Does template specialization work with alias templates?

Question

Does template specialization work with alias templates? Specifically, the following code throws an unexpected error:

$ cat test01.h 
#ifndef TEST01_H
#define TEST01_H

#include <iostream>
#include <typeinfo>
#include <cstdlib>

template <template <typename> class Ops>
double add1(double const & x) {
    std::cerr << "Undefined for type: " << typeid(Ops <double>).name() << std::endl;
    exit(EXIT_FAILURE);
}

template <typename Real>
struct MyOps {
    static Real add(Real const & x,Real const & y) {
        return x+y;
    }
};

template <typename Real> using MyOpsAlias = MyOps <Real>;

#endif

In addition,

$ cat test01.cpp 
#include "test01.h"

template <>
double add1 <MyOps> (double const & x) {
    return MyOps <double>::add(x,1.);
}

int main() {
    std::cout << add1 <MyOps> (2.) << std::endl;
    std::cout << add1 <MyOpsAlias> (2.) << std::endl;
}

After running this code, I receive

$ ./test01 
3
Undefined for type: 5MyOpsIdE

I expected that both answers should return 3 since MyOpsAlias should just be an alias template of MyOps. In case it matters, I'm using GCC 4.7.3.

Perhaps you might be interested in [my recent question](http://stackoverflow.com/questions/22892052/strange-behaviour-of-is-same-template-on-template-aliases). — Constructor, Apr 09 '14 at 08:35
Interesting. That means that gcc typeid's implementation doesn't take care of template aliases. The standard doesn't say how typeid should work (implementation defined). — BЈовић, Apr 09 '14 at 11:11
@BЈовић It is not a bug, it is a standard-compliant behavior. See the reference above. — Constructor, Apr 09 '14 at 16:08
@Constructor Thanks. in that case this question is a duplicate of that question — BЈовић, Apr 10 '14 at 07:06
@Constructor I think you're correct. If you copy that comment into an answer, I'll accept it and close this out. — wyer33, Apr 25 '14 at 16:37
@wyer33 [Done.](http://stackoverflow.com/a/23298947/3043539) — Constructor, Apr 25 '14 at 16:57

score 1 · Accepted Answer · edited May 23 '17 at 10:32

1

As Johannes Schaub - litb explains in his answer to my recent question:

A alias template is not a template alias (despite intended by some to be).

So it is not a bug, it is a standard-compliant behavior.

See also paragraph [temp.alias] 14.5.7/2 of the standard:

[ Note: An alias template name is never deduced. — end note ]

edited May 23 '17 at 10:32

Community

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answered Apr 25 '14 at 16:56

Constructor

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Jarod42 · Answer 2 · 2014-04-09T09:52:19.053

0

You may do the following:

namespace detail
{

template <typename T> struct Adder
{
    double operator() (double) const {
        std::cerr << "Undefined for type: " << typeid(T).name() << std::endl;
        exit(EXIT_FAILURE);
    }
};

template <typename T> struct Adder<MyOps<T>>
{
    double operator() (double x) const {
        return MyOps<T>::add(x, 1.);
    }
};

}

template <template <typename> class T>
double add1(double const & x) {
    return detail::Adder<T<double>>()(x);
}

edited Apr 09 '14 at 09:52

answered Apr 09 '14 at 07:08

Jarod42

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There are a few errors in your code: `template – Constructor Apr 09 '14 at 08:46

Does template specialization work with alias templates?

2 Answers2