my first real perl prog!

[ludeking]

Hey all,
I just finished the first question of one of my perl assignments and I am so happy, I just had to share it with you all.

Also, I wanted to ask if this was good programming style/habit or not??

so yeah, its in perl, it takes a user input of a string of words, then counts how many words in the string: (took out all comments)

#!/usr/bin/perl -w
use strict;

my $text;
my @words;
my $count;

print("\n");
print("Please enter a string of text: ");
$text = <STDIN>;

chomp($text);

@words = split(/ /,$text);


$#words = $#words + 1;

print("The number of words in that string is: $#words \n\n");

output

Please enter a string of text: hello this is my first perl proggie, yay!
The number of words in that string is: 8

YAY!

[ludeking]

only problem i've realised is that two or more white spaces count as a word.... but i think thats ok as the lecturer says this:
"For simplicity lets just call a series of non-white space characters separated by white space (or newlines) a word."

Thats what he means right??
Edit: hmm, maybe not....

[=xer0=]

heheh
good stuff
i just started learning Java so i know how you feel

[ludeking]

Yeah I'm learning Java as well. One of my other courses!

[ludeking]

Anyone else got any feedback for me?

[bigiain]

>Also, I wanted to ask if this was good programming style/habit or not??

A few points, using "-w" and "use strict" is a great habit to get into, but you then throw away some of the benefits by declaring all your variables at the top of the program, effectively globalising them.

Its considered good practice to scope your variables as tightly as possible, ie declare them just before you use them... (you're also declaring $count but never using it, you _sure_ thats not thowing a warning? ;-)

The brackets are unneccessary with print and chomp statements (though the do work OK with them).

Splitting on a literal space is almost certainly not what you _really_ want to do - that "//" in the split statement really is a regex pattern, so you can use \w to mean "any whitespace character", and \w+ to mean "one or more whitespade characters in a row" which is much more likely to "count words" accurately.

Incrementing $#words is actually extending the @words array in memory, not a problem for a simple program like this, but adding 1 to the value when printing (or printing scalar(@words) instead) is argueably "better style".

So, I get:

#!/usr/bin/perl -w
use strict;

print"\nPlease enter a string of text: ";
my $text = <STDIN>;

chomp $text;

my @words = split(/\w+/,$text);

print "The number of words in that string is: ".$#words+1."\n\n";

Having said that, yours works, and working code is much better than fabulously stylish code that doesn't compile to provide the right answer!

;-)

big

[edw]

ludeking, what course are you doing?

[ludeking]

hey guys,
The lecturer told us we need to use the -w flag and the use strict as well so thats why they are there.

Yeah I forgot to delete the $count... whoops....

I'm doing a Masters Degree in InfoTech edw.

Thanks for the revised code, I'll look it over!

[Shalmanese]

Now you need to star learning obfusicated perl which, if I haven't forgotten much is this:

#!/usr/bin/perl -w
use strict;

print"\nPlease enter a string of text: ";
<>;
chomp;
print "The number of words in that string is:" + scalar(split(/\w+/,) + "\n\n";

I think thats it.

or even:

#!/usr/bin/perl -w
use strict;

print"\nPlease enter a string of text: ";
print "The number of words in that string is:" + scalar(split(/\w+/,chomp(<>)) + "\n\n";

I
edit: debugging

edit2: the board sticks a space between the > and the ) for some reason :P.

[ludeking]

I cant get any of your code to compile and run however I thunk I can see what you are getting at.

[ludeking]

Argh,
I tried incorporating this:

@words = split(/\w+/,$text);

and I am still getting more than one white space being a word, plus chars like ? being a word, plus if there is only one word then $#words = 0!!!

Oh well, I'll move on and come back to it. Thanks for all your ideas anyway!!

[Shalmanese]

Originally posted by ludeking
I cant get any of your code to compile and run however I thunk I can see what you are getting at.

Hmm... I have a perl compiler now so I'm playing around with the code now. I can't figure out how to make the <STDIN> implicit in the first example, all the ones I've done were using while(<>) which does make it implicit.

for example:

while(<>) { print;}

is the equivilant of cat (although

print while (<>);

is shorter )

For the second one, chomp can't work on handles since it tries to modify the string directly rather than copy it into another location. A simple but dodgy hack would be to omit the chomp but to minus 1 from the word count.

the debugged code is this:

#!/usr/bin/perl
print"\nPlease enter a string of text: ";
print "The number of words in that string is: ", scalar(split(/\w+/,<>)) - 1, "\n\n";

If you want the one line version, its this:

print "The number of words in that string is: ", scalar(split(/\w+/,<>)) - 1, "\n\n" if print"\nPlease enter a string of text: ";;

ps: was looking through man perl and found this:

BUGS
The -w switch is not mandatory.

ROTFL!

[Shalmanese]

Originally posted by ludeking
Argh,
I tried incorporating this:

@words = split(/\w+/,$text);

and I am still getting more than one white space being a word, plus chars like ? being a word, plus if there is only one word then $#words = 0!!!

Oh well, I'll move on and come back to it. Thanks for all your ideas anyway!!

this worked for me.

stuff like:

   sdfsdfsdf     ???!?!?!   sddsff   ^EOL

was reported as two words

[ludeking]

Originally posted by Shalmanese
this worked for me.

stuff like:

   sdfsdfsdf     ???!?!?!   sddsff   ^EOL

was reported as two words

Yeah but in this assignment, anything that is not white space should count as a word.
I found the correct argument to split on.

split(/\s+/$variable)

will make my proggie do what I want, plus I removed all white space before splitting.
Final code:

#!/usr/bin/perl -w
use strict;

my $text;
my @words;

print("\n");
print("Please enter a string of text: ");
$text = <STDIN>;

$text=~s/^\s+//;

variable
chomp($text);

@words = split(/\s+/,$text);

$#words = $#words + 1;

print("The number of words in that string is: $#words \n\n");

and output

Please enter a string of text: this is a test and        it                    works!
The number of words in that string is: 7

p.s. I just modified it to read lines till EOF and report number of words for each line of input!!