我有此代码,请注意,两行已注释成员变量继承
#include <iostream>
class foo {
public:
foo();
int i;
};
class bar: foo {
public:
bar();
//int i;
};
foo::foo()
{
i = 2;
}
bar::bar()
{
i = 4;
}
int main()
{
bar *b = new(bar);
//std::cout << "bi = " << b->i << std::endl; /*line 28*/
foo *f = (foo*) b;
std::cout << "fi = " << f->i << std::endl;
}
随着注释掉两行,代码编译和输出是
fi = 4
通过这两个代码编译和输出是
bi = 4
fi = 2
只有i wit的声明欣类吧注释掉编译失败
var.cc: In function ‘int main()’:
var.cc:6:7: error: ‘int foo::i’ is inaccessible
var.cc:28:30: error: within this context
我理解前两种情况,但我不明白这个编译错误。为什么
变量“我”可以从酒吧的构造函数,但不是从一个酒吧指针?
尝试从富公开'类酒吧继承:公共foo' – StoryTeller 2013-03-19 11:19:49