Question

GCC inlines a statement — no matter how hard I try to prevent it. I tried

• -fno-inline
• -O0
• __attribute__ ((noinline))
• dummy asm("")

No success!
Here the code:

#include<iostream>

using namespace std;

struct A {
  A() {cout << "A::A()" <<endl; }
  A(const A& a) {cout << "A::A(copy)" <<endl; }
  A& operator=(const A& a) {cout << "A::=()" <<endl; return *this;}
};

A __attribute__ ((noinline)) func() 
{
  cout << "func()" << endl;
  A loc;
  asm("");
  return loc;
}

int main() {
  A a = func();
}

The unfortunate output of this (g++ (Ubuntu/Linaro 4.5.2-8ubuntu4) 4.5.2) is

func()
A::A()

What happened to the statement A a = func(); ??

The reason for this experiment is that I would like to know what happens when execution comes to this statement (because I need control how this is done):

A a = func();

I read that the copy constructor is called when one does

A a = b;

(In this case the copy-constructor is called. But not in the case A a = func();)
The function is inlined instead. I NEED control over this statement since my “struct A”
in real life contains dynamically allocated data that needs to be taken care of.

Am I missing something obvious here ?!

Answer 1

No, this has nothing to do with the function being inlined. Inlining the function would not change observable behaviour.

This is an optimization called copy elision that allows the compiler to avoid a copy by constructing the return value directly at the destination. You can disable it with the g++ flag -fno-elide-constructors.

In any case, the dynamically allocated data should not be a problem. Assuming a sane copy constructor, the only difference you will see will be possibly better performance.