std::add_pointer_t> unexpected behavior

Asked Jul 05 '19 at 17:49

Active Jul 05 '19 at 17:55

Viewed 411 times

Why does the following not compile:

#include <iostream>
#include <cstring>

template <typename U>
void f_broken(std::add_pointer_t<std::remove_pointer_t<U>> p)
{
}

template <typename U>
void f_ok(U p)
{
}


int main() {
    int * ptr;

    std::cout << typeid(ptr).hash_code() << std::endl;
    std::cout << typeid(std::add_pointer_t<std::remove_pointer_t<decltype(ptr)>>).hash_code() << std::endl;

    // f_broken(ptr); Compiler error.
    f_ok(ptr) // Works.
};

The types are the same as expected but when used it is not behaving as expected.

The compiler error is:

<source>:21:17: error: no matching function for call to 'f_broken(int*&)'

     f_broken(ptr);

                 ^

<source>:5:6: note: candidate: 'template<class U> void f_broken(std::add_pointer_t<typename std::remove_pointer<_Tp>::type>)'

 void f_broken(std::add_pointer_t<std::remove_pointer_t<U>> p)

      ^~~~~~~~

<source>:5:6: note:   template argument deduction/substitution failed:

<source>:21:17: note:   couldn't deduce template parameter 'U'

     f_broken(ptr);

Running example: https://godbolt.org/z/UUYHvM

asked Jul 05 '19 at 17:49

Andreas Pasternak

1,250
10
18

See https://stackoverflow.com/q/25245453/1896169 – Justin Jul 05 '19 at 17:51
How could `U` be deduced for `f_broken`? – François Andrieux Jul 05 '19 at 17:52
You might call `f_broken(ptr)` or `f_broken(ptr)`. or to enforce that parameter is a pointer, you can simply do `template void f_ptr(T*p){/*..*/}`. – Jarod42 Jul 05 '19 at 17:59

std::add_pointer_t> unexpected behavior

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