why this simple program gives a "bus error"?

Question

#include <stdio.h>

void squeeze(char s[], int c)
{
int i, j;
for(i = j = 0; s[i]!= '\0'; i++)
{       
    if (s[i] != c)
    {
        s[j++] = s[i];
    }
}   
s[j] = '\0';
printf("%s\n", s);
}


int main(void)
{
squeeze("asfafsdfsaaaasdfsd", 'a');
}

I run this program in mac, and it looks like the problem comes from the line: s[j++] = s[i]

but I don't know why.

Thanks!

You are trying to write to a string literal. – Weather Vane Jul 01 '16 at 22:41 — Weather Vane, Jul 01 '16 at 22:41

score 1 · Answer 1 · answered Jul 01 '16 at 22:42

1

Because string literals are read-only. To make a writable array of characters, try this

int main( void )
{
    char array[] = "asfafsdfsaaaasdfsd";
    squeeze( array, 'a' );
}

answered Jul 01 '16 at 22:42

user3386109

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why this simple program gives a "bus error"?

1 Answers1