Assigning Values to Variables Within Structs Through Pointers in C

Question

Ok so I'm sure there's a simple fix that I'm missing, but right now my code is causing a segment fault on the line "A[i]->key = 0;." The Record* Item part is a necessity for the program, so I need to make it work this way for an assignment I'm working on, however if I do change it so that Item becomes a non-pointer typedef of Record, then I can use A[i].key no problem. I just need a nudge in the right direction so that I can make standInput correctly assign values to an array of pointers to records. Thanks!

Item.h:

#include "stdio.h"
#include "stdlib.h"

typedef int keyType;

struct Record
{
    keyType key;
    int other;
};
typedef struct Record* Item;

void standInput(Item *A, int n)
{
    int i, input;
    for(i = 0; i <= n-1; i++)
    {
        A[i]->key = 0;
        printf("%d ", A[i]->key);
    }
}

Main:

#include "stdio.h"
#include "stdlib.h"
#include "Item.h"

int main()
{
    int n;
    Item *A;
    printf("Enter a length for the array: ");
    scanf("%d", &n);
    A = (Item*)malloc(n * sizeof(Item));
    standInput(A, n);
    return 0;
}

are you sure A is of Item* and not just Item ?

keety
– keety

2012-04-22 22:30:10 +00:00
Commented Apr 22, 2012 at 22:30 — keety
– keety, Commented Apr 22, 2012 at 22:30

hobbs · Accepted Answer · 2012-04-22 22:32:19Z

2

The values in A are all uninitialized, but you're using them as struct Record pointers anyway. If you want to have A continue holding pointers (rather than the structs directly), then you need to allocate space for A and for each item pointed to by A.

answered Apr 22, 2012 at 22:32

hobbs

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1 Comment

Paul Ruiz Over a year ago

Hobbs: Thank you! I would never have gotten that, but it worked perfectly.

Karoly Horvath · Accepted Answer · 2012-04-22 22:35:59Z

2

Note that Item is already a pointer!

You have to allocate space for the struct, not for the pointer:

A = (Item)malloc(n * sizeof(struct Record));

Note: If the typedef for pointer confuses you, don't use it ;)

A[i]->key means that A[i] is a pointer, but you just allocated an array, so use A[i].key.

Note: you have to change the type of A accordingly.

2nd solution: if you want A[i] to be a pointer, you have to fist allocate space for the pointers (as you do now), then for each pointer (in a loop) allocate space for the struct.

edited Apr 22, 2012 at 22:35

answered Apr 22, 2012 at 22:30

Karoly Horvath

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2 Comments

Aslan986 Over a year ago

I'm not sure this is a problem; the standInput function accept a Item pointer and A is declared as Item*. Am i wrong?

Karoly Horvath Over a year ago

It depends on how you want to tackle the problem. A continuous array of elements or an array of pointers?

Aslan986 · Accepted Answer · 2012-04-22 22:32:10Z

0

Your structure name is Record not Item. So you should use sizeof(struct Record).

answered Apr 22, 2012 at 22:32

Aslan986

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3 Comments

Paul Ruiz Over a year ago

Thanks, I switched it out to use sizeof(struct Record), but it's still giving me the same problem.

Aslan986 Over a year ago

I'm not sure i understand what you want your code to do. Are you creating an array of "Item"?

Paul Ruiz Over a year ago

Yeah the assignment asked for an array of Items that are pointers to structs Record. Honestly I'm still a little confused by it as well since the instructors communication skills aren't the greatest, but I've been piecing it together from Robert Sedgewick's book Algorithms in C.

Ehsan Shoja · Accepted Answer · 2012-04-22 23:00:29Z

0

Do it this way:

int main()
{
    int n, i;
    Item *A;
    printf("Enter a length for the array: ");
    scanf("%d", &n);
    A = (Item*)malloc(n * sizeof(Item));
    for(i=0; i<n; i++){
        A[i] = (Item)malloc(sizeof(struct Record));
    }
    standInput(A, n);
    return 0;
}

answered Apr 22, 2012 at 23:00

Ehsan Shoja

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