in reply to Enforcing formal arguments at compile time?

Sure- declare your subs at the beginning of your script. For example, 

#!/usr/bin/perl
use strict;
sub foo($);   #declare foo as a sub that accepts 
              #a single scalar, to be defined later

foo;          #compile time error
foo 1;        #okay - no error

sub foo($) {  #actual foo implementation
  print shift;
}

[download]

Update: tachyon rightly points out that this won't always behave as expected - and gives two good links for reference. Perl tries to figure out the context of your arguments, which may (or may not) DWIM. 

sub foo($$);      #two scalar arguments
foo $bar;         #error
foo @bar;         #error
foo $bar, $baz;   #ok
foo $bar, @baz;   #no error, but does it do what you want?

[download]

Replies are listed 'Best First'.
Re: Re: Enforcing formal arguments at compile time?
by tachyon (Chancellor) on May 03, 2002 at 05:28 UTC

    Actually prototypes don't always work the way you might expect (see links in node below): 

    #!/usr/bin/perl
use strict;
sub foo($);
my @ary = (4,3,2);
foo(@ary);   # no compile time error

sub foo($) { print shift }

    [download]

    Update

    gmax points out I should run my code before posting as I forgot the my to satisfy strict. Ooops :-)

    cheers

    tachyon

    s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

[reply]
[d/l]
      I expect that in the example you give, foo will be called with a single scalar argument which is a reference to the array @ary. Am I wrong? Or isn't that what you expect?

      Remember arrays are not lists. This is one of those occasions when forgetting that fact can leave you vaguely confused. Of course, if you've been following perl 6 you'd want to write that as foo(*@ary).

      Update: Dang! I am wrong. I didn't expect it to be a compile error, but the direction Perl 6 is taking led me astray as to what would actually happen. Ah well.

[reply]

        Am I wrong?

        Yes, you are wrong.

        foo() is called with a single scalar argument which is precisely equal to scalar @ary or in other words the number of elements in @ary rather than a ref to it. I suggest you might like to 1) run the sample code where we print shift in foo and 2) follow the links I presented to learn more. I have a fair handle on scalar and array context BTW

        cheers

        tachyon

        s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print

[reply]

In Section Seekers of Perl Wisdom