Ruby if, case

if-elsif-else-end

How to write an if statement

if conditional expression then
processing
elsif conditional expression then
processing
else
processing
end

Leap year judgment (singular)

First determine if it is a multiple of 4

p year = ARGV[0].to_i

if year%4==0 then
  p true
else
  p false
end

If you enter year, it is true if it is a multiple of 4, otherwise false

$ ruby check_leap.rb 2004
2004
true

Let's try it in 1999

$ ruby check_leap.rb 1999
1999
false

Becomes Leap year conditions are not just once every four years

A year divisible by 100 is not a leap year
A year divisible by 400 is a leap year Let's judge including those two conditions

p year = ARGV[0].to_i
if year%400 == 0
  p true
elsif year%100 == 0
  p false
elsif year%4 == 0
  p true
else
  p false
end

When you do this

$ ruby check_leap.rb 1900
1900
false
$ ruby check_leap.rb 2000
2000
true

Became

Multiple

If you want to test multiple values at once, just make an array and loop

[2004,1999,1900,2000].each do |year|
  p year
  if year%400 == 0
    p true
  elsif year%100 == 0
    p false
  elsif year % 4 == 0
    p true
  else
    p false
  end
end

Execution result

$ ruby check_leap_year.rb
2004
true
1999
false
1900
false
2000
true

It's getting longer, so let's use method

def leap?(year)
  if year % 400 ==0
    p true
  elsif year % 100 ==0
    p false
  elsif year % 4 == 0
    p true
  else
    p false
  end
end

[2004,1999,1900,2000].each do |year|
  p year
  leap?(year)
end

case

How to write a case statement

case A
when 1 then  
processing
when 2 then  
processing
else       
processing
end

Rewrite the previous code using case

def leap?(year)
  case
  when year % 400 ==0 ; true
  when year % 100 ==0 ; false
  when year % 4 ==0 ;   true
  else ;                false
  end
end

[2000, 1900, 2004, 1999].each do |year|
  p year
  p leap?(year)
end

The program has become shorter

source ~/grad_members_20f/members/wjswnsrud12/memo7.org