理解/编写 C 代码，用于单精度浮点数的汇编

Question

I am having trouble deciphering this given Assembly code:

1 unknown5:
2    pxor   %xmm0, %xmm0
3    movl   $0, %eax
4    jmp    .L13
5  .L14:
6    movss  (%rdx,%rax,4), %xmm1
7    mulss  (%rsi,%rax,4), %xmm1
8    addss  %xmm1, %xmm0
9    addq   $1, %rax
10 .L13:
11   cmpq   %rdi, %rax
12   jb .L14
13   rep ret

What I think is going on:

• 1: function named unknown5()
• 2: sets the floating point return register to 0
• 3: sets the 32-bit return register to 0
• 4: jumps to 10:
• 11: if %rdi > %rax, go to 5:, else go to 13:
• 6: %xmm1 = %rdx+%rax+4
• 7: %xmm1 *= %rsi+%rax+4
• 8: %xmm0 += %xmm1
• 9: %rax += 1
• 13: return %eax, %rax, or %xmm0??

What I have no clue of is what the possible parameters are of the function, nor what each register is representing. I know %rdi, %rsi, and %rdx are first, second, and third parameters (that might be floating point pointers due to single precision operations?), and %xmm0, %xmm1 are first and second floating point parameters (OR local variables?). I also know %rax is the 64-bit return register, but am unsure why it is used here as such. Could someone please elaborate and correct my comprehension? Thank you.

Answer 1

A few hints:

• %xmm0, %xmm1 are used to pass floating-point parameters, if the function actually takes any. But are the values they had on entry to the function actually used in any way?

• Don't think about %rax as inherently being "a return register". It's just a register. If the function is meant to return an integer or pointer then yes, %rax (or one of its sub-registers) is where that return value will go, but in the meantime it's just a register that the code can use as it likes, like a local variable as you say. Same goes for %xmm0 .

• Your guess about %rdx and %rsi being pointers is a good one, given how they are used in the movss/mulss memory operands. Where exactly are these instructions loading values from? On the other hand, %rdi does not appear in such a context; how is it used instead?

  Note movss (%rdx,%rax,4), %xmm1 doesn't set %xmm1 equal to the value gotten by multiplying %rax by 4 and adding %rdx . Rather, it uses %rdx+%rax*4 as an address , fetches a single-precision float from that address in memory, and loads it into %xmm1 . That's what the ( ) signify - an indirect memory reference. If you like, it's the rough equivalent of the unary * dereference operator in C.

  Likewise, the value being multiplied in the following mulss is also fetched from memory.

• There's no certain way to tell from the code alone whether it's intended to return an int (return value in %eax ) or long ( %rax ) or float/double ( %xmm0 ). But you can take a very good guess. Think about the values that each of those registers has when the function returns. Which seems more plausible as a value for the function to be computing for its caller?

  Notice there are no instructions in this function that write to memory, so it has no "side effects"; its sole purpose must be to compute and return some value.

• This function appears to be performing a familiar mathematical operation. Can you tell what it is?