How to use idiv assembly. idiv — Integer Division.
- How to use idiv assembly You can use FPU or SSE instructions to calculate non-integer values. It's like C where unsigned x=; x *= y; has the same width for the result as the inputs. You manage to show the quotient via temp, then answer, then temp1 How can I collect the remainders of a div instruction into a register so that it can be converted into a string and displayed to the console. It's going to be integer division though. 0: no remainder, the average is already correctly rounded, i. Interesting. Your current code is already broken because you don't sign-extend indexMat into AX. But we can do better by following the same pattern as for signed division by constant powers of 2. Optimized for code-size at the I'm getting confused on what does pop actually do in assembly. Improve this question. It also has two rounding steps, first to 80-bit (64-bit mantissa, unless precision-control is lowered), then to integer. push ebp mov ebp, esp xor eax,eax ;mov eax,0 mov al, [ebp+0x8] div al, byte ptr [ebp+0xC] xor edx,edx mov dl,ah ;put the remainder in edx xor ah,ah mov esp, ebp pop ebp ret The section ## The high 16 bits of the 32-bit word "b" ## will get lost when doing so! ## ## All operations on the "dx" register above ## are also useless when doing this here! cwd ## "div" will do an unsigned division. This instruction is used to divide unsigned integers. Do simple calculations using the add, sub, mul, and div instructions. IDIV r/m64. ) – If you just divide by 2 the solution is easy. We see that the k-th digit of the result is the sum of all the addends such that i + j = k plus an eventual carry However, the idea always works; use two sets of functions and ID at run time. Algorithm: when operand is a byte: AL = AX / operand AH = remainder (modulus) when operand is a word: AX = (DX AX) / operand DX = remainder idiv divides rdx:rax by the specified divisor. What is happening is that your code is experiencing a Division Overflow. (Dividing in Assembler x86 and Displaying Time in Assembly has an example of using aam in 16-bit mode). data section . use cents instead of dollars), but output with a . However what is needed in this case is to round towards negative infinity, so -10/3 has a quotient of -4 with remainder +2, to get this result, the quotient of -3 is decremented to -4 and the remainder of -1 has 3 added to it to get a remainder of +2. That's opposite of printing order. You can always use inline-assembler to get the instruction if the compiler is not supporting it. Division + Modulo in assembly x86. Old. Q&A for work. Without loss of generality, we can assume base 2 and that the numbers are n + 1 bits long (so that the indices run from 0 to n) - then. What kind of competition is it where fewer but slower instructions is better, even when the program is compiled with -O3 not -Os?-O3 will spend extra code size all over the place to gain speed, e. Jeff Duntemann in his assembly language book has an example assembly code for printing the command line arguments. Follow command idiv operand. The problem is that if the 64-bit signed result is large enough and/or if n3 is small enough, an overflow will result and idiv will throw a #DE exception. To quote from the documentation:. Performs an unsigned division of two operands. 1: the sum was positiv, the Starting to teach myself assembly (NASM) I wanted to know how to divide 2 numbers (For instance on Windows). If idiv simply set #DE on overflow, I could check to confirm that ((n1 * n2) / n3) * n3 + ((n1 * n2) mod n3) = (n1 * n2). For example as a hack it can be used to null edx when you know eax You can use fixed-point arithmetic, calculate the division with integers (e. Originally (8086), there was just cbw (ax = If you don't care too much about performance and want to use the straightforward way, you can use either DIV or IDIV. ) Share. Any more clues ? IDIV operation in assembly (understanding) 1 x86 Assembly instruction. 13 bytes for 64-bit operand-size. In this case, since you are only dividing one digit numbers, the best solution is to simply clear AH:. Division of Signed Number. @PeterCordes Noted. The dividend is implicit, and the divisor is the one explicit operand. You don't need an alternative to imul or mul; the one-operand form was part of 8086, not added later. 2^15-1, not 0 . The fact that you're passing "Your Number Is %d \n" as the format string to scanf is a problem, since it contains a bunch of non-format specifier characters. e. NASM idiv can't divide correctly. Also, your hex digits are backwards: you're getting the low digit but storing in increasing order. Also, nightcracker's statement that A XOR B in english would be translated as "are A and B not equal" is only correct when you're looking at the result from a Boolean zero/nonzero So basically I just figured out how to do those 4 simple operations (adding, diving etc. result in st0) fdivp - divides st1 by st0, then pop from reg stack (again, push the result in st0) In this video, you will learn how to use the DIV and IDIV instructions in x86This video is part of my free Foundations of Assembly Programming with nasm cour In microsoft x86 assembly language, I believe your first example would look like: mov eax, 400 ; 400 / 4 and eax, 3h ; Masking with hex 3 (the IDIV Remainder x86 Assembly language. It can just not be that big that the result