Updating struct array fails in C

Question

I'm looking for a simple solution to this. I only have one element and could easily turn it into a simple array of long integers but in the future I will have a struct of random data types so instead of me declaring a bunch of separate arrays of random types, I want to pack the data into a struct.

In this code the problem lies with the calling of load() but I don't know how to solve it. When I use the &, in front of the struct variable, the compiler reports warning: passing argument 1 of 'load' from incompatible pointer type.

The output I expect instead of errors or a segmentation fault is:

0= 1
1= 11

What am I doing wrong?

and here's the code:

#include <stdio.h>
#include <stdlib.h>

typedef struct{
    long n;
}a;

void load(a** x){
    x[0]->n=1;
    x[1]->n=11;
}

int main(void){
    a** b=malloc(200);
    b[0]->n=2;
    b[1]->n=2;
    load(&b); //trouble starts here
    printf("0= %ld\n",b[0]->n);
    printf("1= %ld\n",b[1]->n);
    free(b);
    return 0;
}

Possible duplicate of [Definitive List of Common Reasons for Segmentation Faults](http://stackoverflow.com/questions/33047452/definitive-list-of-common-reasons-for-segmentation-faults) — CodeMouse92, Oct 15 '15 at 18:05

Giorgi Moniava · Answer 1 · 2015-10-15T19:58:24.777

2

You don't need pointer to pointers. Just use

a* b=malloc(ELEMENTS * sizeof (a)); // essentially array with nr of ELEMENTS of type a

The function

void load(a* x){
    x[0].n=1; // 0th structure object
    x[1].n=11; // 1th structure object .. you can access till ELEMENT-th index
}

You can call it like

load(b);  // you can directly pass the pointer

edited Oct 15 '15 at 19:58

answered Oct 15 '15 at 18:04

Giorgi Moniava

27,046
9
53
90

Also, note that malloc returns void pointer . So, another better practice can be to typecast with the appropriate pointer. – Karthik Balaguru Oct 16 '15 at 16:50
@Karthik Balaguru: not in C, in C you don't cast result of malloc – Giorgi Moniava Oct 16 '15 at 19:31

score 0 · Answer 2 · answered Oct 15 '15 at 18:05

0

Anytime you run malloc, you need to check that it has not returned a NULL pointer (ptr == 0). If you try and access a NULL pointer, it can throw a segmentation fault.

One way to do this would be...

a** b=malloc(200);
if(b != 0)
{
   //The rest of the code.
}

answered Oct 15 '15 at 18:05

CodeMouse92

6,840
14
73
130

That is easy to catch but I just found out my segmentation fault happens when `b[0]->n=2;` is executed if I change `load(&b)` to `load(b)` – Mike -- No longer here Oct 15 '15 at 18:07
If b is a pointer to a pointer of `a` (what IS `a`), then you're getting some funky pointer arithmetic off of that, which is one more way to segfault. @Giorgi is absolutely right in his answer. – CodeMouse92 Oct 15 '15 at 18:09

score 0 · Answer 3 · answered Oct 15 '15 at 18:13

0

Thanks for the help, but I couldn't accept one answer because everyone solved portions of the problem.

Here's what I came up with:

#include <stdio.h>
#include <stdlib.h>

typedef struct{
    long n;
}a;

void load(a* x){
    x[0].n=1;
    x[1].n=11;
}

int main(void){
    a* b=calloc(1,sizeof(a)*100);
    if (!b){printf("memory error\n");return 1;}
    b[0].n=2;
    b[1].n=2;
    load(b);
    printf("0= %ld\n",b[0].n);
    printf("1= %ld\n",b[1].n);
    free(b);
    return 0;
}

answered Oct 15 '15 at 18:13

Mike -- No longer here

2,064
1
15
37

problems solved? though strange why put long inside a struct like that – Giorgi Moniava Oct 15 '15 at 18:16
yes. I was doing this as an example but later I will use more than just a long inside a struct. – Mike -- No longer here Oct 15 '15 at 18:31

Updating struct array fails in C

3 Answers3