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CuriousDeveloper
CuriousDeveloper

Reputation: 77

Date input format as dd/mm/yy

I'm trying to take the date as dd/mm/yy format, but when I write it like this, it gives me an output as 191/033/0 for 19/07/14 input.

#include <stdio.h>

int main(){
    int d1, d2, m1, m2, year;
    
    printf("Enter date (dd/mm/yy): ");
    scanf("%d,%d/%d,&,d/%d", &d1,&d2,&m1,&m2,&year);

    return 0;
}

How can I fix this?

Upvotes: 0

Views: 53283

Answers (2)

Iharob Al Asimi
Iharob Al Asimi

Reputation: 53006

You need the strftime() function, and handle the input correctly so nothing unexpected happens

This is an example of how to do it

#include <stdio.h>
#include <time.h>
#include <string.h>

int main()
{
    char      buffer[100];
    struct tm date;

    memset(&date, 0, sizeof(date));

    printf("Enter date (dd/mm/yy): ");
    if (fgets(buffer, sizeof(buffer), stdin) == NULL)
        return -1;
    if (sscanf(buffer, "%d/%d/%d", &date.tm_mday, &date.tm_mon, &date.tm_year) == 3)
    {
        const char *format;

        format = "Dated %A %dth of %B, %Y";
        if (strftime(buffer, sizeof(buffer), format, &date) > sizeof(buffer))
            fprintf(stderr, "there was a problem converting the string\n");
        else
            fprintf(stdout, "%s\n", buffer);
    }
    return 0;
}

Upvotes: 1

ForceBru
ForceBru

Reputation: 44838

# include <stdio.h>
int main(){
    int d, m, year;
    printf("Enter date (dd/mm/yy): ");
    scanf("%d/%d/%d", &d,&m,&year);
    if (d%10==1 && d!=11) printf("%d st",d);
    else if (d%10==2 && d!=12) printf("%d nd",d);
    else if (d%10==3 && d!=13) printf("%d rd",d);
    else printf("%d th",d);
    return 0;
}

By the way, dd/mm/yy does not mean 'two days, two months and two years'. This means 'two digits for day, month and year'. That's why you don't need so many variables.

Upvotes: 3

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