current = current * 10 + (c - '0');

Question

I\'m trying to read unknown number of integers by this piece of code:

while (1) {
        int c = getchar ();
        if (c == EOF)
            break;
        el         


        

Answer 1

The character '0' does not have ASCII value 0. However, fortunately the character '1' does have ASCII value '0' + 1, and so forth through 2, 3, 4, 5, 6, 7, 8, and 9. Therefore (pseudo code) 'n' - '0' == n holds for each digit.

Answer 2

C specifies the 10 decimal digits, not surprisingly, as

0 1 2 3 4 5 6 7 8 9

C11 §5.2.1 further states "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.

Thus when assessing a string for characters that are digits, the language guarantees that subtracting '0' from a decimal digit char will result in its integer value.

if (isdigit (c))
  int value = c - '0';

This is not dependent on char using ASCII.

Answer 3

The digits are represented in ASCII by increasing values: 0 is 48, 1 is 49, and so on. Since characters in C are just integers in disguise, you can subtract the ASCII value of 0 from any other digit to obtain the numerical value of the digit. For example, '1' - '0' is the same as 49 - 48 which is 1.

Answer 4

c - '0' is a rudimentary way of turning a single ASCII digit into an integer.

For example, if c is equal to '9', then its integer value is 0x39. The ASCII value for '0' is 0x30, so, 0x39-0x30 == 0x09 which is equal to the integer value 9.

Here is the ASCII table for digits:

chr  hex   dec
'0'  0x30  48
'1'  0x31  49
'2'  0x32  50
'3'  0x33  51
'4'  0x34  52
'5'  0x35  53
'6'  0x36  54
'7'  0x37  55
'8'  0x38  56
'9'  0x39  57