causes an overflow. It is unusual to use idiv just to check if divisible by 2, it is not done in real-life code. Using div is masochistic. Or if you can't use 386 instructions like shrd, you can emulate SHRD with a left and right shift and an OR. Q&A for Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Visit the blog What is IDIV in assembly language? Description. Ex: MOV AL,31h If you want to deal with numbers larger than 2 32-1, you'll need to do multiple-precision arithmetic. IDIV gives you in this case only three possible remainders: {-1,0. You would have better shown us more from your code. idiv/div leave all flags undefined. That might be ok for most use cases. Every time you use idiv with a power of 2, an electron cries. Signed divide RDX:RAX by r/m64, with result stored in RAX ← Quotient, RDX ← Remainder. DIV or IDIV takes only one operand where it divides a certain register with this operand, the operand can be register or memory location only. Signed divide EDX:EAX by r/m32, with result stored in EAX ← Quotient, EDX ← Remainder. mov eax, -4 xor edx Thanks to not knowing Assembly, I don't know how to use this to change the code correctly. I do, however, know some x86 assembly languages including the one used by GCC and NASM. Follow answered Mar 19, 2011 at 1:04. IDIV instruction is a divide operation. Use movsx eax, byte [indexMax] before byte-size idiv. I'd like to optimize the core using inline assembly, but I'm completely new to this. Implement unsigned division first and The mul instruction has 2 operands: one is specified and the other one is implicit. There are other optional instructions that you need to take care of that fast_idiv. From Intel's manual: "The CWD instruction can be used to produce a doubleword dividend from a word before word division. 0 for negative inputs anyway. This means that the two's-complement value of AX will be the same, but the binary representation will be different. i got into truble in assembly x86 assignment. Find centralized, trusted content and collaborate around the technologies you use most. Note, that despite them using the whole AX, it is not a true 16-bit divide. Your question is tagged Appendix A. Also, do you want signed or unsigned division? div is unsigned, sar is signed (and rounds differently than idiv for negative numbers) – I'm trying divide two numbers in assembly[Irvine-Intel x86 processor]. If you want to divide with byte values, you should use. I'm writing a performance-critical, number-crunching C++ project where 70% of the time is used by the 200 line core module. For example: (if memory serves me right this is correct :-) In a typical assembly language, the integer divide instruction will also give you the remainder. I am developing an assembly language program that will check whether an inputted number is divisible by 3 or not, now I am struggling to get things correct but my code is running. The source The IDIV (signed divide) instruction performs signed integer division, using the same operands as the DIV instruction. When the dividend is positive and the divider is negative you can negate the divider, use div as before, but negate the quotient. My code looks like this but it crashes. Especially, there are arch and tune/cpu flags. So here's the recap of all the pain I've been For a variable count, return x / (1<<n), compilers naively just materialize the 1<<n and use idiv. 7k 5 div bl divides ax by bl and stores the quotient in al, and the remainder in ah. Learn more about Collectives Teams. MUL r8 wil multiply r8 with al and store the result in ax. Using 2x idiv has a fundamental problem: we need the 2nd division to produce the low half of the quotient, which is unsigned and can be anything from 0 to 0xffff. Recall that \(\dfrac{dividend}{divisor} = quotient\) Guide to Using Assembly in Visual Studio — a tutorial on building and debugging assembly code in Visual Studio Intel x86 Instruction Set Reference; idiv — Integer Division The idiv instruction divides the contents of the 64 bit integer EDX:EAX (constructed by viewing EDX as the most significant four bytes and EAX as the least Division without DIV and IDIV Using AND to compute remainders to assembly language is easy, just use the assembly language statement: mov variable, constant This move immediate instruction copies the constant into the variable. Only the highest word of a multi-word integer contains the sign bit, all bits below that have positive place-value. Also JMP @WITHNEG on this line of code @SUB : MOV AL, NUM1 CMP AL, NUM2 JG @WITHOUTNEG JMP @WITHNEG is totally unnecessary as the next line is @WITHNEG:. It compiles and gives me no warnings, but when I try to run it, it gives me a segmentation fault. Therefore the value of DX prior to the IDIV instruction matters, and you should either sign-extend AX into DX using the CWD instruction (before IDIV), or clear DX using e. default rel global WinMain section . To change from divisible by 4 to divisible by 3, you're supposed to use idiv in a sequence that ends with a conditional branch that will have the same effect as the brclr — which is to branch to label yes when the condition of interest holds, You put the high bits into rdx and the low bits into rax, as you probably know. But we can do better by following the same pattern as Division without DIV and IDIV Using AND to compute remainders to assembly language is easy, just use the assembly language statement: mov variable, constant This move immediate instruction copies the constant into the variable. in between. model flat, stdcall option casemap :none include \masm32\include\windows. 