This is because the entire expression is evaluated before being passed as arguments to print. print doesn't get called iteratively for each argument. It gets its arguments, and then iterates.

So that means $a = 2 is evaluated, then ' ', then $a = 3, and then the values of $a, ' ', and $a are passed to print. By the time the expression has been evaluated, $a contains 3 no matter how many times it appears in @_.

Well, I would characterize the reason as having more to do with the fact that passing $a = 2 as an argument causes an alias to $a to be put into @_. Had a copy of $a's value been placed into @_ instead, then subsequently changing $a to 3 would not impact that copy.

print $a = 1, 0+$a, $a = 2, 0+$a, $a = 3, 0+$a;
# prints "313233"
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The reason that $a = 2 passes an alias is because it is an "lvalue" expression, which just means that you can do previously mentioned things like ( $a = 2 ) = 3. And the reason that lvalue expressions cause aliases to be passed is two-fold: So you can modify arguments via code like $_[0] = 3 in your sub and because it is more efficient than creating a copy.

- tye        

This is because the entire expression is evaluated before being passed as arguments to print

No, the arguments are placed on the stack as they are evaluated.

1. $a = 2 is evaluated and the result ($a) is placed on the stack.
2. ' ' is evaluated and the result is placed on the stack.
3. $a = 3 is evaluated and the result ($a) is placed on the stack.
4. "\n" is evaluated and the result is placed on the stack.
5. print is evaluated.

@_ isn't even constructed here since print is an operator, not a sub.

The reason you get the observed behaviour is that $a itself (not a copy of it) is placed on the stack.