Question

Assuming shorts are 2 bytes long, and pointers are 8 bytes long, how many bytes does this program need? short a = 3; short b = a; short * c = &a; *c = b++; short & d = a; short & e = b; short * f = c; e = *f + d; (a) 0 (b) 4 (c) 8 (d) 12 (e) 16 (f) 20 (g) 24 (h) 28

Please confirm. My answer: (f) 20

short a = 3; //2 bytes

short b = a; // +2

short * c = &a; //pointer variable +8

*c = b++; short & d = a;

//reference variable +0 bytes

short & e = b;//reference variable +0

bytes short * f = c;//pointer variable +8

e = *f + d;

Answer 1

Answer: (f) 20 bytes

short a = 3; //2 bytes

short b = a; //2 bytes

short * c = &a; // pointer variable = 8 bytes

*c = b++; //computation = 0 bytes

short & d = a; //reference variable just aliases the name, it doesn't takes space. so 0 bytes

short & e = b; //Reference variable. so 0 bytes.

short * f = c; //pointer variable = 8 bytes

e = *f + d; //it is same as b = *f + a, so the memory for them is already allocated and it is just computation. So 0 bytes.

Total memory needed = 2 + 2 + 8 + 0 + 0 + 0 + 8 + 0 = 20 bytes.

Therefore, The given program needs 20 bytes of memory.