Chapter 2, Problem 2.61HW

A.

Explanation of Solution

C expression for given statement:

The C expression for any bit of “x” equals “1” is shown below.

Expression: !~x

Program:

For the above expression “!~x”, a complete program has been developed in the below section:

//Header file

#include <stdio.h>

#include <assert.h>

//Function definition "A" statement

int partA(int x)

{

/* Returns "1" if any bits of "x" equals "1" by using expression "!~x" */

return !~x;

}

//Main function

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

{

//Assign all bit is "1"

int allBit1 = ~0;

//Assign all bit is "0"

int allBit0 = 0;

/* Call partA function with "alllBit1" using "assert" function */

assert(partA(allBit1));

/* Call partA function with "alllBit0" using "assert" function */

assert(!partA(allBit0));

/* Call partA function with number "0x1234FF" using "assert" function */

assert(!partA(0x1234FF));

return 0;

}

From the given code,

• Define the function for “partA()”
  • In this function, returns “1” if any bits of “x” equals “1” by using the expression “!~x”...

B.

Explanation of Solution

C expression for given statement:

The C expression for any bit of “x” equals “0” is shown below.

Expression: !x

Program:

For the above expression “!x”, a complete program has been developed in the below section:

//Header file

#include <stdio.h>

#include <assert.h>

//Function definition "B" statement

int partB(int x)

{

/* Returns "1" if any bits of "x" equals "0" by using expression "!x" */

return !x;

}

//Main function

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

{

//Assign all bit is "1"

int allBit1 = ~0;

//Assign all bit is "0"

int allBit0 = 0;

/* Call partB function with "alllBit1" using "assert" function */

assert(!partB(allBit1));

/* Call partB function with "alllBit0" using "assert" function */

assert(partB(allBit0));

/* Call partB function with number "0x1234FF" using "assert" function */

assert(!partB(0x1234FF));

return 0;

}

From the given code,

• Define the function for “partB()”
  • In this function, returns “1” if any bits of “x” equals “0” by using the expression “!x”.
• Define the main function
  • In this function, assign “~0” in variable “allBit1”...

C.

Explanation of Solution

C expression for given statement:

The C expression for any bit in the least significant byte of “x” equals “1” is shown below.

Expression: !~(x | ~0xff)

Program:

For the above expression “!~(x | ~0xff)”, a complete program has been developed in the below section:

//Header file

#include <stdio.h>

#include <assert.h>

//Function definition "A" statement

int partA(int x)

{

/* Returns "1" if any bits of "x" equals "1" by using expression "!~x" */

return !~x;

}

//Function definition "C" statement

int partC(int x)

{

/* Returns "1" if any bits in the least significant byte of "x" equals "1" by calling the function "partA" with "x | ~0xff" */

return partA(x | ~0xff);

}

//Main function

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

{

//Assign all bit is "1"

int allBit1 = ~0;

//Assign all bit is "0"

int allBit0 = 0;

/* Call partC function with "alllBit1" using "assert" function */

assert(partC(allBit1));

/* Call partC function with "alllBit0" using "assert" function */

assert(!partC(allBit0));

/* Call partC function with number "0x1234FF" using "assert" function */

assert(partC(0x1234ff));

return 0;

}

From the given code,

• Define the function for “partA()”
  • In this function, returns “1” if any bits of “x” equals “1” by using the expression “!~x”...

D.

Explanation of Solution

C expression for given statement:

The C expression for any bit in the least significant byte of “x” equals “0” is shown below.

Expression: !((x >> ((sizeof(int)-1) << 3)) & 0xff)

Program:

For the above expression “!((x >> ((sizeof(int)-1) << 3)) & 0xff)”, a complete program has been developed in the below section:

//Header file

#include <stdio.h>

#include <assert.h>

//Function definition "B" statement

int partB(int x)

{

/* Returns "1" if any bits of "x" equals "0" by using expression "!x" */

return !x;

}

//Function definition "D" statement

int partD(int x)

{

/* Returns "1" if any bits in the least significant byte of "x" equals "0" by calling the function "partB" with "x >> ((sizeof(int)-1) << 3)) & 0xff" */

return partB((x >> ((sizeof(int)-1) << 3)) & 0xff);

}

//Main function

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

{

//Assign all bit is "1"

int allBit1 = ~0;

//Assign all bit is "0"

int allBit0 = 0;

/* Call partB function with "alllBit1" using "assert" function */

assert(!partD(allBit1));

/* Call partB function with "alllBit0" using "assert" function */

assert(partD(allBit0));

/* Call partB function with number "0x1234FF" using "assert" function */

assert(partD(0x1234ff));

return 0;

}

From the given code,

• Define the function for “partB()”
  • In this function, returns “1” if any bits of “x” equals “0” by using the expression “!x”...