How to use idiv assembly Dividing and rounding in assembly This video is on the DIV instruction. 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 IDIV CX divides the 32-bit value DX:AX by CX, and stores the quotient in AX and the remainder in DX. Follow edited May 23, 2017 at 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. print the division result and the leftovers from the result. Kerrek SB thanks for explaining, could 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. For output you can divide by 100/1000 with remainder, and put a decimal point in between. Examples The assembly instructions don't follow wishes or logic, i. 8086 assembler,INT 16,2. DivNeg: neg ebx cdq div ebx neg eax jmp Done1 Integer overflow exception on idiv instruction occurs when result of n-bit division can not fit into n-bit register. When the dividend is positive and the divider is negative you can negate the divider, use div as before, but negate the quotient. Related. It is interesting because it computes not only the division, but The code you suggested uses fld and fst, which both operate on floating point values, so the result you get won't be what you expect it to be. The idiv instruction divides the contents of the 64-bit integer EDX:EAX by the specified operand value. 386 . Now, if any of the numbers is signed but shorter than idiv expects, you need to sign-extend that number. It compiles and gives me no warnings, but when I try to run it, it gives me a segmentation fault. 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. The quotient goes in AL, and the remainder goes in AH. 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. inc includelib \masm32\lib\kernel32. IDIV gives you in this case only three possible remainders: {-1,0. For Everything you did in your program is right except one. Or for (unsigned long)-1 / 1 with signed division like you're doing by zero-extending into RDX:RAX Find centralized, trusted content and collaborate around the technologies you use most. 7k 5 You would have better shown us more from your code. The Kite plugin integrates with all the top editors and IDEs to give 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). ; fine. mul source If the operand is byte sized, it is multiplied by the byte in the AL register and the result is stored in the 16 bits of AX. 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. 8. Learn more about Collectives The reason to do this is because, even though there is an integer division instruction div/idiv in the I'm limited to using the . auto-vectorizing. imul ecx, esi does ecx *= esi like you'd expect, without touching EAX or EDX. Myth busting. 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. avg = avg + (-1). (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. XOR DX,DX (before DIV). – Michael. According to line by line execution scheme, if JG is not met, the next line will be ⭐ Kite is a free AI-powered coding assistant that will help you code faster and smarter. Connect and share knowledge within a single location that is structured and easy to search. because they require a lot of CPU cycles 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. So fdiv should devide st(0)/size and the result should be stored in st(0), right? Because Intel uses little endian, it does not matter that you're only using a single byte, the mov al,[some_address_as_before] will still work. NASM on linux: Using This should probably use idiv, or cast the inputs to unsigned. The rdx:rax is fixed, idiv always uses that pair of registers for the dividend. . Note, that despite them using the whole AX, it is not a true 16-bit divide. broadcasts the sign bit of eax into every bit of edx. BMI2 mulx uses an implicit rdx input operand because its purpose is to allow multiple dep chains of add-with-carry for extended-precision multiply. Commented The cbw instruction sign-extends a byte into a word. This means that the two's-complement value of AX will be the same, but the binary representation will be different. As such, different instructions are used for unsigned division (div) and signed division (idiv). 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 The idiv instruction takes 2 arguments in 3 registers. I want to divide the whole value (64-bits) by 3, not only 32-bits, and store it in 2 registers. c doesn't emit. (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. However, the idea always works; use two sets of functions and ID at run time. I admit I didn't read the whole edit history. Probably the cortex-a5 is the best one to use as it supports idiv and is a lowest common denominator. 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. To quote from the documentation:. While you're at it, I'd suggest a comment to point out that it's intentional that you don't zero EDX inside the loop; that the remainder from the previous division is the upper half of the input to the next limb. lib includelib On a IA-32 architecture how can i divide a signed number by 3 (e. 81. Not to be confused with cdqe, the 64-bit instruction that is a more compact form of movsxd rax, eax. ) 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. 3) How To Use Add, Sub, Mul, Div Instructions. – Johan. 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. What's the difference between `lodsl` and `movl (%esi), %eax; addl 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. The last thing you set ah to before the division is 9. MUL r8 wil multiply r8 with al and store the result in ax. The question should be tag with x86 – phuclv. 386 with my current process. Use shifts and adds/subs instead of multiplication. should retain content across function calls) in the x86 ABI. 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),. kquinn kquinn. 