Help with a script

c3Rhcno · June 7th, 2010, 02:09 AM

I'm really new to perl, as in I started reading this book on it 2 weeks ago, so forgive me if I'm asking very basic questions.

I'm trying to get this perl calculator working and I've only just started it, here's the code for it;

Code:

#!perl
print "What would you like to do? \n.1 Add \n.2 Subtract \n.3 Multiply \n.4 Divide";
$choice = <STDIN>;
print "\n"
if ($choice = 1) {
print "What two numbers would you like to add?\n";
print "First: ";
$first = <STDIN>;
print "$first times: ";
$second = <STDIN>;
$result = $first x $second;
print "The result of $first times $second is $result";
}

My question is how do I get the if statement to know wether the user input in $choice is 1?

chorny · June 7th, 2010, 03:48 AM

You need chomp here.

c3Rhcno · June 7th, 2010, 04:54 AM

ok I know what chomp does and whatnot but where would I put it and why cant I have an if statement on a variable with user input, for example;

Code:

$choice = <STDIN>;
print "$choice";
if ($choice = 1) ....

c3Rhcno · June 7th, 2010, 05:00 AM

So I've updated my code to say

Code:

#!perl
print "What would you like to do" . "\n1 Add" . "\n2 Subtract" . "\n3 Multiply" . "\n4 Divide\n";

$choice = <STDIN>;
print "$choice\n";
if ($choice = 1) {
print "You chose to Add\n";

}

if ($choice = 2) {
print "You chose to Subtract\n";

}

if ($choice = 3) {
print "You chose to Multiply\n";

}

if ($choice = 4) {
print "You chose to Divide\n"

}

And this is what I get

http://i45.tinypic.com/2dgq5p4.jpg

aurelia · June 7th, 2010, 06:20 AM

In if statement you should use "==" to compare, not assign.

To receive warnings add

Code:

use warnings;

in the beginning of program, that would help to prevent some errors.

To remove "\n", You can chomp a variable after the assignment:

Code:

$choice = <STDIN>;
chomp($choice);

c3Rhcno · June 7th, 2010, 05:20 PM

Thank you for that aurelia, I needed that.