在下面的示例中,我们有一个Class
包含一个Bridge
对象的类,该对象为我们处理所有内存处理(三规则)。
class Base {
public:
Base() {};
virtual Base* clone() const = 0;
virtual ~Base() {};
};
class Derived : public Base {
public:
Derived() {};
virtual Base* clone() const {
return new Derived(*this);
}
virtual ~Derived() {}
};
class Bridge {
public:
Bridge(const Bridge& bridge_) {
base = bridge_.base->clone();
}
Bridge(const Base& base_) {
base = base_.clone();
}
~Bridge() { delete base; }
Bridge& operator=(const Bridge& assignFrom) {
if(this != &assignFrom) {
delete base;
base = assignFrom.base->clone();
}
return *this;
}
private:
Base *base;
};
class Class {
public:
Class(const Bridge& bridge_) : bridge(bridge_) {};
private:
Bridge bridge;
};
int main()
{
Derived derived;
Class c(derived);
Class c1(c);
}
现在,我刚刚了解了智能指针,并尝试使用unique_ptr
. 据我了解,我们基本上不需要自己实现 3 规则,因为智能指针已经包含它。为了测试这一点,我做了以下示例:
class BaseSMRT {
public:
BaseSMRT() {};
virtual std::unique_ptr<BaseSMRT> clone() const = 0;
virtual ~BaseSMRT() {};
};
class DerivedSMRT : public BaseSMRT {
public:
DerivedSMRT() {};
virtual std::unique_ptr<BaseSMRT> clone() const {
return std::make_unique<DerivedSMRT>(*this);
}
virtual ~DerivedSMRT() {}
};
class ClassSMRT {
public:
ClassSMRT(const BaseSMRT& base) {
ptr = base.clone();
};
private:
std::unique_ptr<BaseSMRT> ptr;
};
int main()
{
DerivedSMRT derivedSMRT;
ClassSMRT cSMRT(derivedSMRT);
ClassSMRT cSMRT2(cSMRT); // error: Call to implicitly-deleted copy constructor of 'ClassSMRT'
}
正如您在上面的示例中所看到的,cSMRT2
通过cSMRT
复制构造函数进行初始化不起作用并给出了上述错误。
我不明白:为什么我可以Class
像这样调用 的默认复制构造函数Class c1(c);
,但不能调用ClassSMRT
默认的复制构造函数ClassSMRT cSMRT2(cSMRT);
?
这表明我们在使用unique_ptr
.