Chapter 2, Problem 2.82HW

A.

Explanation of Solution

For “C” expression “$\left(\text{x<y}\right)\text{==}\text{(-x>-y)}$”:

• The given “C” expression does not always produce “1”.
• When the value of “x” is “INT_MIN”, the given expression producing the result of “0”.

Program:

A complete program has been developed for the above expression when producing result “0” is shown below.

#include <stdio.h>

#include <assert.h>

#include <limits.h>

#include <stdlib.h>

//Function definition for part A

int partA(int x, int y)

{

  //Returns the value

  return (x < y) == (-x > -y);

}

//Main function

int main(int argc, char* argv[])

{

/* Call function "partA" with checking value using

B.

Explanation of Solution

For “C” expression “$\left(\left(\text{x + y}\right)\text{<< 4}\right)\text{+ y - x}\text{=}\text{17 * y + 15 * x}$”:

The given “C” expression is always producing the result as “1”. The mathematical principle for given expression shown below:

$\begin{array}{c}\text{=}\left(\left(\text{x + y}\right)\text{ << 4}\right)\text{+}\text{y - x}\text{=>}\text{Leftside}\text{of}\text{given}\text{expression}\\ \text{=}\text{x << 4}\text{+}\text{y << 4}\text{+}\text{y - x}\text{=>}\text{By (a + b) * c}\text{=}\text{(a * c)}\text{+}\text{(b * c)}\\ \text{=}\text{x << 4}\text{+ -x}\text{+}\text{y << 4}\text{+}\text{y}\text{=>}\text{By rearranging equation}\\ \text{=}\text{x}\text{*}\text{16}\text{-}\text{x}\text{+}\text{y}\text{* 16}\text{+ y}\text{=>}\text{By}\text{x << 4}\text{=}\text{x}\text{*}\text{16}\end{array}$

C.

Explanation of Solution

For “C” expression “$\sim \text{x+}\sim \text{y+1}\text{==}\sim \left(\text{x+y}\right)$”:

The given “C” expression is always producing the result as “1”. The mathematical principle for given expression shown below:

$\begin{array}{c}\text{=}\sim \text{x + }\sim \text{y + 1}\text{=> From the left side of given expression}\\ \text{=}\sim \text{x + }\sim \text{y + 1 + 1 - 1 => The equation is added and subtracted by 1}\\ \text{= }\sim \text{x + 1 + }\sim \text{y + 1 - 1 => Rearranging equaition}\\ \text{=}\left(\sim \text{x + 1}\right)\text{ + }\left(\sim \text{y + 1}\right)\text{ -1}\end{array}$

$\begin{array}{c}\text{= -x + -y -1 => By -x = :x + 1}\\ \text{=}\text{-}\left(\text{x + y}\right)\text{ -1}\\ \text{=}\sim \left(\text{x + y}\right)\text{ +1 -1}\text{=> By -x = }\sim \text{x + 1}\\ \text{=}\sim \left(\text{x +y}\right)\\ \text{=}\text{Right}\text{of}\text{expression}\end{array}$

Program:

A complete program has been developed for the above expression when producing result “1” is shown below

D.

Explanation of Solution

For “C” expression “$\left(\text{ux-uy}\right)\text{==}\text{-(unsigned)}\left(\text{y-x}\right)$”:

The given “C” expression is always producing the result as “1”. The mathematical principle for given expression shown below:

$\begin{array}{c}\text{=}\text{ux}\text{-}\text{uy}\\ \text{=}\left(\text{unsigned}\right)\text{x}\text{-}\left(\text{unsigned}\right)\text{y}\\ \text{=}\left(\text{unsigned}\right)\left(\text{x}\text{-}\text{y}\right)\\ \text{=}\text{-}\left(\text{unsigned}\right)\left(\text{y}\text{-}\text{x}\right)\end{array}$

Program:

A complete program has been developed for the above expression when producing result “1” is shown below.

//Header file

#include <stdio.h>

#include <assert.h>

#include <limits.h>

#include <stdlib

E.

Explanation of Solution

For “C” expression “$\left(\left(\text{x>>2}\right)\text{<<2}\right)\text{<=}\text{x}$”:

The given “C” expression is always producing the result as “1”. The mathematical principle for given expression shown below:

$\begin{array}{l}\left(\left(\text{x>>2}\right)\text{<<2}\right)\text{<=}\text{x}\\ \text{x}\text{\& }\sim \text{ 0x3}\\ \text{x - num(00/01/10/11)}\\ \text{Hence}\left(\left(\text{x>>2}\right)\text{<<2}\right)\text{<=}\text{x}\end{array}$

For proving $\left(\left(\text{x>>2}\right)\text{<<2}\right)\text{=}\text{x}$is given below

$\begin{array}{c}\text{=}\left(\text{x>>2}\right)\text{<<2}\text{=> Left side of given expression }\\ \text{=}\left(\frac{\text{x}}{{\text{2}}^{\text{2}}}\right)\text{<<}\text{2}\text{=> By x}\text{>>k}\text{=}\frac{\text{x}}{{\text{2}}^{\text{k}}}\\ \text{= }\left(\frac{\text{x}}{\text{4}}\right)\text{*}{\text{2}}^{\text{2}}\text{=>}\text{By}\text{x}\end{array}$