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templates

Started by Xtreme Jul 15, 2004 at 10:25 PM 18 replies 1k views

Original Post

Jul 15, 2004 at 10:25 PM

In C++, do the following templates have the same meaning?


//1. 

template<> void Swap(float x, float y);

//2. or:
template<float> void Swap(float x, float y);

//3. or:
template void Swap<float>(float x, float y);

//4. or:
void Swap(float x, float y);

Whenever I declare it and run the function, i see that 4. is called no matter what. Whats the point of writing unneccessary template keyword in front when you know that 4. will be called all the time instead of the rest?

SiCrane

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473

SiCrane

Jul 15, 2004 at 10:29 PM

No they don't have the same meaning. Templates have a strict order of application and qualification. This is a truly obnoxious and arcane corner of the C++ language, so I'll just refer you to this article. Hopefully that will explain the differences.

Fruny

203

Fruny

Jul 15, 2004 at 10:29 PM

When the compiler has the choice between a template version and a non-template version of an overloaded function, it always chooses the non-template version (13.3.3 p220): "... a viable function F1 is defined to be a better function than another viable function F2 if ... F1 is a non-template function and F2 is a template function specialization ..."

Xtreme

Author

Xtreme

Jul 15, 2004 at 10:37 PM

the anon poster was me :)

Fruny

203

Fruny

Jul 15, 2004 at 10:43 PM

Quote:
Original post by Anonymous Poster
IE, whats the point of specialization templates when you know that non-template version will be called instead of specialization templates in overloaded function cases?

But why would you both provide a templated and a non-templated version of the same function, with the same parameters?

Xtreme

Author

Xtreme

Jul 15, 2004 at 10:47 PM

Quote:
Original post by Fruny
Quote:
Original post by Anonymous Poster
IE, whats the point of specialization templates when you know that non-template version will be called instead of specialization templates in overloaded function cases?

But why would you both provide a templated and a non-templated version of the same function, with the same parameters?

to check to see which function gets called. My point is it seems if you have a same function with same parameters, it is a waste of declaring template<> in front of the function, since declaring it without template <> is still a specialization!

SiCrane

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473

SiCrane

Jul 15, 2004 at 10:50 PM

No, it's not. It's an overload. There's a difference.

Shannon Barber

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368

Shannon Barber

Jul 15, 2004 at 10:51 PM

You generally shouldn't overload template functions. You define a generic one and then handle special cases as, well, specializations.

#4 isn't a template

Xtreme

Author

Xtreme

Jul 15, 2004 at 10:54 PM

Quote:
Original post by SiCrane
No, it's not. It's an overload. There's a difference.

So, what you are saying is that say you have the following template declared:

template <typename T> void Swap(T x, T y);

Is the following not an overload?

template <> void Swap(float x, float y);

And the following is?

void Swap(float x, float y);

Xtreme

Author

Xtreme

Jul 15, 2004 at 10:56 PM

Quote:
Original post by Magmai Kai Holmlor
You generally shouldn't overload template functions. You define a generic one and then handle special cases as, well, specializations.

#4 isn't a template

Yep. I know #4 isnt but i dont know whats the point of having specialization templates when you can define it like #4.

SiCrane

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SiCrane

Jul 15, 2004 at 11:07 PM

Quote:
Original post by Xtreme
So, what you are saying is that say you have the following template declared: /snip/

Pretty much. If you call the function through explicit template arguments rather than you'll call the specialization, but if you call it without explicit template arguments then you'll call the overload. Read the article I posted up at the beginning of the thread.

Xtreme

Author

Xtreme

Jul 15, 2004 at 11:15 PM

Quote:
Original post by SiCrane
Quote:
Original post by Xtreme
So, what you are saying is that say you have the following template declared: /snip/

Pretty much. If you call the function through explicit template arguments rather than you'll call the specialization, but if you call it without explicit template arguments then you'll call the overload. Read the article I posted up at the beginning of the thread.

thanks. I'm just confused because I do not see the advantages of declaring a specialization template.

Fruny

203

Fruny

Jul 15, 2004 at 11:17 PM

Quote:
Original post by Xtreme
thanks. I'm just confused because I do not see the advantages of declaring a specialization template.

Well, for one thing, you can have partial specializations...

Xtreme

Author

Xtreme

Jul 15, 2004 at 11:23 PM

Quote:
Original post by Fruny
Quote:
Original post by Xtreme
thanks. I'm just confused because I do not see the advantages of declaring a specialization template.

Well, for one thing, you can have partial specializations...

and what exactly are partial specializations?

Fruny

203

Fruny

Jul 15, 2004 at 11:40 PM

Quote:
Original post by Xtreme
[nd what exactly are partial specializations?

template<class T, class U> void Foo(T,   U) {}template<class U>          void Foo(int, U) {}

Warning: VC6 doesn't support partial template specialization.

fallenang3l

Jul 15, 2004 at 11:42 PM

Quote:
Original post by Fruny
template<class T, class U> void Foo(T,   U) {}template<class U>          void Foo(int, U) {}
Warning: VC6 doesn't support partial template specialization.

The C++ standard doesn't support partial specializations for functions.

Teknofreek

Jul 15, 2004 at 11:59 PM

Quote:
Original post by Xtreme
thanks. I'm just confused because I do not see the advantages of declaring a specialization template.

One advantage appears to be that if the caller specifies the template parameter explicitly then the overload will not get called. Consider the following:

template< typename T > T Sum( T a, T b ){    return a + b;}template<> float Sum( float a, float b ){    return a + b;}float Sum( float a, float b ){    return a + b;}int main( int argc, char* argv[] ){    int   i = Sum( 1, 2 );    float x = Sum( 1.0f, 2.0f );    float y = Sum<float>( 1.0f, 2.0f );    float z = Sum<float>( 1.0f, 2 );}

I compiled this with VS2003 and found that the last two calls will end up in the template specialization, not the overloaded version. Furthermore, if you comment out the specialized template function, the calls will end up in the vanilla template function, not the overload.

So, in the end, I think it's safer to do the template specialization instead of an overload because it encompasses all calling cases.

-John

Xtreme

Author

Xtreme

Jul 16, 2004 at 12:22 AM

i get the idea. Thanks for enlightening me on specialization and partial templates!

Qw3r7yU10p!

Jul 16, 2004 at 3:27 AM

Also, practically, with most compilers, the overload doesn't have to be defined in the header whereas all templates do.

I know its not a big deal but it is a difference.

What about this, gurus?

template<typename T>T Sum(T a, T b) {    return a + b;}float Sum(float, float);//defined in a .cpp somewheretemplate<> float Sum(float a, float b) {    return Sum(a, b);//calls non-templated version?}

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