Question

A one dimensional array A has indices 0....54.Each element is a string and takes up four memory words. The array is stored starting at location 1000 in decimal form. The starting address of 41th element of A is

Answer 1

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Answer 2

A one-dimensional array A has indices 0....54. Each element is a string and takes up four memory words. The array is stored starting at location 1000 in decimal form. The starting address of the $41^t^h$ element of A is 1164.

Explanation:

• If an array starts from zero then the formula will be:

             Address of array + $index \times size$ of an element.

• The array is stored starting at the location is considered as the address of the array.

• An address of an array is determined as the first element of an array.

• The first byte is a address of the element.

• So, the address of the array = 1000.

       Here the Index = 41 .

       The size of the element is 4.

       Therefore, Total address occupied till the 40th element = 41x 4 =  164.  

            $1000 + 41 \times 4$ = 1164.

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