Does constexpr strictly require any sub-expression involved being constant expression?

Example:

int main(int argc, char**)
{
    constexpr int a = argc * 0;
    (void)a;
    constexpr int b = argc - argc;
    (void)b;
    return 0;
}

arg is not a constant expression, but the compiler is still able to compute the results of a and b in compile time (i.e. 0) in both cases.

g++ accepts the code above, while clang and MSVC14 reject it.

Does the standard allows the compiler being as smart as g++ with regard to constexpr?