ISRO CSE 2020 | Question: 74

Question

Following declaration of an array of struct, assumes size of byte, short, int and long are $1,2,3$ and $4$ respectively. Alignment rule stipulates that $n$ – byte field must be located at an address divisible by $n$, the fields in the struct are not rearranged, padding is used to ensure alignment. All elements of array should be of same size.

Struct complx
       Short s
       Byte b
       Long l
       Int i
End Complx
Complx C[10]

Assuming $C$ is located at an address divisble by $8$, what is the total size of $C$, in bytes?

• A. $150$
• B. $160$
• C. $200$
• D. $240$

Comments

None are correct. Question cancelled by ISRO.

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Correct question must be,

“Assuming elements of C is located at an address divisible by 8, what is the total size of C, in bytes?”

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Ans

Answer 1

I believe the options are incorrect

Alignment rule stipulates that $n$−byte field must be located at an address divisible by $n$.

So,

• $Short$ must be in an address divisible by $2$
• $Byte$ must be in an address divisible by $1$
• $Long$ must be in an address divisible by $4$
• $Int$ must be in an address divisible by $3$

 

Following the same (assuming Base Address to be $0$),

$Short$	$Byte$	-	$Long$	-	$Int$
$0$	$2$	$3$	$4$	$8$	$9$

 

So the data type is $12$ bytes after padding. Since it's an array of $10$ elements, total size = $120$ bytes

Comments

That's why ISRO dropped this question

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great explanation.(good job man).

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I have checked many resources and found that Int size is 4 bytes

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@Dhainik Read the first line of the question again :)

Answer 2

$\underline{\mathbf{Answer:}\Rightarrow}$

$\underline{\mathbf{Explanation:}\Rightarrow}$

$\underline{\mathrm{Padding:}\Rightarrow}$

For aligning the data in the memory, some empty bytes are inserted between the memfory addresses which are assigned for other structure members during memdory allocation.

This is called the structure padding.

Size of the complex data type will be $2+1+4+3 = 10\;\text{Bytes}$

But because of padding, size will become $\mathbf{16}-\text{Bytes}\;\;\text{Since, $6$ Bytes are added for padding}]$

$\therefore $ Total size $=16\times 10 = 160 \;\text{Bytes}$

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http://www.firmcodes.com/structure-padding-and-packing-in-c-example/

Answer 3

Size of Byte = 1

Size of Short= 1

Size of Int= 1

Size of Long= 1

Total Size of Complex Structure = 10

Alignment Rule for N byte :- N Byte mod N = 0

This C code is located at an address which is divisible by N = 8

So, how many more(x) Bytes should we add to get (10 + x) mod 8 = 0

Here Size of Byte is 1 so keep add Bytes until we get (10 + x) mod 8 = 0

(10+1) mod 8 = 3

(10+2) mod 8 = 4

(10+3) mod 8 = 5

(10+4) mod 8 = 6

(10+5) mod 8 = 7

(10+6) mod 8 = 0

So, we need 16 Bytes ( 6 Bytes Padding ) 

Total 16*10 = 160 Bytes

Answer 4

1. Place short s (2 bytes)

• Struct starts at address 0 (divisible by 8 already given).

• Needs multiple of 2 → address 0 is fine.

0 1 [s][s]

We’re at offset 2 now.

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2. Place byte b (1 byte)

• Next free address = 2 → works (1-byte alignment).

2 [b]

We’re at offset 3 now.

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3. Place long l (4 bytes)

• Needs address divisible by 4.

• Next free = 3 → NOT divisible by 4.

• Add 1 byte padding to reach 4.

3 <- pad 4 5 6 7 [l][l][l][l]

We’re at offset 8 now.

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4. Place int i (3 bytes)

• Needs address divisible by 3.

• Next free = 8 → NOT divisible by 3.

• Add 1 byte padding to reach 9.

8 <- pad 9 10 11 [i][i][i]

We’re at offset 12 now.

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5. End padding for struct size

• We want every struct in an array to start at an address divisible by 8 (biggest alignment requirement given).

• Next multiple of 8 after 12 is 16.

• So add 4 bytes padding at end.

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0-1 : s 2 : b 3 : pad 4-7 : l 8 : pad 9-11 : i 12-15 : pad (end)

Size of one struct = 16 bytes.

we have 10 structs:

10 × 16 = 160 bytes total.

So the correct answer is B. 160 bytes.

Answer 5

Ans