I was always under the impression that a std::unique_ptr
had no overhead compared to using a raw pointer. However, compiling the following code
#include <memory>
void raw_pointer() {
int* p = new int[100];
delete[] p;
}
void smart_pointer() {
auto p = std::make_unique<int[]>(100);
}
with g++ -std=c++14 -O3
produces the following assembly:
raw_pointer():
sub rsp, 8
mov edi, 400
call operator new[](unsigned long)
add rsp, 8
mov rdi, rax
jmp operator delete[](void*)
smart_pointer():
sub rsp, 8
mov edi, 400
call operator new[](unsigned long)
lea rdi, [rax+8]
mov rcx, rax
mov QWORD PTR [rax], 0
mov QWORD PTR [rax+392], 0
mov rdx, rax
xor eax, eax
and rdi, -8
sub rcx, rdi
add ecx, 400
shr ecx, 3
rep stosq
mov rdi, rdx
add rsp, 8
jmp operator delete[](void*)
Why is the output for smart_pointer()
almost three times as large as raw_pointer()
?
question from:
https://stackoverflow.com/questions/40635107/why-does-unique-ptr-instantiation-compile-to-larger-binary-than-raw-pointer 与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…