unexpected output in perl

doughyi8u · 12-14-2013, 08:18 PM

Code:

#!/usr/bin/perl -w

@var = (0 .. 10);
$value = 1;
$a = 0;
$var[$a++] = $var[$a++] + $value;
foreach (@var) {
    print $_ . " ";
}
print "\n";

I'm expecting this to output:

Code:

2 1 2 3 4 5 6 7 8 9 10

but I'm getting:

Code:

0 1 2 3 4 5 6 7 8 9 10

When I change the line:

Code:

$var[$a++] = $var[$a++] + $value;

to:

Code:

$var[0] = $var[1] + $value;

I get the expected out put:

Code:

2 1 2 3 4 5 6 7 8 9 10

What's the problem?

druuna · 12-15-2013, 04:35 AM

Using multiple increments on the same variable in one line leads to undefined behaviour. This from the perlop document:

Quote:

Note that just as in C, Perl doesn't define when the variable is incremented or decremented. You just know it will be done sometime before or after the value is returned. This also means that modifying a variable twice in the same statement will lead to undefined behavior. Avoid statements like:

$i = $i ++;
print ++ $i + $i ++;

Perl will not guarantee what the result of the above statements is.

You can change your code to something like this:

Code:

#!/usr/bin/perl -w

@var   = ( 0 .. 10 ) ;
$value = 1 ;
$a     = 0 ;
$b     = 0 ;

$var[ $a ] = $var[ ++$b ] + $value ;

foreach ( @var ) {
  print $_ . " " ;
}

print "\n" ;

Which will result in:

Code:

$ ./foo.pl 
2 1 2 3 4 5 6 7 8 9 10

mina86 · 12-15-2013, 02:50 PM

By the way, what's likely happening is that “$var[$a++] + $value” is evaluated fully with its side effects before the rest of the assignment. This leads to accessing “$var[0]”, which is zero, adding value of “$value”, which is one, to get the result one, followed by incrementing “$a”. At this point, the rest of the statement is evaluated: “$var[$a++] = 1”. This time, “$a” is one so “$var[1]” is assigned the value of one (which it already has).

Of course, as druuna described, this is not reliable behaviour and may change at any time theoretically even within the same run of perl interpreter.

chrism01 · 12-22-2013, 05:21 PM

Any time I see the phrase 'undefined behaviour' it reminds me of a line in the Advanced C book by Peter Van der Linden (http://www.amazon.com/Expert-Program.../dp/0131774298).
It went something along these lines

Quote:

.. undefined behaviour (or implementation defined) means anything can happen as desired by the author (he's talking about the C compiler writer here..).
This may range from doing that calcn all the way to launching nuclear missiles, assuming you have the relevant Hardware module installed...

That last definitely helps me to treat it as a red flag

Sergei Steshenko · 12-26-2013, 10:02 PM

Quote:

Originally Posted by doughyi8u

... What's the problem?

If you have an expectation, you should first verify that the expectation is correct according to the official documentation.