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

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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> > >

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最佳答案

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.

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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

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Here's a typeid version that uses boost::core::demangle to get the type name at runtime.

#include <string>
#include <iostream>
#include <typeinfo>
#include <vector>
using namespace std::literals;


#include <boost/core/demangle.hpp>


template<typename T>
std::string type_str(){ return boost::core::demangle(typeid(T).name()); }


auto main() -> int{
auto make_vector = [](auto head, auto ... tail) -> std::vector<decltype(head)>{
return {head, tail...};
};


auto i = 1;
auto f = 1.f;
auto d = 1.0;
auto s = "1.0"s;
auto v = make_vector(1, 2, 3, 4, 5);


std::cout
<< "typeof(i) = " << type_str<decltype(i)>() << '\n'
<< "typeof(f) = " << type_str<decltype(f)>() << '\n'
<< "typeof(d) = " << type_str<decltype(d)>() << '\n'
<< "typeof(s) = " << type_str<decltype(s)>() << '\n'
<< "typeof(v) = " << type_str<decltype(v)>() << '\n'
<< std::endl;
}

Which prints this on my system:

typeof(i) = int
typeof(f) = float
typeof(d) = double
typeof(s) = std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >
typeof(v) = std::vector<int, std::allocator<int> >

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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.

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This SO answer gives a nice function for printing out the name of a type (actually a couple of implementations).

Additionally this free, open-source, header-only library gives a nice way to print out the value and type of chrono::durations.

Putting these two utilities together:

#include "chrono_io.h"
#include "type_name.h"
#include <iomanip>
#include <iostream>


int
main()
{
using namespace date;
typedef std::ratio<1, 1> sec;
std::chrono::duration<double, sec > timePerTick2{0.001};
auto nextTickTime = std::chrono::high_resolution_clock::now() + timePerTick2;
std::cout << type_name<decltype(nextTickTime)>() << '\n';
std::cout << std::setprecision(12) << nextTickTime.time_since_epoch() << '\n';
}

This output for me:

std::__1::chrono::time_point<std::__1::chrono::steady_clock, std::__1::chrono::duration<double, std::__1::ratio<1, 1000000000> > >
4.8530542088e+14ns

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As Daniel Jour said, read the error message:

... _Tp = std::chrono::time_point<
std::chrono::_V2::system_clock,
std::chrono::duration<
double, std::ratio<1l, 1000000000l> > > ...

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A lo-fi trick that doesn't require any prior helper definitions is:

typename decltype(nextTickTime)::_

The compiler will complain that _ isn't a member of whatever type nextTickTime is.

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Here is a way to force a compile error, which shows the type of tickTime:

struct {} baD = tickTime;

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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.

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@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)