C++ functions can have return type auto and it is not always obvious what the return type of a function is. The return type can be dependent on template parameters as well. In this article, we’ll go over finding the return type of a function for compile time use. I will also go over the tricks to get the compiler to tell us the type so it’s not a big guessing game.

Return type using decltype

Suppose we have the function foo() as follows:

int foo()
  return 123;

To get the return type of foo, we simply use the following:

using t = decltype(foo());

t now contains the return type.

We can use t as a type declaration as follows,

t i = foo();

Note, that decltype is all done at compile time and does not call the method decltype.

Getting the compiler to tell us type

The C++ compiler knows the type of each and every variable during compile time. However, it is not trivial to get the type of a variable from the compiler.

One of the tricks to get it to tell us the type involves producing a compiler error so that outputs the type in the compiler error.

template<typename T>
class error;

and using the above to produce the error during compilation.

using t = decltype(foo());

The output of compilation is following error that tells the type of t.

error: invalid use of incomplete type class error<int>

We thus know that the type of t is int.

Printing out the type

The first instinct is to print out the type and C++ does provide the function typeid and the name() function to convert it to string. However, it strips out things like const and references &.

The other trick is to use the __PRETTY_FUNCTION__ which prints out the contents of pretty version of a function at compile time. This also prints out the template type.

For example, using the function

template <class T>
auto type_name()
  return __PRETTY_FUNCTION__;

and calling the function in this manner,

using t = decltype(foo());
std::cout << type_name<t>() << std::endl;

the output in my system with gcc with ubuntu produces the following output:

auto type_name() [with T = int]

and so, we know the type of t is int.

Functions with Paramters

Suppose we change our function foo in the following way,

int foo(int i)
  return 123+i;

and our method of using just decltype(foo()) does not work since foo now has parameters. We can use decltype(foo(2)). However, if we don’t really want to specify that we are going to use 2, we can use std::declval to not have to use a parameter.

using t = decltype(foo(std::declval<int>()));

If we had two parameters for foo and looks like the following function,

int foo(int i, double j)
  return 123+i+j;

we can use the following:

using t = decltype(foo(std::declval<int>(), std::declval<int>()));


If we template foo, it just involves using decltype with how we would call foo.

template<typename T>
T foo(int i, double j)
  return 123+i+j;

The way to call the decltype is as follows.

using t = decltype(foo<double>(std::declval<int>(), std::declval<int>()));


Getting the return type of a function just involves using decltype and declval. It is also useful to use some tricks to get the compiler to tell us what the actual types are before we use them.