为什么VS2013将一个函数调用编译成两个指令而不是一个？

Question

Here is a simple program:

void func()
{
    printf("hello");
}

int main()
{
    printf("%p",func);
    func();
    return 0;
}

Stepping over the line printf("%p",func) , I get 00F811AE printed on the console.

Disassembling the line func() , gives me call _func (0F811AEh) - so far so good.

But disassembling the contents of func , the first instruction appears at address 00F813C0 .

So I "went to see" what's on address 00F811AE , and there I found jmp func (0F813C0h) .

To summarize this, it appears that the function-call is compiled as two instructions:

call _func (0F811AEh)
jmp   func (0F813C0h)

Why does the VS2013 compiler use two instructions instead of just one?

It appears that a single jmp would do the the job. I am asking even this because I have a feeling that the other compilers do it in a similar manner (depending on the underlying HW architecture of course).

Thanks

Answer 1

Learn about "thunking": http://en.wikipedia.org/wiki/Thunk

One benefit with "thunking" in your example is that the rest of your code will always call func , but any function performing the same role could be injected into the call at address 0x00F811AE.

Try making func a static one and find out if anything changes.