Returning From a Void Function in C++

Question

Consider the following snippet:

void Foo()
{
  // ...
}

void Bar()
{
  return Foo();
}

What is a legitimate reason to use the above in C++

Answer 1

You would use it in generic code, where the return value of Foo() is unknown or subject to change. Consider:

template T Bar(Foo f) {
    return f();
}

In this case, Bar is valid for void, but is also valid should the return type change. However, if it merely called f, then this code would break if T was non-void. Using the return f(); syntax guarantees preservation of the return value of Foo() if one exists, AND allows for void().

In addition, explicitly returning is a good habit to get into.