Conversion of hexadecimal string constant to decimal value using C - c

Question

The following program gives the result as 0 instead of the expected decimal equivalent of the hexadecimal string constant.
#include <stdio.h>
int my_htoi(char[]);
int main(void) {
printf("%d", my_htoi("0xABC"));
return 0;
}
int my_htoi(char str[]) {
int i, num = 0;
for (i = 0; i != '\0'; ++i) {
if (str[i+1] == 'x' || str[i+1] == 'X') {
i = i + 1;
continue;
}
if (str[i] >= '0' && str[i] <= '9') {
num = num * 16 + (str[i] - '0');
} else if (str[i] >= 'a' && str[i] <= 'f') {
num = num * 16 + (str[i] - 'a' + 10);
} else if (str[i] >= 'A' && str[i] <= 'F') {
num = num * 16 + (str[i] - 'A' + 10);
}
}
return num;
}
While the following program runs fine and outputs the correct decimal equivalent of the hexadecimal string constant.
#include <stdio.h>
#include <string.h>
int my_htoi(char[]);
int main(void) {
printf("%d", my_htoi("0xABC"));
return 0;
}
int my_htoi(char str[]) {
int i, num = 0;
for (i = 0; i < strlen(str); ++i) {
if (str[i+1] == 'x' || str[i+1] == 'X') {
i = i + 1;
continue;
}
if (str[i] >= '0' && str[i] <= '9') {
num = num * 16 + (str[i] - '0');
} else if (str[i] >= 'a' && str[i] <= 'f') {
num = num * 16 + (str[i] - 'a' + 10);
} else if (str[i] >= 'A' && str[i] <= 'F') {
num = num * 16 + (str[i] - 'A' + 10);
}
}
return num;
}
The only difference is in the way we find the qualifying condition for the loop. Why does it not work with the null byte checking?

Answer 1

Wrong code: i != '\0' checks if the index is 0.
for(i = 0; i != '\0'; ++i) {
Should be the below to check if the element str[i] is the null character.
for(i = 0; str[i] != '\0'; ++i) {
Other issues exists unneeded increment, int overflow (better to use unsigned here), wrong x detection - consider "0x0x0x1", leading - or +, char str[] --> const char str[], ...

Answer 2

There are some problems in your code:
the loop index i is compared to '\0' instead of str[i], causing immediate termination of the loop with a return value of 0.
the test for x is incorrect: it would cause "1x2" to convert to 2 instead of 1.
you accept letters beyond f and convert them to digits. The function should instead stop parsing at the first character that is not a hex digit.
Here is a corrected version:
#include <stdio.h>
int my_htoi(const char[]);
int main(void) {
printf("%d", my_htoi("0xABC"));
return 0;
}
int my_htoi(const char str[]) {
int i = 0, num = 0;
if (str[0] == '0' && (str[1] == 'x' || str[1] == 'X'))
i += 2;
for (; str[i] != '\0'; ++i) {
if (str[i] >= '0' && str[i] <= '9') {
num = num * 16 + (str[i] - '0');
} else if (str[i] >= 'a' && str[i] <= 'f') {
num = num * 16 + (str[i] - 'a' + 10);
} else if (str[i] >= 'A' && str[i] <= 'F') {
num = num * 16 + (str[i] - 'A' + 10);
} else {
break;
}
}
return num;
}