In the first line, def is a statement to define a function
(or, more precisely, a method; we'll talk more about what a
method is in a later chapter). Here, it
specifies that the function fact takes a single argument,
referred to as n.
if n == 0
The if is for checking a condition. When the condition
holds, the next bit of code is evaluated; otherwise whatever follows the
else is evaluated.
1
The value of if is 1 if the condition holds.
else
If the condition does not hold, the code from here to end is
evaluated.
n * fact(n-1)
If the condition is not satisfied, the value of if is
the result of n times fact(n-1).
end
The first end closes the if statement.
end
The second end closes the def statement.
print fact(ARGV[0].to_i), "\n"
This invokes our fact() function using a value specified
from the command line, and prints the result.
ARGV is an array which contains command line arguments.
The members of ARGV are strings, so we must convert this into a
integral number by to_i. Ruby does not convert strings into
integers automatically like perl does.
Hmmm... what would happen if we fed this program a negative
number? Do you see the problem? Can you fix it?
Next we examine the puzzle program from the chapter
on strings. As this is somewhat longer,
we number the lines for reference.
01 words = ['foobar', 'baz', 'quux']
02 secret = words[rand(3)]
03
04 print "guess? "
05 while guess = STDIN.gets
06 guess.chop!
07 if guess == secret
08 print "you win\n"
09 break
10 else
11 print "you lose.\n"
12 end
13 print "guess? "
14 end
15 print "the word is ", secret, ".\n"
In this program, a new control structure, while, is
used. The code between while and its corresponding
end will execute repeatedly as long as some specified
condition remains true.
rand(3) in line 2 returns a random number
in the range 0 to 2. This random number is used to extract one
of the members of the array words.
In line 5 we read one line from standard input by the method
STDIN.gets. If EOF (end of file) occurs
while getting the line, gets returns
nil. So the code associated with this
while will repeat until it sees ^D (or
^Z under DOS), signifying the end of input.
guess.chop! in line 6 removes the last character from
guess; in this case it will always be a newline
character.
In line 15 we print the secret word. We have written this as
a print statement with three arguments (which are printed one after
the other), but it would have been equally effective to do it with a
single argument, writing secret as #{secret}
to make it clear that it is a variable to be evaluated, not a literal
word to be printed:
01 st = "\033[7m"
02 en = "\033[m"
03
04 while TRUE
05 print "str> "
06 STDOUT.flush
07 str = gets
08 break if not str
09 str.chop!
10 print "pat> "
11 STDOUT.flush
12 re = gets
13 break if not re
14 re.chop!
15 str.gsub! re, "#{st}\\&#{en}"
16 print str, "\n"
17 end
18 print "\n"
In line 4, the condition for while is hardwired to
true, so it forms what looks like an infinite loop.
However we put break statements in the 8th and 13th lines to
escape the loop. These two breaks are also an example
of if modifiers. An "if modifier" executes the
statement on its left hand side if and only if the specified
condition is satisfied.
There is more to say about chop! (see lines 9 and
14). In ruby, we conventionally attach '!' or
'?' to the end of certain method names. The
exclamation point (!, sometimes pronounced aloud as
"bang!") indicates something potentially destructive, that is to say,
something that can change the value of what it touches.
chop! affects a string directly, but chop
with no exclamation point works on a copy. Here is an
illustration of the difference.
ruby> s1 = "forth"
"forth"
ruby> s1.chop! # This changes s1.
"fort"
ruby> s2 = s1.chop # This puts a changed copy in s2,
"for"
ruby> s1 # ... without disturbing s1.
"fort"
You will later come across method names that end in a question mark
(?, sometimes pronounced aloud as "huh?"); this indicates
a "predicate" method, one that can return either true or
false.
Line 15 deserves some careful attention. First, notice that
gsub! is another so-called destructive method. It
changes str by replacing everything matching the pattern
re (sub means substitute, and the
leading g means global, i.e., replace all
matching parts in the string, not just the first one found). So
far, so good; but what are we replacing the matched parts of the text
with? st and en were defined in lines 1-2
as the ANSI sequences that make text color-inverted and normal,
respectively. In line 15 they are enclosed in #{}
to ensure that they are actually interpreted as such (and we do not
see the variable names printed instead). Between these
we see "\\&". This is a little tricky.
Since the replacement string is in double quotes, the pair of
backslashes will be interpreted as a single backslash; what
gsub! actually sees will be "\&", and
that happens to be a special code that refers to whatever matched the
pattern in the first place. So the new string, when printed,
looks just like the old one, except that the parts that matched the
given pattern are highlighted in inverse video.