Build x86-64: why offset 25 bytes?

Question

I started to study the x86 assembly today by analyzing the assembly code corresponding to this C ++ example (I know there is something like atoi, but I wanted this example to be minimal):

#include <vector>

std::vector<int> range(int N) {
    std::vector<int> v(N);
    for (unsigned int i = 0; i < N; ++i)
        v[i] = i;
    return v;
}

int main() {
    return range(100).back();
}

If compiled with g++ -O0 -S -fno-stack-protector return_by_value.cpp, this will result in this excerpt:

... <snip>
_Z5rangei:
.LFB509:
    .cfi_startproc
    .cfi_personality 0x3,__gxx_personality_v0
    .cfi_lsda 0x3,.LLSDA509
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    pushq   %rbx
    subq    $40, %rsp
    .cfi_offset 3, -24
    movq    %rdi, -40(%rbp)
    movl    %esi, -44(%rbp)
    leaq    -25(%rbp), %rax
    movq    %rax, %rdi
    call    _ZNSaIiEC1Ev
    movl    $0, -24(%rbp)
    movl    -44(%rbp), %eax
    movslq  %eax, %rsi
    leaq    -25(%rbp), %rcx
    leaq    -24(%rbp), %rdx
    ... <snip>

I was surprised to see an odd (i.e. not a multiple of 8) offset:, leaq -25(%rbp), %raxmoreover, this is an instruction q, and we also have it -24(%rbp). For what reason does the compiler read across 8-byte boundaries?

Answer 1

Looking at this snippet:

leaq    -25(%rbp), %rax
movq    %rax, %rdi
call    _ZNSaIiEC1Ev

_ZNSaIiEC1Ev std::allocator<int>::allocator(), -25(%rbp) - allocator<int>, . sizeof GCC, 1. 1, 8 .

-24(%rbp), , , 8- .

, lea - ​​ . , q, , 8 .