问题在理解MUL和放大器;汇编语言的指令IMUL

问题描述

我是从

学习80386

  MUL源
 

fine.

Q1: Why DX:AX ? Why can't it store in EAX / EDX?

imul is really confusing

imul dest, source1
imul dest, source1, source2

I've problem in understanding the table.

Q2: in the 2nd entry of the table. Again, why DX:AX. Why not EAX or EDX?

Now consider following code snippet:

imul eax ; edx:eax = eax * eax
mov ebx, eax ; save answer in ebx
mov eax, square_msg ; square_msg db "Square of input is ", 0
call print_string ; prints the string eax
mov eax, ebx 
call print_int ;  prints the int stored in eax
call print_nl ; prints new line

Q3: Its previsously said that The notation EDX:EAX means to think of the EDX and EAX registers as one 64 bit register with the upper32 bits in EDX and the lower bits in EAX. So the answer is also stored in edx, right? in the above code we didn't consider any EDX we are just referring to EAXHow is this still working?

Q4: I've problem with rest of all entries in the table. worst case multiplication result of two n bit numbers(n = 8/16/32 bits) is 2n bits. How come its storing the result of two 16/32 bit multiplication result in register of same size itself?

解决方案

Q1/Q2: The x86 instruction set maintains its 16-bit history. When doing a 16-bit multiply, the answer is stored in DX:AX. That's just the way it is, because that's how it was in 16-bit land.

Q3: The code you showed has a bug if you try to compute the square of a number larger than 2^16, because the code ignores the high 32 bits of the result stored in edx.

Q4: I think you may be misreading the table. 8-bit multiplications are stored in a 16-bit result; 16-bit multiplications are stored in a 32-bit result; 32-bit multiplications are stored in a 64-bit result. Which line are you referring to specifically?

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