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 incis a digit. If it is, the numeric value if that digit isc - '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).
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.