lambda returns '1' all time

Question

Have code like this

#include <iostream>
using namespace std;
int main() 
{ 
  cout<<[](){ return 0;};
  cout<<[](){ return 3.2;};
  cout<<[](){ return true;};
  cout<<[](){ return false;};
  cout<<[](){ return "Hello world!";};
  cout<<[]()->int{ return 0;};
  cout<<[]()->double{ return 3.2;};
  cout<<[]()->bool{ return true;};
  cout<<[]()->bool{ return false;};
  cout<<[]()->const char*{ return "Hello world!";};
  return 0;
}

Compile it with gcc version 4.8.2 and my output is only 1111111111. Why only "1"?

Answer 1

When a lambda expression has no capture, it is implicitly convertible to a function pointer.

A function pointer, in turn, is implicitly convertible to bool, yielding true if the pointer is not null, which gets printed.

If you cout << std::boolalpha before your outputs, you'll see truetruetrue.... printed instead.

If you capture something in your lambda, then it is no longer convertible to a function pointer, and you'd get a compiler error.

If you want to print the result returned by calling the lambda, then you need (), as others have pointed out.

Answer 2

What if you did this:

#include <iostream>
using namespace std;
void foo()
{
}
int main() 
{ 
  cout<<foo;
  return 0;
}

You're not calling the method, but attempting to print its address. The overload operator <<(bool) for cout is chosen, thus for any valid function you attempt to print, you get a 1.

To actually, call the function (or lambadas), add ().

Answer 3

What you say:

cout<<[](){ return 0;};

What you want to say:

cout<<[](){ return 0;}();

See the bracket?