什么是双星(例如 NSError **)? [英] What is double star (eg. NSError **)?

问题描述

所以，我看到了:

error:(NSError **)error

在苹果文档中.为什么是两星?有什么意义?

in the apple doc's. Why two stars? What is the significance?

推荐答案

双星"是指向指针的指针.所以NSError ** 是一个指向NSError 类型的对象的指针.它基本上允许您从函数返回一个错误对象.您可以在函数中创建一个指向 NSError 对象的指针(称为 *myError)，然后执行以下操作:

A "double star" is a pointer to a pointer. So NSError ** is a pointer to a pointer to an object of type NSError. It basically allows you to return an error object from the function. You can create a pointer to an NSError object in your function (call it *myError), and then do something like this:

*error = myError;

将该错误返回"给调用者.

to "return" that error to the caller.

回复下面发表的评论:

你不能简单地使用NSError *，因为在C中，函数参数是按值传递的——也就是说，值被复制 传递给函数时.为了说明这一点，请考虑以下 C 代码片段:

You can't simply use an NSError * because in C, function parameters are passed by value—that is, the values are copied when passed to a function. To illustrate, consider this snippet of C code:

void f(int x)
{
    x = 4;
}

void g(void)
{
    int y = 10;
    f(y);
    printf("%d
", y);    // Will output "10"
}

f() 中 x 的重新赋值不会影响 f() 之外的参数值(在 g()，例如)

The reassignment of x in f() does not affect the argument's value outside of f() (in g(), for example).

同样，当一个指针传入函数时，它的值被复制，重新赋值不会影响函数外的值.

Likewise, when a pointer is passed into a function, its value is copied, and re-assigning will not affect the value outside of the function.

void f(int *x)
{
    x = 10;
}

void g(void)
{
    int y = 10;
    int *z = &y;
    printf("%p
", z);    // Will print the value of z, which is the address of y
    f(z);
    printf("%p
", z);    // The value of z has not changed!
}

当然，我们知道我们可以很容易地改变 z 指向的值:

Of course, we know that we can change the value of what z points to fairly easily:

void f(int *x)
{
    *x = 20;
}

void g(void)
{
    int y = 10;
    int *z = &y;
    printf("%d
", y);    // Will print "10"
    f(z);
    printf("%d
", y);    // Will print "20"
}

所以按理说，要改变 NSError * 指向的值，我们还必须传递一个指向该指针的指针.

So it stands to reason that, to change the value of what an NSError * points to, we also have to pass a pointer to the pointer.

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