clang is correct, note the HEAD revision of gcc accepts also accepts this code. This is a well-formed constexpr function, as long as there is value for the argument(s) that allows the function to be evaluated as a core constant expression. In your case 1
is such a value.
This is covered in the draft C++14 standard section 7.1.5
The constexpr specifier [dcl.constexpr] which tells us what is allowed in a constexpr function:
The definition of a constexpr function shall satisfy the following constraints:
it shall not be virtual (10.3);
its return type shall be a literal type;
each of its parameter types shall be a literal type;
its function-body shall be = delete, = default, or a compound-statement that does not contain
no restriction on throw
and it also says (emphasis mine):
For a non-template, non-defaulted constexpr function or a non-template, non-defaulted, non-inheriting
constexpr constructor, if no argument values exist such that an invocation of the function or constructor
could be an evaluated subexpression of a core constant expression (5.19), the program is ill-formed; no
diagnostic required.
and below this paragraph we have the following example, similar to yours:
constexpr int f(bool b)
{ return b ? throw 0 : 0; } // OK
constexpr int f() { return f(true); } // ill-formed, no diagnostic required
throw
is not allowed in a core constant expression, which is covered in section 5.19
[expr.const] paragraph 2
which says:
A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the
abstract machine (1.9), would evaluate one of the following expressions
and includes the following bullet:
- a throw-expression (15.1).
and so f
would not be usable in a core constant expression when n <= 0
.
Update
As TemplateRex points out, there are two gcc bugs reports for this:
TemplateRex also notes the fixes are not applied to to 5.3.0
and are only in trunk. No, work arounds are provided.
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