while (($input = <STDIN>) < 7)
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?

And the reason it is exiting is that you are using if, not elsif. So this bit:

if($input == 6) {
print "You have chosen option 6\n";
  } else {
exit;
  }
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always exits unless you choose option 6. You could have found out how your program was working with the debugger: perl -d program.pl, then use 's' or 'n' to step through each line of your program.

dave hj~

use strict;

my $input;
my @choice = (
   'Cancel FINAL Schedule for machine A',
   'Submit new FINAL schedule',
   'Ignore Wrkld',
   'Run around the block',
   'Buy a new car',
   'order Pizza',
   'exit',
);

while (1) {
   menu();
   chomp ($input = <>);
   last if $input == 7;
   print "You have chosen option $input\n";
}

sub menu {
   print "Enter a number for the option you want\n";
   print map { ($_+1).": $choice[$_]\n" } (0..$#choice);
}
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jeffa

L-LL-L--L-LL-L--L-LL-L--
-R--R-RR-R--R-RR-R--R-RR
B--B--B--B--B--B--B--B--
H---H---H---H---H---H---
(the triplet paradiddle with high-hat)

okay i understand the array part, but what is
- while (1) doing (is that saying "while True"?)
- menu() what is this doing?
- what is this line doing:
print map { ($_+1).": $choice$_\n" } (0..$#choice);
Thanks
Bob

my $input;
my @choice = (
   'Cancel FINAL Schedule for machine A',
   'Submit new FINAL schedule',
   'Ignore Wrkld',
   'Run around the block',
   'Buy a new car',
   'order Pizza',
   'exit',
);

while (1) {
   menu();
   chomp ($input = <>);
   last if $input == 7;
   print "You have chosen option $input\n";
}

sub menu {
   print "Enter a number for the option you want\n";
   print map { ($_+1).": $choice[$_]\n" } (0..$#choice);
}
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Hey ddrumguy! (got your email BTW ;))
1. Yes. while(1) is saying while TRUE ... the reason i use that is because Perl has the last token to exit. I used to be afraid of while(1) loops, but i like them now.
2. menu() is just calling the menu subroutine. tye gets on to me for calling subs this way ... a more proper way would be to use the ampersand: &menu(). The main reason i can think of is ... what if a new built-in function named menu is introduced? Then menu() would call it, not the one you declared - but &menu() will call yours.
3. Last item ... hehehe sorry about that. How about this instead:

my $total_choices = $#choice;
for my $index (0..$total_choices) {
   print $index + 1, ": $choice[$index]\n";
}
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Same thing ;)

Keep it up, you'll be coding map's and grep's and Schwartzian Transforms in no time. ;)

Still get good ddrum offers? ;)

jeffa

L-LL-L--L-LL-L--L-LL-L--
-R--R-RR-R--R-RR-R--R-RR
B--B--B--B--B--B--B--B--
H---H---H---H---H---H---
(the triplet paradiddle with high-hat)

There is no loop block here for redo to operate on - I am surprised it doesn't give an error actually. You should place make your outer block a loop like:

#!/usr/bin/perl -w

use strict;

my $input = 0;

START: 
while($input != 7 )
{
   menu();
   chomp($input = <STDIN>);

   if ($input == 1) {
      print "You have chosen option 1\n";
      next START;   
   }
   if($input == 2) {
      print "You have chosen option 2\n";
      next START;
   }
   if ($input == 3) {
      print "You have chosen option 3\n";
      next START;   
   }
   if($input == 4) {
      print "You have chosen option 4\n";
      next START;
   }
   if ($input == 5) {
      print "You have chosen option 5\n";
      next START;
   }
   if($input == 6) {
      print "You have chosen option 6\n";
      next START;
   } 
   else {
      last START;
  }
}

# omitted sub ...
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Also bear in mind that nice indentation is next to godliness :)

/J\

#!/usr/bin/perl -w

use strict;

check_input();

sub check_input {

    CHECK : while (1) {
        &menu;
        my $input= (<STDIN>);
        chomp $input;

        print "You have chosen option 1\n" if ($input == 1);
        print "You have chosen option 2\n" if ($input == 2);
        print "You have chosen option 3\n" if ($input == 3);
        print "You have chosen option 4\n" if ($input == 4);
        print "You have chosen option 5\n" if ($input == 5);
        print "You have chosen option 6\n" if ($input == 6);
        last CHECK if ($input == 7);
    }
print "Finished\n";
}

sub menu {
print "Enter a number for the option you want\n";
print "1) Cancel FINAL Schedule for machine A:   \n";
print "2) Submit new FINAL schedule:    \n";
print "3) Ignore Wrkld:    \n";
print "4) Run around the block:   \n";
print "5) Buy a new car:    \n";
print "6) order Pizza:   \n\n";
print "7) exit:   \n\n";
}
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save yourself some typing, ropey!

print "You have chosen option $input\n" if ($input >= 1 && $input <= 6
+);
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~Particle ;Þ

I would agree however .... what happens if the code changes to call different functions instead of printing ??? I was presuming the code supplied was an example ! :)

#!/usr/bin/perl -w
 
use strict;
my $read;
sub menu {
    print "1) Cancel FINAL Schedule for machine A\n";
    print "2) Submit new FINAL schedule\n";
    print "3) Ignore Wrkld\n";
    print "4) Run around the block\n";
    print "5) Buy a new car\n";
    print "6) order Pizza\n";
    print "7) exit\n";
}
for (;;) {
    &menu;
    chomp($read = <STDIN>);
    exit 0 if $read=~ /7/;
    print "You have chosen option $read\n";
    print "Enter a number for the option you want\n";
}
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