Suppose we number the bytes in a w-bit word from 0 (less significant) to w/8-1 (most significant). write code for the followign c function, which will return an unsigned value in which byte i of argument x has been replaced by byte b:unsigned replace_byte (unsigned x, int i, unsigned char b);

Solution:

typedef  unsigned  char  *byte_pointer;

static int  int_of_bit  (byte_pointer x,  int  loc)

{

                  return(x[loc] << loc*8);

}  

static int  replace_byte(unsigned int a,  int loc,  unsigned int  b)

    unsigned int a_loc = int_of_bit((byte_pointer) &a ,  loc);

    unsigned int b_loc = (b  <<  loc*8);

     a  -=  a_loc;

      a  += b_loc;

      return a;

}

Explanation:

This takes two ints in hex format, one with 8 bits (0x00000000) and one with 2 bits (0x00) then places the 2 bit hex into the 8 bit at a given location.

EX:  replace_byte(0x00000000, 1, 0xFF) return 0x0000FF00

First thing first, it looks like you have some mixup:

" one with 8 bits (0x00000000) and one with 2 bits (0x00)*- what you mean is nibble not bits. Each hex character (0-9, A-F) represent 4 bits (16 possible combination), and is called a "nibble".

0x0000000 is 4 bytes. 0x00 is 2 bytes.

Next, you say "takes two ints in hex format" - your function takes two ints in any format. You're just choosing tp express them in hexidecimal. C++ doesn't care if you specify numbers in hex, decimal, octal, binary, etc.

answer:

true

explanation: