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论坛 >编程语言 >C++ new A 和 new A() 的区别详解

C++ new A 和 new A() 的区别详解

希尔瓦娜斯发布于 2017-07-24 09:23查看:8760回复:1

        我们在C++程序中经常看到两种new的使用方式:new A以及new A()。那么这两种究竟有什么区别呢?

        调用new分配的内存有时候会被初始化,而有时候不会,这依赖于A的类型是否是POD(Plain old data)类型,或者它是否是包含POD成员、使用编译器生成默认构造函数的类。

        附:POD类型

        POD是Plain old data的缩写,它是一个struct或者类,且不包含构造函数、析构函数以及虚函数。

        维基百科给出了更加详细的解释:

        C++的POD类型或者是一个标量值,或者是一个POD类型的类。POD class没有用户定义的析构函数、拷贝构造函数和非静态的非POD类型的数据成员。而且,POD class必须是一个aggregate,没有用户定义的构造函数,没有私有的或者保护的非静态数据,没有基类或虚函数。它只是一些字段值的集合,没有使用任何封装以及多态特性。

        附:aggregate的定义:

        An aggregate is an array or a class (clause 9) with no user-declared constructors (12.1), no private or protected non-static data members (clause 11), no base classes (clause 10), and no virtual functions (10.3).

        接着介绍一下C++中的三种初始化方式:

        zero-initialization,default-initialization,value-initialization。

        首先需要注意的是value-initialization是在C++2003标准中新引入的,在原来的1998标准中并不存在。

        C++03标准中针对这三种方式的说明:

        To zero-initialize an object of type T means:

        — if T is a scalar type (3.9), the object is set to the value of 0 (zero) converted to T;
        — if T is a non-union class type, each nonstatic data member and each base-class subobject is zero-initialized;
        — if T is a union type, the object’s first named data member is zero-initialized;
        — if T is an array type, each element is zero-initialized;
        — if T is a reference type, no initialization is performed.

        To default-initialize an object of type T means:

        — if T is a non-POD class type (clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
        — if T is an array type, each element is default-initialized;
        — otherwise, the object is zero-initialized.

        To value-initialize an object of type T means:

        — if T is a class type (clause 9) with a user-declared constructor (12.1), then the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
        — if T is a non-union class type without a user-declared constructor, then every non-static data member and base-class component of T is value-initialized;
        — if T is an array type, then each element is value-initialized;
        — otherwise, the object is zero-initialized

        A program that calls for default-initialization or value-initialization of an entity of reference type is ill-formed. If T is a cv-qualified type, the cv-unqualified version of T is used for these definitions of zero-initialization, default-initialization, and value-initialization.

        注意:VS2008遵循的是98标准,而GCC3.4.5遵循的是03标准。

        采用如下代码可以验证编译器遵循的到底是哪一种标准:

1500859298972928.jpg

        在VS008中输出就不是0,说明遵循的是98标准。

        下面先看一段C++示例代码:

1500859334805510.jpg

        运行结果:

1500859365376412.jpg

        上述测试平台是VS2008.需要注意的是,VS08只支持C++98。

        在这种情况下:

        new A:不确定的值

        new A():zero-initialize

        new B:默认构造(B::m未被初始化)

        new B():默认构造(B::m未被初始化)

        new C:默认构造(C::m被zero-initialize)

        new C():默认构造(C::m被zero-initialize)

        如果用兼容C++03的编译器,比如G++结果:

1500859402740535.jpg

        new A:不确定的值

        new A():value-initialize A,由于是POD类型所以是zero initialization

        new B:默认构造(B::m未被初始化)

        new B():value-initialize B,zero-initialize所有字段,因为使用的默认构造函数

        new C:default-initialize C,调用默认构造函数

        new C():value-initialize C,调用默认构造函数

        在所有C++版本中,只有当A是POD类型的时候,new A和new A()才会有区别。而且,C++98和C++03会有区别。

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    • 不看不知道,我一直以为一样的

      2017-07-24 09:44赞 (0)回复沙发
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