String to binary

[zero-impact]

Hi I'm making a program just for experimentation that takes a string and converts it into binary.
This is what I have so far.

Turing:

fcn strBin (input : string) : string     var output : string := ""     for i : 1 .. length (input)         var dividend : int := ord (input (i))         var quotient : int := dividend div 2         var remainder : string := ""         for k : 1 .. 8             quotient := dividend div 2             remainder := remainder + intstr (dividend mod 2) %binary digit             dividend := quotient         end for         for decreasing k : 8 .. 1             output := output + remainder (k)         end for     end for     result output end strBin put strBin ("A B C D E F G")

Before I go on to make a function to decode the binary I would like to know if there is any easy way to get around the string length limit in Turing. And if there isn't any ideas on how to do it? It has never really bothered me before but with this function converting each character into eight characters it severely limits the string length.

This question also applies to another problem I'm working on Project Euler problem 25http://projecteuler.net/index.php?section=problems&id=25

The function lrgAdd is a function I made for an assignment in my ICS3M course.

Turing:

fcn lrgAdd (numa : string, numb : string) : string %Function adds two positive integers together no matter how large.     var num1 := numa                               %Numbers must be passed as string the sum is returned as a string.     var num2 := numb     var sum : flexible array 1 .. 0 of char %An array to hold the working sum of the numbers.     var rtnSum : string := "" %A string that will later hold the sum of the two numbers.     var carry, smaller, larger, dif : int := 0 %carry is the value of the int carried over to the next column     if length (num1) > length (num2) then %find the difference in length between the two numbers and pad the smaller number with 0's         dif := length (num1) - length (num2)         num2 := (repeat ("0", dif) + num2)     else         dif := length (num2) - length (num1)         num1 := (repeat ("0", dif) + num1)     end if     new sum, length (num1) %+ 1 %make an array of chars 1 bigger than the numbers (in case of a carry on the last number)     for i : 1 .. upper (sum)  %Make the array hold space chars         sum (i) := ' '     end for     for decreasing i : length (num1) .. 1     %go through every number from right to left         var digitTot : string     %The 2 digit number from the addition of the other 2 digits         digitTot := intstr (strint (num1 (i)) + strint (num2 (i)) + carry)      %the total of the two digits and the carry         if length (digitTot) > 1 then     %if the two numbers add up > 9 then make the carry 1             carry := strint (digitTot (1))             sum (i) := digitTot (2)     %if it is greater than 9 the sum in that column is the right number in digitTot         else             carry := 0      %else reset carry to 0 and sum (i) is the sum of the two digits             sum (i) := digitTot         end if     end for     if carry > 0 then      %if there is a carry on the last number add a 1 at the beginning of the number         rtnSum := rtnSum + "1"     end if     for i : 1 .. upper (sum)         rtnSum := rtnSum + sum (i)     end for     result rtnSum end lrgAdd var prev : string := "0" var cur : string := "1" var temp : string var stream : int open : stream, "lrgFib.txt", put setscreen ("text") loop     temp := lrgAdd (cur, prev)     prev := cur     cur := temp     put : stream, cur     exit when length (cur) = 255 end loop

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[The_Bean]

Pass each letter through the function individually, and output

Turing:

var word : string := ("A B C D E F G") for i : 1 .. length (word)     put strBin (word (i)).. end for

or save to an array.

Turing:

var word : string := ("A B C D E F G") var bin : array 1 .. length (word) of string for i : 1 .. length (word)     bin (i) := strBin (word (i)) end for

[zero-impact]

Thanks!
Wow I can't believe I didn't think of something that simple, DOH!
Unfortunately that doesn't help me with the second problem

[The_Bean]

There is an easier way to convert to binary and then you just have to reverse it to decode.

I think I've solved Project Euler problem 25, my answer is F at 4782.
I used a complex array
var fib : array 1 .. 3, 1 .. 1000 of int
so fib(3,i)=fib(1,i)+fib(2,i)
the 1..1000 represents each column in the number, so
189283 would be
fib(1,1)=3 fib(1,2)=8 fib(1,3)=2 fib(1,4)=9 fib(1,5)=8 fib(1,6)=1
reading from right to left
if the sum is greater than 10 then add the corresponding amount to the next column over

Although my algorithm takes 22 - 25 seconds to solve the problem