Programming Forums > Script Programming > Perl

Blackwolf189 · May 8th, 2005, 6:37 PM

I don't know what is wrong with this program. It is supposed to read in a word and definition like so

 word - definition

and store in a hash. I have no clue why it doesn't work. It will open the hash but will not save anything.

(Toggle Plain Text)

###############################################
# Author: Blackwolf189@gmail.com
# Start Date: April 13, 2005
# End Date:
# Description: A program designed to extend 
#	       your vocabulary. 
###############################################

system('cls');
$wordlist = $ENV{"Input"}."WordList.dbm";
dbmopen(%dictionary,$wordlist,0666) || die("Failed To Create Output File");
while($option = printHead())
{
   if($option == 2)
   {
      print "Type each word and definition on a separte line\n";
      print "Type a hyphen between each word and definition\n";
      print "Type :END: to finish\n\n";    
      while($entry = <STDIN>)
      {
	 chomp($entry);
	 if($entry eq ":END:"){last}
	 @entry = split(/-/,$entry);
 	 $dictionary{$entry[0]} = $entry[1] && print "Word Added\n\n" 
	|| warn "Error $!. Word not Added\n\n"; 
      }
   }
	
   elsif ($option == 3)
   {
      print "Type the word you would like to look up:\n";
      lookUp($entry = <STDIN>);
   }
   
   elsif ($option == 4)
   {
      print "Type the definition after each word\n";
      wordtest();
   }
}



sub printHead 
{
print <<HEADING;
Vocab 1 2 3
1. Import Word List
2. Add Words
3. Look Up Word
4. Test on words
5. Random Word
6. Exit
HEADING

   print "Type the number of your selection: ";
   $selection = <STDIN>;
   if($selection != 6){return $selection;}
   exit();
}


sub lookUp
{
   chomp($_[0]);
   if ($dictionary{$_[0]} eq undef) 
     {print "Dictionary does not contain $_[0]\n";}
   else 
     {print "$_[0] = $dictionary{$_[0]}\n";}
}

sub wordtest
{
   while($word,$definiton = each(%dictionary))
   {
       print "$word: ";
       <STDIN>;
       print $definition;
   }
}

spydoor · May 9th, 2005, 8:32 AM

instead of using

(Toggle Plain Text)

$dictionary{$_[0]} eq undef

use

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!defined($dictionary{$_[0]})

unfortunately I can't give you an explanation why.... I don't know

Looks like you have some other bugs..

the way you have this written

(Toggle Plain Text)

$dictionary{$entry[0]} = $entry[1] && print "Word Added\n\n" 
|| warn "Error $!. Word not Added\n\n";

You are assigning the result of an expression (true or false) to $dictionary{$entry[0]}... adding a set of paranthesis should solve that problem

why not add an option to print the entire dictionary (for debugging if nothing else)

(Toggle Plain Text)

   elsif ($option == 7)
   {
        foreach my $word (sort keys %dictionary) {
            print "($word) - ($dictionary{$word})\n";
        }
   }

Blackwolf189 · May 9th, 2005, 8:51 AM

I am unsure of how my code assigns the result of an expression to dictionary. Please explain. i thought that @entry = split(/-/,$entry) would store the word in $entry[0] and the definition in $entry[1]

spydoor · May 9th, 2005, 10:38 AM

The split is working exactly how you think it is.

entry[0] is the word and
entry[1] is the definition

the problem is the assignment.

an example.

(Toggle Plain Text)

$dictionary{test} = 'test's definition' && print "Good" || print "Bad"

the entire right hand side of the assignemnt is being evaluated as an expression.
so in this case it is true (1).

it's like saying

(Toggle Plain Text)

$var = "A" && "B" || "C"
#same as 
$var = ("A" && "B" || "C")
#same as 
$var = (true && true || true) #where true is the boolean result of the statement

literals in Perl are evaluted to true....so $var = 1; after that line.
To make var = "A"; you'd wrap the assignment in paranthesis

(Toggle Plain Text)

($var = "A") && "B" || "C"

if you want to keep similar syntax just wrap the assignment in paranthesis, although I think the test for good or bad is unnecessary and probably not working as you expect. (you need yet another set of parens around the assignment and print "Good" statment for the logic to be correct)

(Toggle Plain Text)

($dictionary{$entry[0]} = $entry[1]) && print "Word Added\n\n" 
|| warn "Error $!. Word not Added\n\n";

I personally find this difficult to read. It's not a bad thing to have more lines of code especially if it makes the logic easier to follow.

