goulashsoup - 1 year ago 51

C Question

I have the following code snippet:

`int ndigit[10] = {0,0,0,0,0,0,0,0,0,0};`

void count() {

char c;

while (cin.get(c)) {

if (c>='0' && c<='9') {

ndigit[c-'0']++;

}

}

So

`c`

`ndigit[c-'0']`

`ndigit[c]`

If I am right, in both cases a

`c`

`ndigit[c-'0']`

I appreciate your wisdom!

Answer Source

"So

`c`

has to be between`0`

and`9`

"

`No!`

`c`

has to be between `'0'`

and `'9'`

, and it makes the difference.

Every character has appropriate numeric value (ASCII code), e. g.:

- 'A' has 65
- 'B' has 66
- 'a' has 98

In C language **characters** are simply **numbers**, e. g. `'A' + 'B'`

is a perfectly valid expression and means `65 + 66`

.

If we want display or read a `digit`

(0, 1, ..., 9), we actually use its symbolic representation, i. e. an `character`

('0', '1', ..., '9'). And these characters amazingly **have not** ASCII values 0, 1, ..., 9, but **48, 49, ... 57** - they are all `shifted by 48`

.

So for converting a digit `symbol`

, e. g. `'7'`

(which has ASCII value `55`

- as `7 + 48`

) into a `number`

which we people see in it (i. e. `7`

- without apostrophes), we need simply subtract this shifting number `48`

from its ASCII value:

```
(7 + 48) - 48
```

which is the same as to subtract **'0'** (= 48) from **'7'** (=55):

```
'7' - '0'
```

and witch is exactly what does the expression

```
c - '0'
```

in your code.