Why does adding a '0' to an int digit allow conversion to a char?

Question

I\'ve seen examples of this all over the place:

int i = 2;
char c = i + \'0\';
string s;
s += char(i + \'0\');

However, I have not yet seen

Answer 1

When ASCII encoding is used, the integer value of '0' is 48.

'0' + 1 = 49 = '1'
'0' + 2 = 50 = '2'

...

'0' + 9 = 57 = '9'

So, if you wanted convert a digit to its corresponding character, just add '0' to it.

Even if the platfrom uses non-ASCII encoding, the lanuage still guarantees that the characters '0' - '9' must be encoded such that:

'1' - '0' = 1
'2' - '0' = 2
'3' - '0' = 3
'4' - '0' = 4
'5' - '0' = 5
'6' - '0' = 6
'7' - '0' = 7
'8' - '0' = 8
'9' - '0' = 9

When ASCII encoding is used, that becomes:

'1' - '0' = 49 - 48 = 1
'2' - '0' = 50 - 48 = 2
'3' - '0' = 51 - 48 = 3
'4' - '0' = 52 - 48 = 4
'5' - '0' = 53 - 48 = 5
'6' - '0' = 54 - 48 = 6
'7' - '0' = 55 - 48 = 7
'8' - '0' = 56 - 48 = 8
'9' - '0' = 57 - 48 = 9

Hence, regardless of the character encoding used by a platform, the lines

int i = 2;
char c = i + '0';

will always result in the value of c being equal to the character '2'.