Explaining C Pointers

Introduction

Pointers are variables that store the address of another variable.

Allow us to indirectly access variables (i.e. we can talk about its address rather than its value)

Importance of Pointers:

More flexible pass-by-reference.
Manipulate complex data structures efficiently, even if their data is scattered in deferent memory locations.
Use polymorphism - calling functions on data without knowing exactly what kind of data it is. (see "pointers and function" section)

Declaring Pointers

Simply <type> *<var_name>;, e.g.

int *ptr;

The pointer can then be initialized to a memory address for a variable, which is found by using &, e.g.

int x = 20;
int *px = &x;

To illustrate this with a diagram and code, consider the following simple program:

// pointers1.c
#include <stdio.h>

int main() {
    int x = 20;
    int *px = &x;
    printf("ptr: %p -> addr %p has: %d\n", &px, px, x);

    return 0;
}

When compiled with gcc pointers1.c and run, you get something similar to this:

ptr: 0x7ffec8780c10 -> addr 0x7ffec8780c0c has: 20

the pointer px is in address 0x7ffec8780c10, pointing 4 bytes away at address 0x7ffec8780c0c that contains x; consider the diagram below

memory_model

ℹ Note
My own preference and the prevalent practice is to put the * just before the name of the variable as opposed to putting it after the type, as some do. The later is also misleading when you have a list of variables in one line, e.g.
int *ptr, x, y vs. int* ptr, x, y.
(x and y are just integers but the later may make it look like all are pointers!)

Dereferencing Pointers

To dereference a pointer is to get the value of what the pointer is pointing to.

We use *<pointer_pointer_var_name>, for example:

  int x = 30;
  int *px = &x;
  *px = 40; // changes x
  // print memory address of where px is pointing 
  // at (px) and the value in the address (*px)
  printf("%p -> %d\n", px, *px); 
  // also note that the pointer also is stored 
  // somewhere in memory and we can get its location 
  // by &px, e.g.
  printf("%p\n", &px);
  // so
  printf("%p stores -> %p (px), which stores -> %d (x)\n", &px, px, *px);

Null Pointers

Any pointer set to 0 is called a null pointer, since there's no memory location 0, it is an invalid pointer.
Dereferencing such a pointer leads to a runtime error. One should check whether the pointer is null before dereferencing it.
```
  int *py = 0; // or int *py = NULL;
  printf("%d\n", *py); // seg-fault!
```
You may ask, what's the point for null pointers? Null pointers are very important for initializing pointers which will point to proper memory addresses later on, but they are not yet determined. If we declared int *pz; without initializing it, the compiler (GCC in my case), will point pz to a random memory address ("allocate"). However, this is not guaranteed, will seg-fault too, sometimes.
```
  int *pz;
  printf("%d\n", *pz);
```

Pointers and Arrays

An array is a list of values arranged sequentially in memory.
The variable name of the array is usually a special kind of pointer, it can decay into a pointer; as we will see below:
```
  int arr[] = { 1, 2, 3 }; // `arr` decays into int* (int pointer)
```
Therefore, arr in the example above is equivalent to an int*. arr is a pointer pointing to the beginning of the array.
To get the first element of the array, we will use *arr.
To get the second element of the array, we use *(arr + 1)

Therefore to get the nth element of the array we will use *(arr + n - 1).

  int arr[] = { 1, 2, 3 };
  printf("%d, %d, %d\n", *arr, *(arr + 1), *(arr + 2));

Let's look at an example for summing up numbers in an array:
```
  int sumArray(int *arr, int sz)
  {
      int sum = 0;
      for (int i = 0; i < sz; ++i)
      {
          sum += *(arr + i); // or sum += arr[i]
      }
      return sum;
  }
```
- sumArray takes in a pointer to an array and the size of that array.
- However, there's not way of telling (AFAIK) that that pointer truly points to an array, it could as well just be an ordinary pointer to an int. For instance of if we gave x (from our example in the beginning) to this function, it will compile correctly and it may even run without a seg-fault!
```
  printf("fake sum -> %d\n", sumArray(px, 3));
  // if you thought that's enough, try this!
  printf("fake sum -> %d\n", sumArray(px, 300));
  // 300 contiguous memory addresses
  // from px are summed up
```
- This is the same reason why strings (array of chars) have a sentinel value \0 at the end that signifies the end of the array (aka, the null terminator).

ℹ Note
In the next section, we will look at pointer-to-pointer. It is worth noting here that &arr in our example above will be an int** (pointer to pointer, or address of a pointer arr), since arr is int*.

Incrementing Pointers

Since pointers point to memory addresses which are contiguous, it is therefore possible to increment a pointer to move to the next address. The pointer will step according to it's size, e.g. int * will be stepping 32 bits (4 bytes) each.

