从重载函数std :: real解析地址<float>

Question

std::vector<std::complex<float> > c;
std::vector<float> d;
std::transform(c.begin(), c.end(), d.begin(), std::real<float>);

Why couldn't the compiler resolve the address from the overloaded function real<float> ?

Which overloaded functions does the compiler mean?

Answer 1

Your library implementation has provided additional overloads for std::real<float> .

Why the overloads?

26.4.9 Additional overloads [cmplx.over]

• 1 The following function templates shall have additional overloads:

   arg norm conj proj imag real 

• 2 The additional overloads shall be sufficient to ensure:
  1. If the argument has type long double , then it is effectively cast to complex<long double> .
  2. Otherwise, if the argument has type double or an integer type, then it is effectively cast to complex<double> .
  3. Otherwise, if the argument has type float , then it is effectively cast to complex<float> .

[...]

Solutions to problem:

You could just use a range based for ...

for (auto v : c) d.push_back(real(v));

... or pack the call to real into a functor or another function ...

struct my_caller {
    template <typename T> T operator() (std::complex<T> const &c) {
        return real(c);
    }
};

... or use the member function ...

std::transform(c.begin(), c.end(), d.begin(), [](std::complex<T> const &c) { 
    return c.real();
});

---

IMPORTANT:

Note that you have to have enough space in the target when using transform :

std::vector<float> d (c.size());

or use a back inserter:

std::transform(c.begin(), c.end(), back_inserter(d), ...);

Otherwise you are iterating over undefined memory, yielding undefined behaviour.

Answer 2

§26.4.9 states that (amongst others), real shall have additional overloads, for arguments of type float, double and long double. It seems your libraray implementation made a template for these overloads, maybe like

template <typename T>
T real(T const& t)
{
  return std::real(std::complex<T>{t});
}

In addition to the solutions phresnel priovided, you could explicitly tell the compiler which kind of function pointer you mean:

std::transform(c.begin(), c.end(), d.begin(), (float(*)(std::complex<float> const&))std::real<float>);

The compiler then looks for a std::real that can be converted into a function pointer of the given type and will find the correct one.

I tell you this only for completeness - I consider this explicit cast ugly and would prefer the ranged based for or transform with a lambda.