1 x86 assembly code confusion Using Skipcond 400 if the number is not divisible by the divider there will be an endless loop as it will go negative not zero. Tried 0 - 80A2 = 7F5E, but didnt help. To multiply by small assemble-time constants, it's often worth it to use a shift and add or subtract instead of multiplying You haven't defined what "it won't work" means, but I'm going to assume that the number displayed by the last printf doesn't match what you inputted. Improve this answer. Hot Network Questions Why are Jersey and Guernsey not considered sovereign states? The apparently random code your debugger seems to jump to is the Arithmetic Exception handler (also the same one as Divide by Zero). Again, using x86 as a demonstration: idiv ebx ; divisor in edx:eax ; quotient in eax, remainder in edx Share. (and then see if the compiler uses the whole instruction set or at least the instructions of interest, if not then make your own library function for division in assembly and call it). Commented May 24, 2017 at 11:30 @Johan idiv has the same This should probably use idiv, or cast the inputs to unsigned. print the division result and the leftovers from the result. The quotient result of the division is stored into EAX; The remainder Divides the (signed) value in the AX, DX:AX, or EDX:EAX (dividend) by the source operand (divisor) and stores the result in the AX (AH:AL), DX:AX, or EDX:EAX registers. See the other question for possible CPUs. Actually it means dx:ax = ax * cx - the high half of the full 32-bit product is always written to dx. (See the x86 tag wiki for a link showing when each form of every instruction was added). mov ecx, 100 / idiv ecx is more like what a compiler would emit (with -Os; otherwise it would replace division by a constant with a multiply by a modular inverse and some shifting. IDIV CX divides the 32-bit value DX:AX by CX, and stores the quotient in AX and the remainder in DX. I'm not the most experienced assembly programmer, and I ran into the "cqo", "cdq" and "cwd" instructions, which are all valid x86_64 assembly. E. The correct setup for signed division is to sign-extend EAX into EDX:EAX, with CDQ. And now the equivalent division code in assembly (NASM syntax) without checking for division by 0:; 64-bit The org directive is mainly used to force a data table or a segment of instructions to start with a certain address. DivNeg: neg ebx cdq div ebx neg eax jmp Done1 With an 8-bit division, the dividend is held in AX, with AH being the high bit and AL being the low bit (naturally enough). When both the dividend and the divider are positive you can safely use div instead of idiv. It's not clear from the question why the code doesn't simply use cdq. IDIV r/m32. Therefore it is up to the programmer to ensure that data types are used correctly. (cdq sign extends eax into edx:eax. Best. Since you did a byte division div bl you obtained a quotient in AL and a remainder in AH. @MichaelPetch you are right, cdq sign extends eax into edx and it is therefore suited for idiv and not div. 3. The rdx:rax is fixed, idiv always uses that pair of registers for the dividend. ) in assembly, because I need to be able to do maths in my own programming language. Assuming the rest of the code does not depend on register values and I haven't messed anything up, that should about work :) I don't know why you use it. -59 can't contain 219 so the quotient is 0, and the remainder is everything that is left which is -59. XOR DX,DX (before DIV). 0xa0000 • Read the wikipedia article on division algorithms, then convert to MIPS assembly. It's not clear what conversion does. n (b j ·2 i+j)] (by the distributive property). If you are using a CPU that does not implement division, use the following code fragment. and it seems like dx register always have content zero even after div has been executed. DIV/IDIV divisor The dividend is idiv executes signed division. From what I understand @TheRookierLearner: A XOR B is a primitive building block for higher-level constructs. This means that you need to have a suitable value in both AH and AL before dividing. asm gcc test. Connect This should probably use idiv, or cast the inputs to unsigned. the request from the program is: get 2 numbers, word size, from the user and divine them. As the assembly code replicating some of the functionality may be tens or hundreds of lines of code (or even thousands for very complex functions), and having it to write every time may be cumbersome. In this case, it'll take the sign bit of AL (which happens to be 1) and copy it into every bit of AH. The code uses esi and edi to store counters as they will be unchanged by the C library function printf. IDIV Remainder x86 Assembly language. Of course, you need 100 as an operand for the second idiv, too. – I am just trying to calculate an average. Original 8086 doesn't have immediate-count shifts I am working on something that will divide two numbers in assembly without using the MUL or DIV operators. Change DIV by IDIV, because DIV is unsigned while IDIV is signed. Posted on January 14, 2017 April 18, 2017 by X86 Assembly. According to line by line execution scheme, if JG is not met, the next