my $i;

$i = 0;
print "\nPost-increment: ", $i++, "\nNow equals: ", $i, "\n\n";

$i = 0;
print "\nPre-increment: ", ++$i, "\nNow equals: ", $i, "\n";
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Post-increment: 0
Now Equals: 1

Pre-increment: 1
Now Equals: 1
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The clearest way of describing the difference I can think of is to fully expand the syntax

print ++$i;
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is equivalent to

$i = $i +1; 

print $i;
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whereas

print $i++;
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is equivalent to

print $i; 

$i = $i + 1;
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which I think emphasises the order in which things occur. The difference only becomes material when you use the pre- or post-increment notation as part of another statement. Ie. ++$i; and $i++; as stand-alone statements have identical effect.

The difference only becomes material when the syntax is used as a part of another statement.

Which can be exemplified by

my @a = (1..10);

my $i = 0; 
print $a[$i++] while $i < @a;

12345678910

$i = 0; 

print $a[++$i] while $i < @a;

2345678910
Use of uninitialized value in print at (eval 8) line 1, <> line 7.
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What is the difference in using post as opposed to pre for increment/decrement.

A post-increment (or decrement) is done after the containing expression is evaluated. A pre-increment (or decrement) is done prior to the containing expression being evaluated. An example will help.

$ perl -le 'my $x = 0; print $x++'
0
$ perl -le 'my $x = 0; print ++$x'
1
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As for when you might want to use one instead of another, the one thing that springs to mind is when you want to increment or decrement an array index within the expression using it. The code below is contrived and I'd never actually do it like this, but it illustrates the point.

# Fill @array with ten random numbers.
my $i = 0; 
$array[$i++] = rand() while $i < 10;
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-sauoq
"My two cents aren't worth a dime.";

Relevant snippet from perlop:

"++" and "--" work as in C. That is, if placed before a variable, they increment or decrement the variable before returning the value, and if placed after, increment or decrement the variable after returning the value.

The only difference is in what value is returned. Maybe the following code will demonstrate the difference more clearly:

$a = 1; print ++$a, ":", $a, "\n";
$a = 1; print $a++, ":", $a, "\n";
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Update: yikes. 2 answers were posted in the time it took me to compose mine. I had a feeling this would happen with a question like this... :)

I fail to see the problem. Pre-increment means that the value will be incremented before the value is returned. Post-increment means the value will be incremented after it is returned. It's very important to realize that it is not defined exactly when the increment occurs. It will be done after execution of the current statement started, and will be done before the statement is finished, but that's all you get.

So, for your first line, you could get "2:2", or "2:1" and for the second line, you could get "1:2" or "1:1". Or some other weird value for the second $a.

As a rule, never use a variable you are pre- or post incrementing or decrementing a second time in the same statement. And most certainly, don't modify the variable a second time. The following is undefined:

    my $i = 0;
    $i = ++ $i;
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meaning that if perl decides to erase your hard disk, it's still well inside the boundaries of its specification.

Abigail

$i = 0;
$j = ++$i;

$i and $j are both 1

$i = 0;
$j = $i++;

$i is now 1 and $j is 0
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Yes, that is correct. One cool use I've seen for post-increment is in object initialization when the Damien Conway prevents multiple inheritance:

return if ($self->{_init}{Object}++);

This returns control to the calling execution if the initializing function has already been called in the inheritance hierarchy. If it hasn't already been initialized, it increments and proceeds. If it has already been initialized, it returns.