Day 9 |
You create a reference by writing the type of the target object, followed by the reference operator (&), followed by the name of the reference. References can use any legal variable name, but for this book we'll prefix all reference names with "r." Thus, if you have an integer variable named someInt, you can make a reference to that variable by writing the following:
int &rSomeRef = someInt;This is read as "rSomeRef is a reference to an integer that is initialized to refer to someInt." Listing 9.1 shows how references are created and used.
Listing 9.1. Creating and using references.
NOTE: Note that the reference operator (&) is the same symbol as the one used for the address of the operator. These are not the same operators, however, though clearly they are related.
1: //Listing 9.1
2: // Demonstrating the use of References
3:
4: #include <iostream.h>
5:
6: int main()
7: {
8: int intOne;
9: int &rSomeRef = intOne;
10:
11: intOne = 5;
12: cout << "intOne: " << intOne << endl;
13: cout << "rSomeRef: " << rSomeRef << endl;
14:
15: rSomeRef = 7;
16: cout << "intOne: " << intOne << endl;
17: cout << "rSomeRef: " << rSomeRef << endl;
18: return 0;
19: }
Output: intOne: 5
rSomeRef: 5
intOne: 7
rSomeRef: 7
Anaylsis: On line 8, a local
int variable, intOne, is declared. On line 9, a reference
to an int,
rSomeRef, is declared and initialized to refer
to intOne. If you declare a reference, but don't initialize it,
you will get a compile-time error. References must be initialized.
On line 15, 7 is assigned to rSomeRef. Since this is a reference, it is an alias for intOne, and thus the 7 is really assigned to intOne, as is shown by the printouts on lines 16 and 17.
Listing 9.2. Taking the address of a reference.
1: //Listing 9.2
2: // Demonstrating the use of References
3:
4: #include <iostream.h>
5:
6: int main()
7: {
8: int intOne;
9: int &rSomeRef = intOne;
10:
11: intOne = 5;
12: cout << "intOne: " << intOne << endl;
13: cout << "rSomeRef: " << rSomeRef << endl;
14:
15: cout << "&intOne: " << &intOne << endl;
16: cout << "&rSomeRef: " << &rSomeRef << endl;
17:
18: return 0;
19: }
Output: intOne: 5
rSomeRef: 5
&intOne: 0x3500
&rSomeRef: 0x3500
NOTE: Your output may differ on the last two lines.
President William_Jefferson_Clinton;You might then declare a reference to President and initialize it with this object:
President &Bill_Clinton = William_Jefferson_Clinton;There is only one President; both identifiers refer to the same object of the same class. Any action you take on Bill_Clinton will be taken on William_Jefferson_Clinton as well.
Be careful to distinguish between the & symbol on line 9 of Listing 9.2, which declares a reference to int named rSomeRef, and the & symbols on lines 15 and 16, which return the addresses of the integer variable intOne and the reference rSomeRef.
Normally, when you use a reference, you do not use the address of operator. You simply use the reference as you would use the target variable. This is shown on line 13.
Even experienced C++ programmers, who know the rule that references cannot be reassigned and are always aliases for their target, can be confused by what happens when you try to reassign a reference. What appears to be a reassignment turns out to be the assignment of a new value to the target. Listing 9.3 illustrates this fact.
Listing 9.3. Assigning to a reference.
1: //Listing 9.3
2: //Reassigning a reference
3:
4: #include <iostream.h>
5:
6: int main()
7: {
8: int intOne;
9: int &rSomeRef = intOne;
10:
11: intOne = 5;
12: cout << "intOne:\t" << intOne << endl;
13: cout << "rSomeRef:\t" << rSomeRef << endl;
14: cout << "&intOne:\t" << &intOne << endl;
15: cout << "&rSomeRef:\t" << &rSomeRef << endl;
16:
17: int intTwo = 8;
18: rSomeRef = intTwo; // not what you think!
19: cout << "\nintOne:\t" << intOne << endl;
20: cout << "intTwo:\t" << intTwo << endl;
21: cout << "rSomeRef:\t" << rSomeRef << endl;
22: cout << "&intOne:\t" << &intOne << endl;
23: cout << "&intTwo:\t" << &intTwo << endl;
24: cout << "&rSomeRef:\t" << &rSomeRef << endl;
25: return 0;
26: }
Output: intOne: 5
rSomeRef: 5
&intOne: 0x213e
&rSomeRef: 0x213e
intOne: 8
intTwo: 8
rSomeRef: 8
&intOne: 0x213e
&intTwo: 0x2130
&rSomeRef: 0x213e
Anaylsis: Once again, an integer
variable and a reference to an integer are declared, on lines 8 and 9.
The integer is assigned the value 5 on line 11, and the values
and their addresses are printed on lines 12-15.
intOne = intTwo;Sure enough, when the values of intOne and rSomeRef are printed (lines 19-21) they are the same as intTwo. In fact, when the addresses are printed on lines 22-24, you see that rSomeRef continues to refer to intOne and not intTwo.
