Skip to content
Related Articles
Get the best out of our app
GeeksforGeeks App
Open App

Related Articles

ISRO | ISRO CS 2020 | Question 12

Improve Article
Save Article
Like Article
Improve Article
Save Article
Like Article

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 a struct are not rearranged, padding is used to ensure alignment. All elements of array should be of same size.

Struct complex
  Short s
  Byte b
  Long l
  Int i
End complex
Complex C[10] 

Assuming C is located at an address divisible by 8, what is the total size of C, in Bytes ?









Answer: (B)


Size of complex data type will be,

= 2 + 1 + 4 + 3 
= 10 Bytes 

But, address divisible by 8, so, it should be minimum,

= 10+6 
= 16 Bytes 

Therefore, total size of such 10 data types,

= 16*10
= 160 Bytes 

So, option (B) is correct.

Quiz of this Question
Please comment below if you find anything wrong in the above post

My Personal Notes arrow_drop_up
Last Updated : 03 Sep, 2020
Like Article
Save Article
Similar Reads