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 Ah, ok, I was wondering how that div eax, ebx got in there in an example based on working code! (and yes, I could tell it must not have been copy/pasted verbatim). Using intel x86 how do I get the decimal places of a division? Basically I want it to output to the hundredth place like: 12. 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). For example:. If AX is positive then DX will get 0, and if AX is negative then DX will get -1. trusted content and collaborate around the technologies you use most. The source In this tutorial, we will see different integer division instructions supported by 8086 microprocessors. ) 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. I wanted to do a simple hybrid code using 16-bit assembly, masm. Snake Assembly 8086: not moving correctly. Posted on January 14, 2017 April 18, 2017 by X86 Assembly. x86 Assembly - idiv to get decimal places. Even if you are using a 64-bit data type, practice shows me that the majority of divisions in a (general purpose) program could still do with just using the built-in div instruction. model flat, stdcall option casemap :none include \masm32\include\windows. However the DIV-Instruction only supports 64/64 Bit divisions. @PeteWilson if you'd like to put your reply in an answer, I'll accept it. use cents instead of dollars), but output with a . My approach was to save rax, the lower Bits of the number, in another Register and left-shift the Bits from the rdx-Register bitwise into rax to Indeed, div and idiv will cause an "integer division exception", not "a floating-point exception". Follow answered Mar 19, 2011 at 1:04. Sorting strings in 8086 Assembly. IDIV operation in assembly When both the dividend and the divider are positive you can safely use div instead of idiv. 6. The quotient must be in the range 0x00. idiv/div leave all flags undefined. The goal was for program to calculate average, but I get very wrong output every single time and I don't know why. DIV/IDIV divisor The dividend is 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 ; The IDIV (signed divide) instruction performs signed integer division, using the same operands as the DIV instruction. mov eax, 174 mov edx, 181 xor eax, edx shr eax, 1 If you do insist on using a div for some reason, you want to use a different register. 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. So here's the recap of all the pain I've been through thanks to intel: This article is useful for people using intel x86_64 architecture with the GNU assembler, gas. CWD is an instruction that will sign-extend the AX register into DX:AX. The size of the divisor (8-, 16- or 32-bit The DIV (Divide) instruction is used for unsigned data and the IDIV (Integer Divide) is used for signed data. See the other question for possible CPUs. Division + Modulo in assembly x86. 8086 assembly on DOSBox: Bug with idiv instruction? Related. For DIV, you set AH=0. Looking back on my book it uses idiv with cdq but uses ecx instead of ebx. and it seems like dx register always have content zero even after div has been executed. It is of course possible to use cqo/idiv from inline asm, you just normally don't want to. Keep the following things in mind: rand() is a standard library function, so you need to link your program with that or the function isn't found / the executable can't run. ) in assembly, because I need to be able to do maths in my own programming language. in between. Assembly - Trying to reverse string, but it adds an extra character on the final string. IDIV operation in assembly (understanding) 2. idiv executes signed division. the request from the program is: get 2 numbers, word size, from the user and divine them. If you use gcc to "compile" your assembly file into an executable, and/or use gcc to link the assembler-generated *. In this case, it'll take the sign bit of AL (which happens to be 1) and copy it into every bit of AH. You use it if (as is usual) the value you want to divide is 32-bit to In this video, you will learn how to use the DIV and IDIV instructions in x86 This video is part of my free Foundations of Assembly Programming with nasm course: httpmore. 5, the average has to be subtracted by 1, i. 0: no remainder, the average is already correctly rounded, i. Signed divide EDX:EAX by r/m32, with result stored in EAX ← Quotient, EDX ← Remainder. MOV ECX,9 CDQ ;this will clear EDX due to the sign extension IDIV ECX MOV EAX,EDX The second way is an optimization used when you modulo using a power of two, in this case you AND by one less than the power of two, eg: eax = eax % 8 becomes AND EAX,7. On a big endian architecture you'd need to adjust the address by 3 bytes. The denominator resides in a source operand and it First, MUL and DIV only take 1 argument. And use movsx rcx, esi , or movsx rsi, esi in your first example. (This is the opposite of AAM, which does accept an immediate operand, but only divides AL, not AX). Or maybe they really do want weird results with negative inputs for some strange reason. Do division by multiplying by the reciprocal value. Intel has many different architectures with corresponding different assembly language. Commented Nov 8, 2013 at 7:45. 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. Not sure I quite understand IDIV and can't really find anything on it. 1. ; cdq sign-extends eax into edx:eax, i. Improve this answer. 0xFF, or you'll get a division overflow. Failure to do so will result in incorrect results, or a division i got into truble in assembly x86 assignment. How to find remainder without using div in Assembly. – 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. The size of the divisor (8-, 16- or 32-bit operand) determines the particular register used as the dividend, quotient, and remainder. Note that the x86 expects the result of the division to fit in one register, so you'll need to zero edx before the division: Find centralized, trusted content and collaborate around the technologies you use most. I am writing this assembly program in 8086, but it is not working properly. Before the division you have to multiply by 100/1000 (see @Chris Halls answer line 3). In the simple/general case: unknown value at runtime. idiv divides a 16-, 32-, or 64-bit register value (dividend) by a register or memory byte, word, or long (divisor). 