A Potential Problem When Passing Objects in .NET framework

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A Potential Problem When Passing Objects
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Even though objects are passed to functions by means of the normal call-by-value parameter-passing mechanism, which, in theory, protects and insulates the calling argument, it is still possible for a side effect to occur that may affect, or even damage, the object used as an argument For example, if an object used as an argument allocates dynamic memory and frees that memory when it is destroyed, then its local copy inside the function will free the same memory when its destructor is called This is a problem
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because the original object is still using the memory This situation will leave the original object damaged and effectively useless Consider this sample program:
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// Demonstrate a problem when passing objects #include <iostream> #include <cstdlib> using namespace std; class myclass { int *p; public: myclass(int i); ~myclass(); int getval() { return *p; } }; myclass::myclass(int i) { cout << "Allocating p\n"; p = new int; *p = i; } myclass::~myclass() { cout << "Freeing p\n"; delete p; } // This will cause a problem void display(myclass ob) { cout << obgetval() << '\n'; } int main() { myclass a(10); display(a); return 0; }
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This program displays the following output:
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Allocating p 10 Freeing p Freeing p
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This program contains a fundamental error Here is why: When a is constructed within main( ), memory is allocated and assigned to ap When a is passed to display( ),
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a is copied into the parameter ob This means that both a and ob will have the same value for p That is, both objects will have their copies of p pointing to the same dynamically allocated memory When display( ) terminates, ob is destroyed, and its destructor is called This causes obp to be freed However, the memory freed by obp is the same memory that is still in use by ap! This is, in itself, a serious bug However, things get even worse When the program ends, a is destroyed, and its dynamically allocated memory is freed a second time The problem is that freeing the same piece of dynamically allocated memory a second time is an undefined operation which could, depending upon how the dynamic allocation system is implemented, cause a fatal error As you might guess, one way around the problem of a parameter s destructor destroying data needed by the calling argument is to pass either a pointer or a reference, instead of the object itself When either a pointer to an object or a reference to an object is passed, no copy is made; thus, no destructor is called when the function returns For example, here is one way to correct the preceding program:
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// One solution to the problem of passing objects #include <iostream> #include <cstdlib> using namespace std; class myclass { int *p; public: myclass(int i); ~myclass(); int getval() { return *p; } }; myclass::myclass(int i) { cout << "Allocating p\n"; p = new int; *p = i; } myclass::~myclass() { cout << "Freeing p\n"; delete p; } /* This will NOT cause a problem Because ob is now passed by reference, no copy of the calling argument is made and thus, no object goes out-of-scope when display() terminates */ void display(myclass &ob) {
cout << obgetval() << '\n'; } int main() { myclass a(10); display(a); return 0; }
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The output from this version of the program is shown here
Allocating p 10 Freeing p
As you can see, only one call to the destructor occurs This is because no copy of a is made when it is passed by reference to display( ) Passing an object by reference is an excellent approach when the situation allows it, but it may not be applicable to all cases Fortunately, a more general solution is available: you can create your own version of the copy constructor Doing so lets you define precisely how a copy of an object is made, allowing you to avoid the type of problems just described Before discussing the copy constructor, let s look at another, related situation that can also benefit from a copy constructor
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