Re: Template instantiation conflict with function definition
Alessandro [AkiRoss] Re wrote:
Hello there!
While developing a policy-based application, I tried a code like this
one (which is real code).
I'm using g++ 4.3.3, which gives me these errors:
templ_error.cpp: In function ?int main()?:
templ_error.cpp:22: error: multiple parameters named ?base?
templ_error.cpp:23: error: request for member ?exec? in ?test?, which
is of non-class type ?Test<Numeric<3>, Numeric<5> > ()(Numeric<3>,
Numeric<5>)?
But I can't understand where is the problem: it seems that g++ read my
declaration of a variable as a function definition. Please tell me if
I'm doing it wrong: I can't see the error.
Here's the code of templ_error.cpp
#include <iostream>
using namespace std;
template <typename A, typename B>
struct Test {
A a;
B b;
Test(A a_, B b_): a(a_), b(b_) {}
void exec() { cout << "Executed!" << a() << "-" << b() <<"\n"; }
};
template <int X>
struct Numeric {
int base;
Numeric(int base_): base(base_) {}
int operator()() { return base + X; }
};
int main() {
int base = 1;
Test<Numeric<3>, Numeric<5> > test(Numeric<3>(base), Numeric<5>
(base));
The line above declares a function 'test' with two arguments, 'base' and
'base'. The first argument is of type 'Numeric<3>', the second is of
type 'Numeric<5>'. If you needed to construct an object 'test' of type
'Test<...>', then you need to move your parentheses a bit:
Test<Numeric<3>, Numeric<5> >
test( ( Numeric<3> ) base, ( Numeric<5> )base );
Notice that I put the types in parentheses making those C-style type
cast expressions.
test.exec();
This line in your original program tries to apply '.' to the function
type, and that's what the second error message is about. Pay attention
to what your compiler is telling you!
return 0;
}
ps. Also with g++ 4.1 gives errors: the only difference is that
"multiple parameters name" is not present as it has introduced in gcc
4.3.
V
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