4. The dividend is implicit, and the divisor is the one explicit operand. 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. I then want to idiv that 64-bit result by n3. 1: the sum was positiv, the You can't use al as divisor, because the command div assumes ax to be the dividend. This shift moves 1 on EAX´s MSB, and you have a positive result 0x80000000 greather than Your assembler code has some bugs: It clobbers EBX which is a saved register (i. Commented May 24, 2017 at 11:30 @Johan idiv has the same 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?). We will also provide assembly program examples of each divide instruction. Use the address calculation options of lea (multiplication only). intel-Based Assembly Language idiv. Using div is masochistic. I am just trying to calculate an average. Therefore you should clear EDX prior to the division if the value you want to divide is 32 bits (fits in EAX alone). After fadd operation the result should be in st(0). 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. Your question is tagged (But yes, with a random integer, not a compile-time-constant, int result = a % b would use idiv. Non-whitespace character, 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). The proposed algorithm will not beat the internal div instruction. Performs an unsigned division of two operands. Your orginal code isn't far off. I am working on something that will divide two numbers in assembly without using the MUL or DIV operators. ; 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. 3681944047089408e-0043 (for the fussy people out there it is actually a long double). Your lines 26 and 27 are already setting up rdx:rax correctly. You still need to setup DX. div bl divides ax by bl and stores the quotient in al, and the remainder in ah. assembly imul and idiv. In this case, since you are only dividing one digit numbers, the best solution is to simply clear AH:. If idiv simply set #DE on overflow, I could check to confirm that ((n1 * n2) / n3) * n3 + ((n1 * n2) mod n3) = (n1 * n2). You don't need an alternative to imul or mul; the one-operand form was part of 8086, not added later. 2. For IDIV, you use CBW instruction to set AH=0 or (-1), depending on the sign of AL. Division operation uses up the AX register. I was wondering if there are any advantages of using cdq or cwd, when operating on smaller values. inc include \masm32\include\kernel32. That is possible because idiv and div divide the whole register pair EDX:EAX. Do simple calculations using the add, sub, mul, and div instructions. avg = avg + 0. Find centralized, trusted content and collaborate around the technologies you use most. @MichaelPetch you are right, cdq sign extends eax into edx and it is therefore suited for idiv and not div. – jfMR. To multiply by small assemble-time constants, it's often worth it to use a shift and add or subtract instead of multiplying 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. Learn more about Collectives Teams. inc include \masm32\include\masm32. e. If AH has a pre value stored in it, AX register will include that value and the final value of AX would be different than what you expect it div / idiv: divides edx:eax by the src. Syntax. And that is Find centralized, trusted content and collaborate around the technologies you use most. But your code looks like you didn't check the instruction manual. – There's no FLAGS / control register / other hardware setting that will make div and idiv not raise #DE divide exception on divisor = 0 or on the quotient not fitting into the operand-size. (See the x86 tag wiki for a link showing when each form of every instruction was added). See What is the difference between 'asm', '__asm' and '__asm__'? When using DIV with a 32-bit operand, it will divide the 64-bit value in EDX:EAX (high DWORD in EDX, low DWORD in EAX) by the operand and put the quotient in EAX and the remainder in EDX. So in the event of there being a remainder, the divider will round up the answer by 1. 3. 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 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. mov ah,01h int16h, how to use it to change the direction of the snake. printing coordinates assembly 8086. I'm assuming this can only be done using shifts operations since imul only works for multiplying 32-bits numbers. The value of AX after the cbw instruction will be FFF0h (a 16-bit -16 value, just like AL was Find centralized, trusted content and collaborate around the technologies you use most. 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 You can use fixed-point arithmetic, calculate the division with integers (e. MOV CL, AL MOV AL, BL MOV AH, 0 ; clear AH ; divide DIV CL MOV Fortunately, to divide by 2, you don't really need to use div at all. I'm learning 80386 from PC Assembly by paul caurter. g. A quick look in the documentation shows two possibilities for division involving 64-bit numbers:. An example? 00 00 00 A9 is the byte representation on 169, but it represents the floating point number +2. Assembly 8086 program. It's going to be integer division though. 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. There are other optional instructions that you need to take care of that fast_idiv. Use movsx eax, byte [indexMax] before byte-size idiv. 10. 