Pointer to multi-dimensional array

Question

I don't know what's the problem with this piece of code, I'm just getting a random result here:

#include <stdio.h>
#include <conio.h>
int main() {
    char arr[3][2] = {{'z','a'},{'e','r'},{'x','v'}};
    int i;
    scanf("%d",&i);
    printf("%c",*(arr+i));
    getch();
}

Thanks

When I give "i" the value 0 it needs to print 'z' , value "1" it needs to print 'a' ...etc — JohnnyCat
– JohnnyCat, Commented Mar 25, 2014 at 0:03

Vlad from Moscow · Accepted Answer · 2014-03-25 00:10:22Z

1

First of all it is not clear what you are trying to do. I can only suggest to change statement

printf("%c",*(arr+i));

the following way

printf("%c",**(arr+i));

In this case if i will be in the range 0 - 2 then this statement will output the first character of the corresponding row. For example for i equal tp 1 the output will be

if you want to output any character in the array using i as an index then use

printf("%c", *( *arr + i ));

Or

for ( int i = 0; i < 6; i++ ) printf("%c", *( *arr + i ));

edited Mar 25, 2014 at 0:10

answered Mar 24, 2014 at 23:59

Vlad from Moscow

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5 Comments

JohnnyCat Over a year ago

What if I want to print 'a' ?

David G Over a year ago

@JohnnyCat Then you would enter the value 0 for i and do *(*(arr + i) + 1) or arr[i][1].

JohnnyCat Over a year ago

I guess arr is a pointer to the whole array and *arr is a pointer to the first row. That's why when we use **(arr+i) it will access the first element of the "i" row.

Vlad from Moscow Over a year ago

@JohnnyCat In expressions arr has type char ( * )[2] that is pointer to char [2]. *arr means the array char[2] In expressions it is converted to char * So the expression *arr + i has type char *

M.M Over a year ago

Maybe you wanted arr[i/2][i%2] . Using the *(+) notation seems more confusing than [][] to me.

achedeuzot · Accepted Answer · 2014-03-25 14:01:46Z

0

If you provide the value 0 then it will give the result assign the address of zero-th item and if you provide the 1 it will print the address of 1st item

zero-th item  = {'z','a'},
first item = {'e','r'}

it will print the address of these above 2 item using format specifier %c , here you are trying the to print the address %c instead of %u.

edited Mar 25, 2014 at 14:01

achedeuzot

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answered Mar 25, 2014 at 13:33

user3459430

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