Calculating and printing factorial at compile time in C++

The factorial can be printed in compiler-generated message as:

template<int x> struct _;
int main() {
        _<Factorial<10>::value> __;
        return 0;
}

Error message:

prog.cpp:14:32: error: aggregate ‘_<3628800> __’ has incomplete type and cannot be defined _::value> __; ^

Here 3628800 is factorial of 10.

See it at ideone : http://ideone.com/094SJz

So are you looking for this?

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EDIT:

Matthieu asked for a clever trick to both print the factorial AND let the compilation continue. Here is one attempt. It doesn't give any error, hence the compilation succeeds with one warning.

template<int factorial> 
struct _{ operator char() { return factorial + 256; } }; //always overflow
int main() {
        char(_<Factorial<5>::value>());
        return 0;
}

It gets compiled with this warning:

main.cpp: In instantiation of '_::operator char() [with int factorial = 120]': main.cpp:16:39: required from here main.cpp:13:48: warning: overflow in implicit constant conversion [-Woverflow] struct _{ operator char() { return factorial + 256; } }; //always overflow

Here 120 is factorial of 5.

Demo at ideone : http://coliru.stacked-crooked.com/a/c4d703a670060545

You could just write a nice macro, and use it instead as:

#define PRINT_AS_WARNING(constant) char(_<constant>())    

int main() 
{
         PRINT_AS_WARNING(Factorial<5>::value);
         return 0;
}

That looks great.

i am learning basics of TMP, and want to know the result at compile to make sure logic is correct.

In that case, what you really want is a static assertion:

static_assert(Factorial<5> ::value ==     120,  "5! should be 120");
static_assert(Factorial<10>::value == 3628800, "10! should be 3628800");

If your compiler does not support static_assert yet, you can use BOOST_STATIC_ASSERT.