Arguments are evaluated from left to right. (For me. I don't know if this is defined or undefined behaviour.) However, everything is passed by reference (or would that be "passed by alias" in Perl terminology). That explains why ++$i affected the $i. It must be this way to allow $_[0] = 4 to work.

The following is the C++ equivalent (ignoring the memory leaks). Try it and you will see similar results.

int func(int &at0, int &at1, int &at2);
func(i, *(new int(++i)), *(new int(i+2)));

[download]

The following is a good example of the execution order in Perl:

my $i = 5;
func($i+0, $i, ++$i, $i+0);  # 5 6 6 6 for me.
#    ^^^^  ^^
#  rvalue  lvalue

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From the above, you can also see the solution to your problem. You can work around the problem by converting lvalues to rvalues:

my $i = 5;
func($i,   ++i, $i+2);  # 6 6 8 for me.
     ^^
   lvalue

my $i = 5;
func($i+0, ++i, $i+2);  # 5 6 8 for me.
     ^^^^
    rvalue

[download]

Of course, you can no longer do $_[0] = 4; from within func if you apply this fix.

In reply to Re: Why is the execution order of subexpressions undefined? by ikegami
in thread Why is the execution order of subexpressions undefined? by BrowserUk

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