line will be imul/mul only set OF and CF, the rest of the flags are undefined (see here or check manual from Intel from which the information was copied). asm. How can I multiply or divide two constants in Assembly? ⭐ Kite is a free AI-powered coding assistant that will help you code faster and smarter. The quotient goes in AL, and the remainder goes in AH. So a check is made for the remainder equalling zero, so X can be incremented when A is no longer positive. So why can't compilers emit this on their own, without inline asm? The cbw instruction sign-extends a byte into a word. printing coordinates assembly 8086. " Note that sarl updates some of the status flags, while cdq doesn't, so movl + sarl won't be exactly the same as cdq. MOV CL, AL MOV AL, BL MOV AH, 0 ; clear AH ; divide DIV CL MOV Use idiv esi like a normal person in your final example. Ask Question Asked 11 years, 3 months ago. Whether C library functions are available depends what libraries your assembly program is linked with. Then you use 64b / 32b => 32b division with the upper half of the dividend = 0. You use it if (as is usual) the value you want to The idiv instruction divides the contents of the 64-bit integer EDX:EAX by the specified operand value. 1. In 64-bit mode, r/m8 can not be encoded to access the following byte registers if a REX prefix is used: AH, BH, CH, DH. You need to use the Floating Point Instruction Set to achieve your goal. What I wanna do is to divide two longs, xor edx,edx before IDIV is usually a bug. imul ecx, esi does ecx *= esi like you'd expect, without touching EAX or EDX. Assembly sarl and idivl - interpreting. avg = avg + (-1). I am supposed to be using long division. In this series of 8086 microprocessor tutorials, we previously discussed; 8086 Microprocessor Addressing Modes , 8086 Data Transfer Instructions , 8086 Integer Arithmetic Instructions , 8086 Integer Multiplication Instructions . The input dividend is in r0 and the divisor is in r1. Another problem with your code is that you seem to assume that int 21h / ah=9 can be used to print numeric values. IDIV will let you handle negative results. I am unsure how to do this. Data in many assembly languages (i. For example:. I don't think you want your remainder to be -9. auto-vectorizing. The idiv instruction divides the contents of the 64 bit integer EDX:EAX (constructed by viewing EDX as the most significant four bytes and EAX as the least significant four bytes) by the specified operand value. idiv — Integer Division. Of course for division by constants idiv and div (and aam) are very inefficient. 1} which means:-1: the sum was negative, the number behind the decimal point is -0. . The last thing you set ah to before the division is 9. 27. How to find remainder without using div in Assembly. For other registers like eax, ecx, edx, there is no guarantee of them not being used by the C library functions. Connect and intel-Based Assembly Language idiv. Controversial. To set up for div instead of idiv, use xor edx,edx) – If the instruction is not in the machine descriptions, then I doubt that gcc will emit code. :. Viewed 2k times 0 . mov ax,12 xor dx,dx div ten (Before signed division with idiv, use cwd to sign-extend AX. However, many OSs will show the message "floating point exception"; POSIX defines SIGFPE as covering any arithmetic exception. The second assignment above is somewhat complicated since the 80x86 doesn’t pro-. inc include \masm32\include\masm32. You can use objdump -drwC -Mintel or gcc -masm=intel -S to get Intel syntax using the mnemonics that Intel and AMD document in their instruction reference manuals (see links in the x86 tag wiki. Your "variable" bNum1 is one byte long. (Unlike with some cases of zero-extension with zero latency (mov / movzx elimination), there's no downside to sign-extending within the same register instead of to a different register. c = a·b = ∑ i=0. If you only want the low 32 bits of the result, use the 2-operand form of imul; it runs faster and doesn't have any implicit operands (so you can use whatever registers are most convenient). The idiv instruction performs a signed integer division. Note2 Since your op-code is fairly rare/machine specific, there is probably not so much effort to get it in the gcc source. (But yes, with a random integer, not a compile-time-constant, int result = a % b would use idiv. To divide something like a BigInteger, which consists of an array of many DWORDS, by 10 (for instance to convert the value to a decimal string), you do Problems with your idiv: There's no form of idiv that takes an immediate operand. Simpler than that, for example: If the bit is less than 0x7F000000, which is less than 127 in decimal, its "positive", is between 0 to 127. Connect and share knowledge within a single location that is structured and easy to search. Since ah is the most significant part of ax that means ax will have the value 0x9XY (where XY are any hexadecimal digits) when you do the division. Now, if any of the numbers is signed but shorter than idiv expects, you need to sign-extend that number. Instruction Operand Encoding ¶ Op/En Operand 1 Operand 2 I am new to Assembly language and I'm getting a problem with the DIV operation. This can be done with the CWD instruction. Likewise, if the dividend is a signed 128-bit value, you don't need to do anything with it, just load the top 64 bits of it into rdx and the low 64 bits into rax. Also, why would you mov rcx,rsi instead of idiv rsi? (Or idiv esi, because you told the compiler you were only going to look at the low 32 bits of input registers by making the return type int. I'm not sure what the easiest way to do this in assembly is I'm afraid. IDIV operation in assembly (understanding) 5. AH must be smaller than a divisor - otherwise "Division overflow" is thrown. – Johan. Does pop move the value PUSHed onto the stack last (meaning it doesn't apply if we MOV a value after the the last element PUSHed) or does it just pop whatever value that's last on the stack (thus, applying to both MOV and PUSH), or does it pop what ever value pointed to by the stack pointer? Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Visit the blog IDIV does signed division, so it expects the dividend in DX:AX to be a signed doubleword. Table 2-7 idiv Register Assignment IDIV r/m64: M: Valid: N. Not sure I quite understand IDIV and can't really find anything on it. You won't receive a result that has a fraction like in 0. ; cdq sign-extends eax into edx:eax, i. n (b j ·2 j) = ∑ i=0. All I know: First, MUL and DIV only take 1 argument. (IDIV, IMUL) and unsigned (DIV, MUL) Share. To multiply by small assemble-time constants, it's often worth it to use a shift and add or subtract instead of multiplying Just to be clear, idiv gives you the signed remainder just as easily, with cdq / idiv for 32-bit operand size for example (When and why do we sign extend and use cdq with mul/div?). b (define constant byte) fcb (form constant byte) Hi I am just playing around in 64 bit assembly but my compiler keeps throwing :(13: Invalid combination of opcode and operands) can anyone explain why this is happening thanks. If the bit is greater than 0x80000000 it's more than 128 in decimal, its "negative", is between -1 and -128. The value of AX after the cbw instruction will be FFF0h (a 16-bit -16 value, just like AL was This video is on the DIV instruction. (Related: see GCD in 9 bytes of x86-64 machine code, or 10 bytes for unsigned. Snake Assembly 8086: not moving correctly. I then want to idiv that 64-bit result by n3. Not to be confused with cdqe, the 64-bit instruction that is a more compact form of movsxd rax, eax. 10: myfunc: 11: mov rax, rcx 12: xor rdx, rdx 13: idiv 10 14: mov rax, rdx 15: ret For people that doesn't read assembly this is similar to: The first two instructions are doing the var <= 5 check, replace everything else with mov ecx, 30; mov eax, r9d; cdq; idiv ecx; test edx, edx; jnz loc_62718B and pad with NOP as needed. where does the remainder goes in div operation (X86 tasm assembler) 6. When operand is a byte: AL = AL / operand, AH = remainder (modulus). Here are the relevant lines of x86 (Intel) assembly code: mov, eax [ebp - 20] cdq idiv ecx Through my input (a network socket) I can control the bytes that go into From what I understand IDIV stores the quotient in the AX register and the remainder in the DX register, but for some reason the value in DX is not the correct value of the remainder. Sort by: Best. The size of the divisor (8-, 16- or 32-bit IDIV Instruction • IDIV (signed divide) performs signed integer division • Uses same operands as DIV Example: 8-bit division of –48 by 5 mov al,-48 cbw ; extend AL into AH mov bl,5 idiv bl ; IDIV - 8086. What is it? It's the value you put there because you want to divide it The CDQ instruction takes a 32-bit value in EAX and converts it into a 64-bit value in EDX:EAX (by copying the sign bit of EAX into every bit of EDX). How would I disable it from assembly? Do I need a syscall or can it be done directly in x86? Reproduction: test. c doesn't emit. How can I get the right answer with idiv? You can't. It is of course possible to use cqo/idiv from inline asm, you just normally don't want to. ;SECOND INPUT CALL SCAN_NUM ;AX/CX = AX IDIV CX The comment is wrong! IDIV CX will do a signed divide of DX:AX by CX. You can easily perform the calculations using any calculator. text WinMain: mov rcx, 0 mov rdx, 0 idiv rcx Commands: nasm -f win64 test. because they require a lot of CPU cycles You don't need an alternative to imul or mul; the one-operand form was part of 8086, not added later. quotient in eax, remainder in edx. 386 . New. It can be used to implement (A != B), but it is a distinctly different operation in its own right. I tried the code in the question (I used NASM so I replaced the writestring and writedec functions with my own), and got the expected output of 4 ((2013 % 10) + 1). See What is the difference between 'asm', '__asm' and '__asm__'? The X86 idiv instruction will return quotient = -3, remainder = -1, since it rounded the quotient towards zero. That is possible because idiv and div divide the whole register pair EDX:EAX. If you're trying to use the number as a constant in your program, then yes, hex is one possible way to split it up. division with a remainders So, I have an assembly function, which is called in C. The size of the divisor (8-, 16- or 32-bit operand) determines the particular register used as the dividend, quotient, and remainder. n [a i ·∑ j=0. kquinn kquinn. If AX is positive then DX will get 0, and if AX is negative then DX will get -1. If you have 32-bit operand size available, you put all of a into eax like you would normally for any other 32-bit operation. When you write mul cx it means something like: ax = ax * cx. inc include \masm32\include\kernel32. Non-whitespace character, There is a way to compute the mod operation, without using DIV or IDIV in assembly x86 language? For instance, one could use DIV in order to take the remainder of the division. 