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c++function-pointers
Zeenobit
Zeenobit

Reputation: 5204

Difference between Functions and Function Pointers as Arguments

This is a rather simple question that is making me a bit curious. Consider the following code snippet:

#include <iostream>

int three()
{
    return 3;
}

void foo(int func(void))
{
    std::cout << func() << std::endl;
}

void bar(int (*func)(void))
{
    std::cout << func() << std::endl;
}

int main()
{
    foo(three);
    bar(three);
    return 0;
}

// output:
// 3
// 3

As you can see, we have two functions that take another function as their only argument. However, the function prototypes for them differ. Mainly, we have void foo(int func(void)) and void bar(int (*func)(void)). At first glance, it looks like foo is taking the function itself, and bar is taking a pointer to a function.

However, they both produce the exact same results, and have the exact same body, and are called in exactly the same manner.

My question is, is there an actually hidden difference between foo and bar? Is this simply an optional syntax in C++? Is one of the two cases considered "bad style" in C++?

If my compiler is a contributing factor, I'm using Visual Studio 2010.

Upvotes: 2

Views: 1536

Answers (1)

James McNellis
James McNellis

Reputation: 355307

There is no difference: the types of foo and bar are the same: int(int(*)()).

There is no such thing as a function-type parameter: when a parameter appears with the function type syntax (like int func(void)), it is transformed into the corresponding pointer-to-function type (like int (*func)(void)).

The transformation is similar to how a parameter declared with array syntax (e.g. int a[]) is transformed into the corresponding pointer type (int* a).

Upvotes: 10

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