Program that tells if a date is valid in C

Question

I'm reading a book about C programming, at the end of each chapter it has some training exercise and one of them was to make a program that tells if a date is valid or not, the code below is what I did and I would like to know if it can be improved?

#include <stdio.h>

int main(){
int dd,mm,yy;

printf("write a date(day/month/year):");
scanf("%d/%d/%d",&dd,&mm,&yy);

int bissextile=(yy%4==0)?1:0;

if (dd>31 || dd<1 || mm>12 || mm<1)
    printf("Invalid date.\n");
    else
        if((mm==2 && dd<30 && bissextile) || (mm==2 && dd<29 && bissextile==0))
            printf("valid date.\n");
            else
                if((mm==4 || mm==6 || mm==9 || mm==11)&& dd<31)
                    printf("valid date.\n");
                    else
                        if(dd<31 && mm!=2)
                            printf("valid date.\n");
                            else
                                printf("Invalid date.\n");
}

Answer 1

Writing a useful function should be the first thing that comes to mind when you have a challenge like this. Functions can be reused, and they can make the logic simpler.

In addition, a return value is easy to apply in a function, and you can return whenever the function value is determined.

In this case, a function like valid_date which returns a conditional-friendly value (int 1 for OK, 0 for not OK).

This function would check for invalid conditions, and return 0 when an invalid condition is found. If all checks pass, though, it returns 1.

Your leap-year check is not accurate though. Your code is:

(yy%4==0)?1:0;

A year is a leap year when:

1. it is a multiple of 4
2. but not if it is also a multiple of 100
3. unless it is also a multiple of 400.

A more accurate check is:

yy % 400 == 0 || (yy % 4 == 0 && yy % 100 != 0)

Putting this advice together, I put up the following method:

int valid_date(int dd, int mm, int yy) {
    if (mm < 1 || mm > 12) {
        return 0;
    }
    if (dd < 1) {
        return 0;
    }

    int days = 31;
    if (mm == 2) {
        days = 28;
        if (yy % 400 == 0 || (yy % 4 == 0 && yy % 100 != 0)) {
            days = 29;
        }
    } else if (mm == 4 || mm == 6 || mm == 9 || mm == 11) {
        days = 30;
    }

    if (dd > days) {
        return 0;
    }
    return 1;
}

Note how I calculate the expected limit for days and then had just one check? That makes the code simpler to read too.

Finally, that can be used like:

int main(){
    int dd,mm,yy;

    printf("write a date(day/month/year):");
    scanf("%d/%d/%d",&dd,&mm,&yy);

    printf("%s Date\n", valid_date(dd, mm, yy) ? "Valid" : "Invalid");
}

Answer 2

Validation is a specific task, distinct from receiving input, so it deserves to have its own function.

The code suffers from excessive nesting, which could be improved by putting else if at the same indentation level. However, another problem is that some of the checks are affirmative and some of them are negative, which makes the code harder to follow.

Since the validation depends mostly on the month, I'd write it this way, with a switch:

int isValid(int dd, int mm, int yy) {
    if (dd <= 0) return 0;
    switch (mm) {
      case 1: case 3: case 5: case 7:
      case 8: case 10: case 12:
        return dd <= 31;
      case 4: case 6: case 9: case 11:
        return dd <= 30;
      case 2:
        return dd <= 28 + (yy % 4 == 0);
      default:
        return 0;  /* invalid month */
    }
}

Then, in main():

if ( 3 == scanf("%d/%d/%d",&dd,&mm,&yy) &&
     isValid(dd, mm, yy) ) {
    puts("Valid date");
} else {
    puts("Invalid date");
}

Note that validating the return value from scanf() is important too.

The code doesn't handle the Gregorian correction properly, such as for year 1900.