跟进： 使用复制构造函数后双释放子对象
我按照建议遵循了 5 规则。但现在似乎移动分配发生在一个未初始化的对象（对象 id 0）上？我预计它会从对象 3 移动到对象 2。
我创建了以下（最少？）示例，这似乎触发了我的问题：
#include <stdio.h>
#include <stdint.h>
class A
{
public:
A()
{
myCtr = ++ctr;
printf("class A default Constructor - object id: %u\n", myCtr);
}
A(const A &a2) {
myCtr = ++ctr;
printf("class A copy constructor - object id: %u\n", myCtr);
}
A(A &&a2) {
myCtr = a2.myCtr;
a2.myCtr = 0;
printf("class A move constructor - object id: %u\n", myCtr);
}
A & operator=(const A &a2) {
myCtr = ++ctr;
printf("class A copy assignment - from object id: %u - to object id: %u\n", a2.myCtr, myCtr);
return *this;
}
A & operator=(A &&a2) {
printf("class A move assignment - from object id: %u - to object id: %u\n", a2.myCtr, myCtr);
if (this != &a2) {
//myCtr = a2.myCtr;
//a2.myCtr = 0;
}
return *this;
}
~A()
{
printf("class A destructor - object id: %u\n", myCtr);
}
private:
uint64_t myCtr;
static uint64_t ctr;
};
class B
{
public:
B() {
}
B(char * input, uint32_t len) {
for (uint32_t i = 0; i < len; i++)
{
/* do something */
}
}
B(const B &b2) {
characters = A(b2.characters);
}
B(B &&b2) {
characters = A(b2.characters);
}
B & operator=(const B &b2) {
characters = A(b2.characters);
}
B & operator=(B &&b2) {
characters = A(b2.characters);
}
~B() {
}
private:
A characters;
};
uint64_t A::ctr = 0;
int main(int argc, char *argv[]) {
B b1 = B((char *)"b1", 2);
B b2 = b1;
return 0;
}
这会产生以下输出：
class A default Constructor - object id: 1
class A default Constructor - object id: 2
class A copy constructor - object id: 3
class A move assignment - from object id: 3 - to object id: 0
class A destructor - object id: 3
class A destructor - object id: 2
class A destructor - object id: 1
这是我没想到的行：
class A move assignment - from object id: 3 - to object id: 0
期待：
class A move assignment - from object id: 3 - to object id: 2
我正在使用以下编译器：Microsoft (R) C/C++ Optimizing Compiler Version 19.14.26429.4
以防你投反对票。请说明原因。我很乐意尝试改进我的问题。
编辑：
似乎只有针对 x86 平台的 Visual C++ 编译器才会出现此问题。它适用于 g++ (x86 & x64)、clang (x86 & x64) 和 msvc (x64)。这让我更难弄清楚这个问题的根源。