K&R Exercise 2-3 "Hex to int converter" Problem - c

Question

The program I wrote works in demographics consisting of only single Hexadecimal values. (Probably not the most elegant solution, but I'm a new programmer) My question is, how would I go about handling of multiple hexadecimal digits, such as 0xAF, or 0xFF, etc? I'm not exactly sure, and I've seemed confuse myself greatly, in the attempt. I'm not asking for someone to hold my hand, but to give me a tip where I've gone wrong in this code and thoughts on how to fix it.
Thanks :)
/* Exercise 2-3. Write the function htoi(s), which converts a string of
* hexadecimal digits (including an optional 0x or 0X) into it's equivalent
* integer value. The allowable digits are 0...9 - A...F and a...f.
*
*/
#include <stdio.h>
#include <string.h>
#define NL '\n'
#define MAX 24
int htoi(char *hexd);
int
main(void)
{
char str[MAX] = {0};
char hex[] = "0123456789ABCDEFabcdef\0";
int c;
int i;
int x = 0;
while((c = getchar()) != EOF) {
for(i = 0; hex[i] != '\0'; i++) {
if(c == hex[i])
str[x++] = c;
}
if(c == NL) {
printf("%d\n", htoi(str));
x = 0, i = x;
}
}
return 0;
}
int
htoi(char *hexd)
{
int i;
int n = 0;
for(i = 0; isdigit(hexd[i]); i++)
n = (16 * i) + (hexd[i] - '0');
for(i = 0; isupper(hexd[i]); i++) /* Let's just deal with lowercase characters */
hexd[i] = hexd[i] + 'a' - 'A';
for(i = 0; islower(hexd[i]); i++) {
hexd[i] = hexd[i] - 'a';
n = (16 + i) + hexd[i] + 10;
n = hexd[i] + 10;
}
return n;
}

Answer 1

Someone has alredy asked this (hex to int, k&r 2.3).
Take a look, there are many good answers, but you have to fill in the blanks.
Hex to Decimal conversion [K&R exercise]
Edit:
in
char hex[] = "0123456789ABCDEFabcdef\0";
The \0 is not necesary. hex is alredy nul terminated. Is len (0...f) + 1 = 17 bytes long.

Answer 2

I'll pick on one loop, and leave it to you to rethink your implementation. Specifically this:
for(i = 0; isdigit(hexd[i]); i++)
n = (16 * i) + (hexd[i] - '0');
doesn't do what you probably think it does...
It only processes the first span of characters where isdigit() is TRUE.
It stops on the first character where isdigit() is FALSE.
It doesn't run past the end because isdigit('\0') is known to be FALSE. I'm concerned that might be accidentally correct, though.
It does correctly convert a hex number that can be expressed solely with digits 0-9.
Things to think about for the whole program:
Generally, prefer to not modify input strings unless the modification is a valuable side effect. In your example code, you are forcing the string to lower case in-place. Modifying the input string in-place means that a user writing htoi("1234") is invoking undefined behavior. You really don't want to do that.
Only one of the loops over digits is going to process a non-zero number of digits.
What happens if I send 0123456789ABCDEF0123456789ABCDEF to stdin?
What do you expect to get for 80000000? What did you get? Are you surprised?
Personally, I wouldn't use NL for '\n'. C usage pretty much expects to see \n in a lot of contexts where the macro is not convenient, so it is better to just get used to it now...

Answer 3

I think that the MAX size of string should be either 10 or 18 instead of 24. (If you have already checked the int on your machine and followed the reasoning bellow, it would be beneficial to include it as a comment in your code.)
10 : since htoi() returns an int , int is usually 4 bytes (check your system's too), so the hexadecimal number can be atmost 8 digits in length (4bits to 1 hex digit, 8 bits to a byte), and we want to allow for the optional 0x or 0X.
18 : would be better if htoi() returned a long and its 8 bytes (again, check your system's), so the hexadecimal number can be atmost 16 digits in length, and we want to allow for the optional 0x or 0X.
Please note that that sizes of int and long are machine dependent, and please look at exercise 2.1 in the K&R book to find them.

Answer 4

Here is my version of a classic htoi() function to convert multiple hexadecimal values into decimal integers. It's a full working program compile it and run.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
int htoi(const char*);
int getRawInt(char);
int main(int argc, char **argv) {
char hex[] = " ";
printf("Enter a hexadecimal number (i.e 33A)\n");
scanf("%s", hex);
printf("Hexedecimal %s in decimal is %d\n", hex, htoi(hex)); // result will be 826
return 0;
}
int htoi(const char *hex) {
const int LEN = strlen(hex) -1;
int power = 1;
int dec = 0;
for(int i = LEN; i >= 0; --i) {
dec += getRawInt(hex[i]) * power;
power *= 16;
}
return dec;
}
int getRawInt(char c) {
if(isalpha(c)) {
return toupper(c) - 'A' + 10;
} return c-'0';
}