Can a template function compare the two typenames?

Question

Possible Duplicate:
Program to implement the is_same_type type trait in c++

I want my template function to do something differently based on whether the two typenames are equal or not:

template <typename T1, typename T2> f()
{
  if (T1==T2) ...;
  else ...;
}

I know "if(T1==T2)" is not gonna working, but, is there a way to do it?

@NPE this is not a duplicate IMHO. Properly applying `is_same` is different from just implementing it. What if he will be doing `T2 t; T1 *u = &t;` in the true branch? Compilation will fail if he uses above `if` for types `char` and `int` respectively. — Johannes Schaub - litb, Nov 30 '12 at 23:16
@JohannesSchaub-litb: You're right. Can't undo the vote, but will cease casting close votes for the rest of the day. :) — NPE, Nov 30 '12 at 23:19

CygnusX1 · Answer 1 · 2012-11-30T23:00:19.260

18

You can check the boost::is_same or std::is_same in C++11.

So, it would be something like this:

template <typename T1, typename T2> f()
{
  if (boost::is_same<T1,T2>::value) ...;
  else ...;
}

edited Nov 30 '12 at 23:00

answered Nov 30 '12 at 22:54

CygnusX1

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Is there a difference between those two that you decided to use `boost` instead? – David G Nov 30 '12 at 23:48
Not really. I am simply more familiar with boost, as it had worked long before C++11. – CygnusX1 Dec 01 '12 at 08:16

score 6 · Answer 2 · answered Nov 30 '12 at 23:52

6

#include <type_traits>

template <typename A, typename B> void f() {

    if ( std::is_same<A, B>::value ) {

        //

    }

}

_{std::is_same returns a typedef of a boolean (true, false) depending on the equlity of the types A and B}

answered Nov 30 '12 at 23:52

template boy

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score 4 · Answer 3 · answered Nov 30 '12 at 22:55

4

If the types can be inferred and are not being explicitly passed, you could make two separate functions:

template<typename SameType>
void f(SameType x, SameType y)
{
    // ...
}

template<typename T1, typename T2>
void f(T1 x, T2 y)
{
    // ...
}

answered Nov 30 '12 at 22:55

jamesdlin

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score 1 · Answer 4 · answered Nov 30 '12 at 22:52

1

Specialize the template

template<typename T1, typename T2>
void f()
{
  //The else part goes here
}

template<typename T>
void f()
{
  //The if part goes here
}

answered Nov 30 '12 at 22:52

StoryTeller - Unslander Monica

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downvoter wants to elaborate on what's wrong? – StoryTeller - Unslander Monica Nov 30 '12 at 22:55
For the second one, don't you mean just template? The version you just posted does not compile. – SplinterOfChaos Nov 30 '12 at 22:55
@SplinterOfChaos, I do... shuold have proof read it one more time – StoryTeller - Unslander Monica Nov 30 '12 at 22:56
It is not a specialization, it is overloading. – demi Nov 30 '12 at 23:34

score 0 · Answer 5 · answered Nov 30 '12 at 22:55

0

You can compare the typeid of T1 and T2

#include <typeinfo>

template <typename T1, typename T2> void f()
{
    bool a;
    if (typeid(T1) == typeid(T2))
        a = true;
    else
        a = false;
}

answered Nov 30 '12 at 22:55

Olaf Dietsche

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1

While it is also a solution, note that typeid is resolved at run time. It introduces performance overhead and both branches will be present in the function. I believe the compile-time evaluation of the condition is preferred. – CygnusX1 Nov 30 '12 at 22:58
2

And it will not work for `f` or `f` since the typeid for all of them is equal. – Johannes Schaub - litb Nov 30 '12 at 23:01

Can a template function compare the two typenames?

5 Answers5

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