Identify if the number is complex C++

Question

How can I identify if the number is complex in C++?

Is there any built in function like this?:

isComplex(1)->false

Answer 1

C++ is a strongly typed language, and the literal 1 is always an int.

The determination you ask about might be relevant when converting text... isComplex("1"), for that you can attempt streaming:

std::istringstream iss(some_text);
std::complex<double> my_complex;
char c;
if (iss >> my_complex &&  // conversion possible...
    !(iss >> c))          // no unconverted non-whitespace characters afterwards
    ...use my_complex...
else
    throw std::runtime_error("input was not a valid complex number");

---

Separately, if you're inside a template and not sure whether a type parameter is std::complex, you can test with e.g. std::is_same<T, std::is_complex<double>>::value, for example:

#include <iostream>
#include <complex>
#include <type_traits>

using namespace std;

double get_real(double n) { return n; }

double get_real(const std::complex<double>& n) { return n.real(); }

template <typename T>
std::complex<double> f(T n)
{
    if (std::is_same<T, std::complex<double>>::value)
        return n * std::complex<double>{1, -1} + get_real(n);
    else
        return -n;
}


int main()
{
    std::cout << f(std::complex<double>{10, 10}) << '\n';
    std::cout << f(10.0) << '\n';
}

Output:

(30,0)
(-10,0)

See the code here.

For more complicated functions, you may want to create separate overloads for e.g. double and std::complex<double>, and/or float and std::complex<float>, long double etc..