address of operator and pointer operator

Question

What is the difference between following two function definitions?

Function declaration:

void fun(int* p);

Function Definition 1:

             void fun (int* p){
                       p += 1;
                      }

Function Definition 1:

                 void fun (*p){
                       p += 1;
                          }

Apart from the first being a function definition and the second being gibberish? — Mike Seymour, Feb 20 '14 at 18:12
`void fun (&p)` is not the definition for `void fun(int* p);`. Even with `int`. — herohuyongtao, Feb 20 '14 at 18:13
This question appears to be off-topic because it is about asking for things that doesn't exist (elves, unicorns, etc.) — πάντα ῥεῖ, Feb 20 '14 at 18:14

πάντα ῥεῖ · Answer 1 · 2014-02-20T18:25:21.607

2

There's only one valid function definition, the 1st one you gave:

Function Definition 1:

 void fun (int* p) {
    p += 1;
 }

Also you probably meant:

    (*p) += 1;

edited Feb 20 '14 at 18:25

answered Feb 20 '14 at 18:17

Sorry for this silly question. I was reading an article for 10 hours straight and in the end I got confused in function call with function definition. – Touseef Ahmed Awan Feb 20 '14 at 20:05

score 1 · Accepted Answer · answered Feb 20 '14 at 18:19

1

Passing an int by pointer:

void fun (int* p) ;

void fun (int* p)
{
    *p += 1 ; // Add 1 to the value pointed by p.
}

Passing an int by reference:

void fun (int& p) ;

void fun (int& p)
{
    p += 1 ; // Add 1 to p.
}

answered Feb 20 '14 at 18:19

jliv902

int *p; p += 1 also works; it just does pointer arithmetic – tenfour Feb 20 '14 at 18:23
@tenfour Judging by the question, I am assuming the OP does not know what pointer arithmetic is. – jliv902 Feb 20 '14 at 18:24
Thank you for your kind response I think I missed the return type of the parameter in 2nd definition. – Touseef Ahmed Awan May 17 '14 at 18:44

2 Answers2