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Define a palindrome operator that duplicates the values on the stack in reverse order in postscript

Define a palindrome operator that duplicates the values on the stack in reverse order.

this is what i have so far and it not doing what i want it to do

/palindrome { 1 dict begin count 1 gt { /first exch def /second exch def temp1 = first temp2 = last first = last last = temp1 } } def

Solution

  • Most of what you have written there doesn't make any sense in PostScript:

    /palindrome
{
  1 dict begin
  count 1 gt 
  {
    /first exch def
    /second exch def
%% The following four lines are not valid PostScript
    temp1 = first
    temp2 = last
    first = last
    last = temp1
%% There is no '=' assignment operator in PostScript, in PS the = operator
%% pops one object from the stack and writes a text representation to stdout.
%% You have not defined any of the keys temp1, temp2 or last
%% in any dictionary. If executed I would expect this program to throw an
%% 'undefined' error in 'temp1'
  }
%% Given the 'count 1 gt' at the opening brace, I would expect this executable
%% array to be followed by a conditional, such as 'if'. Since it isn't this just
%% leaves the executable array '{....}' on the stack
} def

    So overall I would expect this PostScript function to push a boolean on to the operand stack, true or false depending on whether the stack has at least 2 objects on it at the time of execution, and then an executable array on to the operand stack and exit.

    If I were doing this I would store the stack into an array, then unload the array back to the stack and then iterate through the array from end to beginning. Something like:

    %!

/palindrome
{
  count array astore
  aload
  dup length 1 sub -1 0 {
   1 index exch get
    exch
  } for
  pop
} def

(line 1)
2
3

(before palindrome\n) print
pstack
palindrome
(after palindrome\n) print
pstack

    It's also possible (I have a working example here) to do this in a single pass without defining any extra storage objects (dictionaries or arrays) by using a for loop and manipulating the stack. That seems like a more elegant solution to me, and is left as an exercise for the reader :-)