Converting a digit from char to integer in C language

I'm reading the C programming textbook¹ and in 1.6 they're saying the conversion between a char containing a digit and an int can be done like this:

char character = '7';
int integerChar = character - '0';

I'm having trouble understanding what is happening here and why the integer value is equal to the character minus the character '0'.

¹ Brian W Kernighan and Dennis M Ritchie The C Programming Language, 2nd Edn 1988.

Solution

From the C Standard (5.2.1 Character sets)

...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 it means that the Standard guarantees that for example the difference '1' - '0' is equal to 1, or the difference '2' - '0' is equal to 2 and so on.

So independent of the internal representation of characters as for example ASCII or EBCDIC you can get an integer digit that is represented by a character c like c - '0'.

For example in the ASCII character table the characters '0' through '9' have codes from 48 up to 57. In the EBCDIC character table they have codes from 240 up to 249.

Here is a demonstration program.

#include <stdio.h>

int main( void )
{
    for (char c = '0'; c <= '9'; c++)
    {
        printf( "'%c': %d\n", c, c - '0' );
    }
}

The program output is

'0': 0
'1': 1
'2': 2
'3': 3
'4': 4
'5': 5
'6': 6
'7': 7
'8': 8
'9': 9