Are you printing 08 or 09 for the first mm value?
If so, then the problem is: the %i conversion specification takes 0x before the hexadecimal value and 0 to the octal value (see the POSIX specification for scanf() ). If you type 08 or 09 , then only the zero value will be equal to octal, so it leaves 8 or 9 at the input, which is immediately read as the value of day1 . Similarly, you will probably enter 08 or 09 for day2 , and the same failure will occur, so day2 contains 0 and year2 contains 8 or 9.
You can help yourself and everyone else:
- Checking success of calls to
scanf() . - Print values entered from the program.
- Inclusion of a transcript of an executable program.
For instance:
#include <stdio.h> int main(void ) { int day1, day2, month1, month2, year1, year2; printf("Enter date 1 mm\n"); scanf("%i", &month1); printf("Enter date 1 dd\n"); scanf("%i", &day1); printf("Enter date 1 yyyy\n"); scanf("%i", &year1); printf("Enter date 2 mm\n"); scanf("%i", &month2); printf("Enter date 2 dd\n"); scanf("%i", &day2); printf("Enter date 2 yyyy\n"); scanf("%i", &year2); printf("%.4d-%.2d-%.2d to %.4d-%.2d-%.2d\n", year1, month1, day1, year2, month2, day2); return 0; }
Run Example:
$ ./date1 Enter date 1 mm 08 Enter date 1 dd Enter date 1 yyyy 1999 Enter date 2 mm 12 Enter date 2 dd 09 Enter date 2 yyyy 1999-00-08 to 0009-12-00 $
The code does not check scanf() calls, but it illustrates the behavior you saw.
Fix
If you want numbers to be decimal even with leading zeros, use %d instead of %i .
Only the toughest C programmers enter dates in octal or hexadecimal.