CODEWITHSUNDEEP
c++templates
Nikos Athanasiou
Nikos Athanasiou

Reputation: 31499

Short circuit template instantiation

My code has the following layout

// A exists only as a true specialization
template<int Boolean>
struct A;

template<>
struct A<1> {};

// B exists only as a false specialization
template<int Boolean>
struct B;

template<>
struct B<0> {};

// then I use it like so
cond ? A<cond> : B<cond>; // actually I want to access an enumeration member value

Doesn't short circuiting apply to the ternary operator? Why do I get a compiler error?

Upvotes: 1

Views: 258

Answers (2)

Marco A.
Marco A.

Reputation: 43662

The ternary operator requires you to know the types used in the expression in order to form a common type.

In your case you might use something like (C++11, see std::conditional):

#include <iostream>
#include <typeinfo>
using namespace std;

// A exists only as a true specialization
template<int Boolean>
struct A;

template<>
struct A<1> {
    void func() { cout << "A"; }
    };

// B exists only as a false specialization
template<int Boolean>
struct B;

template<>
struct B<0> {
    void func() { cout << "B"; }
};


int main() {

    const bool cond = true;
    // std::conditional can provide a compile-time ternary-like evaluation
    typedef std::conditional<cond ,A<cond>,B<cond>> myType;

    myType::type Obj;
    Obj.func();


    return 0;
}

http://ideone.com/0Yrcsp

Upvotes: 1

Sean
Sean

Reputation: 62472

The ternary operator requires the true and false expressions to either be the same type, or be collapsable to a common type. Your A and B classes have no common type and therefore the expression cannot be resolved to a common type, giving an error.

Upvotes: 1

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