Computation of (a-b): It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”. Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable. In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data. Here, the value of b is negative. So when the computation a-b is done, they both add up. The result is out of range of int value. Hence, the wrapping up takes place. Therefore, the result of a-b is “ -294967296 ”.

Question

Want to see more full solutions like this?

Answer 1

Expert Solution & Answer

Chapter 10, Problem 1RE

Explanation of Solution

Computation of (a-b):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation a-b is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of a-b is “-294967296”.

Explanation of Solution

Computation of (b-a):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation b-a is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of b-a is “294967296”.

Explanation of Solution

Result of Integer.compare(a, b):

The method Integer.compare(a, b) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is greater than the value on the right side.

Therefore, the result of Integer.compare(a, b) is “1”.

Explanation of Solution

Result of Integer.compare(b, a):

The method Integer.compare(b, a) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is less than the value on the right side.

Therefore, the result of Integer.compare(b, a) is “-1”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 2

Expert Solution & Answer

Chapter 10, Problem 1RE

Explanation of Solution

Computation of (a-b):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation a-b is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of a-b is “-294967296”.

Explanation of Solution

Computation of (b-a):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation b-a is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of b-a is “294967296”.

Explanation of Solution

Result of Integer.compare(a, b):

The method Integer.compare(a, b) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is greater than the value on the right side.

Therefore, the result of Integer.compare(a, b) is “1”.

Explanation of Solution

Result of Integer.compare(b, a):

The method Integer.compare(b, a) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is less than the value on the right side.

Therefore, the result of Integer.compare(b, a) is “-1”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Expert Solution & Answer

Chapter 10, Problem 1RE

Explanation of Solution

Computation of (a-b):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation a-b is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of a-b is “-294967296”.

Explanation of Solution

Computation of (b-a):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation b-a is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of b-a is “294967296”.

Explanation of Solution

Result of Integer.compare(a, b):

The method Integer.compare(a, b) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is greater than the value on the right side.

Therefore, the result of Integer.compare(a, b) is “1”.

Explanation of Solution

Result of Integer.compare(b, a):

The method Integer.compare(b, a) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is less than the value on the right side.

Therefore, the result of Integer.compare(b, a) is “-1”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Expert Solution & Answer

Chapter 10, Problem 1RE

Explanation of Solution

Computation of (a-b):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation a-b is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of a-b is “-294967296”.

Explanation of Solution

Computation of (b-a):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation b-a is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of b-a is “294967296”.

Explanation of Solution

Result of Integer.compare(a, b):

The method Integer.compare(a, b) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is greater than the value on the right side.

Therefore, the result of Integer.compare(a, b) is “1”.

Explanation of Solution

Result of Integer.compare(b, a):

The method Integer.compare(b, a) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is less than the value on the right side.

Therefore, the result of Integer.compare(b, a) is “-1”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 10, Problem 1RE

Explanation of Solution

Computation of (a-b):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation a-b is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of a-b is “-294967296”.

Explanation of Solution

Computation of (b-a):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation b-a is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of b-a is “294967296”.

Explanation of Solution

Result of Integer.compare(a, b):

The method Integer.compare(a, b) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is greater than the value on the right side.

Therefore, the result of Integer.compare(a, b) is “1”.

Explanation of Solution

Result of Integer.compare(b, a):

The method Integer.compare(b, a) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is less than the value on the right side.

Therefore, the result of Integer.compare(b, a) is “-1”.

Answer 8

Chapter 10, Problem 1RE

Explanation of Solution

Computation of (a-b):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation a-b is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of a-b is “-294967296”.

Answer 9

Chapter 10, Problem 1RE

Explanation of Solution

Computation of (a-b):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation a-b is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of a-b is “-294967296”.

Answer 10

Explanation of Solution

Computation of (b-a):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation b-a is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of b-a is “294967296”.

Answer 11

Explanation of Solution

Computation of (b-a):

It is given that the value assigned to a is “2000000000” and the value assigned to b is “-a”, which is “-2000000000”.

Whenever the result of an arithmetic operation is out of range, it returns an overflow, which can be stored in the result variable.

In order to store the result, the value is wrapped around and is assigned to the result variable. This wrapping will result in loss of data.

Here, the value of b is negative. So when the computation b-a is done, they both add up.

The result is out of range of int value. Hence, the wrapping up takes place.

Therefore, the result of b-a is “294967296”.

Answer 12

Explanation of Solution

Result of Integer.compare(a, b):

The method Integer.compare(a, b) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is greater than the value on the right side.

Therefore, the result of Integer.compare(a, b) is “1”.

Answer 13

Explanation of Solution

Result of Integer.compare(a, b):

The method Integer.compare(a, b) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is greater than the value on the right side.

Therefore, the result of Integer.compare(a, b) is “1”.

Answer 14

Explanation of Solution

Result of Integer.compare(b, a):

The method Integer.compare(b, a) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is less than the value on the right side.

Therefore, the result of Integer.compare(b, a) is “-1”.

Answer 15

Explanation of Solution

Result of Integer.compare(b, a):

The method Integer.compare(b, a) compares two test values for the order in which they must come.

If the variable on the left side is less than the right side, then “-1” is returned.

Otherwise, if the variable on the left side is greater than the right side, then “1” is returned.

In case, both are equal, then “0” is returned.

Here, the value of a is greater than the value of b. It means that the value on the left side is less than the value on the right side.

Therefore, the result of Integer.compare(b, a) is “-1”.

Answer 16

Ch. 10.1 - Suppose you want to use the average method to...Ch. 10.1 - Prob. 2SC Ch. 10.1 - Prob. 3SC Ch. 10.1 - What is wrong with this code? Measurable meas =...Ch. 10.1 - What is wrong with this code? Measurable meas =...Ch. 10.2 - Can you use a cast (BankAccount) meas to convert...Ch. 10.2 - Prob. 7SC Ch. 10.2 - Why is it impossible to construct a Measurable...Ch. 10.2 - Why can you nevertheless declare a variable whose...Ch. 10.2 - What does this code fragment print? Why is this...

Explanation of Solution

Explanation of Solution

Explanation of Solution

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions