How to call a function using an argument supplied from the template function

Question

EDIT: Just to clarify "t" is successfully called when casted. The compiler knows and does state that it is a function pointer that takes an argument of type int. I supply a null int pointer to break the loop because it is calling itself recursively. It may just be a bug in the compiler.

I am trying to call a function from a template function argument. I would assume that it would be possible to call the function without explicit casting but that does not seem to be the case. Using VC2013.

template<typename T>
void func(T t)
{

    printf("calling func...\n");

    if (t)
    {
        ((void(__cdecl*)(int))t)((int)nullptr);     // explicit casting is successful

        t ((int)nullptr);                           // compile error: ``term does not evaluate to a function taking 1 arguments``

    }

}

void main()
{

    auto pe = func < int > ;
    auto pf = func < void(__cdecl*)(int) >;

    pf(pe);


}

One suggestion concerning casts: Casting a nullptr to int is obviously rubbish. Guessing what you want to achieve though, I would guess that you want a default-initialized T. E.g. if T is some pointer type, you want a null pointer, if it's an int you want a zero. The way to achieve that is simply T(), like T nil = T();. — Ulrich Eckhardt
– Ulrich Eckhardt, Commented Jun 1, 2014 at 14:42

Jarod42 · Accepted Answer · 2014-06-01 13:48:10Z

2

You have the error for func<int> which becomes:

void func(int t)
{
    printf("calling func...\n");

    if (t)
    {
        ((void(__cdecl*)(int))t)((int)nullptr); // bad casting
        t ((int)nullptr);                       // compile error: int is not a callable object
    }
}

answered Jun 1, 2014 at 13:48

Jarod42

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Comments

Ulrich Eckhardt · Accepted Answer · 2014-06-01 13:47:34Z

1

When t is an int, of course you can't treat it like a function. You'll have to specialize the template for ints or use a different function. Also, please forget that there are C-style casts, they only serve to shoot yourself into the foot.

answered Jun 1, 2014 at 13:47

Ulrich Eckhardt

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4 Comments

user3613229 Over a year ago

t is not int, t is of type: void (__cdecl*)(int) at the point of failure.

Ulrich Eckhardt Over a year ago

Actually, I just threw the code at GCC (after fixing some obvious bugs) and you can remove the second and third line of main() and it still doesn't compile. I hereby claim that it is as people have suggested here, that you are trying to call an int as a function. Note that your "works with a cast" is no indication that this isn't the case, as the code never actually gets executed, but at compile time the call that is never made must still be valid.

user3613229 Over a year ago

Ok you just verified that template implementation is broken in GCC as well. The code gets executed trust me. The buggy line should be valid code but these compilers are poorly crafted unfortunately.

Ulrich Eckhardt Over a year ago

func<int> is called with a zero as argument, so the code within if(t) is NOT executed, trust me. However, that code still has to be valid and it isn't, as two compilers and several C++ programmers pointed out to you. You're welcome. ;)

grisha · Accepted Answer · 2014-06-01 13:56:37Z

0

I don't understand what do you want exactly. But maybe something like this ?:

#include <iostream>
#include <type_traits>

template<typename T>
void call_helper(T value, std::true_type) // value is function
{
    std::cout << "Function" << std::endl;
    value(0);
}

template<typename T>
void call_helper(T value, std::false_type) // value is NOT function
{
    std::cout << "Not function" << std::endl;
    std::cout << value << std::endl;
}

template<typename T>
void call(T value)
{
    call_helper(value, std::is_function<typename std::remove_pointer<T>::type>());
}

int main()
{
    void (*f)(int) = call<int>;
    call(f);
}

live example: http://rextester.com/DIYYZ43213

answered Jun 1, 2014 at 13:56

grisha

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