有人知道为什么在分配后,调用Dog :: operator new的Latter语法会调用默认构造函数,最终会调用2个构造函数吗?
我想知道我做错了什么,怎么办:Dog *ptr = new("arg") Dog();
无需调用2个构造函数。而不使用任何技巧,例如检查默认构造函数是否已构造对象。这是代码:
class Dog
{
public:
Dog() // default
{
std::cout << "default Dog constructor [" << this << "]" << std::endl;
}
Dog(int x) // int argument
{
std::cout << "dog constructor int " << x << "[" << this << "]" << std::endl;
}
Dog(const std::string& word) // std::string argument
{
std::cout << "dog constructor std::string: " << word << " ["<< this << "]" << std::endl;
}
Dog(std::string &&word) // rvalue string argument
{
std::cout << "dog constructor std::string&& rvalue: " << word << " [" << this << "]" << std::endl;
}
// custom operator new
static void *operator new(std::size_t size) noexcept // for default constructor
{
Dog *ptr = (Dog*)malloc(size); // allocate memory
if (ptr) // if allocate ok
{
::new(ptr) Dog(); // call default constructor on object in memory
return ptr; // returns
}
else
return nullptr;
}
template<class T>
static void * operator new(std::size_t size, T&& value) noexcept // for argument constructor
{
Dog *ptr = (Dog*) malloc(size); // allocate the memory
if (ptr)
{
::new (ptr) Dog(std::forward<T>(value)); // pass the argument exactly as was passed to operator new,
// using perfect forwarding
return ptr;
}
else
return nullptr;
}
~Dog() { std::cout << "destructor " << std::endl; }
};
int main(void)
{
Dog *d = (Dog*) Dog::operator new(sizeof(Dog), "Const Char * Argument"); // argument version
Dog *d2 = (Dog*)Dog::operator new(sizeof(Dog)); // default constructor argument
//1 this works as expected, do what you specified in the member operator new, everything goes normal.
Dog *d3 = new Dog(); // default constructor
Dog *d4 = new("Const Char * Argument") Dog(); // argument constructor
// this is shorter, goes into your member operator new, BUT when it returns to this scope,
// call the default constructor for *d3, and for *d4 too.
// so this ends up calling constructors twice for both objects.
}
因此,我将分配与构造混合在一起,这里没有理由这样做,也许在操作符new []中有一些用,它使用默认构造函数以外的构造函数构造数组。
但是定义这些成员运算符的最佳方法是:
class Dog {
public:
// .......
// custom operator new
static void *operator new(std::size_t size) noexcept // for default constructor
{
void *memory = malloc(size); // allocate memory
if (memory) // if allocate ok
{
return memory; // returns
}
else
return nullptr;
}
static void *operator new[](std::size_t size) noexcept
{
void *memory = malloc(size); // allocate memory
if (memory) // if allocate ok
{
return memory; // returns
}
else
return nullptr;
}
static void operator delete(void *block) noexcept
{
free(block);
}
static void operator delete[](void *block) noexcept
{
free(block);
}
~Dog() { std::cout << "destructor " << std::endl; }
};
int main(void)
{
// now we can use new operator normaly without complications
Dog *d1 = new Dog[10]; // default constructor on all objects
Dog *d2 = new Dog("const char * argument"); // call std::string&& constructor
delete[] d1;
delete d2;
}
最佳答案
使用(关键字)new:它调用分配运算符new,并且(!)调用构造函数。
注意:向操作员提供新操作时,您也应提供操作员删除操作(在您的情况下,它将免费)。另外,不要忘记阵列版本。
关于c++ - 类成员运算符new,两次调用构造函数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/33451233/