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Home/ Questions/Q 8238611
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T19:52:22+00:00 2026-06-07T19:52:22+00:00

I’d like your help with understand what can be an input from a user

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I’d like your help with understand what can be an input from a user to the following program that can make the output:U%ae'$ffq' ong string

int main(void) {
    int i=0;
    char j[22]="This is a long string", k[3];

    scanf("%2s ", k);
    sprintf(j, k);
    printf("%s", j);
    for (; i< 21; printf("%c", j[i++]))
        ;
    return 1;

}

I don’t understand couple of things:

k can get only two chars from the user- Is this what "%2s" means, no? and then writes into the array pointed by j the content pointed by the array k, so j is not pointed to k, but if we’ll j[5] we’ll still get i. so I don’t understand how can we get this input whatsoever since the input would be chopped to two chars j[0], j[1] would be the two chars from the input and the rest of j[i] would be the original rest of “This is a long string”.

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  1. Editorial Team
    Editorial Team
    2026-06-07T19:52:23+00:00Added an answer on June 7, 2026 at 7:52 pm

    I’m only guessing here, but the problem is probably with the loop. You do not check for the string terminator, but print all of the array regardless of if the string has ended or not.

    If you change the loop to this:

    for (; i < 22 && j[i] != '\0'; printf("%c", j[i++])) ;
    

    You should get the expected output.

    (Note: I also changed 21 to 22 which is the size of the array. You can of course do i <= 21 as that is the same.)

    Edit: Rereading the question after the comment from hmjd.

    If the input as read by scanf contains a percentage ('%') character the call to sprintf afterwards will try to parse it as a formatting code. If I test this program with the input %d123, then k will be "%d" as expected, but the resulting array j will be "192795408\0long string".

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