Possible to both initialize and call a function pointer?

[LB]

Just out of curiosity, is it possible to initialize a temporary function pointer and call it in one go?

void (*)(int, short)(&SomeFunc)(3, 2);
Or would I have to use a typedef?

typedef void (*visf)(int, short);
visf(&SomeFunc)(3, 2);

Obviously it has no practical use except having a lot of parenthesis in one expression...

I'm basing the idea off the fact that you can construct other temporary objects:
std::cout << std::string("text").length() << std::endl;

[webJose]

I would say no. See if you can make sense out of it: What is the data type of SomeFunc? Probably it is a function that matches exactly the type of function pointer you are declaring, and therefore, what is the point anyway since you can call it directly like this SomeFunc(3, 2);?

Now let's imagine SomeFunc is a variable of type void*. Maybe here you can typecast without declaring the typedef. But I think it is as far as you can go.

[Cubbi]

Sure you can, you forgot some parentheses though

Typedef version

#include <iostream>
void SomeFunc(int n1, short n2)
{
    std::cout << n1 << ' ' << n2 << '\n';
}
int main()
{
    typedef void (*visf)(int, short);
    (visf(SomeFunc))(3, 2);
}

demo: http://ideone.com/6YVvv

one-liner version

#include <iostream>
void SomeFunc(int n1, short n2)
{
    std::cout << n1 << ' ' << n2 << '\n';
}
int main()
{
    ((void(*)(int,short))(SomeFunc))(3,2);
}

demo: http://ideone.com/VmmaN

Although it's obviously redundant, you don't need a type to make a function pointer: (&SomeFunc)(3,2); or even (+SomeFunc)(3,2);

To make a better analogy with std::string("text"), try std::function:

#include <iostream>
#include <functional>
void SomeFunc(int n1, short n2)
{
    std::cout << n1 << ' ' << n2 << '\n';
}
int main()
{
    (std::function<void(int,short)>(SomeFunc))(3,2);
}

demo: http://ideone.com/v9OOZ