pointer to array of struct in loop with index

Question

i have code, but it doesn't compile

#include <stdio.h>
#include <stdlib.h>
typedef struct {
    unsigned int ip;
    unsigned int mask;
} nic_t;

typedef struct {
    int nnic; /*number of nic*/
    nic_t *nic;
} host_t;

int main(void)
{
    int i, j;
    host_t *host = malloc(sizeof(host_t) * 5);
    for (i = 0; i < 5; i++) {
        host[i]->nnic = 3;
        host[i]->nic = malloc(sizeof(nic_t) * 3);
        for (j = 0; j < host[i]->nnic; j++) {
            host[i]->nic[j]->ip = 0x7f000001;
            host[i]->nic[j]->mask = 0xffffff00;
        }
    }
    for (i = 0; i < 5; i++) {
        for (j = 0; j < 3; j++) {
            printf("ip = 0x%x, mask = 0x%x\n",
                    host[i]->nic[j]->ip,
                    host[i]->nic[j]->mask);
        }
        printf("\n\n");
    }
    return 0;
}

please explain to me how I need to address an array of structures in the program, that there was no compilation errors. I have the following errors:

test.c: In function ‘main’:
test.c:18:10: error: invalid type argument of ‘->’ (have ‘host_t’)
test.c:19:10: error: invalid type argument of ‘->’ (have ‘host_t’)
test.c:21:11: error: invalid type argument of ‘->’ (have ‘host_t’)
test.c:22:11: error: invalid type argument of ‘->’ (have ‘host_t’)
test.c:28:13: error: invalid type argument of ‘->’ (have ‘host_t’)
test.c:29:13: error: invalid type argument of ‘->’ (have ‘host_t’)

thanks for all. it works.

host[i]->nnic : host[i] is not pointer. change to host[i].nnic — BLUEPIXY
– BLUEPIXY, Commented Oct 15, 2014 at 19:48
No. host is a pointer to host_t, therefore host[i] is a host_t. — Paul Roub
– Paul Roub, Commented Oct 15, 2014 at 19:52
when Type *pointer, pointer[i] meant *(pointer + i). type of pointer dereference is Type. — BLUEPIXY
– BLUEPIXY, Commented Oct 15, 2014 at 19:56

Degustaf · Accepted Answer · 2014-10-15 19:50:02Z

2

You are confusing the -> operator with the . operator. If you have a struct st, then you can access its members by st.member. If you have a struct pointer st_ptr, then st_ptr->member is short for (*st_ptr).member. In your case, you are dereferencing the pointer with the array index.

answered Oct 15, 2014 at 19:50

Degustaf

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Comments

Brian Cain · Accepted Answer · 2014-10-15 19:48:56Z

1

You want to use . and not ->.

host[i]->nnic = 3;

should be

host[i].nnic = 3;

answered Oct 15, 2014 at 19:48

Brian Cain

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Paul Roub · Accepted Answer · 2014-10-15 19:49:07Z

1

You have an array of host_t structs, not host_t * pointers. So you want to say, e.g.:

host[i].nnic = 3;

answered Oct 15, 2014 at 19:49

Paul Roub

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user58697 · Accepted Answer · 2014-10-15 19:49:48Z

1

You already have an array of structures. host[i] is a structure, not a pointer. So . instead of -> is what you need.

answered Oct 15, 2014 at 19:49

user58697

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ryyker · Accepted Answer · 2014-10-15 19:58:04Z

1

host[i] is not a pointer, it represents a single instance of host_t,

therefore you must use the . operator, not the -> operator.

Change: (and all other occurrences)

From:  

host[i]->nnic = 3;  

To:  

host[i].nnic = 3;

edited Oct 15, 2014 at 19:58

answered Oct 15, 2014 at 19:48

ryyker

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