C++ Primer Plus (P72) potx

A: cout << "We have " << cheeses << " varieties of cheese\n"; .9:What does the following function header tell you about the function? A:It tells us that the function froop() expects to be called with one argument, which will be type double, and that the function will return a type int value. .10:When do you not have to use the keyword return when you define a function? A:You don't have to use return in a function when the function has return type void. However, you can use it providing you don't give a return value: return; Chapter 3 .1:Why does C++ have more than one integer type? A:Having more than one integer type lets you choose the type best suited to a particular need. For example, you could use short to conserve space, long to guarantee storage capacity, or find that a particular type speeds up a particular calculation. .2:Define the following: A short integer with the value 80a. An unsigned int integer with the value 42110 b. An integer with the value 3000000000c. This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. A: short rbis = 80; // or short int rbis = 80; unsigned int q = 42110; // or unsigned q = 42110; unsigned long ants = 3000000000; Note Don't count on int being large enough to hold 3000000000. .3:What safeguards does C++ provide to keep you from exceeding the limits of an integer type? A:C++ provides no automatic safeguards to keep you from exceeding integer limits; you can use the climits header file to determine what the limits are. .4:What is the distinction between 33L and 33? A:The constant 33L is type long, whereas the constant 33 is type int. .5:Consider the two C++ statements that follow. Are they equivalent? char grade = 65; char grade = 'A'; A:The two statements are not really equivalent, although they have the same effect on some systems. Most important, the first statement assigns the letter A to grade only on a system using the ASCII code, while the second statement also works for other codes. Second, 65 is a type int constant, while 'A' is a type char constant. .6:How could you use C++ to find out which character the code 88 represents? Come up with at least two ways. A:Here are four ways: This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. char c = 88; cout << c << "\n"; // char type prints as character cout.put(char(88)); // put() prints char as character cout << char(88) << "\n"; // new-style type cast value to char cout << (char)88 << "\n"; // old-style type cast value to char .7:Assigning a long value to a float can result in a round-off error. What about assigning long to double? A:The answer depends on how large the two types are. If long is 4 bytes, there is no loss. That's because the largest long value would be about 2 billion, which is 10 digits. Because double provides at least 13 significant figures, no rounding would be needed. .8:Evaluate the following expressions as C++ would: 8 * 9 + 2a. 6 * 3 / 4b. 3 / 4 * 6c. 6.0 * 3 / 4d. 15 % 4e. A: 8 * 9 + 2 is 72 + 2 is 74a. 6 * 3 / 4 is 18 / 4 is 4b. 3 / 4 * 6 is 0 * 6 is 0c. 6.0 * 3 / 4 is 18.0 / 4 is 4.5d. 15 % 4 is 3e. This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. .9:Suppose x1 and x2 are two type double variables that you want to add as integers and assign to an integer variable. Construct a C++ statement for doing so. A:Either of the following work: int pos = (int) x1 + (int) x2; int pos = int(x1) + int(x2); Chapter 4 .1:How would you declare each of the following? actors is an array of 30 char. a. betsie is an array of 100 short. b. chuck is an array of 13 float. c. dipsea is an array of 64 long double. d. A: char actors[30];a. short betsie[100];b. float chuck[13];c. long double dipsea[64];d. .2:Declare an array of five ints and initialize it to the first five odd positive integers. A: int oddly[5] = {1, 3, 5, 7, 9}; This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. .3:Write a statement that assigns the sum of the first and last elements of the array in question 2 to the variable even. A: int even = oddly[0] + oddly[4]; .4:Write a statement that displays the value of the second element in the float array ideas. A: cout << ideas[1] << "\n"; // or << endl; .5:Declare an array of char and initialize it to the string "cheeseburger". A: char lunch[13] = "cheeseburger"; // number of characters + 1 or char lunch[] = "cheeseburger"; // let the compiler count elements .6:Devise a structure declaration that describes a fish. The structure should include the kind, the weight in whole ounces, and the length in fractional inches. A: struct fish { char kind[20]; int weight; float length; }; .7:Declare a variable of the type defined in question 6 and initialize it. This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. A: fish petes = { "trout", 13, 12.25 }; .8:Use enum to define a type called Response with the possible values of Yes, No, and Maybe. Yes should be 1, No should be 0, and Maybe should be 2. A: enum Response {No, Yes, Maybe}; .9:Suppose ted is a double variable. Declare a pointer that points to ted and use the pointer to display ted's value. A: double * pd = &ted; cout << *pd << "\n"; .10:Suppose treacle is an array of 10 floats. Declare a pointer that points to the first element of treacle and use the pointer to display the first and last elements of the array. A: float * pf = treacle; // or = &treacle[0] cout << pf[0] << " " << pf[9] << "\n"; // or use *pf and *(pf + 9) .11:Write a code fragment that asks the user to enter a positive integer and then creates a dynamic array of that many ints. This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. A: unsigned int