Python backslash in string evaluation

Question

I am debugging my program and I am confused as to why one of my statements is evaluating to false. I am checking if the second to last index in my array (which is a string) starts with ” character. I am writing an interpreter for post script and so this helps me determine whether or not the user is defining a variable ex: x. My if statement that is checking this for me is evaluating to false and I cannot figure out why. Any thoughts?

def psDef():
   if(opstack[-2].startswith('\')):   # this is evaluating to false for some reason
       name = opPop() #pop the name off the operand stack
       value = opPop() #pop the value off the operand stack
       define(name, value)
   else:
       print("Improper use of keyword: def")

def testLookup():
   opPush("n1")
   opPush(3)
   psDef()
   if lookup("n1") != 3:
       return False
   return True

Asked By: CodySig

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Source

Answer 1

Have a look at String literals

The backslash () character is used to escape characters that
otherwise have a special meaning, such as newline, backslash itself,
or the quote character.

n means ASCII Linefeed (LF).So to display a backslash in a string literal you need to escape the backslash with another backslash.

like opPush("\n1")

Hope this helps.

Answered By: McGrady

Answer 2

@McGrady has already pointed out your error. If you change this line:

 opPush("n1")

to something like:

 opPush("\n1")

you’ll probably get what you want.

Answered By: aghast

Answer 3

By default Python accepts backslash escape sequences in strings. n becomes a newline.

To get an actual sequence of and n you can use a double backslash: '\n' or mark the string as “raw”: r'n'.

Answered By: Klaus D.

Python backslash in string evaluation

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