为什么不使用时分配堆栈内存？

Question

Consider the following example:

struct vector {
    int  size() const;
    bool empty() const;
};

bool vector::empty() const
{
    return size() == 0;
}

The generated assembly code for vector::empty (by clang, with optimizations):

push    rax
call    vector::size() const
test    eax, eax
sete    al
pop     rcx
ret

Why does it allocate stack space? It is not used at all. The push and pop could be omitted. Optimized builds of MSVC and gcc also use stack space for this function (see on godbolt ), so there must be a reason.

Answer 1

It allocates stack space, so the stack is 16-byte aligned. It is needed, because the return address takes 8 bytes, so an additional 8-byte space is needed to keep the stack 16-byte aligned.

The alignment of stack frames can be configured with command line arguments for some compilers.

• MSVC : The documentation says that the stack is always 16-byte aligned. No command line argument can change this. The godbolt example shows that 40 bytes are subtracted from rsp at the beginning of the function, which means that something else also affects this.
• clang : The -mstack-alignment option specifies the stack alignment. It seems, that the default is 16, although not documented. If you set it to 8, the stack allocation ( push and pop ) disappears from the generated assembly code.
• gcc : The-mpreferred-stack-boundary option specifies the stack alignment. If the given value is N, it means 2^N bytes of alignment. The default value is 4, which means 16 bytes. If you set it to 3 (ie 8 bytes), the stack allocation ( sub and add for rsp ) disappears from the generated assembly code.

Check out on godbolt .