Pointer and Arrays

Arrays and pointers very close together. To use arrays effectively, you have to know how to use pointers with them. Let's start with a simple example of arrays in C:

#define MAX 10

int main()

{

    int a[MAX];

    int b[MAX];

    int i;

    for(i=0; i<MAX; i++)

        a[i]=i;

    b=a;

    return 0;

}

Now try to compile it. But C will not compile it. If you want to copy a into b, you have to enter something like the following instead:

for (i=0; i<MAX; i++)

    b[i]=a[i];

Or, to put it more briefly:

for (i=0; i<MAX; b[i]=a[i], i++);

Now try to compile it. But C will not compile it. If you want to copy a into b, you have to enter something like the following instead:

for (i=0; i<MAX; i++)

    b[i]=a[i];

Or, to put it more briefly:

for (i=0; i<MAX; b[i]=a[i], i++);

Better yet, use the memcpy utility in string.h.
Arrays in C are unusual in that variables a and b are not. Instead they are permanent pointers to arrays. a and b permanently point to the first elements of their respective arrays -- they hold the addresses of a[0] and b[0] respectively. Since they are permanent pointers you cannot change their addresses. The statement a=b; therefore does not work.
Because a and b are pointers, you can do several interesting things with pointers and arrays.
Example:

#define MAX 10

void main()

{

    int a[MAX];

    int i;

    int *p;

    p=a;

    for(i=0; i<MAX; i++)

        a[i]=i;

    printf("%d\n",*p);

}

The statement p=a; works because a is a pointer. a points to the address of the 0th element of the actual array. This element is an integer, so a is a pointer to a single integer. Therefore, declaring p as a pointer to an integer and setting it equal to a works. Another way to say exactly the same thing would be to replace p=a; with p=&a[0];. Since a contains the address of a[0], a and &a[0] mean the same thing.
Now that p is pointing at the 0th element of a, you can do some rather strange things with it. The a variable is a permanent pointer and can not be changed, but p is not subject to such restrictions. C actually encourages you to move it around using pointer arithmetic . For example, if you say p++;, the compiler knows that p points to an integer, so this statement increments p the appropriate number of bytes to move it to the next element of the array. If p were pointing to an array of 100-byte-long structures, p++; would move p over by 100 bytes. C takes care of the details of element size.
You can copy the array a into b using pointers as well. The following code can replace (for i=0; i<MAX; a[i]=b[i], i++); :

p=a;

q=b;

for (i=0; i<MAX; i++)

{

    *q = *p;

    q++;

    p++;

}

You can abbreviate this code as follows:

p=a;

q=b;

for (i=0; i<MAX; i++)

    *q++ = *p++;

And you can further abbreviate it to:

for (p=a,q=b,i=0; i<MAX; *q++ = *p++, i++);

What if you go beyond the end of the array a or b with the pointers p or q? C does not care -- it carelessly goes along incrementing p and q, copying away over other variables with abandon. You need to be careful when indexing into arrays in C, because C assumes that you know what you are doing.
You can pass an array such as a or b to a function in two different ways. Imagine a function dump that accepts an array of integers as a parameter and prints the contents of the array to stdout. There are two ways to code dump:

void dump(int a[],int nia)

{

    int i;

     for (i=0; i<nia; i++)

        printf("%d\n",a[i]);

}

or:

void dump(int *p,int nia)

{

    int i;

     for (i=0; i<nia; i++)

        printf("%d\n",*p++);

}

The nia (number_in_array) variable is required so that the size of the array is known. Note that only a pointer to the array, rather than the contents of the array, is passed to the function.