For "C" expression "(float)x==(float)dx":

The given "C" expression is always producing the result as "1". From the expression "(float)x==(float)dx",

• All double numbers in the given expression are comes from ints.
• So, when casting from "double" to "float", the value may be rounded because, the precision is lesser.
• Therefore, both side of expression will be equal.

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

For "C" expression "dx-dy=(double)(x-y)":

• The given "C" expression does not always produce "1".
• When the value of "y" 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.

//Header file

#include <stdio.h>

#include <assert.h>

#include <limits.h>

#include <stdlib.h>

//Function definition for part B

int partB(int x, double dx, int y, doubledy)

{

//Return the value

return dx-dy == (double)(x-y);

}

//Main function

int main(intargc, char* argv[])

{

//Assign the value for "x"

int x = rand();

//Assign the value for "y"

int y = rand();

//Converting the value of "x" into double

double dx = (double)x;

//Converting the value of "y" into double

double dy = (double)y;

//Display the value of "x"

printf("%x\n", x);

//Display the value of "y"

printf("%x\n", y);

/* Call function "partB" with checking value using "assert" function */

/* Here producing result as "0", when passing arguments for "x" is "0" and for "y" is "INT_MIN */

assert(!partB(0, (double)(int)0, INT_MIN, (double)(int)INT_MIN));

return 0;

}

The given code is used to return the result for expression "dx-dy=(double)(x-y)"

For "C" expression "(dx+dy)+dz==dx+(dy+dz)":

The given "C" expression is always producing the result as "1". From the given expression, the value of "dx", "dy" and "dz" are all double numbers from "ints".

• So, the result will be "double" value when adding the numbers in both side of given expression.

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.h>

int partC(double dx, double dy, double dz)

{

//Returns the value

return (dx+dy)+dz == dx+(dy+dz);

}

//Main function

int main(intargc, char* argv[])

{

//Assign the value for "x"

int x = rand();

//Assign the value for "y"

int y = rand();

//Assign the value for "z"

int z = rand();

//Converting the value of "x" into double

double dx = (double)x;

//Converting the value of "y" into double

double dy = (double)y;

//Converting the value of "z" into double

double dz = (double)z;

//Display the value of "x"

printf("%x\n", x);

//Display the value of "y"

printf("%x\n", y);

//Display the value of "z"

printf(

For "C" expression "(dx*dy)*dz==dx*(dy*dz)":

• The given "C" expression does not always produce "1".
• When multiplying three "double" numbers, the results from both side of expression is different that is not equal.
  • Since, the precedence of the multiplication.
  • There are two multiplication operation in each side of given expression
    • The multiply operator "*" performs a left to right associativity.
    • Consider the left side of expression "(dx*dy)*dz".
      • In this expression "dx*dy" is calculated first and then multiplied with "dz".
    • Consider the right side of expression "dx*(dy*dz)".
      • In this expression "dx" is multiplied with"dy*dz".
  • So, calculation of any one side of given expressionoccurs to yield a rounding error in result.
    • Because, if a value in it simple, it can’t be denoted by a floating point value.

Program:

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

//Header file

#include <stdio.h>

#include <assert.h>

#include <limits.h>

#include <stdlib

For "C" expression "dx/dx==dz/dz":

• The given "C" expression does not always produce "1".
• When "dx" is not equal to "0" and "dz" value is equal to "0", the given expression producing the result of "0".
  • That is the expression of both side are not equals.

Program:

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

//Header file

#include <stdio.h>

#include <assert.h>

#include <limits.h>

#include <stdlib.h>

//Function definition

int partE(double dx, double dz)

{

//Returns the value

return dx/dx == dz/dz;

}

//Main function

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

{

//Assign the value for "x"

int x = rand();

//Assign the value for "z"

int z = rand();

//Converting the value of "x" into double

double dx = (double)x;

//Converting the value of "z" into double

double dz = (double)z;

//Display the value of "x"

printf("%x\n", x);

//Display the value of "z"

printf("%x\n", z);

/* Call function "partE" with checking value using "assert" function */

assert(partE(dx, (double)(int)0));

return 0;

}

The given code is used to return the value of expression "dx/dx==dz/dz"