Re: template function instantiation
George wrote:
1.
Instantiation of 'f' depends on its _use_ or explicit instantiation
and has NOTHING to do with the actual meaning of the 'g' symbol.
What is the differences between use and explicit instantiation? Could
you show some pseudo code please?
The use causes implicit instantiation. For example, calling it would
be an example of 'use'. Assigning it to a pointer to function would
be another example of 'use'. Explicit instantiation is a declaration
that begins with 'template' without the angle brackets after it.
int main()
{
f(42); // a call
void (*p)(double) = f; // initialising a pointer to function
}
template void f<char>(char); // explicit instantiation.
2.
The existence of 'g' declaration before the 'f' template is needed
to allow proper parsing of the body of the 'f' function. However,
for all we know it could have been an object (and not a function).
Or it could have been a type. Or a macro that expands into someth
acceptable by the compiler. However, it has NOTHING to do with
the instantiation of 'f'. Instantiation may not even exist, yet
the declaration of 'g' is required -- to allow the definition of
'f' to be compiled.
I think MSVC does not parse f to find whether there is g or what is g
before instantiation, agree?
It would seem so. I don't know for sure since I didn't write MSVC.
V
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