09], by subtraction or by the use of a library. The Kite plugin integrates with all the top editors and IDEs to give I'm not quite sure yet how division works in x86 assembly (GAS AT&T syntax). I admit I didn't read the whole edit history. Follow div / idiv: divides edx:eax by the src. Easy to extend to 64-bit input / 16-byte ASCII hex output, with a 64-bit calling I'm writing a program to practice some reverse engineering techniques. – jfMR. g. That is not the case. 6. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company This makes sense for larger extended-precision shifts: you want to use a chain of SHRD instructions to shift bits into the top of each element in turn, and don't want to shift in zeros. Here is my current code:;calculate average : sum / numCount = average calculateAverage: mov edx, 0 mov eax, sum mov ebx, numCount idiv ebx mov average, eax mov edx, OFFSET averageMessage ; "The average is " call writestring mov ebx, average call Introduction to Assembly Language Programming: From Soup to Nuts: ARM Edition (Kann) Instead, the function __aeabi_idiv is used to do division. Or maybe they really do want weird results with negative inputs for some strange reason. (That also lets you keep the sign-extended dividend around in a register other than eax if you want). The quotient must be in the range 0x00. As a general rule, this directive should be used as infrequently as possi-ble. After a division with DIV, the quotient is in EAX and the remainder is in EDX. You are doing a 16-bit/8-bit IDIV. Commented Dec 22, 2014 at 1:41. This will jump if the "equal flag" (also known as the "zero flag") in the FLAGS register is set. For DIV, you set AH=0. Using too many orgs will make your program less reusable. 2. (This is the opposite of AAM, which does accept an immediate operand, but only divides AL, not AX). So in the event of there being a remainder, the divider will round up the answer by 1. It's also incorrect to use div instead of idiv after cqo. The logic is, if n*var < x, where x is the number to divide and var is the number to divide by, increment n by 1 and repeat. I do think though that by your edit you effectively produced another question. number (instead of just 2), the division instruction will usually produce both the quotient and the remainder. Next two links will take you to the procedures to convert from string to number : For assembly language programs the Least Significant Quadword (LSQ) is added with the add instruction and then immediately the Most Significant Quadword division of signed values. (Fun fact: as input, gas accepts either mnemonic in How to find remainder without using div in Assembly. Sorting strings in 8086 Assembly. Q&A. idiv's quotient range is -2^15 . Capture numbers with int=21h ah=0Ah as strings (not as single chars). 3) How To Use Add, Sub, Mul, Div Instructions. Usually one wants to use it to avoid writing the code providing that functionality himself. CWD is an instruction that will sign-extend the AX register into DX:AX. Search for 'intel mul' and 'intel div' to see the instruction details: For 8 bits: Using an 8 bit register r8 as argument (where r8 is one of the 16 8 bit registers),. This step is necesssary for the IDIV CX instruction to work Just like everything else in assembly there are many ways to do multiplication and division. This is because, for instance, multiplying 127 with 127 is bigger than 8 bits (but never more than 16). 65535. Unlike higher-level programming languages (such as C), assembly language is there to tell the CPU directly what to do. If the FP rounding mode is still the default, that will round to nearest, unlike idiv which truncates towards zero. Probably the cortex-a5 is the best one to use as it supports idiv and is a lowest common denominator. How to find remainder without division or modulo operator in MIPS assembly. An exercise in the book is to implement number = result % divisor in assembly. For a fixed (compile-time-constant) divisor, there's a fixed-point trick using multiply to do exact integer division: Why does GCC use multiplication by a strange number in implementing integer division?. Later, you convert the strings into numbers. You still need to setup DX. This assembly will not perform as good as similar C code. The compiler doesn't have to zero or sign-extend inputs to 64 bit for you, in either of the I get a SIGFPE exception when dividing by 0 in x86 assembly using idiv. Instead of clearing DX prior to the division you should therefore sign-extend AX into DX:AX. And use movsx rcx, esi, or movsx rsi, esi in your first example. From the documentation: Since you are using idiv (and not div), 80A2 is interpreted as a negative number – Matteo Italia. This gets set as a result of arithmetic operations, or instructions like TEST and CMP. As such, different instructions are used for unsigned division (div) and signed division (idiv). 0. According to the book I am learning from, the result of the idiv operation is placed in eax and the remainder in edx. How to divide a number by 32 in ARM assembly? Hot Network Questions look for mov dword ptr [ebp-4],eax you have 8 bits of code. How does imul and idiv really work 8086? 