Question

Since my shifting from C to C++ I have a question on STL’s formatting output. How ostreams tell one basic type from another?

In C with its printf and formatting strings it was pretty straightforward, but in C++ ostreams somehow distinguish basic types automatically. It puzzles me.

For example, in the following code,

int i;
float f;

std::cout << i << std::endl;
std::cout << f << std::endl;

how cout “knows” that i is an int and f is a float?

Answer 1

The compiler converts the operators to function calls. So that

std::cout << i

becomes

operator<<(std::cout, i)

Somewhere buried deep in the bowels of the standard library headers there are function declarations (functionally equivalent to):

std::ostream& operator<<(std::ostream& o, int i);
std::ostream& operator<<(std::ostream& o, double d);

That is, operator<< is overloaded. When the function call is made, the compiler chooses the function overload which is the best match to the arguments passed in.

In the case of std::cout << i, the int overload is chosen. In the case of std::cout<<d, the double overload is chosen.

You can see function overloading in action fairly simply with a contrived example:

#include <stdio.h>

void print(int i) {printf("%d\n", i);}
void print(double d) {printf("%f\n", d);}

int main()
{
   int j=5;
   double f=7.7;

   print(j);
   print(f);
}

Producing the output:

5
7.700000

Try it for yourself: http://ideone.com/grlZl.

Edit: As Jesse Good points out, the functions in question are member functions. So really we have:

std::cout << i

becomes

std::cout.operator<<(i)

And in the headers there are declarations (equivalent to):

class ostream {
    ostream& operator<<(int i);
    ostream& operator<<(double d);
    ...
};

The same basic idea holds, however.