Understanding C: Pointers and Structs

I'm trying to better understand c, and I'm having a hard time understanding where I use the * and & characters. And just struct's in general. Here's a bit of code:

void word_not(lc3_word_t *R, lc3_word_t A) {
    int *ptr;
    *ptr = &R;
    &ptr[0] = 1;
    printf("this is R at spot 0: %d", ptr[0]);
}  

lc3_word_t is a struct defined like this:

struct lc3_word_t__ {
  BIT b15;
  BIT b14;
  BIT b13;
  BIT b12;
  BIT b11;
  BIT b10;
  BIT b9;
  BIT b8;
  BIT b7;
  BIT b6;
  BIT b5;
  BIT b4;
  BIT b3;
  BIT b2;
  BIT b1;
  BIT b0;
};

This code doesn't do anything, it compiles but once I run it I get a "Segmentation fault" error. I'm just trying to understand how to read and write to a struct and using pointers. Thanks :)

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New Code:

void word_not(lc3_word_t *R, lc3_word_t A) {
    in开发者_JAVA百科t* ptr;
    ptr = &R;
    ptr->b0 = 1;
    printf("this is: %d", ptr->b0);
}

---

Here's a quick rundown of pointers (as I use them, at least):

int i;
int* p; //I declare pointers with the asterisk next to the type, not the name;
        //it's not conventional, but int* seems like the full data type to me.

i = 17; //i now holds the value 17 (obviously)
p = &i; //p now holds the address of i (&x gives you the address of x)
*p = 3; //the thing pointed to by p (in our case, i) now holds the value 3
        //the *x operator is sort of the reverse of the &x operator
printf("%i\n", i); //this will print 3, cause we changed the value of i (via *p)

And paired with structs:

typedef struct
{
    unsigned char a;
    unsigned char r;
    unsigned char g;
    unsigned char b;
} Color;

Color c;
Color* p;

p = &c;     //just like the last code
p->g = 255; //set the 'g' member of the struct to 255
            //this works because the compiler knows that Color* p points to a Color
            //note that we don't use p[x] to get at the members - that's for arrays

And finally, with arrays:

int a[] = {1, 2, 7, 4};
int* p;

p = a;    //note the lack of the & (address of) operator
          //we don't need it, as arrays behave like pointers internally
          //alternatively, "p = &a[0];" would have given the same result

p[2] = 3; //set that seven back to what it should be
          //note the lack of the * (dereference) operator
          //we don't need it, as the [] operator dereferences for us
          //alternatively, we could have used "*(p+2) = 3;"

Hope this clears some things up - and don't hesitate to ask for more details if there's anything I've left out. Cheers!

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I think you are looking for a general tutorial on C (of which there are many). Just check google. The following site has good info that will explain your questions better.

http://www.cplusplus.com/doc/tutorial/pointers/
http://www.cplusplus.com/doc/tutorial/structures/

They will help you with basic syntax and understanding what the operators are and how they work. Note that the site is C++ but the basics are the same in C.

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First of all, your second line should be giving you some sort of warning about converting a pointer into an int. The third line I'm surprised compiles at all. Compile at your highest warning level, and heed the warnings.

The * does different things depending on whether it is in a declaration or an expression. In a declaration (like int *ptr or lc3_word_t *R) it just means "this is a pointer."

In an expression (like *ptr = &R) it means to dereference the pointer, which is basically to use the pointed-to value like a regular variable.

The & means "take the address of this." If something is not a pointer, you use it to turn it into a pointer. If something is already a pointer (like R or ptr in your function), you don't need to take the address of it again.

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int *ptr;
*ptr = &R;

Here ptr is not initialized. It can point to whatever. Then you dereference it with * and assign it the address of R. That should not compile since &R is of type lc3_word_t** (pointer to pointer), while *ptr is of type int.

&ptr[0] = 1; is not legal either. Here you take the address of ptr[0] and try to assign it 1. This is also illegal since it is an rvalue, but you can think of it that you cannot change the location of the variable ptr[0] since what you're essentially trying to do is changing the address of ptr[0].

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Let's step through the code.

First you declare a pointer to int: int *ptr. By the way I like to write it like this int* ptr (with * next to int instead of ptr) to remind myself that pointer is part of the type, i.e. the type of ptr is pointer to int.

Next you assign the value pointed to by ptr to the address of R. * dereferences the pointer (gets the value pointed to) and & gives the address. This is your problem. You've mixed up the types. Assigning the address of R (lc3_word_t**) to *ptr (int) won't work.

Next is &ptr[0] = 1;. This doesn't make a whole lot of sense either. &ptr[0] is the address of the first element of ptr (as an array). I'm guessing you want just the value at the first address, that is ptr[0] or *ptr.