What's the real use of using n[c-'0']?

Question

I'm a novice in C and I came across the code like this :

int n[10];
if(c>='0' && c<='9')
++n[c-'0']

In if loop why we have to use single quotes around 0, whats the use of it, why we can't define 0 as an integer straight away? And in the second line of code ++n[c-'0'], whats the use of using array like this, in sense why we need to subtract 0(once again why the use of single quotes in this context?) from c in the array index?

If i do like this n[c-'0'], the result of index operation(c-'0') will be an character or integer?

Given that can anyone say me, whats the real use of such array and what are the disadvantages well?

Thanks in advance.

Answer 1

In C, '0' is an integer whose value represents the digit zero as a character, not the value 0, which would be the null character. Other answers have omitted this, but it's important to note that the C language mandates that the decimal digits have consecutive values, so that if c is a digit, c-'0' is the numeric value of that digit, i.e.

'0'-'0' = 0
'1'-'0' = 1
'2'-'0' = 2
.
.
.
'9'-'0' = 9

Answer 2

c is (likely) a char, which also has an integer representation and in C it can be converted implicitly. '0' is the character zero, and a convenient feature of numeric characters is that they are laid out sequentially in their integer representations.

So, now that you know each character has an integer representation and that the number characters are laid out sequentially, you can convert a character to its integer representation using simple subtraction.

'0' - '0' == 0
'1' - '0' == 1
'2' - '0' == 2
/* and so on and so forth */

So if you would like to count the occurrences of digits in a string, you can use this to your advantage:

int n[10]; /* 10 digits */

n['0' - '0'] /* where we store the counts for the character 0, aka n[0] */
n['1' - '0'] /* where we store the counts for the character 1, aka n[1] */