将默认构造函数添加到基类会更改sizeof（）派生类型 [英] Adding a default constructor to a base class changes sizeof() a derived type

问题描述

我倾向于认为我对C ++内部和内存布局有很好的把握，但这个让我感到困惑。我有以下测试代码：

I tend to think I have a pretty good grasp of C++ internals and memory layouts, but this one has me baffled. I have the following test code:

#include <stdio.h>

struct Foo
{
    //Foo() {}
    int x;
    char y;
};

struct Bar : public Foo
{
    char z[3];
};

int main()
{
    printf( "Foo: %u Bar: %u\n", (unsigned)sizeof( Foo ), (unsigned)sizeof( Bar ) );
}

输出合理：

Foo：8 Bar：12

Foo: 8 Bar: 12

然而，这是非常奇怪的部分，如果我取消注释Foo（）上的简单默认构造函数，sizeof（Bar）发生了变化！如何添加ctor可能会改变这些类的内存布局？

However, this is the very odd part, if I uncomment that simple default constructor on Foo(), the sizeof( Bar ) changes! How can the addition of a ctor possibly change the memory layout of these classes?

Foo：8 Bar：8

Foo: 8 Bar: 8

使用gcc-7.2编译

Compiled using gcc-7.2

推荐答案

GCC遵循Itanium ABI for C ++，它阻止POD的尾部填充用于存储派生类数据成员。

GCC follows the Itanium ABI for C++, which prevents the tail-padding of a POD being used for storage of derived class data members.

添加用户提供的构造函数意味着 Foo 不再是POD，因此该限制不适用于 Bar 。

Adding a user-provided constructor means that Foo is no longer POD, so that restriction does not apply to Bar.

有关ABI的详细信息，请参阅此问题。

这篇关于将默认构造函数添加到基类会更改sizeof（）派生类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！