Ian Lance Taylor <[EMAIL PROTECTED]> writes:
> Alan Mackenzie <[EMAIL PROTECTED]> writes:
>
>> I'm thinking of things like
>>
>> foo (a < b, c > d);
>>
>> I think this is unambiguously a function call with 2 parameters, the
>> expressions "a < b" and "c > d". It cannot be be one with 1 parameter
>> beginning with the template invocation "a < b , c >". Or can it?
>
> No, it can't be, because a<b, c> is a type. The result would be
> foo(TYPE d), which can not be a function call. On the other hand, if
> there were a type before foo then this would be a function
> declaration. For example, this is valid C++ code:
>
> template <int a1, int a2> class a;
>
> int fn(int d, int e)
> {
> const int b = 1;
> const int c = 2;
> typedef int f;
> f foo (int, int);
> f foo (a < b, c > d);
> foo (e < b, c > d);
> }
>
> The line "f foo (a < b, c > d);" uses a template, the line "foo (e <
> b, c > d);" does not.
Oh, wait, I forgot about constructors. This is valid C++ code:
template <int a1, int a2> class a;
const int b = 1;
const int c = 2;
int d;
int e;
class foo
{
foo (a < b, c > d);
};
extern void quux (int, int);
void bar()
{
quux (e < b, c > d);
}
Inside the class foo, you've got a template, inside the function bar,
you've got a function call.
I think you're going to have a hard time with this.
Ian
