C++ overloaded function as template argument

Question

the simplified version of my code is here

int foo(int x)
{
  return x;
}

int foo(int x, int y)
{
  return x+y;
}

template
int ba         


        

Answer 1

No, you can't, because you are calling the function always with only one argument, you need a type with only one argument. Instead, you can use template by value (no typename or class)

One argument:

int foo(int x)
{
    return x;
}

int foo(int x, int y)
{
    return x+y;
}

typedef int (*foo_fcn)(int);

template
int bar(int k)
{
    return unary_func(k);
}

int main()
{
    bar(3);
    return 0;
}

Two arguments:

int foo(int x)
{
    return x;
}

int foo(int x, int y)
{
    return x+y;
}

typedef int (*foo_fcn)(int, int);

template
int bar(int k)
{
    return unary_func(k, k);
}

int main()
{
    bar(3);
    return 0;
}

Both:

int foo(int x) // first foo
{
    return x;
}

int foo(int x, int y) // second foo
{
    return x+y;
}

typedef int (*foo_fcn)(int);
typedef int (*foo_fcn_2)(int, int);

template
int bar(int k)
{
    return unary_func(k);
}

template
int bar(int a, int b)
{
    return unary_func(a, b);
}


int main()
{
    bar(3,1); // compiler will choose first foo
    bar(4); // compiler will choose second foo
    return 0;
}