Nowhere does Perl guarantee (or even document) the order in which operands are evaluated. For example, consider 

var() = foo() + bar() * baz();

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Perl is documented to multiply the result of bar() with the result of baz() before adding the product to the result of foo(), but it doesn't document when it calls the functions to get the results.

Can you guess the order?

I was going to say Perl always evaluates operands in left to right order, but it's not the case. 

var() = foo() + bar() * baz();
  6       1       2       3
      7       5       4

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You can see that here: 

my $x;

sub var : lvalue { print 'var'; $x }
sub foo          { print 'foo'; 3 }
sub bar          { print 'bar'; 4 }
sub baz          { print 'baz'; 5 }

local $\ = "\n";
var() = foo() + bar() * baz();

print($x);

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>perl 709887.pl
foo
bar
baz
var
23

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However, the order in which Perl evaluates operands is not platform- or compiler-specific. And some people rely on the order for some operators, particularly the comma operator. So it's not likely to change.

There's a catch where lvalues are involved. Even if you know the order in which operands are evaluated, you could be surprised by the results. The operands of the comma operator are evaluated left to right, but you'd think otherwise by looking at the following: 

>perl -le"$i=3; print 0+$i,$i,++$i,$i"
3444

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In short, you might be able to rely on the order in which operands are evaluated, but I consider it a bad practice. Avoid changing an argument and reading it in the same expression. At the very least, it reduces readability and maintainability. The length of this post is proof of that.

In reply to Re: Will "$_[0]=shift" always resolve shift first? by ikegami
in thread Will "$_[0]=shift" always resolve shift first? by kyle

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