Ruby Convert If Statements to Case
Is it possible to use a case statement to replace these if statements?
if (a%3 == 0) then puts "%3"
elsif (a%4 == 0) then puts "%4"
elsif (a%7 == 0 && a%13 == 0) then put开发者_StackOverflow社区s "%%"
case
when (a % 3).zero? then puts "%3"
when (a % 4).zero? then puts "%4"
when (a % 7).zero? && (a % 13).zero? then puts "%%"
end
Sure:
case
when (a%3 == 0) then puts "%3"
when (a%4 == 0) then puts "%4"
when (a%7 == 0 && a%13 == 0) then puts "%%"
end
It isn't much better, is it? ;-)
puts [3,4,91,10].collect do |a|
case 0
when a % 3 then
"%3"
when a % 4 then
"%4"
when a % 91 then
"%%"
end
end
You should be able to copy that right into irb to see it work. Please forgive the slight 7*13 = 91 hack, but if you're working with actual modulos they should be equivalent.
Using Proc#===
def multiple_of( factor )
lambda{ |number| number.modulo( factor ).zero? }
end
case a
when multiple_of( 3 ): puts( "%3" )
when multiple_of( 4 ): puts( "%4" )
when multiple_of( 7*13 ): puts( "%%" )
end
(a%7 == 0 && a%13 == 0) is equal to (a%7*13 == 0).
in ruby, you can make 1-line if-else statement use && and ||.
puts (a%3 == 0)&&"%3"||(a%4 == 0)&&"%4"||(a%(7*13) == 0)&&"%%"||""
or
log = (a%3 == 0)&&"%3"||(a%4 == 0)&&"%4"||(a%(7*13) == 0)&&"%%"
puts log if log
it's looks agly but short.
精彩评论