How to make a pogram output TeXt LiKe ThIs

[zero44]

Im trying to make a program that can get a String of text then check if it is capital or lowercase for each go then change all the characters so the text come out like AbCdEfGhIJkLmNo.

[BigBear]

In the F10 Help under Turing Language

Keystroke code

how can u tell if a letter is lower case or upper case base on its ordinal value

[zero44]

thx for the help but what i need to knw is what i can do to check if a number is odd or even would help i tried this but it doesn't help

code:

% checks if number is odd var num : int get num if num/2 = int then put "youre number is even" else put "youre number is odd"

[Insectoid]

Every letter has a numerical value: It's ordinal value. This is it's ASCII code. Capital letters have different values than lower case letters. Iterate over each letter (with a for loop I would suggest) and use the ord() command to determine the value of the letter. From there you can figure it out yourself.

[BigBear]

The thing I wanted you to realize was check if it is less than or greater than 96

but to check if it is odd or even

code:

% checks if number is odd var num : int get num if num/2 = int then put "youre number is even" else put "youre number is odd"

First of all there is a few things

code:

if num/2 = int then

num is an integer because when you declared it you put

code:

var num : int

Integer means it can only be whole numbers real numbers can be decimal numbers and whole numbers

for integers instead of using / for divided by you use div

so

code:

if num div 2 = int then

But you cannot say = int because int is a type

You could use

code:

if num div 2 = 0 then

[zero44]

every one thx for your help but these method seems to work perfectly

code:

var num : int var x : int get num x = num mod 2 if x = 1 then put "odd" else put "even" end if

[Insectoid]

So...did we answer your first question right or did you just switch questions? Re-reading the thread it looks like this:

You: Hey I need help with this...
Us: Have a look at this...
You: Thanks, but all along I actually needed help with this other thing.

[zero44]

ahh no i just forget to mention that i already know how to convert a capital to lowercase by adding 32 in ord but i need to figure out how to tell the computer which one to change and which one not to, so i was gonna do it by odds and evens so all the odd counts in string become caps and all the even become lowercase .... i hope i make sense.....having that said if anyone could help change the script so it comes out as boleen so if even true and if odd false?

[BigBear]

you are already checking if even or odd so just after you put "even"

have a variable be assigned to true?

[zero44]

ok this is what i have so far but it doesn't work need to fix 2 things the boleen and how to read what postion in loop the procces is at

code:

var upperCaseString := "a" var lowerCaseString := "" get upperCaseString var num : int var t : int var x : int procedure check     x := num mod 2     if x = 1 then         var t := 1     else         var t := 0     end if     for i : 1 .. length (upperCaseString)         upperCaseString (i) := num         check         if t := 1 then             if ord (upperCaseString (i)) >= 65 and ord (upperCaseString (i)) <= 90 then                 lowerCaseString += upperCaseString (i)             else                 if                         ord (upperCaseString (i)) >= 97 and ord (upperCaseString (i)) <= 122 then                     lowerCaseString += chr (ord (upperCaseString (i)) - 32)                 else                     if t := 0 then                         if ord (upperCaseString (i)) >= 65 and ord (upperCaseString (i)) <= 90 then                             lowerCaseString += chr (ord (upperCaseString (i)) + 32)                         else                             if                                     ord (upperCaseString (i)) >= 97 and ord (upperCaseString (i)) <= 122 then                                 lowerCaseString += upperCaseString (i)                             end if                         end if                     end if                 end if             end if         end if     end for put lowerCaseString

BigBear i think i understand you i will try and reply
EDIT:
thx BigBear i fixed the boolean problem now i have

code:

var upperCaseString := "a" var lowerCaseString := "" get upperCaseString var num : int var t : int var x : int procedure check     x := num mod 2     if x = 1 then         var t := 1     else         var t := 0     end if     for i : 1 .. length (upperCaseString)         upperCaseString (i) := num         check         if t := 1 then             if ord (upperCaseString (i)) >= 65 and ord (upperCaseString (i)) <= 90 then                 lowerCaseString += upperCaseString (i)             else                 if                         ord (upperCaseString (i)) >= 97 and ord (upperCaseString (i)) <= 122 then                     lowerCaseString += chr (ord (upperCaseString (i)) - 32)                 else                     if t := 0 then                         if ord (upperCaseString (i)) >= 65 and ord (upperCaseString (i)) <= 90 then                             lowerCaseString += chr (ord (upperCaseString (i)) + 32)                         else                             if                                     ord (upperCaseString (i)) >= 97 and ord (upperCaseString (i)) <= 122 then                                 lowerCaseString += upperCaseString (i)                             end if                         end if                     end if                 end if             end if         end if     end for put lowerCaseString

[zero44]

sry for double post but my last post was getting too big i have completed the program that i wished to right and it works successfully this is my first ever algorithm and im glad i got the help from you guys so thx
for whoever wants to do something similar or has similar issues here is a script to compare with the old

code:

var upperCaseString := "a" var lowerCaseString := "" get upperCaseString var num : int var t : boolean := true var x : int var a : int procedure check     x := num mod 2     if x = 1 then         t := true     else         t := false     end if end check     procedure tolower     if ord (upperCaseString (a)) >= 65 and ord (upperCaseString (a)) <= 90 then         lowerCaseString += upperCaseString (a)     else         if                 ord (upperCaseString (a)) >= 97 and ord (upperCaseString (a)) <= 122 then             lowerCaseString += chr (ord (upperCaseString (a)) - 32)             else             lowerCaseString += upperCaseString (a)                     end if         end if             end tolower             procedure toup     if ord (upperCaseString (a)) >= 65 and ord (upperCaseString (a)) <= 90 then         lowerCaseString += chr (ord (upperCaseString (a)) + 32)     else         if                 ord (upperCaseString (a)) >= 97 and ord (upperCaseString (a)) <= 122 then             lowerCaseString += upperCaseString (a)             else             lowerCaseString += upperCaseString (a)        end if        end if              end toup for i : 1 .. length (upperCaseString) num := i check     a := i     if t = true then         tolower     end if     if t = false then         toup     end if end for put lowerCaseString

[btiffin]

old guy comment
tests for even and odds;

I don't know Turing syntax

code:

if (var AND 1) == 0 then even

The logical AND is usually a lot cheaper than modulus or division.
Cheers

[Zren]

btiffin @ Tue Sep 15, 2009 10:39 pm wrote:

old guy comment
tests for even and odds;

I don't know Turing syntax

code:

if (var AND 1) == 0 then even

The logical AND is usually a lot cheaper than modulus or division.
Cheers

Mind explaining how the heck that works? or give a link.

[Euphoracle]

Well in binary, if something is odd it _has_ to have 2^0 (1). Therefore, any number represented in binary that has 2^0 as 1 is odd. You can apply this to your current problem.

eg.
7 = 0111 <-- last bit is 2^0 = 1 = odd
6 = 0110 <-- last bit is 2^0 = 0 = even

Rather than using the modulus operator, you can use this knowledge to cheaply check if something is odd.

<var> AND 1 will return 1 if the 2^0 bit of <var> is 1 and 1 is 1 (which it is). If it is zero, it is even because the <var>'s 2^0 bit is not 1.

[Zren]

So if you changed 1 to a three, it would also look at 2^1 and compare that, then use an XAND between the two. So it would be true on 0,4,8,...
I see, so it only looks at the digits in the smallest number. Funky. Thanks Euphoracle.