Why does “&& true” added to a constraint make a function template a better overload?

26

Consider the following two overloads of a function template foo:

template <typename T>
void foo(T) requires std::integral<T> {
    std::cout << "foo requires integral\n";
}

template <typename T>
int foo(T) requires std::integral<T> && true {
    std::cout << "foo requires integral and true\n";
    return 0;
}

Note the difference between the two constraints: the second has an extra && true.

Intuitively speaking, true is redundant in a conjunction (since X && true is just X). However, it looks like this makes a semantic difference, as foo(42) would call the second overload.

Why is this the case? Specifically, why is the second function template a better overload?

c++ language-lawyer c++20 c++-concepts

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edited Apr 8 at 15:06

T.C.

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asked Apr 8 at 14:38

MeowmereMeowmere

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7

• 1
  It's "more specialized". There are more requirements, so it's narrower, so it's "better". – NathanOliver Apr 8 at 14:39
• 6
  So, would adding && !false make it even better? – Adrian Mole Apr 8 at 14:40
• 2
  AFAIK, the standard doesn't really care what it does, just that there are more requirements. – NathanOliver Apr 8 at 14:41
• 1
  @AdrianMole Looks like the answer is no: godbolt.org/z/j6hG98WEv – Meowmere Apr 8 at 14:50
• 3
  It falls out of the rules. There's not really motivation to add a special case for this. – T.C. Apr 8 at 14:56

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