Why doesn't this if statement work (C)?

[Thsharp01]

if (inputAnswer != 'A' || inputAnswer != 'B' || inputAnswer != 'C' || inputAnswer != 'D') { printf("error"); }

Even if 'inputAnswer' equals A, B, C or D it still displays 'error', i can't figure out why it's doing this. If i take out the ||(or) and just have the one inputAnswer !='A' then it works, but not when i add more than one.

Anyone able to shed some light on this please?

Thank you for reading!

 

[Mickey]

Think about what the statement is saying.

A simplified version is:

if (inputAnswer != 'A' || inputAnswer != 'B') { do something }

If inputAnswer is not 'A' OR inputAnswer is not 'B', then do something. Is one of those statements still true if inputAnswer is 'A'?


Mick.

 

[Haircut]

It seems to me that you want to get:

Not (A or B or C or D), correct?

What you are actually doing is (Not A) or (Not B) or (Not C) or (Not D)
and the two are not logically equivalent (see here, http://en.wikipedia.org/wiki/De_Morgan's_laws)

You can either change the parentheses, or change the ORs to ANDs and it should work.

 

[GravyMonster]

*edit* dammit, didnt read the last paragraph above

I'm not a C programmer, but shouldn't it be...

if (inputAnswer != 'A' && inputAnswer != 'B' && inputAnswer != 'C' && inputAnswer != 'D') { printf("error"); }

As if inputAnswer is 'A', the != 'B'/'C'/'D' conditions are true, hence you get an error?

 

[Thsharp01]

Thank you both for your help. Changed them to 'ands' and it does the job i want it to do.

Edit: I should have known this, but it seemed i was getting myself confused