C destructor called timing for returned value-CodePudding

Consider C code as below:

struct V {
  int s;
  V(int s): s(s) {}
  ~V() { cout << "Destructor\n"; }
};

V f() {
  V x(2);
  return x;
}

int main(){
  V a = f();
  cout << "Function End\n";
  return 0;
}

The execution result shows that the destructor is called only once.

Function End
Destructor

However, if I add a meaningless if statement as below, the destructor is called twice.

V f() {
  if(false) return V(3);
  V x(2);
  return x;
}

Destructor
Function End
Destructor

Why can this happen? Is there some points to avoid calling destructor twice?

CodePudding user response：

In the first example, NRVO (Named Return Value Optimization) may kick in and elide the copy in return x;.

When you have two exit paths, not returning the same variable, NRVO is less likely to kick in and you'll actually get a copy, even though if(false) is never going to be true.

The below would most likely also elide the copy because x is returned in all paths that leads to return.

V f() {
    V x(2);
    if (false) x = V(3);
    return x;
}