Using 'auto' type deduction - how to find out what type the compiler deduced?

Question

How can I find out what type the compiler deduced when using the auto keyword?

Example 1: Simpler

auto tickTime = 0.001;

W

Answer 1

@jonathan-oconnor points out that you can use the [[deprecated]] attribute introduced in C++14 to produce a very clean solution:

template<typename T>
[[deprecated]] constexpr void printType(T const&) {}

Unfortunately though, the diagnostic emitted by MSVC doesn't mention the type. (example)

Answer 2

I like to use idea from Effective Modern C++ which uses non-implemented template; the type is output with compiler error:

 template<typename T> struct TD;

Now for auto variable var, after its definition add:

 TD<decltype(var)> td;

And watch error message for your compiler, it will contain type of var.

Answer 3

typeid can be used to get the type of variable most of the time. It is compiler dependent and I've seen it give strange results. g++ has RTTI on by default, not sure on the Windows side.

#include <iostream>
#include <typeinfo>
#include <stdint.h>
#include <chrono>
#include <ctime>

typedef std::ratio<1, 1> sec;
int main()
{
    auto tickTime = .001;
    std::chrono::duration<double, sec > timePerTick2{0.001};
    auto nextTickTime = std::chrono::high_resolution_clock::now() + timePerTick2;
    std::cout << typeid(tickTime).name() << std::endl;
    std::cout << typeid(nextTickTime).name() << std::endl;

    return 0;
}

./a.out | c++filt

double
std::__1::chrono::time_point<std::__1::chrono::steady_clock, std::__1::chrono::duration<long long, std::__1::ratio<1l, 1000000000l> > >

Answer 4

As a side note, to effectively print out the value in nextTickTime you should explicitly convert to a suitable std::chrono::duration and output the result of duration::count.

using std::chrono::duration_cast;
using std::chrono::seconds;

auto baseTime = ...;
std::cout << std::setprecision(12) << duration_cast<seconds>(nextTickTime - baseTime).count()
    << std::endl; // time in seconds

Answer 5

A low tech solution is hover the mouse over nextTickTime which in some GUIs gives the type else set a . after nextTickTime in the cout and select a reasonable looking value or function.

In general if You know what type You get use auto if you don't know it don't use it. Which is a bit counter intuitive.

So if you know its a interator just use auto to reduce the incantations, if the result is some unknown type you have to find out what it is before using auto.

See also Herb, Andrei and Scott discussing auto

Answer 6

The type deduced by the compiler is in the error message:

/usr/include/c++/4.8.2/ostream:602:5: error:   initializing argument 1 of ‘std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>;
 _Tp = std::chrono::time_point<std::chrono::_V2::system_clock, std::chrono::duration<double, std::ratio<1l, 1000000000l> > >]’
  ^^   <-------- the long type name --------------------------------------------------------------------------------------->

It's a complicated type name but it is there in the error message.