为什么不能将前向声明用于 std::vector?
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++
或 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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