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). But only found solutions for dividing by 2^n 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. Originally (8086), there was just cbw (ax = 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 If you want to deal with numbers larger than 2 32-1, you'll need to do multiple-precision arithmetic. just because some instruction is called "DIV", it doesn't mean it works as you expect. AH must be smaller than a divisor - otherwise "Division overflow" is thrown. Your current code is already broken because you don't sign-extend indexMat into AX. I do think though that by your edit you effectively produced another question. into RAX for 64-bit idiv rcx, like for INT64_MIN / -1 which also faults on x86-64. IDIV r/m64. ) Share. How can I multiply or divide two constants in Assembly? 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). This is because, for instance, multiplying 127 with 127 is bigger than 8 bits (but never more than 16). The lower Bits are in raxand the higher ones in rdx. The quotient result of the division is stored into EAX; The remainder is placed in EDX. You manage to show the quotient via temp, then answer, then temp1 I need to divide an unsigned 128-Bit number on a 64-Bit Processor at the Register rdx:rax with a 64-Bit divisor in rdi. It's not clear what conversion does. :) Much appreciated. Sign extension is very simple. There's no form of div / idiv that ignores edx in the input. The quotient result of the division is stored into EAX, while the remainder is placed in EDX. IDIV r/m32. Assembly sarl and idivl - interpreting. This means that you need to have a suitable value in both AH and AL before dividing. If you're clearing a register it's usually better to use xor a,a than using mov reg,0 the former encodes in fewer bytes and runs 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 command idiv operand. e. Inside the CPU, the negative "-2" and the positive "4294967294" are transformed to the same value: 0xFFFFFFFE. 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. Your EDX value is 1 and you divide EDX:EAX by 2, which results into a bit shift to right. But, instead to use DIV, there are other options? 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 This answer will use a technique named Division by Partial Quotients aka Chunking. We would understand better what you need if we were shown the actual text in Danswer and Drem. 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). The format for the DIV/IDIV instruction −. If you are dividing using AL only, It is an utmost necessity that you must first clear the AH register. Since you did a byte division div bl you obtained a quotient in AL and a remainder in AH. Just didn't quite get there. It's like C where unsigned x=; x *= y; has the same width for the result as the inputs. 1} which means:-1: the sum was negative, the number behind the decimal point is -0. Returning result to 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. ;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. @PeterCordes Noted. Syntax idiv idiv . Here is my current code: mov AX, Dividend cwd idiv Divisor outputW AX ;Outputs Quotient (whole number) outputW DX ;Outputs Remainder If you just divide by 2 the solution is easy. 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. But this hypothetical new form of idiv would exist to save code-size and uops around normal uses of idiv that aren't widening. Irvines's WriteDec should be replaced by WriteInt which handles the argument EAX as signed number. If the divisor is a signed 64-bit value, you don't need to do anything with it. Much better to use idiv with all explicit operands. The DIV instruction (and its counterpart IDIV for signed numbers) gives both the quotient and remainder. So basically I just figured out how to do those 4 simple operations (adding, diving etc. Signed divide RDX:RAX by r/m64, with result stored in RAX ← Quotient, RDX ← Remainder. o file into an executable, all necessary dependent libs This is why cdq / cqo is used to set up for idiv, while xor edx,edx is used to set up for div. Use idiv esi like a normal person in your final example. 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 Thanks guys, I was on the right track. Q&A for work. 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:. Recall that \(\dfrac{dividend}{divisor} = quotient\). Otherwise I'll just put in an answer. If the source is 16-bit, it is multiplied by the word in AX and the 32-bit result is stored in DX:AX. ; For the Remainder out parameter, you were not storing to the location it pointed to. What is IDIV in assembly language? Description. This instruction is used to divide unsigned integers. Syntax idiv <reg32> idiv <mem> Examples There's nothing called "Intel-based assembly language". So, I have an assembly function, which is called in C. In this series of 8086 microprocessor tutorials, we previously discussed; IDIV (signed numbers) AAD; 8086 DIV Instruction ( Unsigned Operands) The DIV instruction performs the division of two unsigned operands. For both DIV and IDIV, all of the arithmetic status flags are undefined after idiv divides rdx:rax by the specified divisor. For unsigned, remainder and modulus are the same thing. – user2922055. quotient in eax, remainder in edx. DIV performs the division 6/-2 positively (6/4294967294) and gets the result 0 = 0x00000000, with IDIV the result is correct: -3 = Please provide more details on what do you wish to achieve with this code. The first implicit argument is the dividend, a 64-bit argument in edx:eax See: Can I use Intel syntax of x86 assembly with GCC? Share. However, many OSs will show the message "floating point exception"; POSIX defines SIGFPE as covering any arithmetic exception. ) a value stored in 2 registers, edx:eax (a 64-bit value). MASM - Rounding Integers. The compiler doesn't have to zero or sign-extend inputs to 64 bit for you, in either of the Both 8-bit DIV and IDIV use the whole AX as input, so you need to set AH to a valid value. Problems with your idiv: There's no form of idiv that takes an immediate operand. So it still has to produce 2 outputs. hxetxn pbym mugidk tuewo ihfwm dmsjj zjkghxf kkz kbok wvywd