Help real quick, I gotta go in 4 mins eep

[Flikerator]

Im going to a contest soon and I only have a few minutes before I leave I know its chance someone will answer but I can try. Why doesn't this work, it removes all letters.

code:

var word : string := "hel3lOE1asa5d" var neword : string := "" for i : 1 .. length (word)     if ord (word (i)) < 65 and ord (word (i)) > 90 and ord (word (i)) < 97 and ord (word (i)) > 122 then         neword += word (i)         put ord (word (i))     end if end for put neword

25 bits if I get the answer before I leave

[MysticVegeta]

What do you mean removes all the letters?

[MysticVegeta]

code:

var word : string := "hel3lOE1asa5d" var neword : string := "" for i : 1 .. length (word)     if (ord (word (i)) > 65 and ord (word (i)) < 90) or (ord (word (i)) > 97 and ord (word (i)) < 122) then         neword += word (i)         put ord (word (i))     end if end for put neword

you mean this?

[cornputer]

because of the and at the second part
turing doesnt take the second part as (condition1 +condition2)or (1 + 2 )
it takes condition 1 AND (1 or 2) AND condition 3
not sure if im totally right but ive tried a similar error and i think it was like this

[Flikerator]

Im back, came second I think

There were no questions with that so its all good.

I jsut wanted all the uppercase and lowercase to be taken out of the string leaving the rest. If anyone has an answer to it ill give em 10 bits, it has to use ord.

if word (i) = "a" then

Is something I want to avoid, thats the whole reason for the ord

[MysticVegeta]

code:

var word : string := "ASDasdaASI/././D123JsE" var ne : string := "" for s : 1 .. length (word)     if (word (s) >= "A" and word (s) <= "Z") or (word (s) >= "a" and word (s) <= "z") then         ne := ne     else         ne += word (s)     end if end for put ne

This?

Offtopic: what contest?

[Token]

code:

var word : string := "hel3lOE1asa5d" var neword : string := "" for i : 1 .. length (word)     if ord (word (i)) > 64 and ord (word (i)) < 91 or ord (word (i)) > 96 and ord (word (i)) < 122 then     else         neword += word (i)     end if end for put neword

You had the right idea, just a little off

[Cervantes]

It should be noted that, depending on what you intend to do with the program, the second variable may be unnecessary, for one could just do this:

code:

word := word (1 .. i - 1) + word (i + 1 .. *)

But that brings me to a question:

code:

var word : string := "hel3lOE1asa5d" for i : 1 .. length (word)     if ord (word (i)) > 64 and ord (word (i)) < 91 or ord (word (i)) > 96 and ord (word (i)) < 122 then     else         word := word (1 .. i - 1) + word (i + 1 .. *)     end if end for put word

I'm getting an error: substring index is greater than length of string. Do you guys get the error as well? (Over the years I've sometimes noticed Turing behaving irrationally. I've failed to find anything wrong with my code, so I'm wondering if it's Turing itself. ) Does turing highlight the if statement? It seems to me that if there is an error it should be at the word := word (1 etc.
I get the same error when I rearrange the if statement so we don't have that hanging "else" there:

code:

var word : string := "hel3lOE1asa5d" for i : 1 .. length (word)     %if the character, word (i), is not a letter.     if ord (word (i)) < 65 or ord (word (i)) > 122 or ord (word (i)) > 90 and ord (word (i)) < 97 then             word := word (1 .. i - 1) + word (i + 1 .. *)     end if end for put word

Even the word assignment line should not be a proble. Assuming i = 1:
word (1 .. i - 1) will be word (1 .. 0) which equates to nothing. We are allowed to have the right number one less than the left number.
word (i + 1 .. *) will be word (2 .. *), which should be fine.

Assuming i = length (word):
word (1 .. i - 1) should be fine. word (i + 1 .. *) will be word (* + 1 .. *) which is again a case of the right side being one less than the left side.
When that happens (right side 1 less than left) it just skips over it.

Anyone have any ideas as to why there's an error?

[Token]

yah man thats probibly turing just being screwy again, it has done that to me before too, i made this calculator program and it kept telling me that the square root of 9 was 2.999999 lol it was messssssssssed up.

[MysticVegeta]

Token wrote:

code:

var word : string := "hel3lOE1asa5d" var neword : string := "" for i : 1 .. length (word)     if ord (word (i)) > 64 and ord (word (i)) < 91 or ord (word (i)) > 96 and ord (word (i)) < 122 then     else         neword += word (i)     end if end for put neword

You had the right idea, just a little off

eh? so he needs to use (ord)? because this works too

[Token]

yah, urs is rite, it works, but he needed to use ord, which is the same thing as what your doing becuase it automatically recognises the ascii value in yours.

[MysticVegeta]

why make it so confusing using ord?

[Cervantes]

1.) It wouldn't be confusing if it were commented.
2.) Comparing the ordinal value of each character to a set of numbers (we can use ranges, because of the way the ASCII chart is set up) saves us from having to do 52 (26 * 2) if statements or a case structure with 52 labels or an index comparing the character to a string with 52 characters.

It's a lot easier!

EDIT: Hey, whaddya know, you're code would work too. I missed the >= and <= operators and thought you'd only checked for a's and z's.

In any case, when comparing two characters, the computer converts them to their ordinal values and compares those integers. So, essentially it's the same thing.

[MysticVegeta]

Cervantes wrote:

when comparing two characters, the computer converts them to their ordinal values and compares those integers. So, essentially it's the same thing.

i win? just kidding

[Cervantes]

It's a question of how you like to think about it. Personally, I find the ord method easier to understand. It means you have to open up the ascii chart, but it means you don't have to go the extra step of thinking about how the computer compares characters. Plus, since you've already got the ascii chart open, you can look at the range and everything within it. For things other than the alphabet and numbers, you'd have to open it up regardless. So using the ord method is more coding, but, for me at least, it's easier to understand.