在 new (c++) 的构造函数调用中不使用括号
可能的重复:
在类型名后面加括号make和新的有区别吗?
所以我的主要内容:
Class* pC = new Class;
它的工作原理
Class* pC = new Class();
我今天才意识到我省略了括号(所以我在某种程度上被最烦人的解析的相反"击中了).
I realized just today that I had omitted the parentheses (so I was hit by the "opposite" of the most vexing parse in a way).
我的问题:这两种形式是否等价?
My question: Are these two forms equivalent ?
推荐答案
如果类定义了默认构造函数，则两者是等价的；该对象将通过调用该构造函数来创建.
If the class has a default constructor defined, then both are equivalent; the object will be created by calling that constructor.
如果类只有一个隐式的默认构造函数，那就有区别了.第一个将使 POD 类型的任何成员未初始化；第二个将值初始化它们(即将它们设置为零).
If the class only has an implicit default constructor, then there is a difference. The first will leave any members of POD type uninitialised; the second will value-initialise them (i.e. set them to zero).
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