Convert a character digit to the corresponding integer in C

As per other replies, this is fine:

char c = '5';
int x = c - '0';

Also, for error checking, you may wish to check isdigit(c) is true first. Note that you cannot completely portably do the same for letters, for example:

char c = 'b';
int x = c - 'a'; // x is now not necessarily 1

The standard guarantees that the char values for the digits '0' to '9' are contiguous, but makes no guarantees for other characters like letters of the alphabet.

Subtract '0' like this:

int i = c - '0';

The C Standard guarantees each digit in the range '0'..'9' is one greater than its previous digit (in section 5.2.1/3 of the C99 draft). The same counts for C++.

If, by some crazy coincidence, you want to convert a string of characters to an integer, you can do that too!

char *num = "1024";
int val = atoi(num); // atoi = ASCII TO Int

val is now 1024. Apparently atoi() is fine, and what I said about it earlier only applies to me (on OS X (maybe (insert Lisp joke here))). I have heard it is a macro that maps roughly to the next example, which uses strtol(), a more general-purpose function, to do the conversion instead:

char *num = "1024";
int val = (int)strtol(num, (char **)NULL, 10); // strtol = STRing TO Long

strtol() works like this:

long strtol(const char *str, char **endptr, int base);

It converts *str to a long, treating it as if it were a base base number. If **endptr isn't null, it holds the first non-digit character strtol() found (but who cares about that).

To convert character digit to corresponding integer. Do as shown below:

char c = '8';                    
int i = c - '0';

Logic behind the calculation above is to play with ASCII values. ASCII value of character 8 is 56, ASCII value of character 0 is 48. ASCII value of integer 8 is 8.

If we subtract two characters, subtraction will happen between ASCII of characters.

int i = 56 - 48;   
i = 8;