Let's look at an example:

int arr[] = { 1, 2, 3, 4 };
int *p = arr;   // points to the first element in arr
p++;            // now p is pointing at the 2nd element
printf("p -> %d\n", *p);
printf("p -> %d\n", *(++p)); // pointer now pointing to the 3rd

This should be done strictly for guaranteed contiguous memory addresses, i.e. arrays, and you should know where the end is (where to stop).

Character Pointers

In C, we create a string by using an array of characters (loosely). A pointer pointing to this array is therefore a character pointer.

It will be insane to just have one pointer pointing to one char, what's the point?

Now, how do we tell we have reached the end of our "string"? We use a null terminator \0. That is when it is a proper string, else, it's just an array of characters.

Let's look at an example:

char s[] = { 'h', 'e', 'l', 'l', 'o', '\0' };
char *ps = s;
// notice that the length of the array will always be +1 the length of the
// string, because of the \0
printf("%s, str length = %ld, array length = %d\n", ps, strlen(ps), sizeof(sl));
// a shorter way to initalize this:
char *ps2 = "another hello"; // using double quotes to denote string
printf("%s, length = %ld\n", ps2, strlen(ps2));

As we had mentioned earlier, there is no way to know you have reached the end of an array, unless you put a sentinel value to mark the end. We use \0 to mark the end of a string. For instance, this:

char hackedStr[] = { 'g', 'o', '\0', 'o', 'd'};
printf("%s, str length = %ld, array length = %ld\n", hackedStr, strlen(hackedStr), sizeof(hackedStr));

Pointer to Pointer

We can have a pointer pointing to a pointer, and even another pointer pointing to the/that pointer (pointer -> pointer -> pointer).

Let's look at a simple example:

int y = 10;
int *py = &y;
int **ppy = &py;
int ***pppy = &ppy; // we can go on and on

printf("%p -> %p -> %p -> %d\n", pppy, ppy, py, y);
// we can deference any to get to our int value
// notice the symetry in the *
// as per the declaration
printf("%d, %d, %d\n", ***pppy, **ppy, *py);
// likewise you can modify through indirection
***pppy = 40;
printf("%d\n", y);

Likewise, you can have a pointer that points to the array pointer, eg:

int arr2[] = { 2, 5, 6, 8 };
int *p2 = arr2;
int **ppArr = &p2;
printf("1st element in arr: %d\n", **ppArr);
printf("2nd element in arr: %d\n", *(*ppArr + 1)); // notice the brackets

We will see why this is important when we look at the the next section on passing by value and by reference.

Pointers and Functions

Passing by Value vs. by Reference

A pointer is a value too, only that that value is a reference. Let that sink in.

Therefore you can pass a pointer to a function by value or by reference. Reference here will be a pointer to that pointer.

To illustrate this, let's look at the following example:

void swap1(int *a, int *b)
{
    int *temp = a;
    a = b;
    b = temp;
}

// the above function is for illustration purpose only
// the actual swapping function should be:
/*
void swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}
*/

int m = 30, n = 20;
int *pm = &m, *pn = &n;
swap1(pm, pn); // passed by value
printf("m -> %d, n -> %d\n", *pm, *pn); // no swap done!

Basically, we just passed by value (a copy of the pointers) to the function, and therefore, our original pointers remained untouched.

We have to pass by refeference (a pointer to the pointer):

void swap2(int **a, int **b)
{
    int *temp = *a;
    *a = *b;
    *b = temp;
}

int m = 30, n = 20;
int *pm = &m, *pn = &n;

swap2(&pm, &pn);
printf("m -> %d, n -> %d\n", *pm, *pn); // now swap done

Returning Pointers

As you may have known by now, you can pass pointers to functions and also you can return pointers from a function.

The following example is very buggy but it passes the point across -- that you can return a pointer from a function. I leave the exercise of finding out why it's buggy to you, to save on the space that I'd have to use to explain how the function call-stack works:

We will revisit this example when we look at malloc.

int* return_ptr() {
    int x = 30;
    return &x; // pointer to x (local)
}

Pointer to Functions

I'd initially planned to cover this topic as a sub-section of Pointers and Functions but I think it deserves it's own section.

This concept is not covered in most books but it is such a powerful concept. With pointers to functions, you can now pass functions to other functions (by reference).

