coding a word with a givin factor

funnyf · #1 11-Jan-2004, 07:28

hi, I have the following problem, I need to code a word with a givin factor
for example:
- word = "ABC"
- factor = 5
'A' + 5, 'B' + 5, 'C' + 5
so the coded word should be "FGH"
can someone help me how I can code this?
this is what i've got, so in the for i 've each letter from the givin word, but how can I count the factor so i it counts 5 letters?
thx a lot..

CPP / C++ / C Code:

#include <iostream.h>

void main()
{
	char woord1[20];
        int factor;
	int teller=-1;

	cout<<"Geef een woord in: ";
	cin>>woord1;
        cout<<"Geef een factor: ";
        cin>>factor;

	while(*(woord1+(++teller))!='\0');

	for(int i=0;i<teller;i++)
	{
		cout<<*(woord1+i)<<" ";
	}
}

Garth Farley · #2 13-Jan-2004, 04:36

This shouldn't be too hard, once you know exactly how chars are stored in the computer's memory. They're stored as 8 bit integers, so (unsigned) up to 256 ACSII characters exist. Each character (including punctuation, and system characters) is assigned a number, the ASCII standard is the main. 'A' is represented by 65, 'B' by 66, and so on (check out www.asciitable.com for the rest)

Say you have

CPP / C++ / C Code:

char letter='a';

To "increase" this letter by 5 places, i.e. to count up 5 letters, you can do the following:

CPP / C++ / C Code:

letter += 5; //or letter = letter +5;

Printing the value of letter will give you 'F'

You just gotta be careful when you reach the end of the alphabet, where does 'Z' go to? Does it wrap around?

GF

funnyf · #3 13-Jan-2004, 08:32

thx a lot dude, i've found it.

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