'size`来自哪里？

Question

我知道应该有一个删除运算符,我不会在某个地方关注它.我只是想知道,哇,它有效."大小"的论点来自何处？

#include<iostream>
#include<string>

class Base {
public:
    Base() { }
    void *operator new( unsigned int size, std::string str ) {
    std::cout << "Logging an allocation of ";
    std::cout << size;
    std::cout << " bytes for new object '";
    std::cout << str;
    std::cout << "'";
    std::cout << std::endl;
    return malloc( size );
    }
private:
    int var1;
    double var2;
};

int main(int argc, char** argv){
    Base* b = new ("Base instance 1") Base;
}

结果如下:

为新对象'Base instance 1'记录16字节的分配

Answer 1

它由编译器在编译时提供.当编译器看到:

new ("Base instance 1") Base;

它会添加一个调用:

Base::operator new(sizeof(Base), "Base instance 1");

编辑:编译器当然也会添加一个调用Base::Base()