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- operators - What does =~ do in Perl? - Stack Overflow
man perlop "perlop - Perl operators and precedence" (at least in UNIX-like) answers this question: "Binary "=~" binds a scalar expression to a pattern match " – U Windl Commented Apr 1, 2020 at 21:11
- variables - What is the meaning of @_ in Perl? - Stack Overflow
perldoc perlvar is the first place to check for any special-named Perl variable info Quoting: @_: Within a subroutine the array @_ contains the parameters passed to that subroutine More details can be found in perldoc perlsub (Perl subroutines) linked from the perlvar: Any arguments passed in show up in the array @_
- syntax - What are the differences between $, @, % in a Perl variable . . .
Here the sigil changes to $ to denote that you are accessing a scalar, however the trailing [0] tells perl that it is accessing a scalar element of the array in _ or in other words, @_ – Eric Strom
- What does the - gt; arrow do in Perl? - Stack Overflow
Perl arrow operator has one other use: Class−>method invokes subroutine method in package Class though it's completely different than your code sample Only including it for completeness for the question in the title
- Whats the use of lt; gt; in Perl? - Stack Overflow
So, if the shell is handing you a bunch of file names, and you'd like to go through each one's data in turn, perl's <> operator gives you a nice way of doing that it puts the next line of the next file (or stdin if no files are named) into $_ (the default scalar) Here is a poor man's grep: while(<>) { print if m pattern ; } Running this script:
- operators - What is the difference between || and or in Perl . . .
From Perl documentation: OR List operators On the right side of a list operator, it has very low precedence, such that it controls all comma-separated expressions found there
- What does exactly perl -pi -e do? - Stack Overflow
You can inspect the code actually used by Perl with the core module B::Deparse This compiler backend module is activated with the option -MO=Deparse
- Perl: Use s (replace) and return new string - Stack Overflow
In Perl, the operator s is used to replace parts of a string Now s will alter its parameter (the string) in place I would however like to replace parts of a string befor printing it, as in
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