How does the compiler calculates the address of the last element in the below array?

[radha gogia]

Say I have an array of a[5] , and base address is 3000 now , and I have to get the address of the last element a[4] and I have a pointer b , and I do
b=a+n-1 , so I have one confusion in this that first a+n would be executed like
a+n*sizeof(a) -1 ,now a+n*sizeof(a)= 3010 ,now when we do -1 from it , how does it evaluate to 3008 , the address of last element since contiguous memory cells are from address 3000 to 3008 so then how come 3010-1 evaluates to 3008 ,how does compiler come to know how much bytes it has to move backwards since 3010 is a random location .

[weaknessforcats]

All you need do is:

b = a+4;

In doing pointer arithmetic, the compiler will add 4 times the sizeof the array type. You don't have to worry about that.

In your example, if n is number of elements in the array, and b is a pointer to the array type, then you would:

b =  a + n - 1;
 

You might also read: http://bytes.com/topic/c/insights/77...rrays-revealed

[radha gogia]

Sorry but couldn't get clear with this concept that how 3010 -1 =3008 when that is some random location in the memory as array is contiguously located from 3000 to 3008 only .
As u said that compiler will add 4 times the sizeof the array type but no of elements in the array are 5 and not 4 so then ?

Sorry but seriously I am stucked at this point .

[weaknessforcats]

The first element of an array is element 0. The name of the array is the address of that element. Therefore, that address plus 4 is the address of the fifth element,which is what you asked about.

Did you read the article?

[radha gogia]

Yes , I read and agree to your explanation but what I am saying ,plz see it once, I am confused in that point.

[radha gogia]

Since b=a+n-1 =3010-1=3008 , how's that evaluating to 3008 when 3010 is a random location.

[donbock]

Let's be very specific. Let's assume for a moment that data pointers can be cast to integer type unsigned long.

#define N 5
int main()
{
  int a[N];
  int *b;
  unsigned long p;
 
  b = a;
  p = (unsigned long)a;
  printf("1a. %p\n", b);
  printf("1b. %ld\n", p);
 
  b = a + N;
  p = (unsigned long)a + N*sizeof(int);
  printf("2a. %p\n", b);
  printf("2b. %ld\n", p);
 
  b = a + N - 1;
  p = (unsigned long)a + (N-1)*sizeof(int);
  printf("3a. %p\n", b);
  printf("2b. %ld\n", p);
}

If sizeof(int) is 4, and if array a happens to start at address 3000, then I expect the output to be:

1a. 3000
1b. 3000
2a. 3020
2b. 3020
3a. 3016
3b. 3016

Pointer arithmetic involves an arbitrary but well-defined pointer value and an arbitrary but well-defined integer expression. Nothing is random.

Pointer arithmetic works as if it were implemented in accordance with the p expressions above.