DO use references to create an alias to an object. DO initialize all references. DON'T try to reassign a reference. DON'T confuse the address of operator with the reference operator.
int & rIntRef = int; // wrongYou must initialize rIntRef to a particular integer, such as this:
int howBig = 200; int & rIntRef = howBig;In the same way, you don't initialize a reference to a CAT:
CAT & rCatRef = CAT; // wrongYou must initialize rCatRef to a particular CAT object:
CAT frisky; CAT & rCatRef = frisky;References to objects are used just like the object itself. Member data and methods are accessed using the normal class member access operator (.), and just as with the built-in types, the reference acts as an alias to the object. Listing 9.4 illustrates this.
Listing 9.4. References to objects.
1: // Listing 9.4
2: // References to class objects
3:
4: #include <iostream.h>
5:
6: class SimpleCat
7: {
8: public:
9: SimpleCat (int age, int weight);
10: ~SimpleCat() {}
11: int GetAge() { return itsAge; }
12: int GetWeight() { return itsWeight; }
13: private:
14: int itsAge;
15: int itsWeight;
16: };
17:
18: SimpleCat::SimpleCat(int age, int weight)
19: {
20: itsAge = age;
21: itsWeight = weight;
22: }
23:
24: int main()
25: {
26: SimpleCat Frisky(5,8);
27: SimpleCat & rCat = Frisky;
28:
29: cout << "Frisky is: ";
30: cout << Frisky.GetAge() << " years old. \n";
31: cout << "And Frisky weighs: ";
32: cout << rCat.GetWeight() << " pounds. \n";
33: return 0;
34: }
Output: Frisky is: 5 years old. And Frisky weighs 8 pounds.Anaylsis: On line 26, Frisky is declared to be a SimpleCat object. On line 27, a SimpleCat reference, rCat, is declared and initialized to refer to Frisky. On lines 30 and 32, the SimpleCat accessor methods are accessed by using first the SimpleCat object and then the SimpleCat reference. Note that the access is identical. Again, the reference is an alias for the actual object.
int hisAge; int &rAge = hisAge;Example 2
CAT boots; CAT &rCatRef = boots;
Most compilers will support a null object without much complaint, crashing only if you try to use the object in some way. Taking advantage of this, however, is still not a good idea. When you move your program to another machine or compiler, mysterious bugs may develop if you have null objects.
Passing values to a function by reference can overcome both of these limitations. In C++, passing by reference is accomplished in two ways: using pointers and using references. The syntax is different, but the net effect is the same. Rather than a copy being created within the scope of the function, the actual original object is passed into the function.
Passing an object by reference allows the function to change the object being referred to.
NOTE: If you read the extra credit section after Day 5, you learned that functions are passed their parameters on the stack. When a function is passed a value by reference (either using pointers or references), the address of the object is put on the stack, not the entire object. In fact, on some computers the address is actually held in a register and nothing is put on the stack. In either case, the compiler now knows how to get to the original object, and changes are made there and not in a copy.
Recall that Listing 5.5 in Day 5 demonstrated that a call to the swap() function did not affect the values in the calling function. Listing 5.5 is reproduced here as Listing 9.5, for your convenience.
1: //Listing 9.5 Demonstrates passing by value
2:
3: #include <iostream.h>
4:
5: void swap(int x, int y);
6:
7: int main()
8: {
9: int x = 5, y = 10;
10:
11: cout << "Main. Before swap, x: " << x << " y: " << y << "\n";
12: swap(x,y);
13: cout << "Main. After swap, x: " << x << " y: " << y << "\n";
14: return 0;
15: }
16:
17: void swap (int x, int y)
18: {
19: int temp;
20:
21: cout << "Swap. Before swap, x: " << x << " y: " << y << "\n";
22:
23: temp = x;
24: x = y;
25: y = temp;
26:
27: cout << "Swap. After swap, x: " << x << " y: " << y << "\n";
28:
29: }
Output: Main. Before swap, x: 5 y: 10 Swap. Before swap, x: 5 y: 10 Swap. After swap, x: 10 y: 5 Main. After swap, x: 5 y: 10Anaylsis: This program initializes two variables in main() and then passes them to the swap() function, which appears to swap them. When they are examined again in main(), they are unchanged!
There are two ways to solve this problem in C++: You can make the parameters of swap() pointers to the original values, or you can pass in references to the original values.
Listing 9.6. Passing by reference using pointers.
