Strcmp() function realization on C

Question

I need to make an strcmp function by myself, using operations with pointers. That's what I got:

int mystrcmp(const char *str1, const char *str2) {
    while ('\0' != *str1 && *str1 == *str2) {
        str1 += 1;
        str2++;
    }
    int result1 = (uint8_t)(*str2) - (uint8_t)(*str1); // I need (uint8_t) to use it with Russian symbols.
    return result1;
}

But my tutor told me that there are small mistake in my code. I spend really lot of time making tests, but couldn't find it.

Answer 1

Does this answer the question of what you're doing wrong?

#include <stdio.h>
#include <stdint.h> 
#include <string.h>

int mystrcmp(const char *str1, const char *str2);

int main(void)
{
    char* javascript = "JavaScript";
    char* java = "Java";

    printf("%d\n", mystrcmp(javascript, java));
    printf("%d\n", strcmp(javascript, java));
    return 0;
}

int mystrcmp(const char *str1, const char *str2) {
    while ('\0' != *str1 && *str1 == *str2) {
        str1 += 1;
        str2++;
    }
    int result1 = (uint8_t)(*str2) - (uint8_t)(*str1); // I need (uint8_t) to use it with Russian symbols.
    return result1;
}

Output:

-83
 83

---

I'll propose a quick fix:

Change

int result1 = (uint8_t)(*str2) - (uint8_t)(*str1);

To

int result1 =  (uint8_t)(*str1) - (uint8_t)(*str2);

---

And why you were wrong:

The return values of strcmp() should be:

if Return value < 0 then it indicates str1 is less than str2.

if Return value > 0 then it indicates str2 is less than str1.

if Return value = 0 then it indicates str1 is equal to str2.

And you were doing exactly the opposite.