您的位置:首页 > 其它

new对象时,类名后加括号与不加括号的区别

2015-04-11 15:51 267 查看
#include <iostream>
using namespace std;

// 空类
class empty
{
};

// 一个默认构造函数,一个自定义构造函数
class Base
{
public:
Base()
{
cout << " default Base construct " << endl;
m_nValue = 100;
};
Base(int nValue)
{
cout << " custom Base construct " << endl;
m_nValue = nValue;
};

private:
int m_nValue;
};

// 一个默认复合构造函数
class custom
{
public:
custom(int value = 100)
{
cout << " default && custom construct " << endl;
m_nValue = value;
}

private:
int m_nValue;
};

void main()
{
empty* pEmpty1 = new empty;
empty* pEmpty2 = new empty();

Base* pBase1 =  new Base;
Base* pBase2 = new Base();
Base* pBase3 = new Base(200);

custom* pCustom1 = new custom;
custom* pCustom2 = new custom();

delete pEmpty1;
delete pEmpty2;

delete pBase1;
delete pBase2;
delete pBase3;

delete pCustom1;
delete pCustom2;
}
// Result:
/*
default Base construct
default Base construct
custom Base construct
default && custom construct
default && custom construct
*/


加括号与不加的区别
  (1)加括号
    1. 若括号为空,即无实参项,那么理解为调用默认构造函数;
    2. 若括号非空,即有实参项,可以理解为调用重载构造函数,或默认复合构造函数。
  (2)不加括号
    调用默认构造函数,或默认复合构造函数。
加空括号与不加括号的区别:对于自定义类型,调用的都是默认构造函数,没区别的。

只不过对于内建类型不太一样,加了扩号会做默认值初值化,比如:

int* p = new int;//p指向的int值是不确定的;

int* q = new int();//q指向的int值为0。
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: