Why are double parentheses needed with std::is_same

Question

I was trying to use std::is_same to verify the underlying type of a strongly typed enum and I noticed a strange situation where I needed to use double parentheses, but I don't understand why. I've reduced the example down to as follows:

#include <type_traits>
#include <cassert>
#include <stdint.h>

int main(int argc, char *argv[])
{
    assert((std::is_same<unsigned int,uint32_t>::value == true)); // OK
    assert((std::is_same<unsigned int,uint32_t>::value) == true); // OK
    //assert(std::is_same<unsigned int,uint32_t>::value == true); // Compile error
    static_assert(std::is_same<unsigned int,uint32_t>::value == true, "BAD"); // OK
    return 0;
}

Compile error:

isSameAssert.cpp:9:62: error: macro "assert" passed 2 arguments, but takes just 1
     assert(std::is_same<unsigned int,uint32_t>::value == true); // Compile error
                                                              ^
isSameAssert.cpp: In function ‘int main(int, char**)’:
isSameAssert.cpp:9:5: error: ‘assert’ was not declared in this scope
     assert(std::is_same<unsigned int,uint32_t>::value == true); // Compile error
     ^
make: *** [build/isSameAssert] Error 1

Could anyone explain this or point me to a reference that does?

That's because `assert()` is a macro. – πάντα ῥεῖ Apr 05 '16 at 05:58 — πάντα ῥεῖ, Apr 05 '16 at 05:58

score 4 · Accepted Answer · answered Apr 05 '16 at 06:09

Because assert is a macro, the expression assert(std::is_same<unsigned int,uint32_t>::value == true); seems to call assert with two parameters due to the comma between int and uint32_t so the compiler complains that assert takes only one parameter, but two were supplied.

Indeed, putting it again in parenthesis solves this issue.

Why are double parentheses needed with std::is_same

1 Answers1