Converting from hexadecimal to decimal number in C - c

Question

I am doing the exercise in the C Programming language book, and exercise 2-3 asked us to write a function htoi to convert a hexadecimal number to decimal number.
This is the code I wrote, however when it runs, it always show that my hexadecimal number is illegal.
Please help!
#include<stdio.h>
#define TRUE 1
#define FALSE 0
int htoi (char s[]);
int main() {
printf("The decimal number is %d\n", htoi("0x134"));
return 0;
}
int htoi (char s[]) {
int j; /* counter for the string */
int temp; /* temp number in between conversion */
int number; /* the converted number */
int ishex; /* if the number is a valid hexadecimal number */
char c;
number = 0;
temp = 0;
ishex = FALSE;
if (s[0] == '0' && (s[1] == 'x' || s[1] == 'X')) {
ishex = TRUE;
}
else {
ishex = FALSE;
printf("This is not valid hexadecimal number.\n");
return number = 0;
}
if (ishex == TRUE) {
for (j = 2; (c = s[j]) != EOF; ++j) {
if (c >= '0' && c <= '9')
temp = c - '0';
else if (c >= 'a' && c <= 'f')
temp = 10 + c - 'a';
else if (c >= 'A' && c <= 'F')
temp = 10 + c - 'A';
else {
printf("This is a illegal hexadecimal number.\n");
ishex = FALSE;
return 0;
}
number = number * 16 + temp;
}
}
return number;
}

Answer 1

A string is a sequence of characters that terminates at the first '\0' character. That means "0x134" terminates with a '\0' character value, not an EOF value.
You are operating on a sequence of characters that you expect to be terminated by an EOF value, but that is simply not possible. I'll explain why later... Suffice to say for now, the string "0x134" contains no EOF value.
Your loop reaches the string-terminating '\0', which isn't in the range 0..9, a..f or A..F and so this branch executes:
else {
printf("This is a illegal hexadecimal number.\n");
ishex = FALSE;
return 0;
}
Perhaps you meant to write your loop like so:
for (j = 2; (c = s[j]) != '\0'; ++j) {
/* SNIP */
}
I promised to explain what is wrong with expecting EOF to exist as a character value. Assuming an unsigned char is 8 bits, getchar can return one of 256 character values, and it will return them as a positive unsigned char value... OR it can return the negative int value EOF, corresponding to an error or end-of-file.
Confused? In an empty file, there are no characters... Yet if you try to read a character from the file, you will get EOF every time, in spite of there being no characters. Hence, EOF is not a character value. It's an int value, and should be treated as such before you attempt to convert the value to a character, like so:
int c = getchar();
if (c == EOF) {
/* Here, c is NOT A CHARACTER VALUE! *
* It's more like an error code ... *
* XXX: Break or return or something */
}
else {
/* Here, c IS a character value, ... *
* so the following conversion is ok */
char ch = c;
}
On another note, c >= '0' && c <= '9' will evaluate truthfully when c is one of the digits in the range 0..9... This is a requirement from the C standard
Neither c >= 'a' && c <= 'f' nor c >= 'A' && c <= 'F' are required to evaluate truthfully under any circumstance, however. It happens to work on your system, because you are using ASCII which contains all of the lowercase letters in one contiguous block, and all of the uppercase letters in another contiguous block. C does not require that ASCII be the character set.
If you want this code to work portably, you might consider something like:
char alpha_digit[] = "aAbBcCdDeEfF";
if (c >= '0' && c <= '9') {
c -= '0';
}
else if (strchr(alpha_digit, c)) {
c = 10 + (strchr(alpha_digit, c) - alpha_digit) / 2;
}
else {
/* SNIP... XXX invalid digit */
}