How can I make strftime show year as in 2022 instead of 22? Is there a way?

Mariam Ahmed Elsendiony 0
Viewed 49
time_t rawtime;
time(&rawtime);
struct tm* timeinfo = localtime(&rawtime);
strftime(arr, sizeof(arr)-1 , "%d/%m/%y", timeinfo);
puts(arr);

The output is 15/9/22, I want it 15/9/2022.I need it to be stored in a char array

c
2 Answers

Just use %Y instead of lowercase y. Clear in documentation.

strftime(arr, sizeof(arr)-1 , "%d/%m/%Y", timeinfo);
Hogan0

Based on additional information supplied by the OP under @Hogan's answer, here is the fix:

Change:

strftime(arr, sizeof(arr)-1 , "%d/%m/%y", timeinfo);

to

strftime( arr, sizeof arr, "%d/%m/%Y", timeinfo );

The 'Y' will give a 4 digit year and the OP has stated that the destination array is "minimal" for "dd/mm/yyyy" (11 bytes only) with nothing to spare. By subtracting 1 in the parameter list, strftime() did the best that it could.

Like fgets( ) and unlike the dangerous strncat( ), one uses this function (passing the full buffer size) safe in the knowledge that the function "knows" it is returning a "null terminated string".

EDIT: Credit where credit is due. @Hogan asked the "pivotal" question that led to this answer.

Fe2O30
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