template <class T>
class SmartPtr
{
public:
explicit SmartPtr(T* pointee) : pointee_(pointee);
SmartPtr& operator=(const SmartPtr& other);
~SmartPtr();
T& operator*() const
{
...
return *pointee_;
}
T* operator->() const
{
...
return pointee_;
}
private:
T* pointee_;
...
};
What does this line of C do?
Author
Its from a smart pointer class definition:
explicit SmartPtr(T* pointee) : pointee_(pointee);
It has nothing to do with inheritance. It''s the initializer
list that, well, initializes the member variables.
Kami no Itte ga ore ni zettai naru!
list that, well, initializes the member variables.
Kami no Itte ga ore ni zettai naru!
It''s also C++, not C. They are not the same language, despite a great many similarities in syntax and behaviour.
The C++ FAQ Lite is full of useful information, such as this.
The C++ FAQ Lite is full of useful information, such as this.
A simpler example would be this random class I made:
class CExample{ public: CExample() : mA(0), mB(0.0), mC(''\0'') { } CExample(int a, double b, char c) : mA(a), mB(b), mC(c) { } private: int mA; double mB; char mC;};class CExample{ public: CExample() { mA = 0; mB = 0.0; mC = ''\0''; } CExample(int a, double b, char c) { mA = a; mB = b; mC = c; } private: int mA; double mB; char mC;};
Author
SO if i want to use the smart pointer class i put at teh top, how would i create a new object. Im following a tutorial an dit doesnt work, giving me one of those annoying liner errors?
regards
ace
regards
ace
Say you have an int* p. Then you create a SmartPtr to the same address using SmartPtr s = (SmartPtr) p.
... when I read that I figured out easily enough that it was a ctor with initializer list; but what is the keyword "explicit"? o_O
A normal constructor that takes a single argument is also an implicit conversion; for instance, if your class A has a single-parameter int constructor, it is possible to implicitly convert (cast) an int to an A. If you wish to prevent this, to avoid mishaps, you may declare the constructor explicit so that you cannot construct an A without explicitly calling its constructor.
I''m extremely tired and have a feeling the above doesn''t make much sense, but a kernel of truth is still present.
I''m extremely tired and have a feeling the above doesn''t make much sense, but a kernel of truth is still present.
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