exercise2-3.md

Exercise 2-3

Write the function htoi(s), which converts a string of hexadecimal digits (including an optional 0x or 0X) ito its equivalent integer value. The allowable digits are 0 through 9, a through f, and A through F.

Process

This took me a session. Figuring out how this one worked was kind of cool and maybe slightly annoying...

I just felt an intuitive answer and was able to write the code, without consciously knowing where I was going with it yet. Thats pretty cool, but hard to explain how I arrived at the solution or replicate the process later on. I suppose it happened because I've been working on these exercises everyday and could just feel the answer.

I added length checking to the function because it was continuing past the string, as it only terminates on non valid characters

---

I also stumbled across two strange properties of C:

1. You can add more characters to an array than the limit that you set

Apparently its called 'undefined behavior' and C allows for it due to the old days when people understood the hardware

2. The first time running C program is slightly slower then the other times

Apparently this is because it needs to read from disk the first time, then it is cached

Final Code

#include <stdio.h>

int htoi(char s[], int len);
    
int main(void) {
    char s1[2];
    s1[0] = 'f';
    s1[1] = 'a';
    
    char s2[4];
    s2[0] = '0';
    s2[1] = 'x';
    s2[2] = '5';
    s2[3] = 'b';
        
    printf("\'fa\': %d\n", htoi(s1, 2));
    printf("\'0x5b\': %d", htoi(s2, 4));
    return 0;
}

int htoi(char s[], int len){
    int i, n;
    n = 0;
    
    /* ignore initial 0x or 0X */
    if(s[0] == '0' && (s[1] =='x' || s[1]=='X'))
        i = 2;
    else
        i = 0;
    
    /* ignore characters outside legal hex values */
    for(i; 
        (s[i] >= '0' && s[i] <= '9') || 
        (s[i] >= 'A' && s[i] <= 'F') ||
        (s[i] >= 'a' && s[i] <= 'f') &&
        i < len;
        i++
        )
    {
        /* add value to total integer */
        if(s[i] >= '0' && s[i] <= '9')
            n = 16 * n + (s[i] - '0');   
        
        if(s[i] >= 'A' && s[i] <= 'F')
            n = 16 * n + (s[i] - 'A') + 10; /* add 10 because A is 10 */
        
        if(s[i] >= 'a' && s[i] <= 'f')
            n = 16 * n + (s[i] - 'a') + 10; /* add 10 because a is 10 */
    }
    return n;
}

Output

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