1: //Listing 9.6 Demonstrates passing by reference
2:
3: #include <iostream.h>
4:
5: void swap(int *x, int *y);
6:
7: int main()
8: {
9: int x = 5, y = 10;
10:
11: cout << "Main. Before swap, x: " << x << " y: " << y << "\n";
12: swap(&x,&y);
13: cout << "Main. After swap, x: " << x << " y: " << y << "\n";
14: return 0;
15: }
16
17: void swap (int *px, int *py)
18: {
19: int temp;
20:
21: cout << "Swap. Before swap, *px: " << *px << " *py: " << *py << "\n";
22:
23: temp = *px;
24: *px = *py;
25: *py = temp;
26:
27: cout << "Swap. After swap, *px: " << *px << " *py: " << *py << "\n";
28:
29: }
Output: Main. Before swap, x: 5 y: 10
Swap. Before swap, *px: 5 *py: 10
Swap. After swap, *px: 10 *py: 5
Main. After swap, x: 10 y: 5
Anaylsis: Success! On line
5, the prototype of swap() is changed to indicate that its two
parameters will be pointers to int rather than int variables.
When swap() is called on line 12, the addresses of x
and y are passed as the arguments.
On line 23, temp is assigned the value at px. On line 24, the value at px is assigned to the value at py. On line 25, the value stashed in temp (that is, the original value at px) is put into py.
The net effect of this is that the values in the calling function, whose address was passed to swap(), are, in fact, swapped.
It is a goal of C++ to prevent the user of a function from worrying about how it works. Passing by pointers takes the burden off of the called function, and puts it where it belongs--on the calling function. Listing 9.7 rewrites the swap() function, using references.
Listing 9.7. swap() rewritten with references.
1: //Listing 9.7 Demonstrates passing by reference
2: // using references!
3:
4: #include <iostream.h>
5:
6: void swap(int &x, int &y);
7:
8: int main()
9: {
10: int x = 5, y = 10;
11:
12: cout << "Main. Before swap, x: " << x << " y: " << y << "\n";
13: swap(x,y);
14: cout << "Main. After swap, x: " << x << " y: " << y << "\n";
15: return 0;
16: }
17:
18: void swap (int &rx, int &ry)
19: {
20: int temp;
21:
22: cout << "Swap. Before swap, rx: " << rx << " ry: " << ry << "\n";
23:
24: temp = rx;
25: rx = ry;
26: ry = temp;
27:
28: cout << "Swap. After swap, rx: " << rx << " ry: " << ry << "\n";
29:
30: }
Output: Main. Before swap, x:5 y: 10
Swap. Before swap, rx:5 ry:10
Swap. After swap, rx:10 ry:5
Main. After swap, x:10, y:5
Anaylsis:Just as in the example
with pointers, two variables are declared on line 10 and their values are
printed on line 12. On line 13, the function swap() is called,
but note that x and y, not their addresses, are passed.
The calling function simply passes the variables.
On lines 24-26, the values are swapped, and then they're printed on line 28. Program execution jumps back to the calling function, and on line 14, the values are printed in main(). Because the parameters to swap() are declared to be references, the values from main() are passed by reference, and thus are changed in main() as well.
References provide the convenience and ease of use of normal variables, with the power and pass-by-reference capability of pointers!
This is another use for the function prototype. By examining the parameters declared in the prototype, which is typically in a header file along with all the other prototypes, the programmer knows that the values passed into swap() are passed by reference, and thus will be swapped properly.
If swap() had been a member function of a class, the class declaration, also available in a header file, would have supplied this information.
In C++, clients of classes and functions rely on the header file to tell all that is needed; it acts as the interface to the class or function. The actual implementation is hidden from the client. This allows the programmer to focus on the problem at hand and to use the class or function without concern for how it works.
When Colonel John Roebling designed the Brooklyn Bridge, he worried in detail about how the concrete was poured and how the wire for the bridge was manufactured. He was intimately involved in the mechanical and chemical processes required to create his materials. Today, however, engineers make more efficient use of their time by using well-understood building materials, without regard to how their manufacturer produced them.
It is the goal of C++ to allow programmers to rely on well-understood classes and functions without regard to their internal workings. These "component parts" can be assembled to produce a program, much the same way wires, pipes, clamps, and other parts are assembled to produce buildings and bridges.
In much the same way that an engineer examines the spec sheet for a pipe to determine its load-bearing capacity, volume, fitting size, and so forth, a C++ programmer reads the interface of a function or class to determine what services it provides, what parameters it takes, and what values it returns.
Once again, this can be done with references or pointers. Listing 9.8 demonstrates a function that returns three values: two as pointer parameters and one as the return value of the function.
Listing 9.8. Returning values with pointers.