size; cout << "Enter a positive integer: "; cin >> size; int * dyn = new int [size]; .12:Is the following valid code? If so, what does it print? A:Yes, it is valid. The expression "Home of the jolly bytes" is a string constant, hence it evaluates as the address of the beginning of the string. The cout object interprets the address of a char as an invitation to print a string, but the type cast (int *) converts the address to type pointer-to-int, which is then printed as an address. In short, the statement prints the address of the string. .13:Write a code fragment that dynamically allocates a structure of the type described in question 6 and then reads a value for the kind member of the structure. A: struct fish { char kind[20]; int weight; float length; }; fish * pole = new fish; cout << "Enter kind of fish: "; cin >> pole->kind; This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. .14:Listing 4.6 illustrates a problem with the following numeric input with line-oriented string input. How would replacing cin.getline(address,80); with cin >> address; affect the working of this program? A:Using cin >> address causes a program to skip over whitespace until it finds nonwhitespace. It then reads characters until it encounters whitespace again. Thus, it will skip over the newline following the numeric input, avoiding that problem. On the other hand, it will read just a single word, not an entire line. Chapter 5 .1:What's the difference between an entry-condition loop and an exit-condition loop? Which kind is each of the C++ loops? A:An entry-condition loop evaluates a test expression before entering the body of the loop. If the condition initially is false, the loop never executes its body. An exit-condition loop evaluates a test expression after processing the body of the loop. Thus, the loop body is executed once even if the test expression initially is false. The for and while loops are entry-condition loops, and the do while loop is an exit-condition loop. .2:What would the following code fragment print if it were part of a valid program? int i; for (i = 0; i < 5; i++) cout << i; cout << "\n"; This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. A:It would print the following: 01234 Note that the cout << "\n"; is not part of the loop body (no braces). .3:What would the following code fragment print if it were part of a valid program? int j; for (j = 0; j < 11; j += 3) cout << j; cout << "\n" << j << "\n"; A:It would print the following: 0369 12 .4:What would the following code fragment print if it were part of a valid program? int j = 5; while ( ++j < 9) cout << j++ << "\n"; A:It would print the following: 6 8 .5:What would the following code fragment print if it were part of a valid program? int k = 8; do cout <<" k = " << k << "\n"; while (k++ < 5); This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. A:It would print the following: k = 8 .6:Write a for loop that prints the values 1 2 4 8 16 32 64 by increasing the value of a counting variable by a factor of 2 each cycle. A:It's simplest to use the *= operator: for (int num = 1; num <= 64; num *= 2) cout << num << " "; .7:How do you make a loop body include more than one statement? A:You enclose the statements within paired braces to form a single compound statement, or block. .8:Is the following statement valid? If not, why not? If so, what does it do? int x = (1,024); What about the following? int y; y = 1,024; A:Yes, the first statement is valid. The expression 1,024 consists of two expressions?1 and 024?joined by a comma operator. The value is the value of the right-hand expression. This is 024, which is octal for 20, so the declaration assigns the value 20 to x. The second statement also is valid. However, operator precedence causes it to be evaluated as follows: (y = 1), 024; That is, the left expression sets y to 1, and the value of the entire expression, which isn't used, is 024, or 20. .9:How does cin>>ch differ from cin.get(ch) and ch=cin.get() in how it This document was created by an unregistered ChmMagic, please go to http://www.bisenter.com to register it. Thanks. [...]... Because C++ interprets "pizza" as the address of its first element, applying the * operator yields the value of that first element, which is the character p Because C++ interprets "taco" as the address of its first element, it interprets "taco"[2] as the value of the element two positions down the line, that is, as the character c In other words, the string constant acts the same as an array name .11: C++. .. fundamental type such as an int or double, it passes it by value so that the function works with a copy Thus, the original data is already protected .7: What are the three forms a C-style string can take in a C++ program? A: A string can be stored in a char array, it can be represented by a string constant in double quotation marks, and it can be represented by a pointer pointing to the first character of a... ChmMagic, please go to http://www.bisenter.com to register it T f (ch >= 'a' && ch = 'A' && ch . being large enough to hold 3000000000. .3:What safeguards does C++ provide to keep you from exceeding the limits of an integer type? A :C++ provides no automatic safeguards to keep you from exceeding. "taco"[2]? A:Because C++ interprets "pizza" as the address of its first element, applying the * operator yields the value of that first element, which is the character p. Because C++ interprets. and 33? A:The constant 33L is type long, whereas the constant 33 is type int. .5:Consider the two C++ statements that follow. Are they equivalent? char grade = 65; char grade = 'A'; A:The