Yes that is exactly what I got. "In the beginning there was , in the end there will be ." As to the reason, the only thing I thought at first was that as far as I
have learned "$_" and "@_" are actually similar to 'argv' in C. Since, we had not passed any parameters to the script nothing was printed out. So I tried that passing two parameters to the script when I ran it. However, as I am sure you know, nothing changed. I have considered that maybe the word 'nothing' is similar to declaring something null, however, I can not find anything to back this up. Does not @_ mean the same thing as @ARGV, that these are the arguments passed to a script? So that if I pass two arguments they will reside in $_[0] and $_[1]? So in effect about the only thing I can conclude at the moment (and I don't particularly like my answer) is that when we state @_ = qw(alpha omega); we are overwriting the first two parameters (if any were passed) and then in the second line, $_ = qw(nothing nothing); setting them both to null. So in effect we set two values, set them to null and then print them out. Am I close? | [reply] [Watch: Dir/Any] [d/l] [select] |
You are over complicating some things. First, answer this question and keep your answer as simple as possible:
How do you pass an argument to a subroutine?
Yes, there are a few different ways, but there is one way that is most commonly used nowadays and that's the answer I'm looking for. | [reply] [Watch: Dir/Any] |
Because @_ is localised for the subroutine. Since you called subroutine with no parms, the @_ it's operating from has undef.
As for &subroutine, @_ is not localised and the existing one is passed in. $_ need not apply outside the sub in any event -- the subroutine deals with @_.
--
Me spell chucker work grate. Need grandma chicken. | [reply] [Watch: Dir/Any] [d/l] [select] |
Give the man above a see-gar...well maybe half of one. You explanation is correct for Part B. But Part C yields "In the beginning there was nothing, in the end there will nothing." (At least it does in my version of Perl 5.005.) I'm not sure I quite follow your explanation for this result.
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