Character to Integer in C

Question

Is there a way to convert a character to an integer in C?

for example, '5' -> 5

thanks.

Answer 1

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++.

Answer 2

char numeralChar = '4';
int numeral = (int) (numeralChar - '0');

Answer 3

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.

Answer 4

You would cast it to an int (or float or double or what ever else you want to do with it) and store it in anoter variable.

Answer 5

If it's just a single character 0-9 in ASCII, then subtracting the the value of the ASCII zero character from ASCII value should work fine.

If you want to convert larger numbers then the following will do:

char *string = "24";

int value;

int assigned = sscanf(string, "%d", &value);

** don't forget to check the status (which should be 1 if it worked in the above case).

Paul.

Answer 6

char chVal = '5';
char chIndex;

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

    chIndex = chVal - '0';
}
else 
if ((chVal >= 'a') && (chVal <= 'z')) {

    chIndex = chVal - 'a';
}
else 
if ((chVal >= 'A') && (chVal <= 'Z')) {

    chIndex = chVal - 'A';
}
else {
    chIndex = -1; // Error value !!!
}

Answer 7

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 converstion 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);

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).

Answer 8

Hi!

is there a way that i can check if the input is either a int or a char?

thanks,

Answer 9

When I need to do something like this I prebake an array with the values I want.

const static int lookup[256] = { -1, ..., 0,1,2,3,4,5,6,7,8,9, .... };

Then the conversion is easy

int digit_to_int( unsigned char c ) { return lookup[ static_cast<int>(c) ]; }

This is basically the approach taken by many implementations of the ctype library. You can trivially adapt this to work with hex digits too.