char math?

[mdh]

Can someone enlighten me?

let "s" be a string array and "i, n" be an integers.

Why does the expression:

s= n % 10 + '0' put the "string value" of the modulo in the array.

I thought by placing the 0 in a single quote, you are telling the
compiler to treat 0 as a character.
I fail to see how "n % 10" added to that produces the correct result,
which it does!
Thanks in advance.

Ref to this is page 64 of K&R(2);

 

[Richard Heathfield]

mdh said:

Why does the expression:

s= n % 10 + '0' put the "string value" of the modulo in the array.

Because of 5.2.1(3) which says, in part, "...the value of each character
after 0 in the above list of decimal digits shall be one greater than the
previous".

(It means '0', by the way.)

So - 0 + '0' is '0', obviously. But the above text means that 1 + '0' is
guaranteed to be '1', 2 + '0' is guaranteed to be '2', etc.

 

[mdh]

mdh said:

Because of 5.2.1(3) which says, in part, "...the value of each character
after 0 in the above list of decimal digits shall be one greater than the
previous".

(It means '0', by the way.)

So - 0 + '0' is '0', obviously. But the above text means that 1 + '0' is
guaranteed to be '1', 2 + '0' is guaranteed to be '2', etc.

OK...thanks...
I was trying to make a much bigger deal out of it.