When is a function call not a function call

dpuu has asked for the wisdom of the Perl Monks concerning the following question:

Everyone known why print (1+1)*2; doesn't work as newbies might expect. But there's a simple rule to understand it: if it looks like a funtion call then it is a function call.

But I came across another example today:

sub uniq
{
    local $_;
    my %seen;
    return grep { !$seen{$_}++ } @_;
}

print uniq 0,0,1,2;         # good: prints 0,1,2
print sort uniq 0,0,1,2;    # bad:  prints 0,0,1,2
[download]

It appears that the sort function is seeing "uniq" as the code block that defines sort critera. OK, I thought, lets make it "look like a function call":

print sort uniq(0,0,1,2);
[download]

But nope, this still prints "0,0,1,2". To make it work I need to resort to:

print sort grep {1} uniq 0,0,1,2;
[download]

or some similar intusive builtin.

This example probably just scratches the surface of some parser logic that I haven't correctly groked. Why doesn't the "if it looks like a function" rule work in this case? If a builtin like "grep" (or "map") isn't treated as a sort-criteria code block, then how do I define my own subroutines that similarly are not sort-critera?

--Dave
Opinions my own; statements of fact may be in error.

Comment on When is a function call not a function call Select or Download Code

Replies are listed 'Best First'.
Re: When is a function call not a function call by Roy Johnson (Monsignor) on Mar 29, 2006 at 21:55 UTC
See Re: leaking into sort when sorting an empty set for a recent explanation of this gotcha. Caution: Contents may have been coded under pressure.	[reply]
Re: When is a function call not a function call by duff (Parson) on Mar 30, 2006 at 04:15 UTC
To make it work I need to resort to: print sort grep {1} uniq 0,0,1,2; I know it's mentioned in the node that Roy Johnson linked to, but I think it bears repeating that you don't need to resort to such arcane chicanery. Simply adding parentheses does the trick: `print sort(uniq(0,0,1,2));` [download] duff	[reply] [d/l]
Re^2: When is a function call not a function call by dpuu (Chaplain) on Mar 30, 2006 at 04:52 UTC
That doesn't seem to work: `% perl -we 'sub fn {print "$a$b\n"} print(sort(fn(0,0,1,2)))' Unquoted string "fn" may clash with future reserved word at -e line 1. 00 10 21 0012` [download] I can eliminate the warnings by choosing a function name that isn't all-lowercase but, as the output demonstrates, the function is still called as a comparator, not a modifier of the input list. Update: I was using perl 5.6.1. The parentheses trick works with 5.8. --Dave Opinions my own; statements of fact may be in error.	[reply] [d/l]
Re^3: When is a function call not a function call by duff (Parson) on Mar 30, 2006 at 05:55 UTC
It's good to know that 5.6.x is broken in this regard in case one of us is unlucky enough to be required to use a 5 year old perl distribution. :-) thanks duff	[reply]


The stupid question is the question not asked
	PerlMonks