[c] Char to int conversion in C

If I want to convert a single numeric char to it's numeric value, for example, if:

char c = '5';

and I want c to hold 5 instead of '5', is it 100% portable doing it like this?

c = c - '0';

I heard that all character sets store the numbers in consecutive order so I assume so, but I'd like to know if there is an organized library function to do this conversion, and how it is done conventionally. I'm a real beginner :)

Normally, if there's no guarantee that your input is in the '0'..'9' range, you'd have to perform a check like this:

if (c >= '0' && c <= '9') {
    int v = c - '0';
    // safely use v
}

An alternative is to use a lookup table. You get simple range checking and conversion with less (and possibly faster) code:

// one-time setup of an array of 256 integers;
// all slots set to -1 except for ones corresponding
// to the numeric characters
static const int CHAR_TO_NUMBER[] = {
    -1, -1, -1, ...,
    0, 1, 2, 3, 4, 5, 6, 7, 8, 9, // '0'..'9'
    -1, -1, -1, ...
};

// Now, all you need is:

int v = CHAR_TO_NUMBER[c];

if (v != -1) {
    // safely use v
}

P.S. I know that this is an overkill. I just wanted to present it as an alternative solution that may not be immediately evident.

Since the ASCII codes for '0','1','2'.... are placed from 48 to 57 they are essentially continuous. Now the arithmetic operations require conversion of char datatype to int datatype.Hence what you are basically doing is: 53-48 and hence it stores the value 5 with which you can do any integer operations.Note that while converting back from int to char the compiler gives no error but just performs a modulo 256 operation to put the value in its acceptable range

Yes. This is safe as long as you are using standard ascii characters, like you are in this example.

You can simply use theatol()function:

#include <stdio.h>
#include <stdlib.h>

int main() 
{
    const char *c = "5";
    int d = atol(c);
    printf("%d\n", d);

}

Since you're only converting one character, the function atoi() is overkill. atoi() is useful if you are converting string representations of numbers. The other posts have given examples of this. If I read your post correctly, you are only converting one numeric character. So, you are only going to convert a character that is the range 0 to 9. In the case of only converting one numeric character, your suggestion to subtract '0' will give you the result you want. The reason why this works is because ASCII values are consecutive (like you said). So, subtracting the ASCII value of 0 (ASCII value 48 - see ASCII Table for values) from a numeric character will give the value of the number. So, your example of c = c - '0' where c = '5', what is really happening is 53 (the ASCII value of 5) - 48 (the ASCII value of 0) = 5.

When I first posted this answer, I didn't take into consideration your comment about being 100% portable between different character sets. I did some further looking around around and it seems like your answer is still mostly correct. The problem is that you are using a char which is an 8-bit data type. Which wouldn't work with all character types. Read this article by Joel Spolsky on Unicode for a lot more information on Unicode. In this article, he says that he uses wchar_t for characters. This has worked well for him and he publishes his web site in 29 languages. So, you would need to change your char to a wchar_t. Other than that, he says that the character under value 127 and below are basically the same. This would include characters that represent numbers. This means the basic math you proposed should work for what you were trying to achieve.

You can use atoi, which is part of the standard library.

As others have suggested, but wrapped in a function:

int char_to_digit(char c) {
    return c - '0';
}

Now just use the function. If, down the line, you decide to use a different method, you just need to change the implementation (performance, charset differences, whatever), you wont need to change the callers.

This version assumes that c contains a char which represents a digit. You can check that before calling the function, using ctype.h's isdigit function.

Try this :

char c = '5' - '0';

int i = c - '0';

You should be aware that this doesn't perform any validation against the character - for example, if the character was 'a' then you would get 91 - 48 = 49. Especially if you are dealing with user or network input, you should probably perform validation to avoid bad behavior in your program. Just check the range:

if ('0' <= c &&  c <= '9') {
    i = c - '0';
} else {
    /* handle error */
}

Note that if you want your conversion to handle hex digits you can check the range and perform the appropriate calculation.

if ('0' <= c && c <= '9') {
    i = c - '0';
} else if ('a' <= c && c <= 'f') {
    i = 10 + c - 'a';
} else if ('A' <= c && c <= 'F') {
    i = 10 + c - 'A';
} else {
    /* handle error */
}

That will convert a single hex character, upper or lowercase independent, into an integer.