1: //Listing 9.8
2: // Returning multiple values from a function
3:
4: #include <iostream.h>
5:
6: typedef unsigned short USHORT;
7:
8: short Factor(USHORT, USHORT*, USHORT*);
9:
10: int main()
11: {
12: USHORT number, squared, cubed;
13: short error;
14:
15: cout << "Enter a number (0 - 20): ";
16: cin >> number;
17:
18: error = Factor(number, &squared, &cubed);
19:
20: if (!error)
21: {
22: cout << "number: " << number << "\n";
23: cout << "square: " << squared << "\n";
24: cout << "cubed: " << cubed << "\n";
25: }
26: else
27: cout << "Error encountered!!\n";
28: return 0;
29: }
30:
31: short Factor(USHORT n, USHORT *pSquared, USHORT *pCubed)
32: {
33: short Value = 0;
34: if (n > 20)
35: Value = 1;
36: else
37: {
38: *pSquared = n*n;
39: *pCubed = n*n*n;
40: Value = 0;
41: }
42: return Value;
43: }
Output: Enter a number (0-20): 3
number: 3
square: 9
cubed: 27
Anaylsis: On line 12, number,
squared, and cubed are defined as USHORTs.
number
is assigned a value based on user input. This number and the addresses
of squared and cubed are passed to the function Factor().
The actual values needed, the square and cube of number, are returned not by using the return mechanism, but rather by changing the pointers that were passed into the function.
On lines 38 and 39, the pointers are assigned their return values. On line 40, return Value is assigned a success value. On line 41, return Value is returned.
One improvement to this program might be to declare the following:
enum ERROR_VALUE { SUCCESS, FAILURE};
Then, rather than returning 0 or 1, the program could return SUCCESS
or FAILURE.
Listing 9.9.Listing 9.8 rewritten using references.
1: //Listing 9.9
2: // Returning multiple values from a function
3: // using references
4:
5: #include <iostream.h>
6:
7: typedef unsigned short USHORT;
8: enum ERR_CODE { SUCCESS, ERROR };
9:
10: ERR_CODE Factor(USHORT, USHORT&, USHORT&);
11:
12: int main()
13: {
14: USHORT number, squared, cubed;
15: ERR_CODE result;
16:
17: cout << "Enter a number (0 - 20): ";
18: cin >> number;
19:
20: result = Factor(number, squared, cubed);
21:
22: if (result == SUCCESS)
23: {
24: cout << "number: " << number << "\n";
25: cout << "square: " << squared << "\n";
26: cout << "cubed: " << cubed << "\n";
27: }
28: else
29: cout << "Error encountered!!\n";
30: return 0;
31: }
32:
33: ERR_CODE Factor(USHORT n, USHORT &rSquared, USHORT &rCubed)
34: {
35: if (n > 20)
36: return ERROR; // simple error code
37: else
38: {
39: rSquared = n*n;
40: rCubed = n*n*n;
41: return SUCCESS;
42: }
43: }
Output: Enter a number (0 - 20): 3
number: 3
square: 9
cubed: 27
Anaylsis: Listing 9.9 is identical
to 9.8, with two exceptions. The ERR_CODE enumeration makes the
error reporting a bit more explicit on lines 36 and 41, as well as the
error handling on line 22.
The larger change, however, is that Factor() is now declared to take references to squared and cubed rather than to pointers. This makes the manipulation of these parameters far simpler and easier to understand.
In the "Extra Credit" section at the end of Day 5, you learned that these objects are copied onto the stack. Doing so takes time and memory. For small objects, such as the built-in integer values, this is a trivial cost.
However, with larger, user-created objects, the cost is greater. The size of a user-created object on the stack is the sum of each of its member variables. These, in turn, can each be user-created objects, and passing such a massive structure by copying it onto the stack can be very expensive in performance and memory consumption.
There is another cost as well. With the classes you create, each of these temporary copies is created when the compiler calls a special constructor: the copy constructor. Tomorrow you will learn how copy constructors work and how you can make your own, but for now it is enough to know that the copy constructor is called each time a temporary copy of the object is put on the stack.
When the temporary object is destroyed, which happens when the function returns, the object's destructor is called. If an object is returned by the function by value, a copy of that object must be made and destroyed as well.
With large objects, these constructor and destructor calls can be expensive in speed and use of memory. To illustrate this idea, Listing 9.9 creates a stripped-down user-created object: SimpleCat. A real object would be larger and more expensive, but this is sufficient to show how often the copy constructor and destructor are called.
Listing 9.10 creates the SimpleCat object and then calls two functions. The first function receives the Cat by value and then returns it by value. The second one receives a pointer to the object, rather than the object itself, and returns a pointer to the object.
Listing 9.10. Passing objects by reference.
