Double Pointers

Consider the following C code

int arr[4][3] = { 
	{81, 92, 2}, 
	{30, 99, 37}, 
	{38, 19, 49}, 
	{69, 78, 40} 
};
int *pptr = &arr[2][2];

The base address of arr is 0x0000bb50. What is the value of pptr in hexadecimal

Assume the sizeof(int) = 4

Sol: Each internal array of size 3 has a byte of

$$3\ast sizeof(int)=3\ast 4=12$$

The the difference from the starting address to arr[2][2] is

$$2\ast 12+4+4=24+8=32\to 0x20$$

Hence the address for pptr will be 0xbb70

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Answer

To solve this problem, we note the base address of arr is 0x0000bb50. This is the starting point of the entire 2D array in memory. To find the address of the element at arr[2][2], calculate the offset as follows:

1. Calculate the number of elements before the desired element:
   • Since the array is stored in row-major order (rows are stored consecutively), calculate the number of elements in previous rows as the number of elements in a row * the number of rows we are passing, or 2 * 3.
   • Then, add the number of elements before the desired column within the current row: 2.
   • Thus, the total number of integers before the desired value is 2 * 3 + 2.
   • Multiply this by 4 (size of each integer) to get the byte offset: offset = (2 * 3 + 2) * 4.
2. Add this offset to the base address to get the final pointer address:
   pptr = 0x0000bb50 + offset