C PROGRAMMING

Question

Answer 1

short variable has a 16-bit size. in a little-endian system, 320 (=00000001 01000000b) will be stored as 01000000 in the first location and  00000001 in the next location.

char variable stores 8 bits and points to 01000000 whose equivalent decimal value is 64. Thus output will be 64

Comments

01000000 will be sxtored in the first location or 00000001 will be stored in first location in little-endian?i am confused in this.

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	Big Endian	Little Endian
value 0x01 23 45 67 is represented as	01   23   45  67	67  45   23  01
Lowest Numeric address stores	MSB	LSB
Starting address of data item	Same in both	Same in both
features	String comparison. Ordering of integers and strings. Human readable has to perform addition when it converts a 32-bit integer address to a 16-bit integer address, to use the least significant bytes	Higher-precision arithmetic. Always reads as the same value if reading in the size larger than or equal to the value.
	Intel	ARM, PowerPC (by Motorola) and SPARK (by Sun); Current versions are bi endian

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thankyou :)

Answer 2

 

Here, (C) is the answer.

Here, a = $(320)_{10}$ = $(0000000101000000)_{2}$ = $(0140)_{16}$.

Short data type takes 2 bytes of memory.

*ptr is char pointer which points to the 1 byte memory location of stored value of variable a.

In little-endian representation of stored data is as least significant byte of stored value will be stored in first memory location and second least byte will be stored on subsequent memory locations and so on.

Refer, Endianness for more clarity on little-endian.

Suppose address of variable a is start at location A in byte addressable memory architecture. than $(40)_{16}$ will be store at location A and $(01)_{16}$ will be stored at address A+1 .

 

Little-Endian representation

$(40)_{16}$

$(01)_{16}$

 

Here, ptr points to the memory address A.

Here, *ptr prints the integer value of the byte stored on location A. which is $(40)_{16}$ = $(64)_{10}$.

 

 

 

Answer 3

320 write in binary form since ptr takes 8 bytes it is 64

Answer 4

We know in binary 320 = 00000001 01000000 .

In little – endian we take values from the LSB , char will take 1 byte , so 01000000  will be taken  = 64.

Answer – C :  64