C++ 错误：'.' 之前的预期不合格 ID 令牌
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error: expected unqualified-id before ‘.’ token
提问by Aquarius_Girl
class A
{
private:
A () {}
public:
static A* getInstance ()
{
return new A ();
}
};
int main ()
{
A.getInstance ();
return 0;
}
results in the error stated in the title. I do realize that if I create a variable in class A and instanciate it there and return it directly, the error will vanish.
导致标题中所述的错误。我确实意识到如果我在 A 类中创建一个变量并在那里实例化它并直接返回它，错误就会消失。
But, here I want to understand what is the meaning of this error and why can't I use it this way.
但是，在这里我想了解这个错误的含义是什么，为什么我不能这样使用它。
回答by Luchian Grigore
You need to call the method using the scope resolution operator - :::
您需要使用范围解析运算符 - 调用该方法::：
A::getInstance ();
Also, if this is meant to be a singleton, it's a very bad one. Whenever you call getInstance(), you'll receive a new object, and you'll run into memory leaks if you forget to delete any instances.
此外，如果这是一个单身人士，这是一个非常糟糕的。每当您调用 时getInstance()，您都会收到一个新对象，如果您忘记删除任何实例，就会遇到内存泄漏。
A singleton is usually implemented like so:
单例通常是这样实现的：
class A
{
private:
A () {}
static A* instance;
public:
static A* getInstance ()
{
if ( !instance )
instance = new A ();
return instance;
}
};
//implementation file
A* A::instance = NULL;
回答by hmjd
Use scope resolution operator ::(not .like in Java for example):
使用范围解析运算符::（.例如不像在 Java 中那样）：
A::getInstance();
回答by Aquarius_Girl
getInstanceis a static function of class A. The right form of calling a static function of a class is <class_name>::<static_function_name>.
getInstance是 class 的静态函数A。调用类的静态函数的正确形式是<class_name>::<static_function_name>.
We can also call the static function by creating object of the class and using .operator:
<class_object>.<static_function_name>
我们还可以通过创建类的对象并使用.运算符来调用静态函数：
<class_object>.<static_function_name>
回答by dirkgently
You can call a static member function using either .or ::. However, if you use class name you need to use the latter and an object then use the former.
您可以使用.或调用静态成员函数::。但是，如果您使用类名，则需要使用后者，而对象则使用前者。
回答by Hemant Metalia
use scope Resolution Operator ::
使用范围解析运算符::
e.g.
例如
class::methodName()