(Toggle Plain Text)

(($x="test") && print "Assignment Worked") || print "Assignment Failed";

# same logic, different sytnax..

if($x="test") {
    print "Assignment Worked";
}else{
    print "Assignment Failed";
}

Blackwolf189 · May 9th, 2005, 7:25 PM

Thanks for the indepth explanation. It all makes sense now.

May 8th, 2005, 6:37 PM	#1
Blackwolf189 Newbie Join Date: May 2005 Posts: 10 Rep Power: 0	Dictionary Program I don't know what is wrong with this program. It is supposed to read in a word and definition like so (Toggle Plain Text) word - definition word - definition and store in a hash. I have no clue why it doesn't work. It will open the hash but will not save anything. (Toggle Plain Text) ############################################### # Author: Blackwolf189@gmail.com # Start Date: April 13, 2005 # End Date: # Description: A program designed to extend # your vocabulary. ############################################### system('cls'); $wordlist = $ENV{"Input"}."WordList.dbm"; dbmopen(%dictionary,$wordlist,0666) \|\| die("Failed To Create Output File"); while($option = printHead()) { if($option == 2) { print "Type each word and definition on a separte line\n"; print "Type a hyphen between each word and definition\n"; print "Type :END: to finish\n\n"; while($entry = <STDIN>) { chomp($entry); if($entry eq ":END:"){last} @entry = split(/-/,$entry); $dictionary{$entry[0]} = $entry[1] && print "Word Added\n\n" \|\| warn "Error $!. Word not Added\n\n"; } } elsif ($option == 3) { print "Type the word you would like to look up:\n"; lookUp($entry = <STDIN>); } elsif ($option == 4) { print "Type the definition after each word\n"; wordtest(); } } sub printHead { print <<HEADING; Vocab 1 2 3 1. Import Word List 2. Add Words 3. Look Up Word 4. Test on words 5. Random Word 6. Exit HEADING print "Type the number of your selection: "; $selection = <STDIN>; if($selection != 6){return $selection;} exit(); } sub lookUp { chomp($_[0]); if ($dictionary{$_[0]} eq undef) {print "Dictionary does not contain $_[0]\n";} else {print "$_[0] = $dictionary{$_[0]}\n";} } sub wordtest { while($word,$definiton = each(%dictionary)) { print "$word: "; <STDIN>; print $definition; } } ############################################### # Author: Blackwolf189@gmail.com # Start Date: April 13, 2005 # End Date: # Description: A program designed to extend # your vocabulary. ############################################### system('cls'); $wordlist = $ENV{"Input"}."WordList.dbm"; dbmopen(%dictionary,$wordlist,0666) \|\| die("Failed To Create Output File"); while($option = printHead()) { if($option == 2) { print "Type each word and definition on a separte line\n"; print "Type a hyphen between each word and definition\n"; print "Type :END: to finish\n\n"; while($entry = <STDIN>) { chomp($entry); if($entry eq ":END:"){last} @entry = split(/-/,$entry); $dictionary{$entry[0]} = $entry[1] && print "Word Added\n\n" \|\| warn "Error $!. Word not Added\n\n"; } } elsif ($option == 3) { print "Type the word you would like to look up:\n"; lookUp($entry = <STDIN>); } elsif ($option == 4) { print "Type the definition after each word\n"; wordtest(); } } sub printHead { print <<HEADING; Vocab 1 2 3 1. Import Word List 2. Add Words 3. Look Up Word 4. Test on words 5. Random Word 6. Exit HEADING print "Type the number of your selection: "; $selection = <STDIN>; if($selection != 6){return $selection;} exit(); } sub lookUp { chomp($_[0]); if ($dictionary{$_[0]} eq undef) {print "Dictionary does not contain $_[0]\n";} else {print "$_[0] = $dictionary{$_[0]}\n";} } sub wordtest { while($word,$definiton = each(%dictionary)) { print "$word: "; <STDIN>; print $definition; } }

