使得在C常量数组++ [英] making a constant array in c++

问题描述

是否有任何理由，$ C $个cblocks告诉我，我不能让一个数组？我只是试图做的：

Is there any reason why codeblocks is telling me that I can't make an array? I'm simply trying to do:

const unsigned int ARRAY[10] = {0,1,2,3,4,5,6,7,8,9};

和它给了我

错误：一个大括号内的初始化在这里不允许使用前'{'令牌

error: a brace-enclosed initializer is not allowed here before '{' token

我已经改变了初始的其他部位，但错误总是说同样的事情。这似乎没有什么意义，因为这是第一件事情我在C ++中了解到的。

I have changed other parts of the initializer, but the error is always saying the same thing. This doesn't seem to make sense, since this is one of the first things I learned in c++.

推荐答案

您说，你这样做是一个类中，作为私有变量。

You say that you did this within a class, as a private variable.

回想一下，（目前），成员变量的可能不会被初始化的在声明它们（只有少数例外）一样的地方。

Recall that (at the moment), member variables may not be initialised in the same place where you declare them (with a few exceptions).

struct T {
   std::string str = "lol";
};

也不行。它必须是：

is not ok. It has to be:

struct T {
   std::string str;
   T() : str("lol") {}
};

但是，要雪上加霜，pre-C ++ 0x中，你不能初始化在构造函数初始化值阵列：

struct T {
   const unsigned int array[10];
   T() : array({0,1,2,3,4,5,6,7,8,9}) {} // not possible :(
};

和，因为你的数组的元素是常量，你不能依靠转让或者：

And, because your array's elements are const, you can't rely on assignment either:

struct T {
   const unsigned int array[10];
   T() {
       for (int i = 0; i < 10; i++)
          array[i] = i; // not possible :(
   }
};

然而，其他一些参与者已相当正确地指出的那样，似乎没有点在数组的副本 T的每个实例如果您不能修改它的元素。相反，你可以使用静态成员。

However, as some other contributors have quite rightly pointed out, there seems little point in having a copy of the array for each instance of T if you can't modify its elements. Instead, you could use a static member.

因此​​，下面将最终解决什么＆MDASH你的问题。或许＆MDASH;最好的方式：

So, the following will ultimately solve your problem in what's — probably — the best way:

struct T {
   static const unsigned int array[10];
};

const unsigned int T::array[10] = {0,1,2,3,4,5,6,7,8,9};

希望这有助于。

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