Easy ROT13 Ruby "programme" mystery -

i made simple rot13 programme , don't understand 1 thing:

a = 'abcdefghijklmnopqrstuvwxyz' (a.length+1).times |i|   print  a[i + 13]   if i>13     print a[i %14]   end  end 

outputs:

nopqrstuvwxyzabcdefghijklm 

if don't add +1 after a.length, iteration ends letter l. however, if use print a[i] inside iteration, starts a , ends z no +1 addition needed.

can explain mystery me?

as may know, .times loop invokes block specified number of times, passing each iteration incremented value.

if 26.times {|i| puts i} print values 0 25. to, not including last value.

now let's walk through loop. @ first iteration, i 0. print 14th character of string, "n" (at index 13, zero-based). don't go condition because 0 not greater 13. on second iteration print 15th character, "o". , keep doing this, until reach i=14.

at point, "magic" begins. first, attempt print 27th character of string. there's no such character print literally nothing. condition triggered , go in.

i % 14 equals 0, print zeroth character, "a". next iteration print character @ index 1 (15 % 14) , on, until .times finishes iteration , stops calling block. now, logic work, last value i must 26, 12 in i % 14 , print "m".

length of entire string 26. remember, .times counts not including number? that's why add 1 length, counts 0 26. that's mystery.

there many-many ways of improving code, , you'll learn them in time. :)

update

i knew looked odd code. and, of course, there's bug. when i 13 don't print first time and don't go condition. waste 1 iteration. classic example of "off 1" class of errors. here's fixed version of code doesn't waste iterations , contains no mysteries:

a.length.times |i|   print  a[i + 13]   if > 12     print a[i % 13]   end end