How to display input value for a date in C-CodePudding

The output has to be printed Date: 2021-FEB-01

After receiving the input month and year value, for example, 2018 January should be displayed 2018-JAN-01.

how could I display YYYY-MMM-DD?

I used else if from Jan to Dec. Is there a simpler code?

if (month == 1)
   printf("Date: %d-JAN-01\n", year);
    
else if (month == 2)
   printf("Date: %d-FEB-01\n", year); 

...

CodePudding user response：

The standard library time.h supports what you are trying to achieve (almost). For example:

struct tm date = {.tm_mday = 1, 
                  .tm_mon = month - 1,       // January == 0
                  .tm_year = year - 1900 } ; // Years since 1900
mktime( &date ) ;
char timstr[] = "YYYY-MMM-DD" ;
strftime( timstr, sizeof(timstr), "%Y-%b-%d", &date ) ;

printf( "%s\n", timstr ) ;

However the month will be output Mmm rather then MMM, e.g. Jan rather the JAN. If that is an issue then you might simply coerce the timstr[6] and timstr[7] to upper case thus:

strftime( timstr, sizeof(timstr), "%Y-%b-%d", &date ) ;
timstr[6] = toupper( timstr[6] ) ;
timstr[7] = toupper( timstr[7] ) ;

printf( "%s\n", timstr ) ;

or you might define a more generic conversion function for greater flexibility - should you change the format for example:

#include <ctype.h>

char* strtoupper( char* str )
{
    for( char* cp = str; *cp != 0; cp   )
    {
        *cp = toupper( *cp ) ;
    }
    
    return str ;
}

then for example:

printf( "%s\n", strtoupper( timstr ) ) ;

A complete example:

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

char* strtoupper( char* str )
{
    for( char* cp = str; *cp != 0; cp   )
    {
        *cp = toupper( *cp ) ;
    }
    
    return str ;
}

int main()
{
    int month = 5 ;
    int year = 1967 ;
    
    struct tm date = {.tm_mday = 1, 
                      .tm_mon = month - 1, 
                      .tm_year = year - 1900 } ;
    mktime( &date ) ;
    char timstr[] = "YYYY-MMM-DD" ;
    strftime( timstr, sizeof(timstr), "%Y-%b-%d", &date ) ;
    
    printf( "%s\n", strtoupper( timstr ) ) ;
    
 
    return 0;
}

Outputs:

1967-MAY-01

As does:

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

int main()
{
    int month = 5 ;
    int year = 1967 ;
    
    struct tm date = {.tm_mday = 1, 
                      .tm_mon = month - 1, 
                      .tm_year = year - 1900 } ;
    mktime( &date ) ;
    char timstr[] = "YYYY-MMM-DD" ;
    strftime( timstr, sizeof(timstr), "%Y-%b-%d", &date ) ;
    timstr[6] = toupper( timstr[6] ) ;
    timstr[7] = toupper( timstr[7] ) ;
    printf( "%s\n", timstr ) ;
    
 
    return 0;
}

CodePudding user response：

You can use an array of string literals:

#include <stdio.h>

int main(void)
{
    const char *months[] = {"", "JAN", "FEB", "MAR", "APR", "MAY", "JUN",
                                "JUL", "AUG", "SEP", "OCT", "NOV", "DEC"};

    int year = 2022;
    int month = 2;

    printf("Date: %d-%s-01\n", year, months[month]);
    return 0;
}

Is there any way to express it without using arrays?

Yes, take a look to strftime:

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
 
int main(void)
{
    struct tm *tmp;
    time_t t;

    char str[12];

    time(&t);
     
    tmp = localtime(&t);

    strftime(str, sizeof(str), "%Y-%b-%d\n", tmp);
     
    printf("Date: %s\n", str);
    return(0);
}

The output is:

Date: 2022-Feb-11

Note that time() takes the current time, you can also fill a struct tm by hand:

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
 
int main(void)
{
    struct tm tmp = {
        .tm_year = 2022 - 1900,
        .tm_mon = 2,
        .tm_mday = 11,
        .tm_hour = 0,
        .tm_min = 0,
        .tm_sec= 0
    };

    char str[12];

    strftime(str, sizeof(str), "%Y-%b-%d\n", &tmp);
     
    printf("Date: %s\n", str);
    return(0);
}