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The following C function:

int main(void) { char name[10] = "H"; }

Produces the following (unoptimized) assembly in Compiler Explorer:

main: pushq %rbp movq %rsp, %rbp movq $72, -10(%rbp) movw $0, -2(%rbp) <--------- ?? movl $0, %eax popq %rbp ret

What does the line above do? I would think we would want to null terminal the string by adding a $0 but I don't undertand why it's being added at -2. If helpful, here is a screenshot:

enter image description here

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  • @JosephSible-ReinstateMonica I don't think so. It seems to be more related to alignment, as the offset is basically ARRAY_LEN % 8. Commented Jan 8, 2021 at 3:05
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    Hint: How many bytes long is your array? How many bytes are affected by the instruction you marked and the one immediately before it? Commented Jan 8, 2021 at 3:06
  • Maybe you're mixed up on where exactly things are getting stored. -2(%rbp) and -1(%rbp) are the last 2 bytes of the 10-byte array. The first 8 of which were written by movq. Commented Jan 8, 2021 at 3:37
  • @PeterCordes ah, ok. So it adds in the first 8 bytes with the movq and then pads the last two with zero from the movw $0. Is that correct? Commented Jan 8, 2021 at 4:28
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