C Strings, Char Arrays, Char Pointers - Confused!

[ToniAz]

Hello everyone!

Because some functions in C support only C strings, I am trying to learn about them.
But I am really confused, what's the difference between the following:

char charr1[] = "Hello!";

char * charr2 = "Hello!";

So if the 1st is a char array, and the second is a char pointer, then what is a C String?

Thanks!

[Athar]

A C string is a char array where the last character of the actual string is followed by null ('\0').
Therefore, charr1 is a C string and charr2 points to a C string (or rather, to its first character) (although you need to fix the type, which must be const char*).

[ToniAz]

If it's const then I can't modify it's content, but apparently I do. What confuses me is that shouldn't we write:

char * charr2;
charr2 = new char[6];

???

And I didn't include a null character in either of charr1 or charr2. Does the null character get automatically inserted?

[Athar]

ToniAz wrote:

If it's const then I can't modify it's content, but apparently I do. What confuses me is that shouldn't we write:

char * charr2;
charr2 = new char[6];

No, you're assigning the address of the first character in the string literal "Hello!" (which is stored somewhere, usually in a read-only memory region) to charr2. String literals are constant character arrays, so you shouldn't be able to assign its address to a non-const char pointer. If your compiler does not report an error, it is probably outdated.

ToniAz wrote:

And I didn't include a null character in either of charr1 or charr2. Does the null character get automatically inserted?

Yes. char charr1[] = "Hello!"; is equivalent to:
char charr1[7] = {'H','e','l','l','o','!','\0'};

[Disch]

When declaring variables usage of empty [brackets] automatically sizes the array to the appropriate size. For example:

// this line:
int foo[] = {1,2,3};

// is the same as this line:
int foo[3] = {1,2,3};

With char arrays, the same idea applies:

char foo[] = "Hello!";

// is the same as:
char foo[7] = "Hello!";

// is the same as:
char foo[7] = {'H','e','l','l','o','!', 0};

// (note the null terminator is put there automatically by the "double quotes") 

On the other hand... when you use char* you only have a pointer and not an array. The pointer points to characters that exist somewhere in memory.

If you make a pointer point to a string literal, then the pointer must be const, because you can't modify a string literal. It doesn't make sense to change the literal string "Hello!" to be something else, any more than it makes sense to change the literal number 5 to be a different number:

int* foo = &5;  // conceptual only, not legal C++
*foo = 3;  // this is like saying 5 = 3; ... which is nonsense.

// same idea applies to char arrays:
char* foo = "bar";
foo[0] = 'c';  // is like saying "bar" = "car" ... which makes no sense 

Therefore pointers must be const char:

const char* charr2 = "Hello!";

EDIT: doh, too slow.