Why it is storing just two characters instead of 4?

Question

This is my code:

 #include<stdio.h>
    int main() 
    {

        char a[10][10];

        int i,n,m,j;

        n=2;
        m=2;


        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
            {
                scanf("%c",&a[i][j]);
            }

        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                printf("%c",a[i][j]);
            }
        }

    }

Output:

a b c d

a b

Process returned 0 (0x0) execution time : 3.965 s Press any key to continue.

I don't know what you asking please update your question in clearly. — Kmg Kumar, Dec 09 '17 at 06:17
@RetiredNinja thank you for your help, now i clearly understood it. — Alaukik Tandlekar, Dec 09 '17 at 10:07

user2736738 · Accepted Answer · 2017-12-09T07:01:23.197

2

Because they are storing and two of them is the space. That's why you don't see them.

Try this and you will understand

printf("[%c]",a[i][j]);

To solve the issue you can do this

scanf(" %c",&a[i][j]);

This consumes the white space character between the two character input.

From standard:- 7.21.6.2

Input white-space characters (as specified by the isspace function) are skipped, unless the specification includes a [, c, or n specifier

and this

A directive composed of white-space character(s) is executed by reading input up to the first non-white-space character (which remains unread), or until no more characters can be read. The directive never fails.

edited Dec 09 '17 at 07:01

answered Dec 09 '17 at 06:17

user2736738

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@LalitVerma.: Yes because `' '` matches it. scanf is kind of like parser. Sorry for late reply. – user2736738 Dec 09 '17 at 06:42
@LalitVerma.: I have mentioned the relevant sections of the standard explainign both – user2736738 Dec 09 '17 at 06:44
thank you @coderredoc for clearing this doubt, now i understood. – Alaukik Tandlekar Dec 09 '17 at 10:01

Why it is storing just two characters instead of 4?

1 Answers1