Please help with array addressing Will give good review if a

Please help with array addressing. Will give good review if answered properly.

Assume all variables are signed 32-bit quantities, except for when you can tell they are characters. Topic: Array addressing Memory is reserved for an array of 2048 4-byte integers. How many bytes are reserved? Memory is reserved for an array of 2048 4-byte integers starting at address AC54h. What is the address of the last element of the array. Array arr1 has 512 elements, where each element is a 1-byte character. If the starting address of the array is FB68h, what is the address of arr1[367]? Array arr2 has 512 elements, where each element is a 4-byte integer. If the starting address of the array is FB68h, what is the address of arr1[367]?

Solution

1)answer : 8192 bytes are reserved

explanation:- array contains 2048 integers, for each integers 4bytes of space needed... so 2048*4 bytes of space needed for the array

2)answer:last element address :0CC50h

explanation:- starting address : 0AC54h convert to decimal = 52,308 now add 8188 (leaving last integer)..result is 52,304 convert it hex : 0CC50h

3)answer: address of arr1[367] is 0FCD7h

explanation:- starting address : 0FB68h to decimal value = 64360 now add 367*1 to get address ot 367 index, result is 64727 to hex 0FCD7h

4)answer: address of arr1[367] is 10124h

explanation:- starting address : 0FB68h to decimal value = 64360 now add 367*4 to get address ot 367 index, result is 65828 to hex 10124h