Pointer notation vs array notation

Question

I like pointer notation in C more than I like array notation, but just can't figure it out for some cases. I have the following code, and the body of main

/*converts arguemnt to number using atoi()*/
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv) {
    int i, times;

    if(argc < 2 || (times=atoi(argv[1])) < 1) {
        printf("Usage :%s positive-number\n", argv[0]);
    } else {
        for(i = 0; i < times; i++) {
            puts("Hello");
        }
    }
    return 0;
}

How would I express argv[1] and argv[0] in pointer notation?

Do you mean pointer arithmetic?

haccks
– haccks

2014-07-13 17:10:39 +00:00
Commented Jul 13, 2014 at 17:10 — haccks
– haccks, Commented Jul 13, 2014 at 17:10

Paul R · Accepted Answer · 2014-07-13 17:12:33Z

6

argv[i] is equivalent to *(argv + i)

I really can't imagine why you would prefer the latter over the former.

answered Jul 13, 2014 at 17:12

Paul R

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Stian Svedenborg · Accepted Answer · 2014-07-13 21:04:00Z

2

Just to point out a fun fact:

If you have and array (int arr[5]) or a pointer all of the following are equivalent:

*(arr + i);
*(i + arr);
arr[i];
i[arr]; // This last one is rather interesting now isn't it.

answered Jul 13, 2014 at 21:04

Stian Svedenborg

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1 Comment

user3833610 Over a year ago

That's neat, I never would have thought of the last one actually working

Ed Swangren · Accepted Answer · 2014-07-13 17:11:16Z

0

p[n]

Is equivalent to:

*(p + n)

answered Jul 13, 2014 at 17:11

Ed Swangren

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1 Comment

user3833610 Over a year ago

Boom, thank you. I'll accept once the time runs out.

haccks · Accepted Answer · 2014-07-13 17:12:21Z

0

If you mean pointer arithmetic then argv[1] would be

*(argv + 1)

but I would go with argv[1].

answered Jul 13, 2014 at 17:12

haccks

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