some issue with string in C

Question

i'm reading the C programming language, question 2-4 asks to write a function called squeeze to delete all char in s1 which is in s2, so i write the code, but it can't run at all.

here is my code

#include <stdio.h>

void squeeze(char s1[], char s2[]);

int main()
{
    squeeze("tabcdge", "abc");
}

void squeeze(char s1[], char s2[])
{
    int i, j, k;

    for (i = k = 0; s1[i] != '\0'; i++) {
        for (j = 0; s2[j] != '\0' && s2[j] != s1[i]; j++)
            ;
        if (s2[j] == '\0')
            s1[k++] = s1[i];
    }
    s1[k] = '\0';
    for (i = 0; s1[i] != '\0'; i++)
        printf("%c", s1[i]);
}

Can you give this question a meaningful title? See http://stackoverflow.com/help/how-to-ask for some title ideas. — Jason C, Sep 07 '16 at 05:20
[See **here**](http://ideone.com/yRxumu) for how the duplicate link relates to your code.. — WhozCraig, Sep 07 '16 at 05:29

score 1 · Answer 1 · answered Sep 07 '16 at 05:21

1

In main, you pass string constants to squeeze. But then squeeze tries to modify them with code like s1[k++] = s1[i]. You can't modify a constant -- that's what makes it a constant.

answered Sep 07 '16 at 05:21

David Schwartz

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score 0 · Answer 2 · answered Sep 07 '16 at 05:21

0

String literals (like "tabcdge") are read-only. Attempting to modify one leads to undefined behavior.

answered Sep 07 '16 at 05:21

Some programmer dude

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score 0 · Answer 3 · edited May 23 '17 at 12:32

0

The String constatns are stored in read-only memory. As @Joachim said, modifying them (deleting in your case) is Undefined behaviour.

I suspect your program is crashing. For more info where the variables and constants are stored https://stackoverflow.com/a/18479996/3747770

edited May 23 '17 at 12:32

Community

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answered Sep 07 '16 at 05:27

NJMR

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some issue with string in C

3 Answers3