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To solve the type of problem just described, C++ allows you to create a copy constructor, which the compiler uses when one object is used to initialize another When a copy constructor exists, the bitwise copy is bypassed The general form of a copy constructor is classname (const classname &o) { // body of constructor } Here, o is a reference to the object on the right side of the initialization It is permissible for a copy constructor to have additional parameters as long as they have default arguments defined for them However, in all cases the first parameter must be a reference to the object doing the initializing It is important to understand that C++ defines two distinct types of situations in which the value of one object is given to another The first is assignment The second is initialization, which can occur in three ways: I when one object explicitly initializes another, such as in a declaration I when a copy of an object is made to be passed to a function I when a temporary object is generated (most commonly, as a return value) The copy constructor applies only to initializations For example, assuming a class called myclass, and that y is an object of type myclass, each of the following statements involves initialization
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myclass x = y; // y explicitly initializating x func(y); // y passed as a parameter y = func(); // y receiving a temporary, return object C++
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Following is an example where an explicit copy constructor is needed This program creates a very simple safe integer array type that prevents array boundaries from being overrun Storage for each array is allocated by the use of new, and a pointer to the memory is maintained within each array object
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/* This program creates a "safe" array class Since space for the array is allocated using new, a copy constructor is provided to allocate memory when one array object is used to initialize another */ #include <iostream> #include <new>
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#include <cstdlib> using namespace std; class array { int *p; int size; public: array(int sz) { try { p = new int[sz]; } catch (bad_alloc xa) { cout << "Allocation Failure\n"; exit(EXIT_FAILURE); } size = sz; } ~array() { delete [] p; } // copy constructor array(const array &a);
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// Copy Constructor array::array(const array &a) { int i; try { p = new int[asize]; } catch (bad_alloc xa) { cout << "Allocation Failure\n"; exit(EXIT_FAILURE); } for(i=0; i<asize; i++) p[i] = ap[i]; }
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void put(int i, int j) { if(i>=0 && i<size) p[i] = j; } int get(int i) { return p[i]; }
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Miscellaneous C++ Topics
int main() { array num(10); int i; for(i=0; i<10; i++) numput(i, i); for(i=9; i>=0; i--) cout << numget(i); cout << "\n"; // create another array and initialize with num array x(num); // invokes copy constructor for(i=0; i<10; i++) cout << xget(i); return 0; }
When num is used to initialize x, the copy constructor is called, memory for the new array is allocated and stored in xp, and the contents of num are copied to x s array In this way, x and num have arrays that have the same values, but each array is separate and distinct (That is, nump and xp do not point to the same piece of memory) If the copy constructor had not been created, the default bitwise initialization would have resulted in x and num sharing the same memory for their arrays (That is, nump and xp would have, indeed, pointed to the same location) The copy constructor is called only for initializations For example, this sequence does not call the copy constructor defined in the preceding program:
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