That's a lot of questions! Here goes...

Answer to Question 1

I'll do two examples, $x,$x,$x++: 

{
   local @_;
   alias $_[0] = $x;
   alias $_[1] = $x;
   alias $_[2] = $x++;
   &print;              # @_ = ($x, $x, $anon) = (1, 1, 0)
}

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$x,$x++,$x++: 

{
   local @_;
   alias $_[0] = $x;
   alias $_[1] = $x++;
   alias $_[2] = $x++;
   &print;              # @_ = ($x, $anon1, $anon2) = (2, 1, 0)
}

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$x,$x++,$x++, on a system that does the alias statments in a different order: 

{
   local @_;
   alias $_[2] = $x++;
   alias $_[1] = $x++;
   alias $_[0] = $x;
   &print;              # @_ = ($x, $anon1, $anon2) = (2, 0, 1)
}

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Answer to Question 2

Your compiler probably does something like: 

load x
inc
push
load x
push
call printf

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C doesn't guarantee the order either.

Answer to Question 3

The return value of $x++ is not $x. It's a new variable that contains the old value of $x. $x is incremented before the print call, but you are aren't printing $x. Check perlop.

Answer to Question 4

                        #   $x  |  anon returned  |  $_[0]  |  $_[1]
                        #       |     by $x++     |         |
                        # ------+-----------------+---------+---------
my $x = 0;              #    0  |             --  |     --  |     --
{                       #    0  |             --  |     --  |     --
   local @_;            #    0  |             --  |  undef  |  undef
   alias $_[0] = $x;    #    0  |             --  |      0  |  undef
   alias $_[1] = $x++;  #    1  |              0  |      1  |      0
   &print;
}

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or in even more detail 

                        #   $x  |  anon returned  |  $_[0]  |  $_[1]
                        #       |     by $x++     |         |
                        # ------+-----------------+---------+---------
my $x = 0;              #    0  |             --  |     --  |     --
{                       #    0  |             --  |     --  |     --
   local @_;            #    0  |             --  |  undef  |  undef
   alias $_[0] = $x;    #    0  |             --  |      0  |  undef
   my $anon = $x++;     #    1  |              0  |      1  |  undef
   alias $_[1] = $anon; #    1  |              0  |      1  |      0
   &print;
}

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Answer to Question 5

Guess what the following prints: 

perl -le"$x=3; sub { $_[1]++; $_[2]++; print @_ }->($x+0, ++$x, $x++, 
+$x+0, $x);"

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On my system, it prints 36556. 

my $x = 3;                #  $anon0 | $x    | $anon2 | $anon3 | $x
{                         #  $_[0]  | $_[1] | $_[2]  | $_[3]  | $_[4]
   local @_;              # --------+-------+--------+--------+-------

   # $x+0
   my $anon0 = $x+0;
   alias $_[0] = $anon0;  #       3 |    -- |     -- |     -- |    --

   # ++$x
   $x=$x+1;               #       3 |    -- |     -- |     -- |    --
   alias $_[1] = $x;      #       3 |     4 |     -- |     -- |    --

   # $x++
   my $anon2 = $x;
   alias $_[2] = $anon2;  #       3 |     4 |      4 |     -- |    --
   $x=$x+1;               #       3 |     5 |      4 |     -- |    --

   # $x+0
   my $anon3 = $x+0;
   alias $_[3] = $anon3;  #       3 |     5 |      4 |      5 |    --

   # $x
   alias $_[4] = $x;      #       3 |     5 |      4 |      5 |     5

   &anon_sub;
   #  $_[1]++;            #       3 |     6 |      4 |      5 |     6
   #  $_[2]++;            #       3 |     6 |      5 |      5 |     6
   #  print @_;
}

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Notice how $x and $x+0 are different.
Notice how $_[1]++ affects more than one argument.
Notice how $_[1]++ affects $x and ++$x.
Notice how $_[1]++ doesn't affect $x+0 and $x++.

In reply to Re^3: print behavior by ikegami
in thread print behavior by Anonymous Monk

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