C++ Is the copy constructor called here?

Question

Suppose you have a functions like this:

Foo foo() {
  Foo foo;

  // more lines of code

  return foo; // is the copy constructor called here?
}

Foo bar() {
  // more lines of code

  return Foo(); // is the copy constructor called here?
}

int main() {
  Foo a = foo();
  Foo b = bar();  
}

When any of the functions return, is the copy constructor called (suppose there would be one)?

What do you mean by "suppose there would be one"? Every class has a copy constructor (though it may be `private`, or in C++0x `delete`d). If the class doesn't declare one, it gets an implicitly-declared copy constructor. — aschepler, Mar 02 '11 at 21:07
Also see http://stackoverflow.com/questions/665825/copy-constructor-vs-return-value-optimization — xtofl, Mar 02 '11 at 21:07
It's easy to test something like this. Just implement a copy constructor and print a message in it. — gregg, Mar 02 '11 at 21:15

score 10 · Accepted Answer · answered Mar 02 '11 at 21:05

10

It might be called, or it might not be called. The compiler has the option of using the Return Value Optimization in both cases (though the optimization is a bit easier in bar than in foo).

Even if RVO eliminates the actual calls to the copy constructor, the copy constructor must still be defined and accessible.

answered Mar 02 '11 at 21:05

aschepler

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There is also the NRVO option, that might actually be easier in foo. :-) – Bo Persson Mar 02 '11 at 21:21

score 8 · Answer 2 · answered Mar 02 '11 at 21:05

8

Depends on the Return Value Optimization being applied or not.

answered Mar 02 '11 at 21:05

Nikolai Fetissov

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score 2 · Answer 3 · edited May 23 '17 at 11:55

2

It may be called. It also may be optimized away. See some other question in the same direction.

edited May 23 '17 at 11:55

Community

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answered Mar 02 '11 at 21:06

xtofl

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C++ Is the copy constructor called here?

3 Answers3