scanf will not take &(*(p+i).grade)

Question

void storing(struct student * p, int count)
{
    int i,j;

    for ( i = 0; i < count; i++)
    {
        printf("the %d student's info\n", i+1);

        printf("name:\n");
        scanf("%s", p[i].name); 
        printf("age:\n");
        scanf("%d", &((&p[i])->age));  
        printf("grade:\n");
        scanf("%d", &(*(p+i).grade));        
    }
    return;
}

I don't know why scanf doesn't take &(*(p+i).grade)? I know *(p+i) ≡ p[i], but I don't know why it just doesn't work here. I did similar thing with *(p+i) in another code I written: *p.age = 10; and it successfully writes the value into member age.

It will take `&(*(p + i)).grade`. For the same reason `p->age` is equivalent to `(*p).age` and not `*p.age`. — StoryTeller - Unslander Monica, Dec 09 '17 at 23:11
You need to learn about [*operator precedence*](http://en.cppreference.com/w/c/language/operator_precedence). I suggest you [get a good beginners book or two](http://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list) to read. — Some programmer dude, Dec 09 '17 at 23:12

Sergey Kalinichenko · Accepted Answer · 2017-12-10T03:43:24.513

0

You did the conversion from a[b] to *(a+b) correctly, but you missed the precedence of the dot . operator vs. the dereference * operator.

You need to force the way the operators are evaluated, like this:

scanf("%d", &((*(p+i)).grade)); // Add parentheses and keep dot . operator

or use -> operator, like this:

scanf("%d", &(p+i)->grade); // Replace dot . with -> operator

Note how the use of -> operator lets you reduce the number of parentheses.

edited Dec 10 '17 at 03:43

answered Dec 09 '17 at 23:13

Sergey Kalinichenko

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@aschepler Right, I mixed up the order of parentheses. Thanks! – Sergey Kalinichenko Dec 10 '17 at 03:44

scanf will not take &(*(p+i).grade)

1 Answers1