CODEWITHSUNDEEP
c++templatestemplate-specialization
asv
asv

Reputation: 181

strange behavior of templated type

#include <iostream>

template<typename T, int = 0>
struct test {
    typedef char type[3];
};

template<int N>
struct test<long *, N> {
    typedef char type[7];
};

int main()
{
    std::cout << sizeof( test<int*>::type ) << std::endl; // 3
    std::cout << sizeof( test<long*>::type ) << std::endl; // 7
    return 0;
}

I expected sizeof( test<long*>::type ) == 3. Why it is 7?

Upvotes: 2

Views: 83

Answers (2)

Mark B
Mark B

Reputation: 96311

You specialized your template so that if the type argument is long* then the size is changed. I'm fairly sure you can't mix the test<int*> and test<123> forms that you want to use (what would the meaning be of a parameter that could be either a type OR a value?). The best you can do is something like:

#include <iostream>

template<typename T, int = 0>
struct test {
    typedef char type[3];
};

template<int N>
struct test<int, N> {
    typedef char type[7];
};

int main()
{
    std::cout << sizeof( test<int*>::type ) << std::endl; // 3
    std::cout << sizeof( test<long*>::type ) << std::endl; // 3
    std::cout << sizeof( test<int, 123>::type ) << std::endl; // 7
    return 0;
}

Maybe if you tell us the real problem you're trying to solve that would help.

Upvotes: 1

Dave S
Dave S

Reputation: 21123

What you did is you specialized the template, saying that whenever the type argument is long*, you should use the second form of the template. For other arguments, it will use the initial form of the template (with size 3).

Upvotes: 3

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