CODEWITHSUNDEEP
c++11metaprogrammingboost-mpl
MB.
MB.

Reputation: 1411

Is it possible to have a branch of code conditionally be compiled in C++?

Title says it all really. Sample code which illustrates the spirit of the affaire:

if( std::is_constructible<T, unsigned long>::value )
{
    unsigned long identity = collection.rbegin()->first + 1;
    std::shared_ptr<T> newObject(new T(identity));

    collection.insert( identity , newObject );
    return true;
}
else
{
    return false;
}

Upvotes: 1

Views: 231

Answers (2)

Svalorzen
Svalorzen

Reputation: 5618

Since the if statement can be determined at compile time, I'd expect the compiler to be smart and optimize it directly, just as if you had something like

if ( true ) {
    // Some code
}
else {
    // Anything here any decent compiler will ignore.
}

Another option is to wrap the behaviour you want in a function, and use std::enable_if:

template <typename T, typename = typename enable_if<is_constructible<T, int>::value>::type>
bool foo() {
    return true;   
}

template <typename T, typename = typename enable_if<!is_constructible<T, int>::value>::type>
bool foo() {
    return false;
}

// ...

return foo<T>();

Example: http://ideone.com/sgNVr5

Yet another option is to specialize on the boolean value:

template <bool b> bool foo();

template <>
bool foo<true>(){
    return true;
}

template <>
bool foo<false>() {
    return false;
}

Upvotes: 2

T.C.
T.C.

Reputation: 137425

Tag dispatch.

template<class T>
bool foo_impl(std::true_type){
    unsigned long identity = collection.rbegin()->first + 1;
    std::shared_ptr<T> newObject(new T(identity));
    collection.insert( identity , newObject );
    return true;
}

template<class T>
bool foo_impl(std::false_type){
    return false;
}

template<class T>
bool foo(){
    return foo_impl<T>(std::is_constructible<T, unsigned long>());
}

Upvotes: 3

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