std::is_same_v<> produces wrong insight

[tocic]

The following code produces wrong insight:

#include <iostream>
#include <type_traits>

int main() {
    std::cout << std::is_same_v<void(int), void(const int)> << '\n';
    std::cout << std::is_same_v<void(int*), void(const int*)> << '\n';
}

||
v

#include <iostream>
#include <type_traits>

int main()
{
  std::operator<<(std::cout.operator<<(std::is_same_v), '\n');
  std::operator<<(std::cout.operator<<(std::is_same_v), '\n');
}

While this example is okay:

#include <iostream>
#include <type_traits>

int main() {
    std::cout << std::is_same<void(int), void(const int)>::value << '\n';
    std::cout << std::is_same<void(int*), void(const int*)>::value << '\n';
}

||
v

#include <iostream>
#include <type_traits>

int main()
{
  std::operator<<(std::cout.operator<<(std::integral_constant<bool, true>::value), '\n');
  std::operator<<(std::cout.operator<<(std::integral_constant<bool, false>::value), '\n');
}

It's taken from the 254th C++ Quiz question, btw.

[andreasfertig]

Hello @tocic,

thanks for reporting this! That is an interesting one. On my Mac it works. However, the web site runs Linux so I have to dig a little deeper to figure out what goes wrong there. Anyway, as it is now it is a bug.

Andreas