CODEWITHSUNDEEP
c++c++11constexpr
fedino
fedino

Reputation: 933

constexpr function with no argument

I am getting confused around this snippet:

constexpr int f(bool b) {
  return b ? throw 0 : 0; }
constexpr int f() { return f(true); }

directly from the c++ draft. The point I am stucked with is why the standard defines as ill-formed the case of a constexpr function without arguments (stated in the same link). May anyone clarify?

Upvotes: 2

Views: 194

Answers (1)

davmac
davmac

Reputation: 20631

The key is "if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression". It's not about the function f() taking no arguments; it's about the fact that there's no set of arguments you could give it that would make it return a usable value - it always calls f(true), which throws an exception.

To re-iterate: a constexpr function without arguments can certainly be well-formed. But for the given example, it is not.

Also of note is "diagnostic not required". That means that a compiler is free to accept the construct anyway. Indeed, GCC compiles the example in your question without complaining.

Upvotes: 8

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