Converting hexadecimal digital to integer

[Joe1903]

I have the code below to convert hex digits to integer.But I want a char with all possible digits e.g. char s[]="0123456789abcdef"; and calculate the integer values dynamically: s[2] = 2 or s[12] = 12 and then calculate hexdigit.How this can be done at best?

int htoi(char s[])
{
    int hexdigit,i,inhex,n;
    i = 0;
    if( s[i] == '0')
    {
        ++i;
        if(s[i] == 'x' || s[i] == 'X')
            ++i;
    }
    
    n = 0;
    inhex = YES;

    for(;inhex==YES;++i)
    {
        if(s[i] >='0' && s[i] <='9')
            hexdigit= s[i] - '0';
        else if(s[i] >='a' && s[i] <='f')
            hexdigit= s[i] -'a' + 10;
        else if(s[i] >='A' && s[i] <='F')
            hexdigit= s[i] -'A' + 10;
        else
            inhex = NO;
        
        if(inhex == YES)
            n = 16 * n + hexdigit;
    }
    return n;
}

[kevinkjt2000]

Could you rephrase your question? I can not understand what you are asking. What do you mean by "calcuate integer values dynamically"?

[Joe1903]

Instead of hexdigit = s[i] - '0' or s[i] - 'a' + 10 => n = 16* n+hexdigit
s[0] = 0, s[1] = 1 ... s[15] = 15 (using char s[] defined above in my question)
and then n = 16* n + s[i]

[kevinkjt2000]

Instead of hexdigit = s[i] - '0' or s[i] - 'a' + 10 => n = 16* n+hexdigit

Here s[i] is an individual character from the hex number you are converting

s[0] = 0, s[1] = 1 ... s[15] = 15

But here it seems like you are trying to express something like s[ s[i] ]. I would recommend creating a seperate array if you would like to lookup values, and do something like other_array[ s[i] ]. This would require an array that has 255 locations, because char variables range from 0 to 255. But there is an even better solution than that, and it is what you are already doing abstracted to a function.

//return -1 if the char is not a hex digit
//otherwise, returns a value in [0, 15]
int int_hex_digit_from_char(char ch) {
    int hexdigit = -1;
    if(ch >= '0' && ch <= '9')
        hexdigit = ch - '0';
    else if(ch >= 'a' && ch <= 'f')
        hexdigit = ch - 'a' + 10;
    else if(ch >= 'A' && ch <= 'F')
        hexdigit = ch - 'A' + 10;
    return hexdigit;
}

The function would be used instead of some other array in something like, int hexdigit = int_hex_digit_from_char( s[i] ); followed by an if statement checking for -1.

The function isxdigit may interest you, because you could call this to validate that a char is a hex digit.
http://www.cplusplus.com/reference/cctype/isxdigit/

[Joe1903]

You are Right but what I want is not to be addicted to specific numeral system (+ 10)
How would it look like with something like s[s[i]]?

[kevinkjt2000]

What I was trying to explain briefly about s[ s[i] ] in my last post, is that it makes no sense. Think about it. s[i] is a hex char from a number that you are converting which is then used as an index to the same s array. s[ char that you are converting ] just does not make any logical sense. Which is why I recommended another array, other_array. If you really did want to use another array to store the values, you would need to initialize the locations '0' to '9' with 0 to 9, 'a' to 'f' and 'A' to 'F' with 10 to 15, and all the other locations with some error value. I can assure you that using another array is a bad road to travel down for this problem. A function handles this same mapping of inputs to outputs in a much more readable way than the code for this other array would.

Not being stuck on a specific numeral system is a bit unrelated to this char to hex digit conversion. 'a' should always have a value of 10₁₀, 12₈, 1010₂, etc.. Therefore this hex digit conversion function is correct when it reports 10₁₀ for the input of 'a'. It would also be correct if it reported 12₈. Because 10₁₀ and 12₈ are equivalent. This kind of conversion is happening very frequently within a computer. All numbers are ultimately represented with binary base in memory. When C++ has to display this binary representation in base 10, somewhere down the line is a function that does the math to figure out what ASCII digits need to be displayed.

In C++ it is possible to convert how integers are displayed to a few different bases. http://www.cplusplus.com/reference/ios/oct/

It is also possible to specify literal numbers in bases other than 10.
int sixteen = 0x10; //0x is the prefix to use hexadecimal

[Nico]

If you can use C++11 features, you could try to do it like:

#include <iostream>
#include <string>     // use std::string, std::stoi


int htoi(char s[]) // int htoi(char* s) works equally well and is a bit more correct
{
	int sum = 0;
	std::string theInputString(s);
	for(int index=0; index<theInputString.length();index++)
	{
		// Here you convert the hex value to base 10 and add it to the previous values.
		sum = sum + std::stoi(std::string(1, theInputString[index]), nullptr, 16);
	}
	return sum;
}

int main()
{
    std::cout << htoi("012345") << std::endl;
    std::cout << htoi("6789") << std::endl;
    std::cout << htoi("abcd") << std::endl;
    std::cout << htoi("0123456789abcd") << std::endl;
    return 0;
}

However, given the name of your function I would expect that this is closer to what you need.

#include <iostream>
#include <string>     // use std::string, std::stoi


int htoi(char s[]) // int htoi(char* s) works equally well and is a bit more correct
{
	std::string theInputString(s);
	// Here you convert the hex value to base 10.
	return std::stoi(theInputString, nullptr, 16);
}

int main()
{
    std::cout << htoi("012345") << std::endl;
    std::cout << htoi("6789") << std::endl;
    std::cout << htoi("abcd") << std::endl;
    // make sure you don't enter anything larger than MAXINT
    return 0;
}

Also, note that you are not really passing an array s as the argument, you are actually passing a char* s that points to an array.

Kind regards, Nico

http://www.cplusplus.com/reference/string/stoi/