-
Here's one way to do it:
print "What temperature is it? ";
chomp($temperature = <STDIN>);
if ($temperature > 72) {
print "Too hot!\n";
} else {
print "Too cold!\n";
}
The first line prompts you for the temperature. The second line accepts the temperature for input. The
if
statement on the final five lines selects one of two messages to print, depending on the value of
$temperature
.
-
Here's one way to do it:
print "What temperature is it? ";
chomp($temperature = <STDIN>);
if ($temperature > 75) {
print "Too hot!\n";
}
elsif ($temperature < 68) {
print "Too cold!\n";
} else {
print "Just right!\n";
}
Here, we've modified the program to include a three-way choice. First, the temperature is compared to 75, then to 68. Note that only one of the three choices will be executed each time through the program.
-
Here's one way to do it:
print "Enter a number (999 to quit): ";
chomp($n = <STDIN>);
while ($n != 999) {
$sum
+= $n;
print "Enter another number (999 to quit): ";
chomp($n = <STDIN>);
}
print "the sum is $sum\n";
The first line prompts for the first number. The second line reads the number from the terminal. The
while
loop continues to execute as long as the number is not 999.
The += operator accumulates the numbers into the
$sum
variable. Note that the initial value of
$sum
is
undef
, which makes a nice value for an accumulator, because the first value added in will be effectively added to
0
(remember that
undef
used as a number is 0).
Within the loop, we must prompt for and receive another number, so that the test at the top of the loop is against a newly entered number.
When the loop is exited, the program prints the accumulated results.
Note that if you enter
999
right away, the value of
$sum
is not 0, but an empty string - the value of
undef
when used as a string. If you want to ensure that the program prints 0 in this case, you should initialize the value of
$sum
at the beginning of the program with
$sum = 0
.
-
Here's one way to do it:
print "Enter some strings, end with ^Z:\n";
@strings = <STDIN>;
while (@strings) {
print
pop @strings;
}
First, this program asks for the strings. These strings are saved in the array variable
@strings
, one per element.
The control expression of the
while
loop is
@strings
. The control expression is looking for a single value (
true
or
false
), and is therefore computing the expression in a scalar context. The name of an array (such as
@strings
) when used in a scalar context is the number of elements currently in the array. As long as the array is not empty, this number is non-zero, and therefore true. This is a very common Perl idiom for "do this while the array is non-empty."
The body of the loop prints a value, obtained by
pop
'ing off the rightmost element of the array. Thus, each time through the loop, the array is one element shorter, because that element has been printed.
You may have considered using subscripts for this problem. As we say, there's more than one way to do it. However, you'll rarely see subscripts in a true Perl Hacker's programs because a better way almost always exists.
-
Here's a way to do it
without
a list:
for ($number = 0; $number <= 32; $number++) {
$square = $number * $number;
printf "%5g %8g\n", $number, $square;
}
And here's how to do it
with
a list:
foreach $number (0..32) {
$square = $number * $number;
printf "%5g %8g\n", $number, $square;
}
These solutions both involve loops, using the
for
and
foreach
statements. The body of the loops are identical, because for both solutions, the value of
$number
proceeds from 0 to 32 on each iteration.
The first solution uses a traditional C-like
for
statement. The three expressions respectively: set
$number
to 0, test to see if
$number
is less than 32, and increment
$number
on each iteration.
The second solution uses a
foreach
statement. A list of
33
elements (0 to 32) is created, using the list contructor. The variable
$number
is then set to each element in turn.