What is std::decay and when it should be used?

C++C++11Standard Library

C++ Problem Overview


What are the reasons for the existence of std::decay? In what situations is std::decay useful?

C++ Solutions


Solution 1 - C++

<joke>It's obviously used to decay radioactive std::atomic types into non-radioactive ones.</joke>

N2609 is the paper that proposed std::decay. The paper explains:

> Simply put, decay<T>::type is the identity type-transformation except > if T is an array type or a reference to a function type. In those > cases the decay<T>::type yields a pointer or a pointer to a function, > respectively.

The motivating example is C++03 std::make_pair:

template <class T1, class T2> 
inline pair<T1,T2> make_pair(T1 x, T2 y)
{ 
    return pair<T1,T2>(x, y); 
}

which accepted its parameters by value to make string literals work:

std::pair<std::string, int> p = make_pair("foo", 0);

If it accepted its parameters by reference, then T1 will be deduced as an array type, and then constructing a pair<T1, T2> will be ill-formed.

But obviously this leads to significant inefficiencies. Hence the need for decay, to apply the set of transformations that occurs when pass-by-value occurs, allowing you to get the efficiency of taking the parameters by reference, but still get the type transformations needed for your code to work with string literals, array types, function types and the like:

template <class T1, class T2> 
inline pair< typename decay<T1>::type, typename decay<T2>::type > 
make_pair(T1&& x, T2&& y)
{ 
    return pair< typename decay<T1>::type, 
                 typename decay<T2>::type >(std::forward<T1>(x), 
                                            std::forward<T2>(y)); 
}

Note: this is not the actual C++11 make_pair implementation - the C++11 make_pair also unwraps std::reference_wrappers.

Solution 2 - C++

When dealing with template functions that take parameters of a template type, you often have universal parameters. Universal parameters are almost always references of one sort or another. They're also const-volatile qualified. As such, most type traits don't work on them as you'd expect:

template<class T>
void func(T&& param) {
    if (std::is_same<T,int>::value) 
        std::cout << "param is an int\n";
    else 
        std::cout << "param is not an int\n";
}

int main() {
    int three = 3;
    func(three);  //prints "param is not an int"!!!!
}

http://coliru.stacked-crooked.com/a/24476e60bd906bed

The solution here is to use std::decay:

template<class T>
void func(T&& param) {
    if (std::is_same<typename std::decay<T>::type,int>::value) 
        std::cout << "param is an int\n";
    else 
        std::cout << "param is not an int\n";
}

http://coliru.stacked-crooked.com/a/8cbd0119a28a18bd

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionEric Javier Hernandez SauraView Question on Stackoverflow
Solution 1 - C++T.C.View Answer on Stackoverflow
Solution 2 - C++Mooing DuckView Answer on Stackoverflow