为什么不能在主体或头文件中的类中初始化静态成员？ [英] Why can't initialize the static member in a class in the body or in the header file?

问题描述

任何机构可以向我提供任何理由吗？

Could any body offer me any reason about that?

如果我们这样做，结果是什么？编译错误？

If we do it like that, what's the outcome? Compile error?

推荐答案

问题是静态初始化只是初始化，它也是定义。例如：

The problem is that static initialization isnt just initialization, it is also definition. Take for example:

class Foo
{
public:
    static std::string bar_;
};

std::string Foo::bar_ = "Hello";

std::string GimmeFoo();

main.cpp：

main.cpp :

#include <string>
#include <sstream>
#include <iostream>

#include "hacks.h"

using std::string;
using std::ostringstream;
using std::cout;

int main()
{

    string s = GimmeFoo();
    return 0;
}

foo.cpp：

foo.cpp :

#include <string>
#include <sstream>
#include <iostream>

#include "hacks.h"

using std::string;
using std::ostringstream;
using std::cout;

string GimmeFoo()
{
    Foo foo;
    foo;
    string s = foo.bar_;

    return s;
}

在这种情况下，您不能初始化 Foo :: bar _ ，因为它将在 #include s hacks.h的每个文件中分配。因此，在内存中将有2个实例 Foo.bar _ - 一个在main.cpp中，一个在foo.cpp中。

In this case, you can't initialize Foo::bar_ in the header because it will be allocated in every file that #includes hacks.h. So there will be 2 instances of Foo.bar_ in memory - one in main.cpp, and one in foo.cpp.

解决方案是分配&仅在一个位置初始化：

The solution is to allocate & initialize in just one place:

...
std::string Foo::bar_ = "Hello";
...

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