点符号和 -> 之间的区别在目标 C [英] Difference Between Dot Notation and -> in Objective C

问题描述

我正在尝试在我的代码中使用尽可能少的内存.我尝试了两种将自定义类对象发送到方法的方法.我不确定这两种方法之间是否有任何区别.假设我有 2 个类，Class1 和 Class2，当然每个类都有自己的类变量和方法.

I'm trying to use as little memory as possible in my code. I've tried two ways of sending a custom class object to a method. I'm not sure if there is any difference between these two approaches. Say I have 2 classes, Class1 and Class2 each with their own class variables and methods of course.

所有代码都写在Class1

方法 1:

Class2 *class2Object = [[Class2 alloc] init];

[self doSomething: class2Object];

[class2Object release];

-(void) doSomething: (Class2 *) var {
int a = var.a;
}

方法 2:

Class2 *class2Object = [[Class2 alloc] init];

[self doSomething: &class2Object];

[class2Object release];

-(void) doSomething: (Class2 **) var {
int a = var->a;
}

这两种方法在性能上有什么区别吗?第二种方法完全没有意义吗?为什么我可以在方法1中使用点表示法，而在方法2中必须使用->?

Is there any performance difference between these two methods? Is the second approach completely pointless? Why is it that I can use the dot notation in Approach 1, but have to use -> in Approach 2?

谢谢.

推荐答案

这两种方法在性能上有区别吗?

Is there any performance difference between these two methods?

确实，性能上的差异可以忽略不计，因为在方法 2 中，您还有一个间接寻址(即指针取消引用，另见下文)；因此，方法 1 将为您节省几个时钟周期.

indeed, there is a negligible difference in performance, due to the fact that in approach 2 you have one more indirection (i.e., pointer dereferencing, see also below); so, approach 1 will save you a few clock cycles.

第二种方法完全没有意义吗?

Is the second approach completely pointless?

方法 2 很有用，例如，当您想分配 Class2 类型的新实例并通过相同的参数将其传回给调用者时；说:

approach 2 is useful,e.g., when you would like to allocate a new instance of type Class2 and pass it back to the caller through the same argument; say:

 - (bool)cloneObject:(Class2 **)var;

你传入一个对象；对象被克隆并在 var 中返回；由于是对象本身的地址发生变化，因此您需要有一个指向对象指针的指针才能设置新地址；返回值仅说明操作是否正确执行.

you pass an object in; the object is cloned and returned in var; since it is the address of the object itself which changes, you need to have a pointer to the pointer to the object in order to set the new address; the return value only states if the operation was executed correctly.

当然，在这个例子中，这样做会更自然:

Of course, in this example, it would be more natural doing:

     - (Class2)cloneObject:(Class2*)var;

即，您返回指向新分配对象的指针，但用例仍然成立.

i.e., you return the pointer to the object newly allocated, but the use case still holds.

为什么我可以在方法 1 中使用点表示法，而在方法 2 中必须使用 ->?

Why is it that I can use the dot notation in Approach 1, but have to use -> in Approach 2?

在第二种情况下，您必须使用 -> 因为您没有直接处理指向对象的指针；您正在处理指向对象指针的指针；在这种情况下，您需要做的是，首先，取消引用"您的指针(即，应用运算符 *)以获得指向对象的指针，然后像其他方式一样访问后者；这可以写成:

in the second case you have to use -> because you are not dealing with a pointer to an object directly; you are dealing with a pointer to a pointer to an object; what you need do in such cases is, first of all, "dereferencing" your pointer (i.e., applying operator *) in order to get a pointer to the object, then accessing the latter as you would do otherwise; this could be written like:

 (*var).a

这里，var是一个指向Class2对象的指针；*var 是解引用它的结果，所以你有一个指向 Class2 对象的指针；最后，.a 是访问对象属性的语法.语法:

here, var is a pointer to a pointer to an object of Class2; *var is the result of dereferencing it, so you have a pointer to an object of Class2; finally, .a is the syntax to access an object property. the syntax:

var->a

只是上述操作的简写".

is simply a "shorthand" for the operations described above.

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