1: //Listing 9.10
2: // Passing pointers to objects
3:
4: #include <iostream.h>
5:
6: class SimpleCat
7: {
8: public:
9: SimpleCat (); // constructor
10: SimpleCat(SimpleCat&); // copy constructor
11: ~SimpleCat(); // destructor
12: };
13:
14: SimpleCat::SimpleCat()
15: {
16: cout << "Simple Cat Constructor...\n";
17: }
18:
19: SimpleCat::SimpleCat(SimpleCat&)
20: {
21: cout << "Simple Cat Copy Constructor...\n";
22: }
23:
24: SimpleCat::~SimpleCat()
25: {
26: cout << "Simple Cat Destructor...\n";
27: }
28:
29: SimpleCat FunctionOne (SimpleCat theCat);
30: SimpleCat* FunctionTwo (SimpleCat *theCat);
31:
32: int main()
33: {
34: cout << "Making a cat...\n";
35: SimpleCat Frisky;
36: cout << "Calling FunctionOne...\n";
37: FunctionOne(Frisky);
38: cout << "Calling FunctionTwo...\n";
39: FunctionTwo(&Frisky);
40: return 0;
41: }
42:
43: // FunctionOne, passes by value
44: SimpleCat FunctionOne(SimpleCat theCat)
45: {
46: cout << "Function One. Returning...\n";
47: return theCat;
48: }
49:
50: // functionTwo, passes by reference
51: SimpleCat* FunctionTwo (SimpleCat *theCat)
52: {
53: cout << "Function Two. Returning...\n";
54: return theCat;
55: }
Output: 1: Making a cat...
2: Simple Cat Constructor...
3: Calling FunctionOne...
4: Simple Cat Copy Constructor...
5: Function One. Returning...
6: Simple Cat Copy Constructor...
7: Simple Cat Destructor...
8: Simple Cat Destructor...
9: Calling FunctionTwo...
10: Function Two. Returning...
11: Simple Cat Destructor...
Anaylsis: A very simplified SimpleCat class is declared on lines 6-12. The constructor, copy constructor, and destructor all print an informative message so that you can tell when they've been called.
NOTE: Line numbers will not print. They were added to aid in the analysis.
On line 36, main() reports that it is calling FunctionOne, which creates output line 3. Because FunctionOne() is called passing the SimpleCat object by value, a copy of the SimpleCat object is made on the stack as an object local to the called function. This causes the copy constructor to be called, which creates output line 4.
Program execution jumps to line 46 in the called function, which prints an informative message, output line 5. The function then returns, and returns the SimpleCat object by value. This creates yet another copy of the object, calling the copy constructor and producing line 6.
The return value from FunctionOne() is not assigned to any object, and so the temporary created for the return is thrown away, calling the destructor, which produces output line 7. Since FunctionOne() has ended, its local copy goes out of scope and is destroyed, calling the destructor and producing line 8.
Program execution returns to main(), and FunctionTwo() is called, but the parameter is passed by reference. No copy is produced, so there's no output. FunctionTwo() prints the message that appears as output line 10 and then returns the SimpleCat object, again by reference, and so again produces no calls to the constructor or destructor.
Finally, the program ends and Frisky goes out of scope, causing one final call to the destructor and printing output line 11.
The net effect of this is that the call to FunctionOne(), because it passed the cat by value, produced two calls to the copy constructor and two to the destructor, while the call to FunctionTwo() produced none.
Passing by value is like giving a museum a photograph of your masterpiece instead of the real thing. If vandals mark it up, there is no harm done to the original. Passing by reference is like sending your home address to the museum and inviting guests to come over and look at the real thing.
The solution is to pass a const pointer to SimpleCat. Doing so prevents calling any non-const method on SimpleCat, and thus protects the object from change. Listing 9.11 demonstrates this idea.
Listing 9.11. Passing const pointers.
1: //Listing 9.11
2: // Passing pointers to objects
3:
4: #include <iostream.h>
5:
6: class SimpleCat
7: {
8: public:
9: SimpleCat();
10: SimpleCat(SimpleCat&);
11: ~SimpleCat();
12:
13: int GetAge() const { return itsAge; }
14: void SetAge(int age) { itsAge = age; }
15:
16: private:
17: int itsAge;
18: };
19:
20: SimpleCat::SimpleCat()
21: {
22: cout << "Simple Cat Constructor...\n";
23: itsAge = 1;
24: }
25:
26: SimpleCat::SimpleCat(SimpleCat&)
27: {
28: cout << "Simple Cat Copy Constructor...\n";
29: }
30:
31: SimpleCat::~SimpleCat()
32: {
33: cout << "Simple Cat Destructor...\n";
34: }
35:
36:const SimpleCat * const FunctionTwo (const SimpleCat * const theCat);
37:
38: int main()
39: {
40: cout << "Making a cat...\n";
41: SimpleCat Frisky;
42: cout << "Frisky is " ;
43 cout << Frisky.GetAge();
44: cout << " years _old\n";
45: int age = 5;
46: Frisky.SetAge(age);
47: cout << "Frisky is " ;
48 cout << Frisky.GetAge();
49: cout << " years _old\n";
50: cout << "Calling FunctionTwo...\n";
51: FunctionTwo(&Frisky);
52: cout << "Frisky is " ;
53 cout << Frisky.GetAge();
54: cout << " years _old\n";
55: return 0;
56: }
57:
58: // functionTwo, passes a const pointer
59: const SimpleCat * const FunctionTwo (const SimpleCat * const theCat)
60: {
61: cout << "Function Two. Returning...\n";
62: cout << "Frisky is now " << theCat->GetAge();
63: cout << " years old \n";
64: // theCat->SetAge(8); const!
