size_t参数new运算符 [英] size_t parameter new operator

问题描述

我有一点在我心中，我不能弄清楚新的操作符重载。
假设我有一个类MyClass，但MyClass.h MyClass.cpp和main.cpp文件是类似的;

I have a point in my mind which I can't figure out about new operator overloading. Suppose that, I have a class MyClass yet MyClass.h MyClass.cpp and main.cpp files are like;

//MyClass.h

class MyClass {
   public:
     //Some member functions
     void* operator new (size_t size);
     void operator delete (void* ptr);
     //...
};

//MyClass.cpp

void* MyClass::operator new(size_t size) {
   return malloc(size);
}

void MyClass::operator delete(void* ptr) {
   free(ptr);
}

//main.cpp

//Include files
//...

int main() {
   MyClass* cPtr = new MyClass();
   delete cPtr
} 

这个程序运行正常。然而，我不能理解的事情是，如何来的新操作符可以被调用没有任何参数，而在其定义它有一个函数参数像size_t size。有没有点我在这里失踪？
感谢。

respectively. This program is running just fine. However, the thing I can't manage to understand is, how come new operator can be called without any parameter while in its definition it has a function parameter like "size_t size". Is there a point that I am missing here? Thanks.

推荐答案

不要将新表达式与运算符新分配函数混淆。前者导致后者。当你说 T * p = new T; 时，首先调用分配函数获得内存，然后构造该内存中的对象。该过程松散地等价于以下内容：

Don't confuse the "new expression" with the "operator new" allocation function. The former causes the latter. When you say T * p = new T;, then this calls the allocation function first to obtain memory and then constructs the object in that memory. The process is loosely equivalent to the following:

void * addr = T::operator new(sizeof(T));    //  rough equivalent of what
T * p = ::new (addr) T;                      //  "T * p = new T;" means.

（在构造函数抛出的情况下加上异常处理程序;在这种情况下内存将被释放。）

(Plus an exception handler in the event that the constructor throws; the memory will be deallocated in that case.)

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