If you run your code under a perl built with -DDEBUGGING, which can show each instruction being executed and what's on the stack after each execution, things may become clearer.

Here's the output of perl -Dst -e'my $x=5; $x=$x + ++$x + $x++;' with some annotations. I've skipped the first assign statement. 

The padsv op pushes $x (not a copy) onto the perl stack;
$x is currently an integer value (IV) with value 5:

(-e:1)    padsv($x)
    =>  IV(5)  

$x is pushed onto the stack again:

(-e:1)    padsv($x)
    =>  IV(5)  IV(5)  

preinc increments the variable $x; note that both values on the stack
have now incremented (because they are both $x, not copies)

(-e:1)    preinc
    =>  IV(6)  IV(6)  

the add op takes the top two values on the stack and
replaces them with a new value:

(-e:1)    add
    =>  IV(12)  

push $x again. Since it was preinc()ed earlier,
it now has the value 6:

(-e:1)    padsv($x)
    =>  IV(12)  IV(6)  

postinc takes the variable $x off the stack, makes a copy
and pushes that copy back on the stack, then increments $x.
If we had any other direct $x's on the stack (which we don't),
their values would appear incremented:

(-e:1)    postinc
    =>  IV(12)  IV(6)  

the final add does the obvious thing.

(-e:1)    add
    =>  IV(18)

[download]
Note that this is what currently *happens* to happen. Code should not rely on this behaviour.

Dave.

In reply to Re: question 1st - undefined behaviour by dave_the_m
in thread question 1st - undefined behaviour by rumos2

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