65: return theCat;
66: }
Output: Making a cat...
Simple Cat constructor...
Frisky is 1 years old
Frisky is 5 years old
Calling FunctionTwo...
FunctionTwo. Returning...
Frisky is now 5 years old
Frisky is 5 years old
Simple Cat Destructor...
Anaylsis: SimpleCat
has added two accessor functions, GetAge() on line 13, which is
a const function, and SetAge() on line 14, which is not
a const function. It has also added the member variable itsAge
on line 17.
On line 46, itsAge is set using the accessor SetAge, and the result is printed on line 47. FunctionOne is not used in this program, but FunctionTwo() is called. FunctionTwo() has changed slightly; the parameter and return value are now declared, on line 36, to take a constant pointer to a constant object and to return a constant pointer to a constant object.
Because the parameter and return value are still passed by reference, no copies are made and the copy constructor is not called. The pointer in FunctionTwo(), however, is now constant, and thus cannot call the non-const method, SetAge(). If the call to SetAge() on line 64 was not commented out, the program would not compile.
Note that the object created in main() is not constant, and Frisky can call SetAge(). The address of this non-constant object is passed to FunctionTwo(), but because FunctionTwo()'s declaration declares the pointer to be a constant pointer, the object is treated as if it were constant!
Since you know the object will never be null, it would be easier to work with in the function if a reference were passed in, rather than a pointer. Listing 9.12 illustrates this.
Listing 9.12. Passing references to objects.
1: //Listing 9.12
2: // Passing references to objects
3:
4: #include <iostream.h>
5:
6: class SimpleCat
7: {
8: public:
9: SimpleCat();
10: SimpleCat(SimpleCat&);
11: ~SimpleCat();
12:
13: int GetAge() const { return itsAge; }
14: void SetAge(int age) { itsAge = age; }
15:
16: private:
17: int itsAge;
18: };
19:
20: SimpleCat::SimpleCat()
21: {
22: cout << "Simple Cat Constructor...\n";
23: itsAge = 1;
24: }
25:
26: SimpleCat::SimpleCat(SimpleCat&)
27: {
28: cout << "Simple Cat Copy Constructor...\n";
29: }
30:
31: SimpleCat::~SimpleCat()
32: {
33: cout << "Simple Cat Destructor...\n";
34: }
35:
36: const SimpleCat & FunctionTwo (const SimpleCat & theCat);
37:
38: int main()
39: {
40: cout << "Making a cat...\n";
41: SimpleCat Frisky;
42: cout << "Frisky is " << Frisky.GetAge() << " years old\n";
43: int age = 5;
44: Frisky.SetAge(age);
45: cout << "Frisky is " << Frisky.GetAge() << " years old\n";
46: cout << "Calling FunctionTwo...\n";
47: FunctionTwo(Frisky);
48: cout << "Frisky is " << Frisky.GetAge() << " years old\n";
49: return 0;
50: }
51:
52: // functionTwo, passes a ref to a const object
53: const SimpleCat & FunctionTwo (const SimpleCat & theCat)
54: {
55: cout << "Function Two. Returning...\n";
56: cout << "Frisky is now " << theCat.GetAge();
57: cout << " years old \n";
58: // theCat.SetAge(8); const!
59: return theCat;
60: }
Output: Making a cat...
Simple Cat constructor...
Frisky is 1 years old
Frisky is 5 years old
Calling FunctionTwo...
FunctionTwo. Returning...
Frisky is now 5 years old
Frisky is 5 years old
Simple Cat Destructor...
Analysis: The output is identical
to that produced by Listing 9.11. The only significant change is that
FunctionTwo()
now takes and returns a reference to a constant object. Once again, working
with references is somewhat simpler than working with pointers, and the
same savings and efficiency are achieved, as well as the safety provided
by using
const.
References cannot be reassigned, however. If you need to point first to one object and then another, you must use a pointer. References cannot be null, so if there is any chance that the object in question may be null, you must not use a reference. You must use a pointer.
An example of the latter concern is the operator new. If new cannot allocate memory on the free store, it returns a null pointer. Since a reference can't be null, you must not initialize a reference to this memory until you've checked that it is not null. The following example shows how to handle this:
int *pInt = new int; if (pInt != NULL) int &rInt = *pInt;In this example a pointer to int, pInt, is declared and initialized with the memory returned by the operator new. The address in pInt is tested, and if it is not null, pInt is dereferenced. The result of dereferencing an int variable is an int object, and rInt is initialized to refer to that object. Thus, rInt becomes an alias to the int returned by the operator new.
