pointer to char question

I am trying to cement my knowledge of pointers ,with very little success at the moment . I want to do the same with the second paragraph of code what i managed with the first . I am obviously using pointers in an illegal way .

Can anyone shed some light.

Code:

                // this works
	char word[80] = {0};
	word[0] = 's';
	cout << word[0] << endl;

	// this doesnt
	char* word2 = {0};
	word2[0] = 's';
	cout << word2[0] << endl;

Code:

// this doesnt
	char* word2 = {0};
	word2[0] = 's';
	cout << word2[0] << endl;

No the porper way of doing it is

Code:

char *word2 = "s";
cout << word2[0] << endl;
/*
char *name = "Steve";
cout << name[0] << endl; //prints S
cout << name[1] << endl; //prints t
cout << name[2] << endl; //prints e
cout << name << endl; //prints Steve
*/

When you say:
char word[80] = {0};

You are telling the compiler to allocate 80 characters in an array for you to use.

When you say:
char *words[80] = {0};

You are telling the compiler to allocate 80 pointers to a char type. So in essence you have allocated space for 80 potential words, but there pointers are not pointing anywhere at the moment.

Your array 'words' looks like this:

Code:

words[0] = NULL;
words[1] = NULL;
words[2] = NULL;
...
words[79] = NULL;

To use it you need to allocate space and assign data to it.

For example:

Code:

words[0] = strdup("This");
words[1] = strdup("is");
words[2] = strdup("a");
words[3] = strdup("sentence.");
...

This is an array of pointers to char* which is what char *[80] allocated for you.

When you are done, each of the lements that you allocated MUST be free'd or you will leak memory:

Code:

free(words[0]);
free(words[1]);
free(words[2]);
free(words[3]);

Usually this can be done in a following way:

Code:

for (int i=0; i<80; ++i) {
  if (words[i]) {
    delete words[i];
    words[i] = NULL;
  }
}

Since we are in the C++ world you should be using something akin to:

Code:

std::string words[80];
words[0] = "This";
words[1] = "is";
words[2] = "a";
words[3] = "sentence";

and not worry about deallocation, as the std::string class will delete the data in its destructor when the obect is removed from scope.

i dont want to initialize a value into the char pointer there and then like this

Code:

char *word2 = "s";

i want something like this

Code:

char* word2 = {0};      // this doesnt work

then initialize it later.
Can this be done?

char *word2 is a pointer to an array of char's, so you are really doing:

char *word2 = NULL;

at a later time you must use new/delete to allocate/deallocate the actual string.

Code:

char *word2 = NULL;

// some code here

word2 = new char[20];
strcpy(word2, "something");

// rest of the code

delete []word2;    // release memory
word2 = NULL;     // just in case