Default template parameter based on type of other template parameter

user_1893075 0
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I have a templated function as below:

template<typename T>
std::vector<RT> myFunc(std::vector<T> inVec){
    ...
    std::vector<RT> outVec;
    return outVec

I need the return type RT to be deduced from T with the logic that if T is std::complex then RT will be same as T but if not then RT will be std::complex<T>.

I tried defining a type trait to check if T is std::complex and have a default template parameter based on that as follows:

template<typename T>
struct is_complex : std::false_type {};

template<typename T> 
struct is_complex<std::complex<T>> : std::true_type {};

template<typename T, typename RT = typename is_complex<T>::value ? T : std::complex<T>>
std::vector<RT> myFunc(std::vector<T> inVec){
    ...
    std::vector<RT> outVec;
    return outVec

but this doesn't compile giving me the following error:

error: expected '>' before '?' token template<typename T, typename RT = typename is_complex<T>::value ? T : std::complex<T>>

Is this not legal C++ code?

template-meta-programming templates c++
1 Answers

Is this not legal C++ code?

Yes it's not valid syntax.

You can use std::conditional.

template<typename T, typename RT = std::conditional_t<is_complex<T>::value, T, std::complex<T>>>
std::vector<RT> myFunc(std::vector<T> inVec){
    std::vector<RT> outVec;
    return outVec;
}

Or you can define the return type trait directly.

template <typename T>
struct rt {
    using type = std::complex<T>;
};

template <typename T>
struct rt<std::complex<T>> {
    using type = std::complex<T>;
};

template <typename T>
using rt_t = typename rt<T>::type;

template<typename T, typename RT = rt_t<T>>
std::vector<RT> myFunc(std::vector<T> inVec){
    std::vector<RT> outVec;
    return outVec;
}
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