DO pass parameters by reference whenever possible. DO return by reference whenever possible. DON'T use pointers if references will work. DO use const to protect references and pointers whenever possible. DON'T return a reference to a local object.
CAT * SomeFunction (Person &theOwner, House *theHouse, int age);This declaration says that SomeFunction takes three parameters. The first is a reference to a Person object, the second is a pointer to a house object, and the third is an integer. It returns a pointer to a CAT object.
NOTE: The question of where to put the reference (&) or indirection (*) operator when declaring these variables is a great controversy. You may legally write any of the following:
1: CAT& rFrisky; 2: CAT & rFrisky; 3: CAT &rFrisky;White space is completely ignored, so anywhere you see a space here you may put as many spaces, tabs, and new lines as you like. Setting aside freedom of expression issues, which is best? Here are the arguments for all three: The argument for case 1 is that rFrisky is a variable whose name is rFrisky and whose type can be thought of as "reference to CAT object." Thus, this argument goes, the & should be with the type. The counterargument is that the type is CAT. The & is part of the "declarator," which includes the variable name and the ampersand. More important, having the & near the CAT can lead to the following bug:
CAT& rFrisky, rBoots;Casual examination of this line would lead you to think that both rFrisky and rBoots are references to CAT objects, but you'd be wrong. This really says that rFrisky is a reference to a CAT, and rBoots (despite its name) is not a reference but a plain old CAT variable. This should be rewritten as follows:
CAT &rFrisky, rBoots;The answer to this objection is that declarations of references and variables should never be combined like this. Here's the right answer:
CAT& rFrisky; CAT boots;Finally, many programmers opt out of the argument and go with the middle position, that of putting the & in the middle of the two, as illustrated in case 2. Of course, everything said so far about the reference operator (&) applies equally well to the indirection operator (*). The important thing is to recognize that reasonable people differ in their perceptions of the one true way. Choose a style that works for you, and be consistent within any one program; clarity is, and remains, the goal. This book will adopt two conventions when declaring references and pointers:
2. Never declare references, pointers, and variables all
on the same line.
Listing 9.13 illustrates the danger of returning a reference to an object that no longer exists.
Listing 9.13. Returning a reference to a non-existent object.
1: // Listing 9.13
2: // Returning a reference to an object
3: // which no longer exists
4:
5: #include <iostream.h>
6:
7: class SimpleCat
8: {
9: public:
10: SimpleCat (int age, int weight);
11: ~SimpleCat() {}
12: int GetAge() { return itsAge; }
13: int GetWeight() { return itsWeight; }
14: private:
15: int itsAge;
16: int itsWeight;
17: };
18:
19: SimpleCat::SimpleCat(int age, int weight):
20: itsAge(age), itsWeight(weight) {}
21:
22: SimpleCat &TheFunction();
23:
24: int main()
25: {
26: SimpleCat &rCat = TheFunction();
27: int age = rCat.GetAge();
28: cout << "rCat is " << age << " years old!\n";
29: return 0;
30: }
31:
32: SimpleCat &TheFunction()
33: {
34: SimpleCat Frisky(5,9);
35: return Frisky;
36: }
Output: Compile error: Attempting to return a reference to a local object!
Anaylsis: On lines 7-17, SimpleCat is declared. On line 26, a reference to a SimpleCat is initialized with the results of calling TheFunction(), which is declared on line 22 to return a reference to a SimpleCat.
WARNING: This program won't compile on the Borland compiler. It will compile on Microsoft compilers; however, it should be noted that it is a bad coding practice.
The body of TheFunction() declares a local object of type SimpleCat and initializes its age and weight. It then returns that local object by reference. Some compilers are smart enough to catch this error and won't let you run the program. Others will let you run the program, with unpredictable results.
When TheFunction() returns, the local object, Frisky, will be destroyed (painlessly, I assure you). The reference returned by this function will be an alias to a non-existent object, and this is a bad thing.
The problem with this approach is: What do you do with the memory allocated for Frisky when you are done with it? Listing 9.14 illustrates this problem.
1: // Listing 9.14
2: // Resolving memory leaks
3: #include <iostream.h>
4:
5: class SimpleCat
6: {
7: public:
8: SimpleCat (int age, int weight);
9: ~SimpleCat() {}
10: int GetAge() { return itsAge; }
11: int GetWeight() { return itsWeight; }
12:
13 private:
14: int itsAge;
15: int itsWeight;
16: };
17:
18: SimpleCat::SimpleCat(int age, int weight):
19: itsAge(age), itsWeight(weight) {}
20:
21: SimpleCat & TheFunction();
22:
23: int main()
24: {
25: SimpleCat & rCat = TheFunction();
26: int age = rCat.GetAge();
27: cout << "rCat is " << age << " years old!\n";
28: cout << "&rCat: " << &rCat << endl;
29: // How do you get rid of that memory?
