Question

Consider the following variable definitions:

char a, *b, *c; int d[2], *e; int i, *j;

1. How many total bytes does this code allocate for variables?  Assume a 32-bit representation for integer and pointer values.

a     sizeof(char)

b     sizeof(char *)

c     sizeof(char *)

d    2*sizeof(int)

e     sizeof(int *)

i     sizeof(int)

j     sizeof(int *)

Total number of bytes ?

2. What is the output of the following piece of code? (Use the above variable definitions).

j = &i;

*j = 50;                      /* i = 50 */

e = &d[0];

*e = 20;                      /* d[0] = 20 */

e[1] = 30;                    /* d[1] = 30 */

j = &d[1];                    

*j = i+d[1];                  /* d[1] = 80 */

printf(“d = [%d, %d]\n”, d[0], d[1]);

Answer 1

size of a sizeof(char) is 1 byte // because character is a data type which requires 1 byte to store data

size of b sizeof(char*) is 4 byte // because it is given that the representation is of 32 bits or 4 bytes

size of c sizeof(char*) is 4 byte

size of d 2*sizeof(int) is 8 byte

size of e sizeof(int*) is 4 byte

size of i sizeof(int) is 4 byte

size of j sizeof(int *) is 4 byte

Therefore the total number of bytes will be (1+4+4+8+4+4+4)bytes=29 bytes.

A. The output of the code will 20,80

Note: If you run the code it will give error . Try  printf(“d = %d, %d”, d[0], d[1]); insteads of

printf(“d = [%d, %d]\n”, d[0], d[1]);.