warning: string literal in condition

Question

Using the first bit of code below I receive two warning messages: warning: string literal in condition x2

if input == \"N\" || \"n\"
  #do this
else         


        

Answer 1

I was also looking for the answer to this question, and thanks to a friend found a few other solutions.

1) Change the case for the input so you only have to make one test:

if input.downcase == "n"

2) Use a more sophisticated check on the input data:

if %w{n N}.include?(input)     or
if ['n', 'N'].include?(input)

The second one allows for far more flexibility with your checking , especially if there are groups of entry you are looking for.

Hope what I found helps others.

Answer 2

When you are writing input == "N" || "n"( internally Ruby sees it (input == "N") || "n"), it means "n" string object is always a truth value. Because in Ruby every object is true, except nil and false. Ruby interpreter is warned you there is not point to put ever true value is conditional checking. Conditional check statement always expect equality/un-equality test kind of expression. Now you can go ahead this way or re-think again. if input == "N" || input == "n" is not throwing any warning, as it obeys the norm of conditional test.

else input == "L" || "l" is wrong, as else statement don't expect any conditional test expression. Change it to elseif input == "L" || "l"

Answer 3

change input == "N" || "n"

to

input == "N" || input == "n"

You must also use else if instead of else

The warning is saying that instead of a boolean or test, you have a string literal, ' n', which always evaluates to true.

Answer 4

I have the same error as you but a different problem, found this through Google might as well post my solution for the next Googler.

I had a typo in my code which also gives the same warning:

if input =! "N"

of course the right way:

if input != "N"