Function template parameter compilation error [duplicate] - c++

Question

This question already has answers here:
Where and why do I have to put the "template" and "typename" keywords?
(8 answers)
Closed 9 years ago.
I'm trying to do something like this:
#include <iostream>
#include <array>
using namespace std;
template <size_t A>
class Test {
public:
typedef array<int, A> TestType;
};
template <size_t A>
void foo(Test<A>::TestType t) {
cout << "test\n";
}
int main() {
Test<5>::TestType q;
foo(q);
return 0;
}
but foo doesn't compile. In gcc I get
prog.cpp:12:19: error: variable or field ‘foo’ declared void
void foo(Test<A>::TestType t) {
^
prog.cpp:12:28: error: expected ‘)’ before ‘t’
void foo(Test<A>::TestType t) {
while in Visual Studio 2010 I get
error C2975: 'Test' : invalid template argument for 'A', expected compile-time constant expression
I don't understand what I am doing wrong, as A is a compile-time constant. What should I change?

Answer 1

If you would add typename like this:
template <size_t A>
void foo(typename Test<A>::TestType t) {
cout << "test\n";
}
the only effect is a better error message. The problem is that you still can not deduce template parameters like that.
When you declare q
Test<5>::TestType q;
q's type is std::array<int,5> and the compiler doesn't know how this type was connected to Test<5>. In the call to foo(q) it would need a much deeper analysis of the code which is not standardized to figure out that there is only one possible match for A. You need to call
foo<5>(q);
to specify it explicitly or change the definition of foo:
template <size_t A>
void foo(std::array<int,A> t) {
cout << "test\n";
}