Calling pointer of structs from nested functions

Question

I have a problem with calling structs, putting them as argument in a void function, and then call another void function, with argument taking the same struct, so I put them both pointers and the result is not showing up. If I call them separately, it works, but with the nested functions, it's not working.

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

typedef struct date {
        int day;
        int month;
        int year;
} date;

void next_day(date *d,int days)
{
        if(d->day < days) {
                int m = d->day + 1;
                d->day = m;
        }
        else
        {
                int m = d->year + 1;
                d->year = m;
                d->month = 1;
        }
}

void tom(date *d)
{
        switch(d->month)
        {
                case 1: case 3: case 5: case 7: case 8: case 10: case 12:
                        next_day(&d,31);
                        break;
                case 4: case 6: case 9: case 11:
                        next_day(&d,30);
                        break;
                case 2:
                        if(d->year % 4 == 0)
                                next_day(&d,29);
                        else
                                next_day(&d,28);
                        break;
        }
}

int main(void)
{
        //date d = malloc(sizeof(date));
        date d;
        printf("Enter day, month, year: ");
        scanf("%d %d %d",&(d.day),&(d.month),&(d.year));
        tom(&d);
        printf("The next day is: %d %d %d",d.day,d.month,d.year);
        return 0;
}

What do you mean by the result is not showing up?

Ed Swangren
– Ed Swangren

2011-11-27 19:50:21 +00:00
Commented Nov 27, 2011 at 19:50 — Ed Swangren
– Ed Swangren, Commented Nov 27, 2011 at 19:50

Mahesh · Accepted Answer · 2011-11-27 19:51:46Z

1

next_day(&d,31);   // At this point `d` is a pointer to a struct type.

You are sending the address of pointer, which means at the receiving end the function argument needs to be a pointer to a pointer ( i.e., date ** ). But in your case, it is not.

void next_day(date *d,int days)

So, just do -

next_day(d,31);  // Notice the removal of &

Now d is of type date * and the receiving end also the argument is of type date *.

answered Nov 27, 2011 at 19:51

Mahesh

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Comments

score 1 · Accepted Answer · 2011-11-27 19:59:50Z

1

I am fairly sure your problem is with next_day(&d,31); You already have the address of the structure, you do not need to use the & operator again. Try calling it with next_day(d,31);.

This should have shown up in the compiler warning; always take care to read those warning.

edited Nov 27, 2011 at 19:59

answered Nov 27, 2011 at 19:51

user406009

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