Understanding char array[] and string

Question

I am new to programming. I am learning C as my first programming language. I found something strange to understand.

I have learnt that in C we can represent a String as a sequence of characters like this (using a char array):

char status[10] = "Married";   

I have learnt that the problem of this approach is that we have to tell the size of the status array during compilation.

But now I have learned we can use a char pointer to denote an string like -

char status[10] = "Married";
char *strPtr;
strPtr = status;

I don't understand it properly. My questions are -

1. How can I get char at index 4 (that is i in Married) using the strPtr?

2. In status there is a null character (\0) at the end of the string represented by the char array - M-a-r-r-i-e-d-\0. So by using the null character (\0) we can understand the end of the string. When we use strPtr, how can we understand the end of the string?

Answer 1

char *strPtr;
strPtr = status;

Now your pointer strPtr is pointing to the first character in the array and you can do

int i =0;
while( strPtr[i] != '\0')
{
  printf("%c ",strPtr[i]);
  i++;
}

*strPtr is called dereferencing the pointer to get the value stored in the location the pointer is pointing to.

Make a note that

strPtr[4] = *(strPtr +4); 

Both will get you the value stored at the index 4 of the array.

Note the difference between a pointer and a array name:

----------------------------------
| s  | t  | r  | i  | n | g | \0 |
----------------------------------
  |
strPtr
status

strPtr ++ will make your pointer point to the next element in the array.

| s  | t  | r  | i  | n | g | \0 |
----------------------------------
       |
      strPtr

Whereas you can't do this for the array name

status++ is not allowed because an array is not a modifiable lvalue.