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3 years ago

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Comment on the following 2 arrays. int a1[8]; int (a2[8]); a1 is pointer to an array; a2 is array of pointers a1 is pointer to

an array; a2 is pointer to an array a1 is array of pointers; a2 is pointer to an array a1 is array of pointers; a2 is array of pointers

Computers and Technology

1 answer:

Hitman42 [59]3 years ago

5 0

Answer:

The answer is "a1 and a2 is an array of pointers".

Explanation:

In this question, A collection of pointers refers to an array of elements where each pointer array element points to a data array element. In the above-given statement, the two-pointer type array "a1 and a2" is declared that holds the same size "8" elements in the array, and each element points towards the array's first element of the array, therefore, both a1 and a2 are pointer arrays.

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Drag each statement to the correct location.

VMariaS [17]

Answer:

$(a)\ 222_{10} = DE_{16}$ --- True

$(b)\ D7_{16} = 11010011_2$ --- False

$(c)\ 13_{16} = 19_{10}$ --- True

Explanation:

Required

Determine if the statements are true or not.

$(a)\ 222_{10} = DE_{16}$

To do this, we convert DE from base 16 to base 10 using product rule.

So, we have:

$DE_{16} = D * 16^1 + E * 16^0$

In hexadecimal.

$D =13 \\E = 14$

So, we have:

$DE_{16} = 13 * 16^1 + 14 * 16^0$

$DE_{16} = 222_{10}$

Hence:

(a) is true

$(b)\ D7_{16} = 11010011_2$

First, convert D7 to base 10 using product rule

$D7_{16} = D * 16^1 + 7 * 16^0$

$D = 13$

So, we have:

$D7_{16} = 13 * 16^1 + 7 * 16^0$

$D7_{16} = 215_{10}$

Next convert 215 to base 2, using division rule

$215 / 2 = 107 R 1$

$107/2 =53 R 1$

$53/2 =26 R1$

$26/2 = 13 R 0$

$13/2 = 6 R 1$

$6/2 = 3 R 0$

$3/2 = 1 R 1$

$1/2 = 0 R1$

Write the remainders from bottom to top.

$D7_{16} = 11010111_2$

<em>Hence (b) is false</em>

$(c)\ 13_{16} = 19_{10}$

Convert 13 to base 10 using product rule

$13_{16} = 1 * 16^1 + 3 * 16^0$

$13_{16} = 19$

Hence; (c) is true

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3 years ago