Very cool. My only question is why, in the
for(@q){...}, do you have to set
$a = $_? I tried to do it with just
$_, and sure enough, it didn't work, but I have no idea why that is.
Spoilers below!
#!/usr/bin/perl -w
sub q {
$; = 1; # $; stores how many tries you've h
+ad; start at 1
for (0..3) {
$;[$_] = int(rand(9)) # @; stores the correct four number
+s; set it up here
}
}
sub _ {
print @_
}
&q;
for (;;) {
$: = $~ = 0; # set $: (right number in the right spot) and
+$~ (right number) to 0
_ "$/$; "; # prints a newline then the number of tries yo
+u've had
$_ = <>; # read a line into $_
if (/^\d{4}$/) { # if the line is exactly 4 digits...
@q = split //; # @q holds the 4 digits
--$#q; # set $#q = 3, ignoring the newline at $q[4]
$@ = -1; # set $@ to -1
# This loops through the input digits and checks for correctne
+ss
for (@q) { # set $_ to each of the 4 i
+nput digits
$a = $_; # set $a to the current dig
+it (why not leave it $_?)
($a == $;[++$@]) && ++$: # increment $: if you have
+a right number in the right spot
|| (grep /$a/, @;) && ++$~ # else increment $~ if you
+have a right number in the wrong spot
}
_ "$~W$:B "; # print out the total of right/wrong ($~W) and r
+ight/right ($:B)
++$;; # increment number of tries
$: == 4 # if we have four right numbers all in the right
+ spots
| $; > 10 # or we have already had 10 tries
&& _ (@;) # print out the correct answer
&& &q # and start the game over
}
}