无法在std :: map成员变量中分配具有正向声明值的类 [英] Can't allocate class with forward declared value in std::map member variable

问题描述

在 test.h ：

#ifndef TEST_H
#define TEST_H

#include <map>
struct Incomplete;

class Test {
     std::map<int, Incomplete> member;
public:
    Test();
    int foo() { return 0; }
};

#endif

在 test.cpp ：

#include "test.h"
struct Incomplete {};
Test::Test() {}

在 main中。 cpp

#include "test.h"

int main() {
    Test test;
    return test.foo();
}

g ++ 4.7给我一个错误，指出 struct Incomplete 在我编写 g ++ main.cpp test.h -o main.o 时被向前声明。

g++ 4.7 gives me an error that struct Incomplete is forward declared when I write g++ main.cpp test.h -o main.o.

但是，如果更改 std :: map< int，Incomplete>成员到 std :: map< int，Incomplete *>成员， main.o 进行编译。为什么是这样？

However, if I change std::map<int, Incomplete> member to std::map<int, Incomplete*> member, main.o compiles. Why is this?

推荐答案

这是为什么？

Why is this?

因为未定义C ++标准库容器以使用不完整的成员类型。 这是设计使然 ¹ –但这可以说是一个错误（在将来的C ++版本中可能会更改）。 Boost.Containers 库对此进行了修复。

Because the C++ standard library containers are not defined to work with incomplete member types. This is by design¹ – but it’s arguably an error (and might be changed in future versions of C++). The Boost.Containers library fixes this.

您的带指针代码可以工作，因为指向不完整类型的指针本身就是完整类型。但是，这显然会极大地改变您类型的语义（特别是谁来管理内存？），通常不建议使用它来代替。

Your code with pointers works because a pointer to an incomplete type is itself a complete type. However, this obviously changes the semantics of your type quite drastically (in particular, who manages the memory?) and it’s generally not a good idea to use this as a replacement.

¹ 值得指出的是，本文声称您在技术上不能实施 std :: map 使用不完整的类型。但是，这种说法是错误的。

¹ It’s worth pointing out that the article claims that you technically cannot implement std::map to work with incomplete types. However, this claim is wrong.

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