前瞻性声明的危险是什么？ [英] What are the dangers of forward declarations?

问题描述

我刚刚接受采访。我被问什么是前进声明。然后我问他们是否是与远期申报相关的危险。

I just had an interview. I was asked what is a "forward declaration". I was then asked if they were dangers associated with forward declarations.

我无法回答第二个问题。在网上搜索没有显示任何有趣的结果。

I could not answer to the second question. A search on the net didn't show up any interesting result.

那么，有人知道与使用前瞻性声明相关的任何危险吗？

So, do someone know any dangers associated with the use of forward declarations ?

推荐答案

很好，除了有关重复的问题...

Well, apart from the issues about duplication...

...

如果在指向不完整类型的指针上调用 delete 得到未定义的行为。在实践中，析构函数可能不会被调用。

If you call delete on a pointer to an incomplete type, you get undefined behavior. In practice, the destructor may not get called.

我们可以看到，在 LiveWorkSpace 使用以下命令和示例：

We can see that on LiveWorkSpace using the following command and sample:

// -std=c++11 -Wall -W -pedantic -O2

#include <iostream>

struct ForwardDeclared;

void throw_away(ForwardDeclared* fd) {
   delete fd;
}

struct ForwardDeclared {
   ~ForwardDeclared() {
      std::cout << "Hello, World!\n";
   }
};

int main() {
   ForwardDeclared* fd = new ForwardDeclared();
   throw_away(fd);
}

诊断：

Compilation finished with warnings:
 source.cpp: In function 'void throw_away(ForwardDeclared*)':
 source.cpp:6:11: warning: possible problem detected in invocation of delete operator: [enabled by default]
 source.cpp:5:6: warning: 'fd' has incomplete type [enabled by default] 
 source.cpp:3:8: warning: forward declaration of 'struct ForwardDeclared' [enabled by default]
 source.cpp:6:11: note: neither the destructor nor the class-specific operator delete will be called, even if they are declared when the class is defined

不要感谢你的编译器警告你;）

Don't you want to thank your compiler for warning you ;) ?

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