#!/usr/bin/perl
open (T, ">out") or die "Can't open out $!\n";
printf(
T "hello\n");
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Prints the warning message:

String found where operator expected at ./t.pl line 4, near "T "hello\
+n""
        (Do you need to predeclare T?)
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If I add use diagnostics I get this message in addition:

    (S) The Perl lexer knows whether to expect a term or an operator. 
+ If it
    sees what it knows to be a term when it was expecting to see an
    operator, it gives you this warning.  Usually it indicates that an
    operator or delimiter was omitted, such as a semicolon.
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I expect whitespace to be benign here, and I don't expect this message. If I wanted this much meaning in the whitespace, I'd be using python! My code these days doesn't look much like this, but perltidy formatted it this way.

Next I tried adding a space like this:

#!/home/toma/perl58i/bin/perl
use strict;
use warnings;
use diagnostics;
open (T, ">out") or die "Can't open out $!\n";
printf (
T "hello\n");
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and I got this additional diagnostic:

printf (...) interpreted as function at ./t.pl line 6 (#1)
    (W syntax) You've run afoul of the rule that says that any list op
+erator
    followed by parentheses turns into a function, with all the list
    operators arguments found inside the parentheses.  See
    perlop/Terms and List Operators (Leftward).
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Removing the parenthesis like this:

#!/usr/bin/perl
open (T, ">out") or die "Can't open out $!\n";
printf
T "hello\n";
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also fixes it. Why?

It should work perfectly the first time! - toma