I have a class which overloads the function call operator with a template function, like so:
class Test
{
public:
template<class T>
void operator()(T t)
{
std::cout<<(&t)<<std::endl;
};
};
I'd like to call it with a reference argument, however when trying to do so, it passes the argument as a value instead. Here's my test setup:
template<class T>
void test(T t) {std::cout<<(&t)<<std::endl;}
int main(int argc,char *argv[])
{
Test t;
int i = 5;
std::cout<<(&i)<<std::endl;
t((int&)i); // Passes the argument as a value/copy?
test<int&>(i); // Passes the argument as a reference
while(true);
return 0;
}
The output is:
0110F738 -- Output of the address of 'i'
0110F664 -- Output of the address of the argument in the template overload
0110F738 -- Output of the address of the argument through 'test'
The template function 'test' is merely for validation.
The visual studio debugger confirms that it's using 'int' instead of 'int&' for the template overload:
test_function_call.exe!Test::operator()(int t) Line 9 C++
How can I force it to use a reference instead? Is there a way to specify the types using <> on a template function call operator?
That's because in your case the cv-qualifiers and the reference-ness of the parameter are discarded when performing template type deduction. Pass via a std::ref
wrapper instead
t(std::ref(i));
Simple example:
#include <iostream>
#include <functional>
template<typename T>
void f(T param)
{
++param;
}
int main()
{
int i = 0;
f(std::ref(i));
std::cout << i << std::endl; // i is modified here, displays 1
}