Checking my intuition/understanding here: 

@a = (1, 2, 3); # works as expected
@a = 1, 2, 3;   # Equivalent to (@a = 1), 2, 3;

[download]

The assignment operator (equals) binds tighter than the comma operator, so in the second case, the assignment is done, then the rest of the commas are processsed. Thus 

$d = ( @a = 1, 2, 3 );
print "$d\n";
print "@a\n";

[download]

prints "3" and "1". $d gets the last value in the comma list, which is being evaluated in scalar context because of the scalar assignment.

Thus in the first example, () doesn't create a list, it just changes the precedence so that all the comma operators evaluate together (and in list context because of the context of the assignment, thus returning a list).

Wow. Never really thought about that before. Thanks, chromatic. I'm not entirely sure why parentheses create list context for lvalues, but I guess that's further homework and grepping through perldoc.

-xdg

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In reply to Re: Re: Re: Re: difference between \() and [] by xdg
in thread difference between \() and [] by eXile

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