2. – Michael. The format for the DIV/IDIV instruction −. But yes, using idiv for constant divisors is well known to be sub-optimal – Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company mov eax, [a] / cdq / mov ecx, 10 / idiv ecx. Irvines's WriteDec should be replaced by WriteInt which handles the argument EAX as signed number. But if you want modulo in the sense of Euclidean or floored division, not truncating (quotient rounds toward zero) where -1 % 2 == -1, then you'd want more I am trying to run the following code in assembly: mov %si, %ax mov $15, %si div %si mov %eax, %esi When I make my program, it compiles, but at runtime it gives me a floating point exception. The CDQ instruction can be used to produce a quadword dividend from a doubleword before doubleword division. Note1. unsigned int i=999; unsigned int j=i The reason to do this is because, even though there is an integer division instruction div/idiv in the instruction set, it's typically very slow, several times slower 8086 assembly on DOSBox: Bug with idiv instruction? and Why should EDX be 0 before using the DIV instruction? explain how to use signed or unsigned division instructions. What is wrong with my code? assembly; Share. The second assignment above is somewhat complicated since the 80x86 doesn’t pro- The function signature takes int args; the caller is allowed to leave garbage in the high 32 bits of rdi and rsi, so it's incorrect as well as much slower to use 64-bit div. Use shifts and adds/subs instead of multiplication. Top. I was wondering if there are any advantages of using cdq or cwd, when operating on smaller values. Open comment sort options. This shift moves 1 on EAX´s MSB, and you have a positive result 0x80000000 greather than But EDX does not have always have to be 0. DIV performs the division 6/-2 positively (6/4294967294) and gets the result 0 = 0x00000000, with IDIV the result is correct: -3 = If the divisor is a signed 64-bit value, you don't need to do anything with it. Any way would do. idiv executes signed division. We would understand better what you need if we were shown the actual text in Danswer and Drem. AT&T syntax (used by GNU as / objdump) uses different mnemonics than Intel for some instructions (see the official docs). The __aeabi_idiv function does integer division. For IDIV, you use CBW instruction to set AH=0 or (-1), depending on the sign of AL. dx will be zero for small products where the result "fits" in ax. It is interesting because it computes not only the division, but I found the assembly instruction cltd by disassembling code on an Intel architecture. 0xFF, or you'll get a division overflow. Use -mtune=haswell, or better -march=haswell. Here is my code: mov eax, 4 mov edx, 0 mov ebx, 2 div ebx I get the correct answer for the division, but when i change the You should use IDIV. avg = avg + 0. Use shifts for powers of 2, or a multiplicative inverse otherwise, unless you're optimizing for code-size instead of performance. But, instead to use DIV, there are other options? I am trying to run the following code in assembly: mov %si, %ax mov $15, %si div %si mov %eax, %esi When I make my program, it compiles, but at runtime it give Skip to main Find centralized, trusted content and collaborate around the technologies you use most. I am learning about division in assembly language. Much better to use idiv with all explicit operands. This should output the quotient and remainder if a 1-digit number is divided to a 1-digit number. you can get it easily here is the procedure first six bits are given or should be memorized for mov command and then add on LSB the destination bit(D) where d=1 when there is a register in the destination or d=0 when the register is in source. I would have thought this would be equivalent to a normal divide operation except edx would be the You can't use al as divisor, because the command div assumes ax to be the dividend. If you only want the low 16 bits of the result, you can just think of it A quick look in the documentation shows two possibilities for division involving 64-bit numbers:. broadcasts the sign bit of eax into every bit of edx. I am writing this assembly program in 8086, but it is not working properly. Do division by multiplying by the reciprocal value. Some instructions you might find useful are: fild <int> - loads and integer into st0 (not an immediate) faddp - adds st0 to st1, and pop from reg stack (i. Modified 8 years, 9 months ago. Currently I am doing division using subtraction using a loop like this but I loose the decimals: Indeed, div and idiv will cause an "integer division exception", not "a floating-point exception". See How to convert a binary integer number to a hex string? for int->hex that doesn't suck. I think it's because I can't move a constant into a register, but to use the mul/div command it requires a value to be in EAX register. should retain content across function calls) in the x86 ABI. Your EDX value is 1 and you divide EDX:EAX by 2, which results into a bit shift to right. idiv divides a 16-, 32-, or 64-bit register value (dividend) by a register or memory byte, word, or long (divisor). 