Why does this algorithm convert decimal to hexadecimal numbers?

Question

I can't quite seem to figure out how does such algorithm work...

Code:

#include <stdio.h>

int main(void){


int number = 0;

while(number < 16){

  if(number < 10){
    printf("decimal: \t%d\n", number);
    printf("hexadec.: \t%c\n", number  + '0');
  }else{
    printf("decimal: \t%d\n", number);
    printf("hexadec.: \t%c\n", number  - 10 + 'A');
  }
  
  printf("\n");
  number++;
  }

  return 1;
}

Output:

decimal:    0
hexadec.:   0

decimal:    1
hexadec.:   1

decimal:    2
hexadec.:   2

decimal:    3
hexadec.:   3

decimal:    4
hexadec.:   4

decimal:    5
hexadec.:   5

decimal:    6
hexadec.:   6

decimal:    7
hexadec.:   7

decimal:    8
hexadec.:   8

decimal:    9
hexadec.:   9

decimal:    10
hexadec.:   A

decimal:    11
hexadec.:   B

decimal:    12
hexadec.:   C

decimal:    13
hexadec.:   D

decimal:    14
hexadec.:   E

decimal:    15
hexadec.:   F

It prints out exactly what I want but I don't understand the process behind it. If I am not mistaken integer + 'character' converts the number to a character by the ascii table.

Let's say we have number 11, then 11 - 10 = 1 which is 49 in ascii, A is 65 in ascii. So how does 49 + 65 = 66. I am sure this thinking is completely wrong, I just wanted to show you what I think it does in the background.

Answer 1

When you change

printf("hexadec.: \t%c\n", number  - 10 + 'A');

to this:

printf("hexadec.: \t%c\n", number  - 9 + 'A');

the first letter you get when 'number' is over 10 is 'B'.

Thats because you count up the Letters from the current value of 'number'.

so when number has the Value 12 you subtract it with 10 and you have the value 2 stored in number 'A' + 2 = 'C'.

you just count up the letters.