asked on

How to convert date to ISO 1806 format ?

/* A 16-bit date is read over modbus. Date fields are as follows:
Bits 0 - 4 Day of Month
Bits 5 - 8 Month of Year
Bits 9 - 15 Year ( 0 to 128 == 2000 to 2128 ) */

unsigned short responseBuffer[128];

/* responseBuffer[0] contains the 16-bit date read over modbus */
dateFunc ( (char *) &responseBuffer[0], 2 );

Result  dateFunc (const char * val, unsigned short size)
{
    /* convert date read over modbus to ISO 8601 format */
   /* enter code here */


}

Open in new window

naseeam

ASKER

ISO 8601 date format is: "2014-02-05" or more generically "YYYY-MM-DD"

jkr

You could do that using bitwise operators, e.g.

bool dateFunc (const char val, char* iso8601_buf, unsigned short size) // pass 'val' as a single byte
{
    /* convert date read over modbus to ISO 8601 format */
/* A 16-bit date is read over modbus.  Date fields are as follows:
Bits 0 - 4      Day of Month
Bits 5 - 8      Month of Year
Bits 9 - 15    Year   ( 0 to 128 == 2000 to 2128 )  */

  unsigned short usDay =   val & 0x0F; // isolate bits 0-4
  unsigned short usMon = (val >> 4) & 0x07; // isolate bits 5-8
  unsigned short usYear = (val >> 9) & 0x3F; // isolate bits 9-15 (bitwise AND is not strictly necessary

  usYear += 2000; // add offset

  if ( size < 11) return false; // buffer not large enough;

  sprintf(iso8601_buf,"%4.4d-%2.2d-%2.2d",usYear,usMon,usDay);

  return true;
}

Open in new window

ASKER CERTIFIED SOLUTION

jkr

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial