英文:
Pointer to the same instance used in a different class
问题
在这个测试中，值是不同的，不是同一个实例，我想在*From `test()`*中得到55的结果。怎么做到这一点？
英文:
How to solve this situation: I have three classes, to call them A, B and C. In C I have members of A and B. How do I set a pointer in B to have the same instance from C to A ?
#include <stdio.h>
class A {
public:
int x; // no init, random to can test
A() {
printf("From A, x=%d\n", x);
}
void getP(A *ptr) {
ptr = this;
}
};
class B {
public:
A *a;
B() {
a->getP(a);
printf("From B, x=%d\n", a->x);
}
void test() {
printf("From test(), x=%d\n", a->x);
}
};
class C {
public:
A a; B b;
C() {
a.x=55;
printf("From C, x=%d\n", a.x);
b.test();
}
};
int main () {
C c;
return 0;
}
In this test, the values are different, is not the same instance, I want to have 55 result in From test() . How to make this?
答案1
得分: 3
**代码中的问题：**
您当前的代码在`B`构造函数的这一行存在未定义行为（[undefined behavior][1]）：
a->getP(a);
这是因为在这一点上`a`尚未初始化，因此使用`->`对其进行解引用是未定义行为。
此外，`A::getP(A *ptr)`通过值获取`ptr`，因此它不会更新来自调用者的指针。
为了实现这一点，您需要通过引用获取它（`void getP(A * & ptr)`）。但实际上，这不再需要 - 请看下文。
**解决方案：**
为了实现您想要的效果，您可以在`B`的构造函数中添加一个`A*`参数，用于初始化`B::a`成员。
然后，在从`C`的构造函数中构造`B`时，传递`C::a`成员的地址。
现在不再需要`A::getP`。
以下是您的完整示例，带有这些修改。
（修改部分用“VVVVV”标记）。
#include <stdio.h>
class A {
public:
int x{ 0 };
A() {
printf("来自A，x=%d\n", x);
}
};
class B {
public:
A* a;
//----VVVVVV----VVVV---
B(A * a_) : a(a_) {
printf("来自B，x=%d\n", a->x);
}
void test() {
printf("从test()，x=%d\n", a->x);
}
};
class C {
public:
A a; B b;
//--------VVVVV--
C() : b(&a) {
a.x = 55;
printf("来自C，x=%d\n", a.x);
b.test();
}
};
int main() {
C c;
return 0;
}
输出：
来自A，x=0
来自B，x=0
来自C，x=55
从test()，x=55
请注意，有未初始化的数据是不好的实践，因此我将`A::x`初始化为0。
[1]:https://en.wikipedia.org/wiki/Undefined_behavior
英文:
Issues in you code:
Your current code has UB (undefined behavior) in this line in B constructor:
a->getP(a);
This is because a is not initialized at this point and therefore it is UB to dereference it (with ->).
In addition, A::getP(A *ptr) is getting ptr by value, and therefore it does not update the pointer from the caller.
In order to do so you need to get it by reference (void getP(A * & ptr)). However it is not needed at all - see below.
Solution:
In order to achieve what you want, you can add a A* parameter to the constructor of B, that will be used to initialized B::a member.
Then pass the address of C::a member, when B is constructed from C's constructor.
There is also no need for A::getP anymore.
Below is your complete example with this modification.
(the changes are marked with "VVVV").
#include <stdio.h>
class A {
public:
int x{ 0 };
A() {
printf("From A, x=%d\n", x);
}
};
class B {
public:
A* a;
//----VVVVVV----VVVV---
B(A * a_) : a(a_) {
printf("From B, x=%d\n", a->x);
}
void test() {
printf("From test(), x=%d\n", a->x);
}
};
class C {
public:
A a; B b;
//--------VVVVV--
C() : b(&a) {
a.x = 55;
printf("From C, x=%d\n", a.x);
b.test();
}
};
int main() {
C c;
return 0;
}
Output:
From A, x=0
From B, x=0
From C, x=55
From test(), x=55
Note that it's not a good practice to have uninitialized data, and therefore I initialized A::x to 0.
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