Ykhjuy

Still learning more C and am a little confused. In my references I find cautions about assigning a pointer that has not been initialized. They go on to give examples. Great answers yesterday by the way from folks helping me with pointers, here:

Precedence, Parentheses, Pointers with iterative array functions

On follow up I briefly asked about the last iteration of the loop and potentially pointing the pointer to a non-existent place (i.e. because of my references cautioning against it). So I went back and looked more and find this:

If you have a pointer

int *pt;

then use it without initializing it (i.e. I take this to mean without a statement like *pt= &myVariable):

*pt = 606;

you could end up with a real bad day depending on where in memory this pointer has been assigned to. The part I'm having trouble with is when working with a string of characters something like this would be ok:

char *str = "Sometimes I feel like I'm going crazy.";

Where the reference says, "Don't worry about where in the memory the string is allocated; it's handled automatically by the compiler". So no need to say initialize *str = &str[0]; or *str = str;. Meaning, the compiler is automatically char str[n]; in the background?

Why is it that this is handled differently? Or, am I completely misunderstanding?

In this case:

char *str = "Sometimes I feel like I'm going crazy.";

You're initializing str to contain the address of the given string literal. You're not actually dereferencing anything at this point.

This is also fine:

char *str;

str = "Sometimes I feel like I'm going crazy.";

Because you're assigning to str and not actually dereferencing it.

This is a problem:

int *pt;

*pt = 606;

Because pt is not initialized and then it is dereferenced.

You also can't do this for the same reason (plus the types don't match):

*pt= &myVariable;

But you can do this:

pt= &myVariable;

After which you can freely use *pt.

When you write sometype *p = something;, it's equivalent to sometype *p; p = something;, not sometype *p; *p = something;. That means when you use a string literal like that, the compiler figures out where to put it and then puts its address there.

The statement

char *str = "Sometimes I feel like I'm going crazy.";

is equivalent to

char *str;

str = "Sometimes I feel like I'm going crazy.";

Simplifying the string literal can be expressed as:

const char literal = "Sometimes I feel like I'm going crazy."

so the expression

char *str = "Sometimes I feel like I'm going crazy.";

is logically equivalent to:

const char literal = "Sometimes I feel like I'm going crazy."

char *str = literal;

of course literals do not have the names

But you can't deference the char pointer which does not have allocated memory for the actual object

/* Wrong */

char *c;

*c = 'a';

/* Wrong  - you assign the pointer with the integer value */ 

char *d = 'a';



/* Correct  */

char *d = malloc(1);

*d = 'a';



/* Correct */

char x

char *e = &x;

*e = 'b';

The last example.

/* Wrong - you assign the pointer with the integer value */

int *p = 666;



/* Wrong oyu dereference the pointer which references to the not allocated space */

int *r;

*r = 666;



/* Correct */

int *s = malloc(sizeof(*s));

*s = 666;



/* Correct */

int t;

int *u = &t;

*u = 666;

And the last one - something similar to the string literals = the compound literals

/* Correct */

int *z = (int){666,567,234};

z[2] = 0;

*z = 5;



/* Correct */

int *z = (const int){666,567,234}; 

Doing char *str = "I am a string"; means declaring and initializing an array of char. There's no problem since you're giving all the content of the array in the declaration.

Doing int *pt = 606; is the same as doing char *str = 'a'; there's a problem: you're not giving enough information while declaring your variable.

You need to do int nb = 606; int *pt = &nb; this way you give you pointer the address of the variable. Or if you want: you pass the int as an array of int.