Question

Consider the following declaration of a two-dimensional
array in c:
char A[100][100];
Assuming that the main memory is byte-
addressable and that the array is stored starting from
memory address 0, the address of A[40][50] is​

Answer 1

Answer:

3950

Explanation:

The size of char is 1 byte.

since the index is [40][50], you have to move 39 rows each having 100 columns. so, the total memory moved in bytes = 39*100*1 = 3900

At 40th row you have to move till column 50

Total meory moved = 3900 + 50 = 3950

Answer 2

Answer:

4050

Explanation:

Address of a[40][50]

= Base address + 40*100*element_size   +50*element_size

= 0 + 4000*1 + 50*1

= 4050