Retrieving the type of auto in C++11 without executing the program

Question

I have some C++11 code using the auto inferred type that I have to convert to C++98. How would I go about converting the code, substituting in the actual type f

Answer 1

You could use typeid and std::type_info::name();

#include 
#include 
#include 

int
main()
{
  using namespace std::literals::complex_literals;

  auto x = 3.1415F;
  std::cout << typeid(x).name() << '\n';
  auto z = 1.0 + 1.0i;
  std::cout << typeid(z).name() << '\n';
}

$ /home/ed/bin_concepts/bin/g++ -std=c++14 typeid.cpp 

$ ./a.out
f
St7complexIdE

The names aren't beautiful but you can at least translate them. These names are got from g++. The name is compiler dependent. There is some movement to standardize a pretty_name(). Here is a non-standard way to unmangle the names.