C++ pointers objects difference

Question

Can somebody explain me the difference between the following?

Those two

float *pointer[10];

float *(pointer[10]);

and those two

int(*pointer_function)();

int *pointer_function();

FWIW: [here](https://cdecl.org/) is a handy website to figure out some of the gibberish — NathanOliver, May 28 '19 at 13:14
Me thinks 1. is a dupe of https://stackoverflow.com/questions/45991094/which-part-of-the-c-standard-allow-to-declare-variable-in-parenthesis and 2. is easy becuase those are quite different things, "pointer to a function" and "a function returning pointer (`int*`)". — luk32, May 28 '19 at 13:18
Would highlight this is why many people prefer `float* pointer[10]` rather than `float *pointer[10]` because the `float*` is the type — UKMonkey, May 28 '19 at 14:00
@VladfromMoscow it's better than `float *pointer[10]` when someone's trying to line up their variable names! — UKMonkey, May 28 '19 at 14:51

Vlad from Moscow · Answer 1 · 2019-05-28T13:45:11.610

These two declarations

float *pointer[10];

float *(pointer[10]);

are equivalent. You can write even the following way

float *( ( pointer )[10] );

That is it is a declaration of an array of 10 pointers to float.

These declarations

int(*pointer_function)();

int *pointer_function();

are different. The first one declares a pointer to a function of type int(). The second one declares a function (not a pointer to a function) with type int *().

Here is a demonstrative program

#include <iostream>

int * pointer_function() // #1
{
    static int x = 1'000'000;

    return &x;
};

int f1() // #2
{
    return 2'000'000;
}

int main()
{
    std::cout << *pointer_function() /* calling #1 */<< '\n';

    int( *pointer_function )() = f1;

    std::cout << pointer_function() /* calling #2 */ << '\n';


    return 0;
}

Its output is

1000000
2000000

To make more obvious the difference between the declarations you can rewrite the second declaration (that is the declaration of function) the following way

int * ( pointer_function )();

Compare it with the pointer to function declaration

int ( *pointer_function )();

Take into account that a declarator may be enclosed in parentheses.

score 0 · Answer 2 · answered May 28 '19 at 16:15

Always keep in mind the place of Parentheses. Parentheses do not alter the type of the object, but they can alter the binding of complex declarators.

From above code snippet:

float *pointer[10];// Array of 10 pointer to float

float *(pointer[10]); //bracket around array of 10 pointer to float, So same as above

int(*pointer_function)(); // function pointer to achieve callbacks

int * pointer_function(); // function to integer pointer

C++ pointers objects difference

2 Answers2