Question 1

This question was intended to be a quick way to pick up some points,
but it turned out to be the hardest question on the test.  I ignored
the first incorrect answer given by everyone when grading. 

# print 1
# variable's value is modified in:
$var1 = "15th " + "St" . "8th " . "St";
# + and . have the same precedence, so we simply
# operate from left to right
# "15th " + "St" 
# + is a numeric operator, so Perl converts both operands to numbers
# "15th " becomes 15 (makes as much of string as possible numeric.
#                     ignores the rest)
# "St" becomes 0 (since it doesn't start with a valid number)
# thus we have   15 . "8th " . "St"
# both periods are concatenation operators (expect strings)
# 15 converted to "15" and concatenated with "8th "
# we then have "158th " . "St", so the print gives us
158th St<n>

# print 2
# variable's value is modified in:
$var2 = 2;
# prints
2<n>

# print 3
# variable's value is modified in:
$var3 = "N = " . 13 x $var2;
# x has higher precedence than . 
# 13 x $var2 is done first
# x is repetition (expects a string on the left and a number on the
# right), so we effectively have
# "13" x 2 (this expression means repeat "13" twice)
# now $var3 = "N = " . "1313", so the print gives us
N = 1313<n>

# print 4
# variable's value is modified in:
$var4 = 'camel\n';
chomp($var4);
$var5 = chop($var4);
# note first of all that the assignment involved a single-quoted
# string; thus, \n is not newline it is simply '\n'
# Because of this chomp does nothing to $var4; however chop always
# removes a single character, in this case, the 'n' at the end
# Hence, the print gives us
camel\<n>

# print 5
# variable's value is modified in:
$var5 = chop($var4);
# as stated in print 4, the 'n' is removed from the string stored in
# $var4 and this is what is returned by chop and stored in $var5
# The print gives us
n<n>

# prints 6, 7, and 8
# variables' values determined by:
@a1 = (1, 2, 3, 4);
@a2 = reverse(@a1); # @a2 is (4, 3, 2, 1)
($var6, $var7, $var8) = @a2[0,1];
# the reverse saves the reverse of array @a1 in @a2 as shown above
# @a2[0,1] is an array slice which equals (4, 3)
# since we have a two element array slice and three scalar variables
# to assign to, the last variable is given the value undef
# The three prints thus give us
4<n>
3<n>
<n>

# print 9
# print's value determined by:
@a1 = (1, 2, 3, 4);
@a2 = reverse(@a1); # @a2 is (4, 3, 2, 1)
# $a1[$a2[$#a1]] may be broken down as follows
# $#a1 gives the last INDEX of @a1 which is 3 because @a1 has 4
# elements and indexing starts at 0
# $a2[$#a1] = $a2[3] which is the last element of @a2 -- 1
# $a1[$a2[3]] = $a1[1] which is the second element of @a1 -- 2
# The print thus gives us
2<n>

# print 10
# print's value determined by:
$a1 = 1;
@a1 = (1, 2, 3, 4);
@a2 = reverse(@a1);
# ${a1}[$a2[-2]]
# ${a1} tells Perl that this involves a simple scalar variable, not an
# array access; thus it uses the value of $a1 and treats [ as a
# literal square bracket
# so far we have 1[$a2[-2]]
# -2 tells Perl to access the second element from the end of @a2 which
# is 2
# The final result is then
1[2]<n>