Templated class: Check if complex at compile time

Question

I have a class A templated with a Scalar which can be real- or complex-valued. It has a method realPart which is supposed to return the real part of the number. If Scalar is real-valued, it should just return the original Scalar, and .real() if it's of complex type.

When writing

#include <complex> #include <iostream> template<class Scalar> class A { public: A (const Scalar z): z_(z) { } Scalar realPart() { return z_.real(); } private: Scalar z_; }; int main() { A<std::complex<double>> z0((1.0, 2.0)); std::cout << z0.realPart() << std::endl; // check A<double> z1(1.0); std::cout << z1.realPart() << std::endl; // mööp }

the compiler will complain at z1.realPart() because double doesn't know .real().

How can I guard .real() at compile time?

@DavidHaim In case of double, just return z_, like you would expect mathematically. — Nico Schlömer
– Nico Schlömer, Commented Jun 9, 2015 at 15:53
so like @AkashPradhan said create a template specialization for it. — David Haim
– David Haim, Commented Jun 9, 2015 at 15:56
See stackoverflow template conditionals with type traits stackoverflow.com/questions/6251889/type-condition-in-template — Richard Chambers
– Richard Chambers, Commented Jun 9, 2015 at 15:58

David G · Accepted Answer · 2015-06-09 15:55:42Z

7

This can be done with a simple is_complex trait and SFINAE:

template<class T> struct is_complex : std::false_type {}; template<class T> struct is_complex<std::complex<T>> : std::true_type {}; template<class Scalar> class A { public: A(const Scalar z) : z_(z) { } template<class S = Scalar, std::enable_if_t<is_complex<S>{}>* = nullptr> Scalar realPart() { return z_.real(); } template<class S = Scalar, std::enable_if_t<!is_complex<S>{}>* = nullptr> Scalar realPart() { return z_; } private: Scalar z_; };

answered Jun 9, 2015 at 15:55

David G

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4 Comments

Eugene Over a year ago

Why is class S template argument needed? Could not you use is_complex<Scalar>?

David G Over a year ago

@Eugene The substitution failure has to occur within the immediate context of the template. The "outer" template of the entire class is not the immediate context of the member function template. So we use a dummy parameter to alias it.

Nico Schlömer Over a year ago

After adding #include <type_traits> and #include <complex>, I'm getting ‘std::enable_if_t’ has not been declared. This is with GCC 5.1.1. Is there something I forgot to add?

Richard Hodges Over a year ago

enable_if_t is c++14. it's equivalent to typename std::enable_if<T>::type - see en.cppreference.com/w/cpp/types/enable_if

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