constexpr question

[Andy Gibbs]

On Thursday, June 09, 2011 5:08 PM, Jonathan Wakely wrote:

>>> The question I have is: why doesn't (or can't) the second case work? In
>>> function 'test2' the compiler believes 'a' is a constant expression and
>>> the
>
> I don't think that's how it works.
>
> In test2 the expression test1(a) is just a normal expression, calling
> a function that returns an int. 'a' is not a constant expression in
> the body of test2.  In the right context it can be substituted with a
> constant expression, but is not actually a constant expression.
>
> In the definition of the constexpr variable 'two' the expression
> test(2) is a call to a constexpr function, so "function invocation
> substitution" takes place. That substitutes the constant 2 in place of
> 'a' in the body of test2, producing test1(2), which is also a call to
> a constexpr function, so function invocation substitution occurs
> again, substituting 2 for 'a' in the body of test1, which produces
> 'return 2' which is a constant expression.
>
> Similarly, in test3 'a' is not a constant expression (it's just
> something which could be substituted for one in the right context) and
> so it can't be used in a way which *requires* a constant expression,
> such as a template parameter.
>
> Does that help?

Hmm, I hadn't thought about it that way round.  You've given me an idea
though, so I'm going to go off and try something, to see if I can "trick"
the compiler into doing what I want!!

Cheers
Andy