NIELIT 2017 July Scientist B (IT) - Section B: 37

Question

Consider the following C declaration 

struct 
{ 
    short s[5]; 
    union 
    { 
        float y; 
        long z; 
    }u; 
}t;

Assume that objects of the type short, float and long occupy $2$ bytes, $4$ bytes and $8$ bytes respectively. The memory requirement for variable $t$ ignoring alignment considerations, is 

• A. $22$ bytes
• B. $14$ bytes
• C. $18$ bytes
• D. $10$ bytes

Answer 1

 In union, all members share the same memory location.

How is the size of union decided by compiler?

Size of a union is taken according the size of largest member in union.

Size $= 8 + 2*5 = 18$ Bytes

C is correct. 

Answer could have been different if "memory requirement for variable t ignoring alignment consideration" wasn't mentioned.

Comments

What could be the answer if we consider alignment ?

If machine architecture has word size = 32 bits, then every piece of addressable memory must be in multiples of a word. Or equivalently 32 bits. Or, equivalently 4B.

Hence for:

short x[5];

 Memory requirement $= 5*2 = 10B + 2B$ padding $= 12B$ (Multiple of 4B)

And for

union {

Memory requirement $= 8B$

So, Total $= 20B$

---

If machine architecture has word size = 64 bits, then every piece of addressable memory must be in multiples of a word. Or equivalently 64 bits. Or, equivalently 8B.

Hence for

short x[5];

 Memory requirement $= 5*2 = 10B + 6B$ padding $= 16B$ (Multiple of 8B)

And for

union {

Memory requirement $= 8B$

So, Total $= 24B$

Src : https://gateoverflow.in/640/gate-cse-2000-question-1-17-isro2015-79?show=328119#a328119