C++ pointer function and non-pointer function

Question

Given a print function:

void print(int i)
{
    cout << i << endl;
}

Why are we allowed to do this in the main function :

void (*bar)(int);
bar = &print;

But not this:

void fizz(int);
fizz = print;

But when it comes to function parameters, we can pass a pointer to a function or a copy of the function:

void foo(void (*f)(int))
{
    (*f)(1);
}

void test(void f(int))
{
    f(1);
} 

Anyone know the reason for these differences?

Answer 1

A function pointer can point to any function that has the same parameters and return type. In your codes above, "bar" is a pointer, and bar=&print means let bar point to print . However, a function cannot = a function.
What's good for a function pointer is that it can point to ANY function as long as the function has the same parameters and return type. This proves quite useful when a function pointer serves as a parameter of another function.
When function is used as an argument, the real thing that is passed is the ADDRESS of the function. A function cannot be passed dereferenced (ie it cannot be passed AS a function). I'm not quite clear, but it seems your implementations of foo and test are identical.

Answer 2

I would recommend reading up on std::function from c++11: http://en.cppreference.com/w/cpp/utility/functional/function

This will allow you to do more what you intend, where void(int) refers to the return type and list of arguments for your function:

void fizz( std::function<void(int)> f )
{
    f(1);
}

Functions are now first class data types. You can even use function for class methods without any painful syntax. You can also make the lazy/dynamic replacement using auto for many situations. Also you can use the same setup to do interesting things like launching functions asynchronously.

auto used_to_call_fizz = fizz;
used_to_call_fizz( f );
std::async( used_to_call_fizz, f );

Answer 3

Every function have its own address, you can see it as an array . The functions void fizz(int) and void print(int i) have different memory address, so the fizz = print; will get an error. But you use void (*bar)(int); , it shows a function's pointer which is used for pointing to a function. When you use bar=&print , it will initialize the bar pointer and then get the correct answer. By the way, you can use bar=print , as they have the same memory address which is same as the array . I hope this can help you.