typedef int array [3]和typedef int（array）[3]有什么区别？

I recently came across this unorthodox way to define a int array type:

我最近遇到了这种非正统的方式来定义一个int数组类型：

typedef int(array)[3];

At first I thought it was an array of function pointers but then I realized that the * and the () were missing, so by looking into the code I deduced the type array was a int[3] type instead. I normally would declare this type as:

起初我以为它是一个函数指针数组但后来我意识到缺少了*和（），所以通过查看代码我推断出类型数组是一个int [3]类型。我通常会将此类型声明为：

typedef int array[3];

Unless I'm mistaken that they are not the same thing, what is the advantage of doing so in the former way other than to make them look similar to a function pointer?

除非我误以为它们不是同一个东西，除了让它们看起来类似于函数指针之外，以前的方式这样做有什么好处呢？

3 个解决方案

#1

What is the difference between typedef int array[3] and typedef int(array)[3]?

typedef int array [3]和typedef int（array）[3]有什么区别？

They are the same.

他们是一样的。

Parentheses could be used when a pointer is being declared, with *, and result in different types. In that case, parentheses could affect the precedence of [] or int. However, this is not your case here.

当声明指针时，可以使用括号，使用*，并产生不同的类型。在这种情况下，括号可能会影响[]或int的优先级。但是，这不是你的情况。

#2

These are both equivalent. The parentheses do not alter the precedence of [] or int in this case.

这些都是等价的。在这种情况下，括号不会改变[]或int的优先级。

The tool cdecl helps to confirm this:

工具cdecl有助于确认这一点：

int (a)[3] gives "declare a as array 3 of int"
int（a）[3]给出“声明一个int的数组3”
int a[3] gives "declare a as array 3 of int"
int a [3]给出“声明一个int的数组3”

#3

-1

If you run this code like this:

如果你运行这样的代码：

typedef int array[6];

array arr={1,2,3,4,5,6};
for(int i=0; i<6; i++)
     cout<<arr[i]<<" ";

And now you run this code like this:

现在你运行这样的代码：

typedef int (array)[6];
array arr={1,2,3,4,5,6};
for(int i=0; i<6; i++)
     cout<<arr[i]<<" ";

Both of those two types of code it generates the same output.This proves that both are same and the parentheses have no effect.

这两种类型的代码都生成相同的输出。这证明两者都是相同的，括号没有效果。

#1