为什么不能将前向声明用于 std::vector?

2022-01-07 00:00:00 c++ stl

如果我像这样创建一个类:

If I create a class like so:

// B.h
#ifndef _B_H_
#define _B_H_

class B
{
private:
    int x;
    int y;
};

#endif // _B_H_

并像这样使用它:

// main.cpp
#include <iostream>
#include <vector>

class B; // Forward declaration.

class A
{
public:
    A() {
        std::cout << v.size() << std::endl;
    }

private:
    std::vector<B> v;
};

int main()
{
    A a;
}

编译main.cpp 时编译器失败.现在我知道的解决方案是 #include "B.h",但我很好奇它为什么会失败.g++cl 的错误信息在这件事上都没有什么启发.

The compiler fails when compiling main.cpp. Now the solution I know is to #include "B.h", but I'm curious as to why it fails. Neither g++ or cl's error messages were very enlightening in this matter.

推荐答案

编译器需要知道B"有多大,才能生成合适的布局信息.相反,如果您说的是 std::vector,那么编译器就不需要知道 B 有多大,因为它知道一个指针有多大.

The compiler needs to know how big "B" is before it can generate the appropriate layout information. If instead, you said std::vector<B*>, then the compiler wouldn't need to know how big B is because it knows how big a pointer is.

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