Re: What's the difference between stat and sort?

Huh? Seems to work here. Both of these work for me:

perl -e 'my $lf = (sort(<*>))[-1];print $lf'
[download]

and

perl -e 'my $foo = (stat("myfile"))[7];print $foo'
[download]

Comment on Re: What's the difference between stat and sort? Select or Download Code

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Re^2: What's the difference between stat and sort? by PetaMem (Priest) on Jul 15, 2004 at 10:03 UTC
Aaah. As the germans say: "I get the crisis!" `my $lf = (sort(<>))[-1]; print $lf,"\n";` [download] works. While the (what I tested) `print (sort(<>))[-1], "\n";` [download] throws a print (...) interpreted as function at ./template.pl line 6. syntax error at ./template.pl line 6, near ")[" Execution of ./template.pl aborted due to compilation errors. Bye PetaMem All Perl: MT, NLP, NLU	[reply] [d/l] [select]
Re^3: What's the difference between stat and sort? by Chady (Priest) on Jul 15, 2004 at 10:12 UTC
print (...) interpreted as function... possible solutions: `print ((sort(<>))[-1]), "\n"; # or print+(sort(<>))[-1], "\n";` [download] see also print (...) interpreted as function He who asks will be a fool for five minutes, but he who doesn't ask will remain a fool for life. Chady \| http://chady.net/	[reply] [d/l]
Re^3: What's the difference between stat and sort? by beable (Friar) on Jul 15, 2004 at 10:17 UTC
Oh that's easy, parenthesisation. Try this: `perl -e 'print((sort(<*>))[1]);'` [download] Once the expression is fully inside parentheses, there is no syntax error. With the square brackets outside the parens, it must be trying to get the first element of the array returned by print(), which doesn't return an array.	[reply] [d/l]