Question

static inline void *__memset(void *s, char c, size_t n) {
int d0, d1;
asm volatile (
    "rep; stosb;"
    : "=&c" (d0), "=&D" (d1)
    : "0" (n), "a" (c), "1" (s)
    : "memory");
return s;
}

What are “d0” and “d1” used for? Could you please explain all the code completely?Thank you!

Answer 1

What are “d0” and “d1” used for?

In effect, it says that the final values of %ecx, %edi (assuming 32-bit) are stored in d0, d1 respectively. This serves a couple of purposes:

It lets the compiler know that, as outputs, these registers are effectively clobbered. By assigning them to temporary variables, an optimizing compiler also knows that there is no need to actually perform the ‘store’ operation.

The “=&” specifies these as early-clobber operands. They may be written to before all the inputs are consumed. So if the compiler is free to choose an input register, it shouldn’t alias these two.

This isn’t technically necessary for %ecx, since it’s explicitly named as an input: "0" (n) – the ‘rep’ count in this case. I’m not sure it’s necessary for %edi either, since it can’t be updated before the input "1" (s) is consumed, and the instruction executed. And again, as it’s explicitly named as an input, the compiler isn’t free to choose another register. In short, “=&” doesn’t hurt here, but it doesn’t do anything.

As "a" (c) specifies an input-only register %eax set to (c), the compiler may assume that %eax still holds this value after the ‘asm’ – which is indeed the case with "rep; stosb;".

"memory" specifies that memory can be modified in a way unknown to the compiler – which is true in this case, it’s setting (n) bytes starting at (r) to the value (c) – assuming the direction flag is cleared, which it should be. This does have the effect of forcing a reload of values, as the compiler can’t assume that registers reflect the memory values they’re supposed to anymore. It doesn’t hurt, and it may be necessary to make it safe for a general case memset, but it’s often overkill.

Edit: Input operands may not overlap clobber operands. It doesn’t make sense to specify something as input-only and clobbered. I don’t think the compiler allows this, and it wouldn’t be wise to use an ambiguous specification even if it did. From the manual:

You may not write a clobber description in a way that overlaps with an input or output operand. For example, you may not have an operand describing a register class with one member if you mention that register in the clobber list.

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Reviewing some old answers, I thought I would add a link to the excellent Lockless GCC inline ASM tutorial. The article builds on prior sections, unlike the gcc manual which is best described as a ‘reference’, and not really suited to any sort of structured learning.