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

The C++ standard says, in its [lex.charset] section “the value of each character after 0 in the above list of decimal digits shall be one greater than the value of the previous” (quoted from draft N4659).

Thus, the value of '1' is '0'+1, and the value of '2' is one more than that, which is '0'+2, and so on. If n has a value from 0 to 9, then '0'+n is the value for the corresponding character from '0' to '9'.

(Unlike the earlier answers, this answer does not assume ASCII and shows how the property derives from the C++ standard.)