I have the following code and have questions about it ..

q1: The variable s is a pointer to an array of characters, so you need to use the pointer type in the parameter field of the function.

q2. That is just a matter of choice, it is not essential to use unsigned.

q3. In this case it does not matter, ++counter or counter++ both do what is needed. You only need to be specific when you are capturing the value at the time that you use the increment operator.

q4. If you wish to use s1 as a pointer to something, then you must declare it as such:

char c1 = 'l'; // c1 is a single character value
char * s1 = "a string"; // s1 is a pointer to an array of characters

q5. The final character in a C character array is '\0' (i.e a zero byte) so that is the way to test if you have reached the end of the string.

See Organization of the C Language Reference | Microsoft Docs[^] for full details on the C language.

[edit]
When creating a 'string' in C you have two possibilities. but remember that there is no string type in C, they are just simple arrays. When you create a string it can be a local array or a pointer to literal:

char s1[] = "this is a local array";
char *s2 = "this is a literal"; 

Variable s1 allocates space for the characters in the normal data space of the application. Variable s2 creates a pointer in the normal data space, but creates the string literal separately. When accessing the data you may use either pointers, or array offsets. Although s1 is an array, its name is treated by the compiler as a pointer to the first character. So when calling the following function to do something with the data you can do :

void function(char *pszText)
{
    while (*s1)
    {
        // do something with each character
    }
}

// and you can call it with:
function(s1); // passes an auto-generated pointer to the array
// or
function(s2); // passes the predefined pointer
// or for some people
function(&s1[0]); // passes the address of the first element

[/edit]

char *s = "Hello"

char *s = (char []){ 'H', 'e', 'l', 'l', 'o', '\0'}

int *x = (int []){ 0, 1, 2, 3 };