How i can separate integer number in 3 "houses"? Hundred, Ten and Unity.

Projects Programming
1

: smiley-neutral:

I have, for example, a variable int number = 300; and i need modify "number" by "number", i wonder if i need separate in 3 variables for hundred, ten, unity or if there a method for divide that enable to me change a unique variable "number' by "number", "house" by "house", a hundred, a ten and unity (3 - 2 - 1).

Example: The user need only change number "2" of 3'2'1, and he want that "2" to being "5", as "321" must become to "351" . In other words, the number 3 and 0 not will modified, only number 2 from 321, turning 3-5-1.

2

It would probably make life easier if you start with the number as text - i.e. 3 or 4 characters. For example

char numberAsText[] = "321";

then you could change the middle character with

numberAsText[1] = '5';

Note that arrays count from 0 and you use single quotes for single characters.

And you can change the text to an int if you need to do maths with it like this

int myNumber = atoi(numberAsText);

The atoi() function converts ASCII text to an integer

...R

3

Using Robins example you can do it in 4 lines including dtostrf() to convert the passed number to a char array,
unfortunately for me i overthought the process in the past and had to divide and multiply the number x times,
I see no need to post it, as quite frankly in comparison to Robins it is blotted but just for giggles here it is.

void setup()
{
  Serial.begin(115200);
  changeDigit(6789, 5, 1);
}

void loop()
{

}

long changeDigit(const long& a_num, const byte& a_newDigit, const byte& a_position) {
  const byte numberOfDigits = countDigits(a_num);
  Serial.print("numberOfDigits = "); Serial.println(numberOfDigits);
  byte digits[numberOfDigits] {0};
  unsigned long division = 1;
  for (byte i = 0; i < numberOfDigits; i++) {
    digits[i] = int(a_num / division) % 10;
    division *= 10;
  }
  // Set new digit at a_position
  if (a_newDigit <= 9 && a_position < numberOfDigits) {
    digits[a_position] = a_newDigit;
  }
  // Add all the digits back into one number
  division = 1;
  long finalNumber = 0;
  for (byte i = 0; i < numberOfDigits; i++) {
    finalNumber += int(digits[i] * division);
    division *= 10;
  }
  Serial.print(F("Final number = ")); Serial.println(finalNumber);
  return finalNumber;
}

byte countDigits(long num) {
  byte count = 0;
  while (num > 0) {
    num /= 10;
    count++;
  }
  return count;
}
4

Your example will be useful to me too, as well as the other rapaze, thanks for your help!