Question

Assume that we are executing the following code on a 32-bit machine using two’s complement arithmetic for signed integers. Which of the following will be printed when the following code is executed (circle those printed, and show work; e.g., how the values are stored):

#include <stdio.h>

int main()

{

char x = 0xF;                // x = ________

char y = -1;                 // y = ________

unsigned char z = 0xFF;      // z = 11111111       

if (x<z)

    printf("performed unsigned compare, x = %u, z = %u\n", x, z);

else

    printf("performed signed compare, x = %d, z = %d\n", x, z);

printf("y>>4 = %d\n", y>>4);    // y>>4 = __________

printf("z>>4 = %d\n", z>>4);    // z>>4 = __________

}

1. performed unsigned compare, x = 15, z = 255
2. performed unsigned compare, x = 15, z = -1
3. performed signed compare, x = 15, z = -1
4. y>>4 = -1
5. y>>4 = 15
6. z>>4 = -1
7. z>>4 = 15

Hint: Recall that the %u specifier is used to display the decimal value as an “unsigned” value

Answer 1

#include <stdio.h>

int main()

{

//0x is used to declare hexadecimal value--

char x = 0xF;                // F in haxadecimal= 1111 in binary so,x = 1111

char y = -1;                 // y = -1

unsigned char z = 0xFF;      // z = 11111111       

if (x<z) //(1111<11111111) = True, so if statement will print

    printf("performed unsigned compare, x = %u, z = %u\n", x, z); // x= decimal value of x and z = decimal value of z

// 1111 = = 8+4+2+1 = 15

//11111111 = = 128+64+32+16+8+4+2+1 = 255

else // it will not be executed

    printf("performed signed compare, x = %d, z = %d\n", x, z);

printf("y>>4 = %d\n", y>>4);    // y>>4 = -1

printf("z>>4 = %d\n", z>>4);    // z>>4= 11111111>>4

// >> means bitwise right shift

z = 0000 0000 1111 1111

z>>1 = 0000 0000 0111 1111

z>>2 = 0000 0000 0011 1111

z>>3 = 0000 0000 0001 1111

z>>4 = 0000 0000 0000 1111

1111 = 15

Hence, the output will be:-

a) performed unsigned compare, x = 15, z = 255

b) y>>4 = -1

c) z>>4 = 15