How to assign an integer array declared inside a structure directly?

Question

struct data
{
    char name;
    int conn[3];
};

typedef struct data unit;
typedef unit *link;

int main()
{
    int i;
    link p[100];
    for(i=0;i<=100;i++)
    {
        p[i]=(link)malloc(sizeof(unit));
        p[i]->name='h';
        p[i]->conn[]=(int*){"1","1","1"};   **// assignment error**
    }
    for(i=0;i<=100;i++)
    {
        printf("%c\t%d\t%d\t%d\n",p[i]->name,p[i]->conn[0],p[i]->conn[1],p[i]->conn[2]); 
    }
    getch();
}

As structures does not support initialization so, is there any way to assign this type of arrays declared inside a structure in a single line of code without using mem allocation functions and all for the sake of simplicity? Please stick to the code.

Unrelated to your question: you declare `conn[2]` but you try to access `conn[2]` which if off by one. This isn't going to end well... — Jabberwocky, Sep 27 '18 at 06:37
@M.M I forgot about structure assignment. Would you make an answer with explanation out of this? — Kami Kaze, Sep 27 '18 at 06:47
You can't assign to arrays directly. You can assign whole structures even if the structure contains arrays, but you can't assign arrays as arrays. — Jonathan Leffler, Sep 27 '18 at 06:55
BTW: both lines `for (i = 0; i <= 100; i++)` are also off by one. It should be: `for (i = 0; i < 100; i++)` — Jabberwocky, Sep 27 '18 at 07:14

score 5 · Answer 1 · answered Sep 27 '18 at 06:49

5

After the malloc line you can write:

*p[i] = (struct data){'h', {1, 1, 1}};

This uses structure assignment with a compound literal as the source.

BTW I would recommend doing the malloc line this way: p[i] = malloc(sizeof *p[i]);, see here for explanation.

answered Sep 27 '18 at 06:49

M.M

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@RahulKumar note the `*` in M.M 's version, it is missing in yours and it is important. – mch Sep 27 '18 at 07:18
Yeah! i missed in excitement – Rahul Kumar Sep 29 '18 at 08:32

How to assign an integer array declared inside a structure directly?

1 Answers1