10. To check if a number is negative either compare it to zero or inspect the most significant bit (it's set if the number is negative). Myth busting. Here the register eax is in source side so should add 0 and then DIV mem16 divides DX:AX by the given operand. This is why cdq / cqo is used to set up for idiv, while xor edx,edx is used to set up for div. This is an example for dividing bp by 7 mov ax,bp // ax is the dividend mov bl,7 // prepare divisor div bl // divide ax by bl We will also provide assembly program examples of each divide instruction. And so, if you want to divide two integers as unsigned, you execute DIV with the values of the integers and if you want to make a signed division, you do the same with the IDIV instruction. mov ah,01h int16h, how to use it to change the direction of the snake. inc includelib \masm32\lib\kernel32. ) The title question is a duplicate of Assembly - How to score a CPU instruction by latency and throughput, the question body is a near-duplicate of Idiomatic way of performance evaluation? and other more specific microbenchmark-methodology questions. The According to my reference, IDIV divides the 64-bit integer EDX:EAX by the provided register's value. lib includelib Both 8-bit DIV and IDIV use the whole AX as input, so you need to set AH to a valid value. Divides (signed) the value in accumulator registers (dividend) by the source operand (divisor) and stores the result in the AX (AH:AL), DX:AX, or EDX:EAX registers. " And use libraries that are likewise compiled for the instruction set of interest and that code stays within the limits of that instruction set. IDIV is for signed numbers wheareas DIV is for unsigned numbers. One example from my BigInteger code:. Your assembler code has some bugs: It clobbers EBX which is a saved register (i. n (a i ·2 i) · ∑ j=0. For both DIV and IDIV, all of the arithmetic status flags are undefined after The DIV (Divide) instruction is used for unsigned data and the IDIV (Integer Divide) is used for signed data. 8086 assembly on DOSBox: Bug with idiv instruction? Related. obj gdb a run How can I divide two numbers in Assembly without using DIV instruction but by using shift and add method? I did that with multiplication and here is my code: mov bl, 56H ;For example mov dl, 79H ;start mov bh, 00H mov dh, 00H xor di, di mov cx, 08H L1: shr dx, 1 ;shifting the multiplier jnc nxt add di, bx ;adding the multiplicand to the result (Note on the above code: it has not been tested, obviously, and I'm a little rusty on my assembly so there is probably something wrong with it, even though it's nice and short. e instruction set architectures) is typeless, I was trying to work out how to calculate modulo 10 in assembly so i compiled the following c code in gcc to see what it came up with. Please provide more details on what do you wish to achieve with this code. Signed divide RDX:RAX by r/m64, with result stored in RAX := Quotient, RDX := Remainder. This would be a variable of the type unsigned char in C programming language. ; You should be using cdq (double to quad word) to sign extend EAX into EDX:EAX instead of cwd (word to double word) to sign extend AX into DX:AX. ; For the Remainder out parameter, you were not storing to the location it pointed to. If that can be done by multiplication of a float number [/9 = *0. Use the address calculation options of lea (multiplication only). Care needs to be taken not to increment X when the remainder is not divisible by B. For example, to add a pair of 64-bit numbers you could do something like this: mov eax, A_1 ; essentially C = A + B mov edx, B_1 add eax, edx ; note that for the low word, we just add mov C_1, eax mov eax, A_2 mov edx, B_2 adc eax, edx ; but for the high word, we I'm writing a small program in assembly and i need to calculate an int that is a power of two and uses the last bit in a register. Inside the CPU, the negative "-2" and the positive "4294967294" are transformed to the same value: 0xFFFFFFFE. For a ## signed division use "idiv" div res1 mov finalres, ax I'm not sure if I found all mistakes in your code. So edx will get 0x00000000 for pos and 0xffffffff for neg. Edit: Has to be long division in binary Share Add a Comment. Assembly Language: How to include the decimal places on the next instruction after dividing? 1. You manage to show the quotient via temp, then answer, then temp1 I m trying to find how to make a division in ARM since there is no DIV command. You could also get faster code if you used unsigned instead of signed division. Your lines 26 and 27 are already setting up rdx:rax correctly. Syntax. ). So you need to zero-extend AX into DX, with MOV DX,0 or XOR DX,DXI. The edit done by the OP (from idiv 10 to idiv ebx) should not have been done either! I do like your suggestion to put a warning comment in the question. The description I found was, Find centralized, trusted content and collaborate around the technologies you use most. db (define byte) dc. 5, the average has to be subtracted by 1, i. @TheFrenchPlaysHdMicraftn you'll need to use a number-to-string conversion function, such as sprintf from the C standard library. If you use IDIV on what you think is usigned integers (or DIV on signed integers), the CPU will happily do that for you and the wrong result (or a division Integer overflow exception on idiv instruction occurs when result of n-bit division can not fit into n-bit register. There's no form of div / idiv that ignores edx in the input. wsq ccldyr bahwi adfnu xspf kwiqj kmd isjqem pceee qyltor
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