"Thoroughly Master C Pointers" Part 3 -- Pointers and Array Names

1. Accessing Array Elements via Array Name

Consider the following code:

int i,a[]={3,4,5,6,7,3,7,4,4,6};
for (i=0;i<=9;i++)
{
   printf ( “%d”, a[i] );
}

Clearly, this displays the values of each element in array a.

We can also access elements like this:

int i,a[]={3,4,5,6,7,3,7,4,4,6};
for (i=0;i<=9;i++)
{
   printf ( “%d”,  *(a+i) );
}

Its result and effect are exactly the same.

2. Accessing Array Elements via Pointers

int i,*pa,a[]={3,4,5,6,7,3,7,4,4,6};
pa =a  ;// Note that array name 'a' is directly assigned to pointer 'pa'
for (i=0;i<=9;i++)
{
   printf ( “%d”, pa[i] );
}

Clearly, this also displays the values of each element in array a.

Additionally, similar to using the array name, it can also be done as follows:

int i,*pa,a[]={3,4,5,6,7,3,7,4,4,6};
pa =a;
for (i=0;i<=9;i++)
{
   printf ( “%d”, *(pa+i) );
}

Looking at pa=a (assigning the array name to a pointer) and the forms of accessing elements via array name and pointer, there seems to be no difference. From this, it appears that an array name is essentially a pointer. Are there really no differences? Yes, please continue.

3. Differences Between Array Names and Pointer Variables Consider the following code:

int i,*pa,a[]={3,4,5,6,7,3,7,4,4,6};
pa =a;
for (i=0;i<=9;i++)
{
   printf ( “%d”, *pa );
    pa++ ;  // Note here, the pointer value is modified
}

It can be seen that this code also outputs the values of each array element. However, try changing pa to a inside the {}. You will find that the program fails to compile. It seems pointers and array names are still different. In fact, the pointer above is a pointer variable, while an array name is merely a pointer constant. The difference between this code and the previous ones is that the value of pointer pa continuously increments throughout the loop; that is, the pointer's value is modified. An array name is a pointer constant, and its value cannot be modified, so operations like a++ are not allowed. In the previous sections (referring to pa[i] and *(pa+i)), the value of pointer pa remained unchanged. Therefore, the pointer variable pa and the array name a can be interchanged in those contexts.

4. Declaring Pointer Constants Now consider the following code:

int i, a[]={3,4,5,6,7,3,7,4,4,6};
int * const pa=a;// Note the position of const: it's not const int * pa,
for (i=0;i<=9;i++)
{
   printf ( “%d”, *pa );
    pa++ ;  // Note here, the pointer value is modified
}

Will this code compile successfully? No. Because the pointer pa has been defined as a constant pointer. At this point, there is no difference from the array name a. This further illustrates that an array name is a constant pointer. However...

int * const a={3,4,5,6,7,3,7,4,4,6};// Not allowed
int a[]={3,4,5,6,7,3,7,4,4,6};// Allowed, so arrays must be initialized this way.

All the above experiments were conducted on VC6.0.