function pointer assignment and call in c++?

Question

I know when we use the name of a function as a value, the function is automatically converted to a pointer. look at following code:

int print(int a)
{
    re         


        

Answer 1

print is not a pointer. Its type is int(int), not int(*)(int). This distinction is especially important in type deduction

auto& f = print;   //type of f is int(&)(int), not int(*(&))(int)
template
foo(Func& f);
foo(print);  //Func is deduced to be int(int), not int(*)(int)

Analogous to arrays, you cannot copy a function "by value", but you can pass around its address. For example,

int arr[4];     //the type of arr is int[4], not int*
int *a = arr;   //automatic array-to-pointer decay
int (*a)[4] = &arr;  //type match 
int (*p)(int) = print;  //automatic function-to-pointer decay
int (*p)(int) = &print; //type match

Now when you call print through p,

p(8)     //automatic dereferencing of p
(*p)(8)  //manual dereferencing of p