C | pointers,arrays and segmentation issue

Question

I have the following snippet

#include <stdio.h>
#include <string.h>
#define SIZE 3
typedef struct node{
    char *name;
    int id;
} Rec;
int main() {
    Rec n[SIZE], *p;
    int i;
    char *s[]= { "one", "two","three"};

    for (i = 0; i < SIZE; i++){
        strcpy(n[i].name, s[i]);
        //n[i].id = i;
    }
    p = n;
    for (i = 0; i < SIZE; i++){
        //printf("%2d %s\n", p->id, p->name);
        p++;
    }
    getchar();
}

I've been sitting on this for like an hour now, and I can't put my finger on the problem. This program gets kicked out when it hits the strcpy(). I do not know how to solve this, tho I get the feeling that there's a problem with n's allocation. I did tried to convert it to malloc, but results stayed the same..

Thanks in advance !

Related: http://stackoverflow.com/q/20997474/694576 if not a duplicate — alk, Jan 25 '15 at 12:28

haccks · Accepted Answer · 2015-01-25T12:57:46.093

1

You have not allocated memory for name member for your structs. You need to allocate memory for them

 for (i = 0; i < SIZE; i++){
    n[i].name = malloc(strlen(s[i]) + 1);
}

edited Jan 25 '15 at 12:57

answered Jan 25 '15 at 12:08

haccks

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Ok, let me see if I got this - So I had the name allocated in my struct as a char and not as an array of chars (string) ? – Aviad Jan 25 '15 at 12:13
1

Yes. Pointer is not array. – haccks Jan 25 '15 at 12:15
1

You are missing a ')' – Spikatrix Jan 25 '15 at 12:42
1

@Aviad: "*So I had the name allocated in my struct as a char*" no, as pointer to `char`. – alk Jan 25 '15 at 12:42
@alk yea, sorry that's what I ment – Aviad Jan 25 '15 at 12:45
@CoolGuy; Fixed now. – haccks Jan 25 '15 at 12:51
Err. Shouldn't the 1 be *outside* the bracket? – Spikatrix Jan 25 '15 at 12:55
@CoolGuy; Yes. It should. – haccks Jan 25 '15 at 12:58

C | pointers,arrays and segmentation issue

1 Answers1