num1 = 35; is a valid assignment as integer 35 is be...

Question

What does the value of the top variable indicate in this ArrayStack implementation? What will happen if we call pop on this stack? What value will be returned, and what changes will occur in the array and the top variable? 3. If we push the value "echo" onto the stack, where will it be stored in the array, and what will be the new value of top? 4. Explain why index 0 contains the string "alpha" even though top is currently 3. 5. What would the state of the stack look like (values in the array and value of top) after two consecutive pop 0 operations?

Answer 1

Expert Solution & Answer

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

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What does the value of the top variable indicate in this ArrayStack implementation? What will happen if we call pop on this stack? What value will be returned, and what changes will occur in the array and the top variable? 3. If we push the value "echo" onto the stack, where will it be stored in the array, and what will be the new value of top? 4. Explain why index 0 contains the string "alpha" even though top is currently 3. 5. What would the state of the stack look like (values in the array and value of top) after two consecutive pop 0 operations?

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Answer 2

Expert Solution & Answer

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

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Answer 3

Expert Solution & Answer

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

Expert Solution & Answer

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Check out sample textbook solution

Answer 4

Expert Solution & Answer

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

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See solution

Check out sample textbook solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

Answer 8

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

Answer 9

Chapter 2, Problem 8CP

a. num1 = 35;

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

Answer 10

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

Answer 11

b. newNum = num1 – num2;

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

Answer 12

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

Answer 13

c. num1 = 5; num2 = 2 + num1; num1 = num2 / 3;

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

Answer 14

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 15

d. num1 * num2 = newNum;

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 16

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

Answer 17

e. x = 12 * num1 - 15.3;

Explanation of Solution

x = 12 * num1 - 15.3; is a valid assignment although the variable num1 in the expression on the right-hand side has been declared but not initialized before being used in the expression. If a variable is used in an expression, the expression would yield a meaningful value only if the variable has first been initialized...

Answer 18

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 19

f. num1 * 2 = newNum + num2;

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 20

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 21

g. x / y = x * y;

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Answer 22

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

Answer 23

h. num2 = num1 % 2.0;

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

Answer 24

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

Answer 25

i. newNum = static_cast<int> (x) % 5;

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

Answer 26

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

Answer 27

j. x = x + y - 5;

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

Answer 28

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

Answer 29

k. newNum = num1 + static_cast<int> (4.6 / 2);

Explanation of Solution

newNum = num1 + static_cast<int> (4.6 / 2); is a valid assignment. The expression on the right-hand side evaluates an unpredictable but int type value. The expression inside the parentheses is evaluated first where the denominator is implicitly casted to floating-point type (as the numerator is floating-point type)...

Answer 30

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Answer 31

Expert Solution & Answer

Answer 32

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Answer 33

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Answer 34

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Answer 35

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Answer 36

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Answer 37

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Answer 38

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Answer 39

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Answer 40

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Answer 41

Ch. 2 - 1. Mark the following statements as true or...Ch. 2 - Prob. 2MC Ch. 2 - Which of the following is not a reserved word in...Ch. 2 - Prob. 4SA Ch. 2 - 5. Are the identifiers quizNo1 and quiznol the...Ch. 2 - 6. Evaluate the following expressions. (3,...Ch. 2 - If int x = 10;, int y = 7;, double z = 4.5;, and...Ch. 2 - Prob. 8CP Ch. 2 - 9. Suppose that x, y, z, and w are int variables....Ch. 2 - Prob. 10SA

num1 = 35; is a valid assignment as integer 35 is be...

Explanation of Solution

num1 = 35; is a valid assignment as integer 35 is be...

Explanation of Solution

newNum = num1 – num2; is a valid assignment although the variables num1 and num2 have not been initialized. If a variable is used in an expression, the expression would yield a meaningf...

Explanation of Solution

num1 = 5; num2 = 2 + num1; num1 = num2 / 3; are valid assignments. num1 =5; declares and initializes the variable num1. num2 = 2 + num1; assigns the value of the expression 2 + num1 to num2. Since num1 has been declared and assigned a value, the right-hand expression value is 2 + 5 = 7...

Explanation of Solution

num1 * num2 = newNum; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Explanation of Solution

Explanation of Solution

num1 * 2 = newNum + num2; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Explanation of Solution

x / y = x * y; is not a valid assignment as the left hand of a valid assignment is a variable and not an expression...

Explanation of Solution

num2 = num1 % 2.0; is not a valid assignment as the expression on the right-hand side would not evaluate. The expression on the right-hand si...

Explanation of Solution

newNum = static_cast<int> (x) % 5; is a valid assignment. The double variable is explicitly cast to an int type before it participates as an oper...

Explanation of Solution

x = x + y - 5; is a valid assignment although the variables x and y have not been initialized before being used in the expression. The expression on the right side...

Explanation of Solution

Chapter 2 Solutions