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Consider the following C declaration
struct {
short s[5];
union {
float y;
long z;
}u;
}t;
Assume that objects of 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)
    18 bytes
  • c)
    14 bytes
  • d)
    10 bytes
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Consider the following C declarationstruct {short s[5];union {float y;...
CONCEPT:
Structure in C is a user-defined data type that is used to store the collection of different data types.
The total size of the structure is the sum of the size of every data member.
Union is a user-defined data type that is used to store different data types in the same memory location.
The total size of the union is the size of the largest data member.
 
Given the size of short, float and long is 2 bytes, 4 bytes, and 8 bytes, respectively.
Therefore,

Size of Structure → size of ( short s[5] ) + size of Union  
Here, Size of Union = 8 bytes   { largest data member is long }.
Size of short s[5]  2×5  10 bytes
Size of Structure → 10+8 →18 bytes.
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Consider the following C declarationstruct {short s[5];union {float y;long z;}u;}t;Assume that objects of type short, float and long occupy 2 bytes, 4 bytes and 8 bytes, respectively. The memory requirement for variable t, ignoring alignment considerations, isa)22 bytesb)18 bytesc)14 bytesd)10 bytesCorrect answer is option 'B'. Can you explain this answer?
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