May 9th, 2005, 8:32 AM	#2
spydoor Programmer Join Date: Feb 2005 Posts: 64 Rep Power: 4	instead of using (Toggle Plain Text) $dictionary{$_[0]} eq undef $dictionary{$_[0]} eq undef use (Toggle Plain Text) !defined($dictionary{$_[0]}) !defined($dictionary{$_[0]}) unfortunately I can't give you an explanation why.... I don't know Looks like you have some other bugs.. the way you have this written (Toggle Plain Text) $dictionary{$entry[0]} = $entry[1] && print "Word Added\n\n" \|\| warn "Error $!. Word not Added\n\n"; $dictionary{$entry[0]} = $entry[1] && print "Word Added\n\n" \|\| warn "Error $!. Word not Added\n\n"; You are assigning the result of an expression (true or false) to $dictionary{$entry[0]}... adding a set of paranthesis should solve that problem why not add an option to print the entire dictionary (for debugging if nothing else) (Toggle Plain Text) elsif ($option == 7) { foreach my $word (sort keys %dictionary) { print "($word) - ($dictionary{$word})\n"; } } elsif ($option == 7) { foreach my $word (sort keys %dictionary) { print "($word) - ($dictionary{$word})\n"; } } Last edited by spydoor; May 9th, 2005 at 8:46 AM.

May 9th, 2005, 8:51 AM	#3
Blackwolf189 Newbie Join Date: May 2005 Posts: 10 Rep Power: 0	I am unsure of how my code assigns the result of an expression to dictionary. Please explain. i thought that @entry = split(/-/,$entry) would store the word in $entry[0] and the definition in $entry[1] Last edited by Blackwolf189; May 9th, 2005 at 8:53 AM.

May 9th, 2005, 10:38 AM	#4
spydoor Programmer Join Date: Feb 2005 Posts: 64 Rep Power: 4	The split is working exactly how you think it is. entry[0] is the word and entry[1] is the definition the problem is the assignment. an example. (Toggle Plain Text) $dictionary{test} = 'test's definition' && print "Good" \|\| print "Bad" $dictionary{test} = 'test's definition' && print "Good" \|\| print "Bad" the entire right hand side of the assignemnt is being evaluated as an expression. so in this case it is true (1). it's like saying (Toggle Plain Text) $var = "A" && "B" \|\| "C" #same as $var = ("A" && "B" \|\| "C") #same as $var = (true && true \|\| true) #where true is the boolean result of the statement $var = "A" && "B" \|\| "C" #same as $var = ("A" && "B" \|\| "C") #same as $var = (true && true \|\| true) #where true is the boolean result of the statement literals in Perl are evaluted to true....so $var = 1; after that line. To make var = "A"; you'd wrap the assignment in paranthesis (Toggle Plain Text) ($var = "A") && "B" \|\| "C" ($var = "A") && "B" \|\| "C" if you want to keep similar syntax just wrap the assignment in paranthesis, although I think the test for good or bad is unnecessary and probably not working as you expect. (you need yet another set of parens around the assignment and print "Good" statment for the logic to be correct) (Toggle Plain Text) ($dictionary{$entry[0]} = $entry[1]) && print "Word Added\n\n" \|\| warn "Error $!. Word not Added\n\n"; ($dictionary{$entry[0]} = $entry[1]) && print "Word Added\n\n" \|\| warn "Error $!. Word not Added\n\n"; I personally find this difficult to read. It's not a bad thing to have more lines of code especially if it makes the logic easier to follow. (Toggle Plain Text) (($x="test") && print "Assignment Worked") \|\| print "Assignment Failed"; # same logic, different sytnax.. if($x="test") { print "Assignment Worked"; }else{ print "Assignment Failed"; } (($x="test") && print "Assignment Worked") \|\| print "Assignment Failed"; # same logic, different sytnax.. if($x="test") { print "Assignment Worked"; }else{ print "Assignment Failed"; } Last edited by spydoor; May 9th, 2005 at 11:28 AM.

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May 9th, 2005, 7:25 PM	#5
Blackwolf189 Newbie Join Date: May 2005 Posts: 10 Rep Power: 0	Thanks for the indepth explanation. It all makes sense now.

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Programming Forums > Script Programming > Perl

Dictionary Program