Dear gcc
gcc 15.2 at -O2
Accumulating iota [1,10] in a vector
https://godbolt.org/z/Pz7GTqP1e
The optimizer somehow has a case where it doesn't see the allocation can
be skipped, even though it successfully pre-calculates the sum without
the allocation.
There's no allocation when
* the sum is returned from main
* the sum is returned from not_main
* the sum is printed from not_main
But if you write the code in main, and you print the sum instead of
returning it, there's an allocation.
Best regards
Søren Rune Nissen
////////////////////////////////////////////////////////////////////////////////
int main(){
std::vector<int>test(10, 0);
std::iota(test.begin(), test.end(), 1);
int sum = std::accumulate(test.begin(), test.end(), 0);
return sum;
}
main:
mov eax, 55
ret
////////////////////////////////////////////////////////////////////////////////
void not_main(){
std::vector<int>test(10, 0);
std::iota(test.begin(), test.end(), 1);
int sum = std::accumulate(test.begin(), test.end(), 0);
printf("Sum is %d\n", sum);
}
.LC0:
.string "Sum is %d\n"
not_main():
sub rsp, 8
mov esi, 55
mov edi, OFFSET FLAT:.LC0
xor eax, eax
call printf
add rsp, 8
////////////////////////////////////////////////////////////////////////////////
int main(){
std::vector<int>test(10, 0);
std::iota(test.begin(), test.end(), 1);
int sum = std::accumulate(test.begin(), test.end(), 0);
printf("Sum is %d\n", sum);
}
.LC3:
.string "Sum is %d\n"
main:
push rbx
mov edi, 40
sub rsp, 32
call operator new(unsigned long)
mov esi, 55
mov edi, OFFSET FLAT:.LC3
movdqa xmm0, XMMWORD PTR .LC0[rip]
lea rdx, [rax+40]
mov QWORD PTR [rsp], rax
mov QWORD PTR [rsp+16], rdx
mov QWORD PTR [rsp+8], rdx
mov rdx, QWORD PTR .LC2[rip]
movups XMMWORD PTR [rax], xmm0
movdqa xmm0, XMMWORD PTR .LC1[rip]
mov QWORD PTR [rax+32], rdx
movups XMMWORD PTR [rax+16], xmm0
xor eax, eax
call printf
mov rdi, rsp
call std::_Vector_base<int,std::allocator<int>>::~_Vector_base() [base
object destructor]
add rsp, 32
xor eax, eax
pop rbx
ret
mov rbx, rax
jmp .L5
