Constexpr if with a non-bool condition

Clang is diagnosing under these paragraphs

[stmt.if] (emphasis mine)

2 If the if statement is of the form if constexpr, the value of the condition shall be a contextually converted constant expression of type bool; this form is called a constexpr if statement.

[expr.const]

4 A converted constant expression of type T is an expression, implicitly converted to type T, where the converted expression is a constant expression and the implicit conversion sequence contains only

• integral conversions other than narrowing conversions,

Now, when it comes to integral conversions, a conversion to bool is listed as an integral conversion. And it is narrowing, in the strictest sense of the word, since a bool cannot represent all the values of an int. So the diagnostic is not without grounds.

But I think it's also quite reasonable to consider the fact a conversion to bool is usually intended to check for "truthiness", and so the narrowing nature of it shouldn't matter. It looks like a minor bug in the standard¹, with GCC taking the common-sense route, and Clang adhering to the dry letter of the law in the strictest sense.

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_{¹ - And a proposal exists to change it.}

We say it, but it's hidden. "contextually converted constant expression of type bool" is a standard term-of-art that excludes narrowing conversions.

Clang is correct.