I am using GCC 6.3 and to my surprise the following code fragment did compile.
auto foo(auto x) { return 2.0 * x; }
...
foo(5);
AFAIK it is GCC extension. Compare to the following:
template <typename T, typename R>
R foo(T x) { return 2.0 * x; }
Besides return type deduction are the above declaration equivalent?
Using the same GCC (6.3) with the -Wpedantic flag will generate the following warning:
warning: ISO C++ forbids use of 'auto' in parameter declaration [-Wpedantic]
auto foo(auto x)
^~~~
While compiling this in newer versions of GCC, even without -Wpedantic, will generate this warning, reminding you about the -fconcepts flag:
warning: use of 'auto' in parameter declaration only available with -fconcepts
auto foo(auto x)
^~~~
Compiler returned: 0
And indeed, concepts make this:
void foo(auto x)
{
auto y = 2.0*x;
}
equivalent to this:
template<class T>
void foo(T x)
{
auto y = 2.0*x;
}
See here: "If any of the function parameters uses a placeholder (either auto or a constrained type), the function declaration is instead an abbreviated function template declaration: [...] (concepts TS)" -- emphasis mine.
