Passing function pointers as arguments

Question

I was revisiting function pointers in C using the following simple code:

unsigned TestFn(unsigned arg)
{   
    return arg+7;
}

unsigned Caller(unsigned (*FuncPtr)(unsigned), unsigned arg)
{
    return (*FuncPtr)(arg);
}

I called it using

Caller(TestFn, 7)  //and
Caller(&TestFn, 7)

both gave the same output : 14. What's the explanation of this. I had been using the second way of calling earlier.

Solution 1

Functions work kind of like arrays here. Suppose you have:

char hello[5];

You can refer to the address of this variable directly as hello - the variable name "decays" into a pointer - or as &hello, which explicitly obtains the address.

Functions are the same: if you write TestFn it decays into a function pointer. But you can also use &TestFn. The latter form might be better because it is familiar to more people.

Solution 2

Just like you pass the address of string without using the ampersand '&' sign, you don't need to use the ampersand sign to say you are passing a function pointer . You can search the book by K & R . It contains a good explaination