This is the general format on how you declare such a pointer:

<return_type> (*<name_of_ptr>)(<type_of_params,...>) = &<the_function_pointed_to>

For example, for a function with a signature like int sum(int x, int y), this is how we will write its pointer:

int sum(int x, int y);
int (*sum_ptr)(int, int) = &sum;

// and then calling
int z = (*sum_ptr)(30, 50);

And thefore you can pass it to another function thus:


void do_op(int (*fn_ptr)(int, int)) {
    int x = 30, y = 40;
    printf("%d + %d = %d\n", x, y, (*fn_ptr)(x, y));
}

int main() {
    // ...
    do_op(&sum);
    // ...
    return 0;
}

Let's look at another example of a callback function print(int) passed to another function mul which multiplies two numbers and then calls the print function which prints out the results. So we leave the caller decide on how they want to print, formatting, etc.

#include <stdio.h>

void print(int prod) {
    printf("The product = %d\n", prod);
}

void mul(int x, int y, void (*print_fn)(int)) {
    int prod = x * y;
    (*print_fn)(prod);
}

int main() {
    mul(20, 30, &print);
    return 0;
}

`malloc`, `calloc` and `free`

`malloc`

malloc() is the function used for dynamic allocation. This allocation is done on the heap. All the allocations we've seen so far up to this point, have been on the stack, and therefore deallocation is done immediately the function returns/exits.

The syntax for malloc is void *malloc(size_t size) -- basically it takes in the size (in bytes) of the memory allocation to be done and returns and void* pointer. You can then type cast void * to the relevant type of the allocation. However, as far as malloc is concerned, all it does is get you a contiguous memory on the heap of size size and returns a pointer pointing to the first byte of that location.

The typical convention is to use number_of_elements * sizeof(type) in the place of `size):

int *m_ptr = (int *)malloc(10 * sizeof(int));
if (m_ptr == NULL) {
    // allocation faield
    // handle error
}

There is no guarantees that the allocation will always succeed, it may fail, especially due to lack of enough memory. Therefore, it is important to check that the pointer returned is not NULL.

malloc does not zero out the memory, so, you might end up with garbage initially in the memory location. Try it out to see for yourself.

`calloc`

calloc is more of syntactic sugar on top of malloc, it dynamically allocates memory and initializes all bytes in the allocated memory to zero; and also provides for num_of_elements in it's signature as you will see shortly.

Basically calloc uses malloc to allocate memory, but it does a step further by also initializing all the allocated memory to zero.

The syntax is void *calloc(size_t num_of_elements, size_t element_size)

int *c_ptr = (int *)calloc(10, sizeof(int));
if (c_ptr == NULL) {
    // allocation failed
    // handle error
}

free

The previous two code-snippets are buggy as stand-alone (has memory leaks). Whenever you allocate memory, it is your responsibility to free it. There's no free lunch as what we get with allocations on the stack which are cleared by the operating systems as soon as the function returns/exits. This is where free() comes in.

free is used to deallocate memory previously allocated by malloc or calloc. It basically releases the memory pointed to by the given pointer. Syntax: void free(void *ptr)

// for the previous allocation done by malloc and calloc
// we can free them once we're done with them
free(m_ptr);
free(c_ptr);

Memory Leaks

Memory leaks occur when you (unintentionally) fail to release or deallocate memory. As a result, the allocated memory remains inaccessible and unusable even after the program (or the program portion) has finished executing.

For a long-running user-space process for example, leaks can keep growing until the memory is exhausted at last, where the operating system will can then step in and kill the process. When it comes to kernel-space processes, it's a different story, leaks can essentially choke the whole system where you end with a panic in Linux for example or a BSOD (blue screen of death) in Windows.

For a short-running user-space processes, leaks can go undetected since the side-effects may not be that visible.

Let's look at a simple example here:

#include <stdlib.h>

int main() {
    // allocate memory for an integer
    int *ptr = (int*)malloc(sizeof(int));
    // check if memory allocation succeeded
    if (ptr == NULL) {
        return 1;
    }

    // proceed to use the allocated memory in the program

    // but at the end, forget to free the memory
    // missing: free(ptr);
}

Luckily now, we have very good static code analysis tools that can help detect such defects like memory leaks.

C Pointers Demystified

Introduction

Importance of Pointers:

Declaring Pointers

Dereferencing Pointers

Null Pointers

Pointers and Arrays

Incrementing Pointers

Character Pointers

Pointer to Pointer

Pointers and Functions

Passing by Value vs. by Reference

Returning Pointers

Pointer to Functions

`malloc`, `calloc` and `free`

`malloc`

`calloc`

Memory Leaks

Further Reading

C Pointers Demystified

Introduction

Importance of Pointers:

Declaring Pointers

Dereferencing Pointers

Null Pointers

Pointers and Arrays

Incrementing Pointers

Character Pointers

Pointer to Pointer

Pointers and Functions

Passing by Value vs. by Reference

Returning Pointers

Pointer to Functions

malloc, calloc and free

malloc

calloc

Memory Leaks

Further Reading

`malloc`, `calloc` and `free`

`malloc`

`calloc`