K&R 1.6 Arrays // Digit representation in an array construct - c

Question

I found this example code on using arrays in the C language.
#include <stdio.h>
main () {
int c, i;
int ndigit[10];
for (i = 0; i < 10; ++i)
ndigit[i]=0;
while ((c = getchar()) != EOF)
if (c >= '0' && c <= '9')
++ndigit[c - '0'];
printf("digits =");
for (i = 0; i < 10; ++i)
printf(" %d", ndigit[i]);
}
I never saw arrays before, but I think I got it.
Still, I'm not sure on why the digit values have to be inserted in '..' nor why the assignement of i has to be expressed as c-'0'.
This is a passage of the book that should clarify my doubts:
This particolar program relies on the properties of the character representation of the digits. For example the text if (c >= '0' && c <= '9') determines whether the characters in c is a digit. If it is, the numeric value if that digit is c - '0'.
I don't understand how can these values be used in arithmetical expressions if they are characters, is it because they are mapped to numerical values?
Then why the whole program just doesn't work if they are written as numbers as in if (c >= 0 && c <= 9) nor it works if c isn't written in that way (which to my understanding is just "whatever number c is minus 0).

Answer 1

TL;DR: a "char" is just a one-byte-long integer.
I don't understand how can these values be used in arithmetical expressions if they are characters, is it because they are mapped to numerical values?
In C, a char is the "smallest addressable unit of the machine that can contain basic character set. It is an integer type." [1]. Normally, char is equivalent to "a one-byte-long integer", so they can hold values from 0 to (2^8)-1, or [0,255].
That being said, when you write
char c = '9';
You are saying "c is a one-byte-long integer whose value is the character-set representation of the character 9". By looking at the most common character set, the ASCII table [2], we see that the character 9 has an integer value of 57, so the above expression is equivalent to
char c = 57;
To convert a digit's character-set value to the digit itself (e.g. '9' to 9, or 57 to 9), you can rely on a property of character sets that digits are always stored sequentially and increasingly, and just subtract by the value of '0', which in ASCII is 48, so:
char c;
c = '9' - '0'; /* = 9 In any character set */
c = 57 - 48; /* = 9 */
c = '9' - 48; /* = 9 In ASCII */
c = 57 - '0'; /* = 9 In ASCII */
Keep in mind that while ASCII is the most common character set, this is actually machine-dependent.
[1] http://en.wikipedia.org/wiki/C_data_types#Basic_types
[2] http://www.asciitable.com/

Answer 2

if you see the man page of getchar() it says
....reads the next character from stdin and returns it as an unsigned char cast to an int....
So, an input of a digit [example, 9] is treated as a char input and the corresponding encoded [Usually ASCII] value is returned by getchar().
Now coming to your question(s),
why the digit values have to be inserted in '..'
A digit [or any other character, for that matter], written as '.', represents the corresponding ASCII value of the same. Check the ASCII table here.
For understanding, a 9 is a 9 whereas a '9' represents the correcsponding ASCII 57.
why the assignment of i has to be expressed as c-'0'.
If you notice the ASCII table closely, you can see, the corresponding values of 0 to 9 are in sequence. So, to get the particular digit as an int value, we can do c - '0' which is same as c - 48 which will give us the digit as an int.

Answer 3

I don't understand how can these values be used in arithmetical
expressions if they are characters, is it because they are mapped to
numerical values?
getchar() returns the character read.Prototype for it is
int getchar(void)
When a character is read getchar() returns the ASCII value of the char read.
The ASCII value for char's 0 to 9 are contiguous. So just making use of it if we have
char ch = '5';
int i = ch - '0'; /* 53 - 48 = 5 */
will give you the integer value 5. Converting character to integer. The arithmetic is performed by implicit conversion.
If you have a character '8' then this doesn't give you the integer value 8 but retuns ASCII value 56. So during arithmetic ch - '0' since both are char's the respective ASCII values are used and the arithmetic operation is performed