🚀 go-pugleaf

Paul Long (plong@perf.com) wrote:

: Is an implementation required to generate a diagnostic for this
: declaration?
: 
:    static int i = 1 ? 3 : (1, 2);

It seems quite clear that it must generate a diagnostic, as you say.

: A constraint in section 6.4 says "Constant expressions shall not
: contain ...  comma operators..." However, a 6.4 Semantics passage
: says "An implementation may accept other forms of constant
: expressions."
: 
: This rather large Semantics loophole does not remove the requirement
: to generate a diagnostic for the constraint violation though, right?
: IOW, an implementation may "accept" a comma operator in a constant
: expression, but it must also generate a diagnostic.

An implementation would be free to generate a diagnostic and then
accept a comma operator even without this `loophole'.  Its purpose is
presumably to allow an implementation to accept, without a diagnostic,
constant expressions that satisfy the restrictions in the Constraints
section but do not satisfy the restrictions in the Semantics section.
For example, it allows an implementation to accept
`(size_t)&((struct s *)0)->m' (which might be generated by the
offsetof macro) as a constant expression.

But does anyone know what the reason for disallowing comma operators
was?  I can see an argument that they are unlikely to be useful in
constant expressions (though I can find a more or less plausible
example of why one _might_ want to use them), but banning them for
that reason isn't in the spirit of C.

[Quibble:  The constraint continues `... except when they are
contained within the operand of a sizeof operator'.  Does it follow
that the expression `1,2' is a valid constant expression in
`sizeof(int[1,2])'?  Or should `... within the constant expression' be
added at the end of the constraint?]

-- 
John Rickard