Why this program showing segmentation fault?

Question

Instead of pointer char *s if I use array char s[100] (in line 4) the code gets executed. Why is that?

#include<stdio.h>
int main(){
    int l=0;
    char *s;
    printf("enter string : \n");
    fgets(s,100,stdin);
    puts(s);
    while(s[l]!='\0'){
        l++;
    }
    printf("length : %d",l-1);
    return 0;
}

Because s is an uninizialized pointer. You must allocate memory for string: s=malloc(someSize); — ulix, Jul 25 '22 at 08:49

score 2 · Answer 1 · answered Jul 25 '22 at 08:48

Because you haven't allocated any memory for s, nor even initialized it to a known value.

That means fgets() is scribbling onto some random address, and your program crashes.

With a local char s[100]; you allocate 100 bytes on the stack, and the value of s is a pointer to the zeroth byte of that allocation.

Why this program showing segmentation fault?

1 Answers1