You're comparing a pointer, i.e. a memory address, against what you're expecting to see when dereferencing the pointer, i.e. the number 0 or '\0'. You need to dereference the pointer.

Also, your if statement is wrong.

And, you're wasting your time duplicating a standard language function.

10 minutes ago, Kylotan said:

You're comparing a pointer, i.e. a memory address, against what you're expecting to see when dereferencing the pointer, i.e. the number 0 or '\0'. You need to dereference the pointer.

Also, your if statement is wrong.

And, you're wasting your time duplicating a standard language function.

ops, missed that dereference in the while, I knew I was tired x_x

Fixed my >= and <= , thanks!

I disagree on the last point though, I don't think this time is wasted, we are probably assigning value to different things 

Though you wouldn't do this in production code, I agree you should know/understand how to handle pointers.

9 minutes ago, Alpha_ProgDes said:

Though you wouldn't do this in production code, I agree you should know/understand how to handle pointers.

Exactly my goal here

What I'm currently writing is probably ugly and a bit buggy, but I'm getting useful practice, today first time I use const_cast and write my template, so...IMPROVEMENT (for better or worse...)

char* findx(const char* s, const char* x)
{
	if(s == nullptr || x == nullptr)
	{ return nullptr; }

	char* Tmp = const_cast<char*>(x);
	int FindSize = 0;
	while (*Tmp++ != '\0') { FindSize++; };

	int ID = 0;
	int firstOccurrence = 0;
	bool isMatch = false;
	
	while (s[ID] != '\0')
	{
		if (s[ID] == x[0])
		{
			firstOccurrence = ID;
			for (size_t i = 0; i < FindSize; i++)
			{
				if (s[ID + i] != x[i]) { break; }
				if (i == FindSize - 1) { isMatch = true; }
			}
			if (isMatch) { break; }
		}
		ID++;
	}
	return (isMatch ? const_cast<char*>(s) + firstOccurrence : nullptr);
}

Using const cast looks horribly wrong, "find" shouldn't change the string!

You have "char *tmp" and return a "char *",  change that to "const char *" too.

At a higher level, you are accepting const char * values as parameters, users don't expect that you break your promise, and sneakily return a modifiable point in their const string as return value.

The Exercise in the book I'm following gave as only instructions:

Quote

Write a function, char* findx(const char* s, const char* x), that finds the first occurrence of the C-style string x in s.

I don't know how to return a char* out of a const char*, so I tried removing the const, just to fullfill the assignment

Book is just wrong

Here you go friend

//take two char* (strings) and return char* to first occurence of second string
char* findx(char* str, char* strx ) {
	
	if (!str || !strx) return nullptr;

	char* found = nullptr;

	for (int s = 0, x = 0; str[s] != '\0' && strx[x] != '\0'; s++) {

		if (str[s] == strx[x]) {
			if( !found ) found = &str[s];
			x++;
		} else {
			found = nullptr;
			x = 0;
		}
	}

	return found;
}

@h8CplusplusGuru much more clean/compact than mine, regardless if I had done it without the const char required by the exercise, thanks for showing me