Character to Integer in C

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

for example, '5' -> 5

thanks.

From stackoverflow

• 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++.

  Arafangion : This answer would be better if you mentioned that a char is /already/ effectively an integer, albiet of implementation defined sign (ie, might be signed or unsigned), and is CHAR_BITS long.

  Johannes Schaub - litb : i wasn't sure knowing that really helps him. but Chris made a good point with a..z being not contiguous necessarily. i should have mentioned that instead. now he's won the race :)

  Arafangion : Yours is better because you've qualified your answer, though - Chris could very well have just made his stuff up. :)

  Johannes Schaub - litb : thanks for the appreciation. i admit i wasn't sure about the state for C. so i looked up and 'cause i already was at it i pasted the reference into the answer :)

  Arafangion : And that, sir, is why you've got a couple more points than me. :)

  Chris Young : I don't make stuff up. :p

  Johannes Schaub - litb : i believe you mate. i often find myself not pasting the reference to some standard section when giving answers either. but this time i looked up after answering and had the section numbers at hand, so it was easy to change my answer and add it in :) cheers

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

  Alok : `numeralChar - '0'` is already of type `int`, so you don't need the cast. Even if it wasn't, the cast is not needed.

  Kevin Conner : Huh, yeah that would coerce, wouldn't it? Maybe Java has colored my perceptions. Or maybe I'm completely mistaken and it would even coerce in Java. :)

• 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.

  Paul Tomblin : Will this work with EBCDIC or other non-ASCII encodings?

  Chris Young : @Paul Tomblin: Yes for digits not letters, because, as I said in the answer, the standard guarantees '0' to '9' are contiguous but does not make such guarantees for other characters such as 'a' to 'z'.

• 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.

  SeanJA : Maybe I shouldn't have answered this one...

• 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.

• 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 !!!
  }
  

• 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).

  Evan Teran : There are no thread issues with atoi (it has no static data) and it is not deprecated. In fact it is functionally equivalent to: #define atoi(x) (int)(strtol((x), NULL, 10)

  Chris Lutz : Well damn. My manpages are old.

  David Thornley : The issue with atoi is that it uses 0 as a "can't find a number here" value, so atoi("0") and atoi("one") both return 0. If that doesn't work for what you're using it for, look for strtol() or sscanf().

• Hi!

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

  thanks,

• 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.