Re^2: Test if a string is a sub name

in reply to Re: Test if a string is a sub name
in thread Test if a string is a sub name

I do not understand how your expression &$string works. The documentation for exists tells us that the expression must return the name of a subroutine. It appears to me that this would only be true for your expression if $string is a symbolic reference (which is prohibited by use strict). I would expect you need string interpolation (perhaps "&$string") but this syntax throws an error that the argument of exists is not a subroutine.

Bill

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Re^3: Test if a string is a sub name by choroba (Cardinal) on Oct 06, 2022 at 15:22 UTC
It seems the documentation uses the term "subroutine name" in quite a confusing way. Check the examples: Use of a subroutine call, rather than a subroutine name, as an argument to `exists` is an error. `exists ⊂ # OK exists &sub(); # Error` [download] So, in this context, "subroutine name" includes the ampersand. Considering strict, note the exceptions to `strict 'refs'`: `$bar = \&{'foo'}; &$bar;` [download] We are not calling the subroutine, we're just checking it exists/is defined. `map{substr$_->[0],$_->[1]\|\|0,1}[\\|\|{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^ARGV,3]`	[reply] [d/l] [select]

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Re^3: Test if a string is a sub name
by choroba (Cardinal) on Oct 06, 2022 at 15:22 UTC

Use of a subroutine call, rather than a subroutine name, as an argument to exists is an error.
exists &sub;    # OK
exists &sub();  # Error
[download]

So, in this context, "subroutine name" includes the ampersand.

Considering strict, note the exceptions to strict 'refs':

$bar = \&{'foo'};
&$bar;
[download]

map{substr$_->[0],$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]

[reply]
[d/l]
[select]

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