2-d Char array in C Unexpected output

Question

I have this Piece of COde that store a string in a 2-d char array.IN my code i am using a 2x6 2-D char array.On providing a 12digit string LIKE > "COMEHOMEARUN". It should store it as
C O M E H O
M E A R U N
but I am getting the output as
C O M E H M
M E A R U N ...i.e the value at [0]6] automatically gets the value of [1][0].

here's the code

#include<stdio.h>
#include<conio.h>
void main()
{
    char string[20];
    char aray[1][5];
    int i,j,k=0;
    gets(string);
    //storing the individual characters in the string in the form of 2x6 char array
    for(i=0;i<=1;i++)
    {
        for(j=0;j<=5;j++)
        {
            aray[i][j]=string[k];
            k++;
        }
    }
    //displaying the array Generated
    for(i=0;i<=1;i++)
    {
        for(j=0;j<=5;j++)
        {
            printf("%c ", aray[i][j]);
        }
        printf("\n");
    }
    getch();
}

Does anybody know where I am going wrong?

unwind · Accepted Answer · 2012-11-20 14:53:49Z

4

In a C array declaration like char array[N][M], the N and M values are not "the highest valid index"; they mean "the number of values".

So your declaration

char aray[1][5];

defines an array sized 1x5, not 2x6 as you intend.

You need:

char aray[2][6];

But of course, the actual indexing is 0-based so for char aray[2][6], the "last" element is at aray[1][5].

edited Nov 20, 2012 at 14:53

answered Nov 20, 2012 at 14:39

unwind

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Comments

Rahul Tripathi · Accepted Answer · 2012-11-20 14:36:31Z

0

You can try by changing char aray[1][5] to char aray[2][6]

answered Nov 20, 2012 at 14:36

Rahul Tripathi

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Comments

andrewsi · Accepted Answer · 2013-06-12 18:25:41Z

0

Your indexing is not correct.

When you declare any char array you have to give sufficient length. In your code you give the dimension 1 but it required 2.

Declare the array as:

   char array[2][6];

That will work.

edited Jun 12, 2013 at 18:25

andrewsi

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answered Jun 12, 2013 at 18:03

Varun Chhangani

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