30: SimpleCat * pCat = &rCat;
31: delete pCat;
32: // Uh oh, rCat now refers to ??
33: return 0;
34: }
35:
36: SimpleCat &TheFunction()
37: {
38: SimpleCat * pFrisky = new SimpleCat(5,9);
39: cout << "pFrisky: " << pFrisky << endl;
40: return *pFrisky;
41: }
Output: pFrisky: 0x2bf4
rCat is 5 years old!
&rCat: 0x2bf4
Anaylss: TheFunction() has been changed so that it no longer returns a reference to a local variable. Memory is allocated on the free store and assigned to a pointer on line 38. The address that pointer holds is printed, and then the pointer is dereferenced and the SimpleCat object is returned by reference.
WARNING: This compiles, links, and appears to work. But it is a time bomb waiting to go off.
To prove that the reference declared in main() is referring to the object put on the free store in TheFunction(), the address of operator is applied to rCat. Sure enough, it displays the address of the object it refers to and this matches the address of the object on the free store.
So far, so good. But how will that memory be freed? You can't call delete on the reference. One clever solution is to create another pointer and initialize it with the address obtained from rCat. This does delete the memory, and plugs the memory leak. One small problem, though: What is rCat referring to after line 31? As stated earlier, a reference must always alias an actual object; if it references a null object (as this does now), the program is invalid.
There are actually three solutions to this problem. The first is to declare a SimpleCat object on line 25, and to return that cat from TheFunction by value. The second is to go ahead and declare the SimpleCat on the free store in TheFunction(), but have TheFunction() return a pointer to that memory. Then the calling function can delete the pointer when it is done.
NOTE: It cannot be overemphasized that a program with a reference to a null object may compile, but it is invalid and its performance is unpredictable.
As you pass this block of memory between functions, someone will "own" the pointer. Typically the value in the block will be passed using references, and the function that created the memory is the one that deletes it. But this is a general rule, not an ironclad one.
It is dangerous for one function to create memory and another to free it, however. Ambiguity about who owns the pointer can lead to one of two problems: forgetting to delete a pointer or deleting it twice. Either one can cause serious problems in your program. It is safer to build your functions so that they delete the memory they create.
If you are writing a function that needs to create memory and then pass it back to the calling function, consider changing your interface. Have the calling function allocate the memory and then pass it into your function by reference. This moves all memory management out of your program and back to the function that is prepared to delete it.
DO pass parameters by value when you must. DO return by value when you must. DON'T pass by reference if the item referred to may go out of scope. DON'T use references to null objects.
You saw that passing objects by reference can be more efficient than passing by value. Passing by reference also allows the called function to change the value in the arguments back in the calling function.
You saw that arguments to functions and values returned from functions can be passed by reference, and that this can be implemented with pointers or with references.
You saw how to use const pointers and const references to safely pass values between functions while achieving the efficiency of passing by reference.
A. References are easier to use and understand. The indirection
is hidden, and there is no need to repeatedly dereference the variable.
Q. Why have pointers if references are easier?
A. References cannot be null, and they cannot be reassigned. Pointers offer greater flexibility, but are slightly more difficult to use.
Q. Why would you ever return by value from a function?
A. If the object being returned is local, you must return by value or you will be returning a reference to a non-existent object.
Q. Given the danger in returning by reference, why not always return by value?
A. There is far greater efficiency in returning by reference. Memory is saved and the program runs faster.
2. When must you use a pointer rather than a reference?
3. What does new return if there is insufficient memory to make your new object?
4. What is a constant reference?
5. What is the difference between passing by reference and passing a reference?
2. Write a program that declares a constant pointer to a
constant integer. Initialize the pointer to an integer variable, varOne.
Assign 6 to varOne. Use the pointer to assign 7 to varOne.
Create a second integer variable, varTwo. Reassign the pointer
to varTwo.
3. Compile the program in Exercise 2. What produces errors? What produces warnings?
4. Write a program that produces a stray pointer.
5. Fix the program from Exercise 4.
6. Write a program that produces a memory leak.
7. Fix the program from Exercise 6.
8. BUG BUSTERS: What is wrong with this program?
1: #include <iostream.h>
2:
3: class CAT
4: {
5: public:
6: CAT(int age) { itsAge = age; }
7: ~CAT(){}
8: int GetAge() const { return itsAge;}
9: private:
10: int itsAge;
11: };
12:
13: CAT & MakeCat(int age);
14: int main()
15: {
16: int age = 7;
17: CAT Boots = MakeCat(age);
18: cout << "Boots is " << Boots.GetAge() << " years old\n";
19: }
20:
21: CAT & MakeCat(int age)
22: {
23: CAT * pCat = new CAT(age);
24: return *pCat;
25: }