Thread: Convert integer to character

It solves for getting a single character only. For instance you have number 62

the You would call the function as byteTochar(62,1), which would return 6+48=54 or "6"
Again if byteTochar(62,2) then it would return 2+48=50 or "2"

Oh I made a simple mistake by the way ,
In line 10, it should be

Code:

anybyte%10

instead of

Code:

10%anybyte

EDIT: byte data type can be obtained by including windows.h

Originally Posted by Swoorup

It solves for getting a single character only. For instance you have number 62

the You would call the function as byteTochar(62,1), which would return 6+48=54 or "6"
Again if byteTochar(62,2) then it would return 2+48=50 or "2"

In that case, your function is misnamed, and the parameters should not be of type byte.

As for improvement: implement the function that I proposed.

Why do you simply not use the function manasij7479 have already suggested? Why do you have to make it so hard on yourself?
Furthermore, why are you using printf?

Originally Posted by Swoorup

Because I have to limit the memory as far as possible. Only a byte is to be passed and output of two characters are needed.
(...)
Furthermore the numbers are only 0 to 63.

In that case, you should do something like this:

Code:

/* Converts integer number in range [0, 63] to its numeric decimal string
 * representation, storing the result in a char array without null termination.
 */
void integerToCharArray(byte number, char* result) {
    result[0] = (number / 10) + '0';
    result[1] = (number % 10) + '0';
}