Question

Consider the following C program: int maino short S; printf("Address of S is %d\n", &S); // printf() Statement 1 printf("%d", S); // printf( Statement 2 return 0; Assume variable "S" is a TWO BYTE, TWO'S COMPLEMENT (SIGNED) "short integer" stored in memory in LITTLE-ENDIAN format (.e. least significant byte at the lowest address). When the program is run, printfo Statement 1 produces the following output: Address of S is 5865 Given the contents of the portion of memory shown in the table below, what is the output (numerical value only) of printf( Statement 2? 80 Memory Contents (all values in decimal) Address Data 5864 212 5865 182 5866 191 5867 21 5869 5870 253 5871 111 5873 5874 122 5868 167 127 5872 78

Answer 1

Little Endian and Big Endian are used to storing multibyte data type.In Little Endian, least significant byte is read while in Big Endian, most significant byte is read.As we know "short" data type takes 2 Bytes but here we are using "Little Endian" hence only 1 least significant byte will be read.

Here "s" is stored on address 5865 and data is 182.

182 in 2 Byte binary= 00000000 10110110

Now take least significant 1 Byte(8 bits)= 10110110

But "s" is 2 byte 2's compliment short integer. Hence decimal value of 10110110 (2's compliment representation) = -74

"%d" in statement 2 shows that it will print a integer value.

Hence statement 2 will print -74 (negative 74)