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Hi,

If I compile the following program, it compiles successfully.
As well as it runs very well. Once I run it there is no output.

Anyone please explain
1) How this program compiles ?
2) How does template work in this case ?

template <int iValue>
void Add(int mValue)
{
    printf("I am in Add");
}

int _tmain(int argc, _TCHAR* argv[])
{
      Add<"sss">;
      return 0;
}

Thanks in advance.

Posted 27-Jul-10 1:59am

san123pune

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Aescleal · Answer 1 · 2010-07-27T02:23:00

Solution 1

The line...:

C++

Add<"sss">;

doesn't actually cause any code to be generated. There's no function call, no object definition. You're not telling the compiler to do anything. If you try actually doing something with your function:

Add<"sss">( 128 );

then your compiler's going to throw a fit and not build the program.

Cheers,

Ash

Posted 27-Jul-10 2:23am

Aescleal

Comments

san123pune 27-Jul-10 8:28am

But how compiler handles this internally ?
Why compiler accept "sss" instead of int?

Aescleal 27-Jul-10 11:12am

It doesn't accept it. Templates expansion errors only occur when they're first used. You're not using the template, you're just making a half-arsed expansion.

As I said most compilers will give you a warning if you do what you have. If yours doesn't then it might be an idea to change it.

Ash

PS: Look up SFINAE (Substituion Failiure Is Not An Error) online. It'll point you to the section of the C++ standard that describes how templates are expanded.

PPS: Edited to make the language more precise and use the same terms as the standard

Emilio Garavaglia · Answer 2 · 2010-07-27T03:09:00

if I do

void func(int x)
{ cout << "in func" << endl; }

int main()
{
   func;
   return 0;
}

No outoput will be produced, since func is not a function call.
Now, in I do

C++

template<int N>
void func(int x)
{ cout << "in func" << endl; }

int main()
{
   func<"hhh">;
   return 0;
}

The compiler will even NOT produce the code for "func", since code is produce at the time it is required to be used.

If I do

C++

int main()
{
   func<"hhh">(128);
   return 0;
}

The compiler attempt to generate func<"hhh">(int) and have no way to do, hence will generate an error.
If I do

C++

int main()
{
    func<10>(100);
    return 0;
}

The compiler generates func<10>(int) and calls it passing 100 as a parameter.

Now, you understand all this correctly, it should be clear to you that
func<10> and func<15> are two different and distinct functions.