“静态变量"和“静态变量"之间有什么区别并在“匿名命名空间"中?

问题描述

extern int iNSIntl; //defined in other cpp
static int iNSIntl=2;
namespace{
    int iNSIntl=0;
}
void main ( int argc, char** argv )
{
    cout<<::iNSIntl<<endl;
}

在main()中，::: iNSIntl引用静态的，我如何引用外部的或在命名空间中呢?

In the main(),::iNSIntl refer to static one,how Can I refer to the extern one or within the namespace?

推荐答案

namespace
{
	void Foo()
	{
	}
}

void Foo()
{
}

现在，如果您调用Foo()，编译器将会很困惑.它们在Foo的相等可调用范围内.您将需要显式调用::Foo，后者将调用全局方法(第二个方法).您现在不能在这里呼叫匿名Foo .现在，如果您注释掉全局Foo，则对Foo 的调用(无::那里)将起作用，因为编译器在首次查找全局版本之后将在匿名名称空间中进行查找.

Now if you call Foo(), the compiler will be confused. They are both under equal callable scope for Foo. You will need to explicitly call ::Foo which calls the global method (the 2nd one). You cannot call the anonymous Foo here now. Now if you comment out the global Foo, then your call to Foo (no :: there) will work since the compiler will look in the anonymous namespace after it first looks for a global version.

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