Char Subtraction in c++

Question

I am fairly new to C++ and i have some trouble in understanding character subtraction in c++.

I had this code intially

char x='2';
x-='0';
if(x) cout << "More than Zero" << endl;

This returned More than Zero as output so to know the value of x i tried this code.

char x='2';
x-='0';
if(x) cout << x << endl;

And i am getting null character(or new line) as output.

Any help is appreciated.

Answer 1

According to the C++ Standard (2.3 Character sets)

3. ...In both the source and execution basic character sets, the value of each character after 0 in the above list of decimal digits shall be one greater than the value of the previous.

So the codes of adjacent digits in any character set differ by 1.

Thus in this code snippet

char x='2';
x-='0';
if(x) cout << x << endl;

the difference between '2' and '0' (the difference between codes that represent these characters; for example in ASCII these codes are 0x32 and 0x30 while in EBCDIC they are 0xF2 and 0xF0 correspondingly) is equal to 2.

You can check this for example the following way

if(x) cout << ( int )x << endl;

or

if(x) cout << static_cast<int>( x ) << endl;

If you just write

if(x) cout << x << endl;

then the operator << tries to output x as a printable character image of the value 2 because x is of type char.

Answer 2

In C/C++ characters are stored as 8-bit integers with ASCII encoding. So when you do x-='0'; you're subtracting the ASCII value of '0' which is 48 from the ASCII value of '2' which is 50. x is then equal to 2 which is a special control character stating STX (start of text), which is not printable.

If you want to perform arithmetic on characters it's better to subtract '0' from every character before any operation and adding '0' to the result. To avoid problems like running over the range of the 8bit value I'd suggest to cast them on ints or longs.

char x = '2';
int tempVal = x - '0';
/*
Some operations are performed here
*/
x = tempValue % 10 + '0'; 
// % 10 - in case it excedes the range reserved for numbers in ASCII
cout << x << endl;

It's much safer to perform these operations on larger value types, and subtracting the '0' character allows you to perform operations independent on the ASCII encoding like you'd do with casual integers. Then you add '0' to go back to the ASCII encoding, which alows you to print a number.