Question

#include <iostream>
using namespace std;

class boiler
{
private:
    static boiler uniqueInstance;

    bool boilerEmpty;
    bool mixtureBoiled;

    boiler()
    {
        boilerEmpty = true;
        mixtureBoiled = false;
    }

public:
    static boiler getInstance()
    {
        if(uniqueInstance == NULL)
        {
            uniqueInstance = new boiler();
        }

        return uniqueInstance;
    }
};

以上法典重述标题中的错误。

anisha@linux-y3pi:~> g++ -Wall test.cpp
test.cpp: In static member function ‘static boiler boiler::getInstance()’:
test.cpp:22:26: error: no match for ‘operator==’ in ‘boiler::uniqueInstance == 0l’
test.cpp:24:34: error: no match for ‘operator=’ in ‘boiler::uniqueInstance = (operator new(2u), (<statement>, ((boiler*)<anonymous>)))’
test.cpp:5:1: note: candidate is: boiler& boiler::operator=(const boiler&)

为什么? 我们能否将“目标”与民族解放军进行比较? 是否存在一些同yn问题?

Answer 1

也许需要一点点:

static boiler* uniqueInstance;

自那时起,你开始使用<代码>新。这里:

uniqueInstance = new boiler ();

汇编者告诉你,它无法将<编码>锅炉的事例与斜体(实际时间很长)进行比较。这一比较没有。点数可与整体类型相比较,后者可与0进行比较。

这里,与<代码>NUL的比较是检查贵点是否已经启动的一种手段。在实践中如何做到这一点并不明显,不存在无效或未经推定的事例。您可以通过界定适当的<代码>operator=而比较一个物体,但这一比较可能没有意义,因为<代码>NUL常常只是<0>的另一个名称。

Answer 2

页: 1

 static boiler uniqueInstance;

but sometimes treat it as a pointer:

uniqueInstance = new boiler ();

我认为,你应当宣布它为点人,以便你能够积极分配这样的情况:

static boiler* uniqueInstance;

You ll also want to change getInstance() to return a pointer:

static boiler* getInstance() //...

Finally make sure you actually have a definition of uniqueInstance somewhere outside of the class declaration. In one spot in a .cpp file (not a header):

boiler* boiler::uniqueInstance;

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