pointer to string and char catch 22

Question

I'm studying on pointers and I'm stuck when I see char *p[10]. Because something is misunderstood. Can someone explain step-by-step and blow-by-blow why my logic is wrong and what the mistakes are and where did I think wrong and how should I think. Because I want to learn exactly. Also what about int *p[10]; ? Besides, for example x is a pointer to char but just char not chars. But how come char *x = "possible";

I think above one should be right but, I have seen for char *name[] = { "no month","jan","feb" }; I am really confused.

Answer 1

char c = 'a';

Here, c is a char, typically a single byte of ASCII encoded data.

char* ptr = &c;

ptr is a char pointer. In C, all it does is point to a memory location and doesn't make any guarantees about what is at that location. You could use a char* to pass a char to a function to allow the function to allow the function to make changes to that char (pass by reference).

A common C convention is for a char* to point to a memory location where several characters are stored in sequence followed by the null character \0. This convention is called a C string:

 char const* cstr = "hello";

cstr points to a block of memory 6 bytes long, ending with a null character. The data itself cannot be modified, though the pointer can be changed to point to something else.

An array of chars looks similar, but behaves slightly differently.

 char arr[] = "hello";

Here arr IS a memory block of 6 chars. Since arr represents the memory itself, it cannot be changed to point to another location. The data can be modified though.

Now,

char const* name[] = { "Jan", " Feb"..., "Dec"};

is an array of pointer to characters. name is a block of memory, each containing a pointer to a null-terminated string.

In the diagram, I think string* was accidentally used instead of char*. The difference between the left and the right, is not a technical difference really, but a difference in the way a char* is used. On the left each char* points to a single character, whereas in the one on the right, each char* points to a null-terminated block of characters.

Answer 2

Let’s go back to basics.

First, char *p is simply a pointer. p contains nothing more than a memory address. That memory address can point to anything, anywhere. By convention, we have always used NULL (or, I hate this method, assigning it to zero – yeah, they are the same “thing”, but NULL has traditionally been used in conjunction with pointers, so when you’re eyes flit across the code, you see NULL – you think “pointer”).

Anyway, that memory address being pointed to can contain anything. So, to use within the language, we type it, in this case it is a pointer to a character (char *p). This can be overridden by type casting, but that’s for a later time.

Second, we know anytime we see p[10], that we are dealing with an array. Again, the array can be an array of characters, an array of ints, etc. – but it’s still an array.

Your example: char *p[10], is then nothing more than an array of 10 character pointers. Nothing more, nothing less. Your problem comes in because you are trying to force the “string” concept onto this. There ain’t no strings in C. There ain’t no objects in C. The concept of a NULL-terminated string can most certainly be used. But a “string” in C is nothing more than an array of characters, terminated by a NULL (or, if you use some of the appropriate functions, you can use a specific number of characters – strncpy instead of strcpy, etc.). But, for all its appearance, and apparent use, there are no strings in C. They are nothing more than arrays of characters, with a few supporting functions that happen to stop going through the array when a NULL is encountered.

So – char a[10] – is simply an array of characters that is 10 characters long. You can fill it with any characters you wish. If one of those is the NULL character, then that terminates what is typically called a “C-style string”. There are functions that support this type of character array (i.e. “string”), but it is still a use of a character array.

Your confusion comes in because you are trying to mix C++ string objects, and forcing that concept onto C arrays of characters. As ugoren noted – your examples are both correct – because you are dealing with arrays of character pointers, NOT strings. Again, putting a NULL somewhere in that character array is happily supported by several C functions that give you the ability to work with a “string-like” concept – but they are not truly strings. Unless of course, you want to phrase it that a string is nothing more than one character following another – an array.

Answer 3

Both are right.

A pointer in C or C++ may point either to a single item (a single char) or to the first in an array of items (char[]).
So a char *p[10]; definition may point to 10 single characters or 10 arrays (i.e. 10 strings).

Answer 4

Your char *p[10] diagram shows an array where each element points to a character.
You could construct it like this:

char f = 'f';
char i = 'i';
char l1 = 'l';
char l2 = 'l';
char a1 = 'a';
char r1 = 'r';
char r2 = 'r';
char a2 = 'a';
char y = 'y';
char nul = '\0';
char *p[10] = { &f, &i, &l1, &l2, &a1, &r1, &r2, &a2, &y, &nul };

This is very different from the array

char p[10] = {'f', 'i', 'l', 'l', 'a', 'r', 'r', 'a', 'y', '\0'};

or

char p[10] = "fillarray";

which are arrays of characters, not pointers.

A pointer can equally well point to the first element of an array, as you've probably seen in constructions like

const char *p = "fillarray";

where p holds the address of the first element of an array defined by the literal.

This works because an array can decay into a pointer to its first element.

The same thing happens if you make an array of pointers:

/* Each element is a pointer to the first element of the corresponding string in the initialiser. */
const char *name[] = { "no month","jan","feb" };

You would get the same results with

const char* name[3];
name[0] = "no month";
name[1] = "jan";
name[2] = "feb";