静态类数据成员和构造函数

发布于 2024-09-05 01:16:56 字数 570 浏览 3 评论 0原文

如何使用所有静态方法访问类中的静态成员？

我想要一组相关的函数，但在调用这些函数之前还需要初始化一些重要的数据成员。我认为只有静态成员的类才是正确的选择。 VS2008 中的编译器不喜欢我尝试访问“a”。

当然，我错过了一些小事，但仍然很困惑。：P （即使没有“a”的无效访问，当从 main 调用 testMethod() 时，也不会调用构造函数。

class IPAddressResolver
{
private:

public:
    static int a;
    IPAddressResolver();
    static void TestMethod();
};


IPAddressResolver::IPAddressResolver()
{
    IPAddressResolver::a = 0;
    cout << "Creating IPAddressResolver" << endl;
}

void IPAddressResolver::TestMethod()
{
    cout << "testMethod" << endl;
}

原文

How do I access a static member in a class with all static methods?

I want to have a group of related functions but also have some important data members initialized before any of these functions are called. I thought a class with only static members would be the way to go. Compiler in VS2008 doesn't like me trying to access "a".

Surely I'm missing something small but still very confused. :P
(Even without the invalid access of "a" the constructor isn't called when calling testMethod() from main.

class IPAddressResolver
{
private:

public:
    static int a;
    IPAddressResolver();
    static void TestMethod();
};


IPAddressResolver::IPAddressResolver()
{
    IPAddressResolver::a = 0;
    cout << "Creating IPAddressResolver" << endl;
}

void IPAddressResolver::TestMethod()
{
    cout << "testMethod" << endl;
}

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你的往事 2024-09-12 01:16:56

您需要在函数外部定义静态数据成员，例如

class IPAddressResolver
{
private:
    static int a;
    IPAddressResolver();
public:
    static void TestMethod();
};

int IPAddressResolver::a = 0;

void IPAddressResolver::TestMethod()
{
    cout << "testMethod" << endl;
}

您的构造函数不会被调用，因为您没有创建该类的新实例。对于静态实用程序类，您不需要实例，因此可以完全省略构造函数。或者，您可能希望将其声明为private，以明确该类不应被实例化（参见上文）。

注意：

不建议在类中使用 public 字段，所以我将 a 改为 private，
静态实用程序类通常是无状态的，所以如果你的类中需要有字段，这可能表明该类最好是 Singleton 。

You need to define your static data member outside of the function, like

class IPAddressResolver
{
private:
    static int a;
    IPAddressResolver();
public:
    static void TestMethod();
};

int IPAddressResolver::a = 0;

void IPAddressResolver::TestMethod()
{
    cout << "testMethod" << endl;
}

Your constructor is not called, since you don't create a new instance of the class. For a static utility class, you don't need instances, so you can omit the constructor altogether. Alternatively, you might want to declare it private to make it explicit that the class shall not be instantiated (see above).

Notes:

it is not recommended to use public fields in classes, so I turned a into private,
static utility classes are usually stateless, so if you need to have fields within your class, this maybe a sign that the class would better be a Singleton.

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