Leap year C shell script

[jrw133]

Hello. So im a little bit stuck on one of my homework questions.
heres the question:

Write a C Shell Script that performs the following functions. Asks the user to input a year. The script should then calculate whether or not the year is a leap year. The script should caculate whether or not it is a leap year using the following algorithm. A leap year is defined as any year divisible by 4 but not by 100. However, if a year is divisible by 4 and by 100, it is not a leap year unless it is also divisible by 400. The script should implement the math to produce this algoritm and tell the user whether or not the year entered is a leap year.

heres what i have so far:
*************** *************** *************** *************** *
#!/bin/csh

echo Please input a year
set year = $<

if ( $year / 4 ) then
echo this is a leap year
else
echo this is not a leap year
endif

*************** *************** *************** ***************

i know its not exactly what the teacher wanted but i was just messing around with it to see if this could tell me whether or not it was a leap year and for every year i input it tells me its a leap year. i dont know what im doing wrong. any help would be helpful thanks

[oberon]

The problem here is the expression if ( $year / 4 ). This test will return true as long as it is a valid division, which in this case is true for all values of $year. What you want is a test that returns true when the division does not give any remainder. You can use the modulo operator for this:

if ( $year % 4 == 0)

[Nepomuk]

Also, your algorithm isn't quite correct - check http://en.wikipedia.org/wiki/Leap_year#Algorithm for a correct algorithm in pseudocode.

Greetings,
Nepomuk

[afixibiranchi]

Calculating leap year

if ((( $year % 4 == 0) && ( $year % 100 != 0)) || $year % 400 == 0 )
echo "Leap Year";
else
echo "Not a leap year";

[Laharl]

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