零在类构造函数C ++中初始化多维数组

Is there a way (with C++11) to zero-initialize a multidimensional array in a class constructor?

有没有办法(使用C ++ 11)在类构造函数中初始化多维数组?

Does this initialize the whole array to zero or just the first dimension?

这会将整个数组初始化为零还是仅初始化第一维?

class Example
{
public:
    int array[100][100];

    Example():array{}
    {
    };
};

Lastly, will this work with enumerated types?

最后,这将使用枚举类型吗?

2 个解决方案

#1

Value-initialization for an array will value-initialize each element in the array. You have an array of arrays, so each element (an array) will be value-initialized. From there, return to the first sentence of this answer for what happens to each array.

数组的值初始化将对数组中的每个元素进行值初始化。您有一个数组数组,因此每个元素(一个数组)都将进行值初始化。从那里,返回到每个数组发生的事情的答案的第一句话。

The answer to your second question, regular enumerations are scalars, and as such are value-initialized, just like your int values in your array, which is to say they're zero-initialized. That may be problematic if you have an enum without a 0-value representation:

对于第二个问题的答案,常规枚举是标量,因此是值初始化的,就像数组中的int值一样,也就是说它们是零初始化的。如果你有一个没有0值表示的枚举,这可能会有问题:

#include <iostream>

enum EnumStuff
{
    one = 1,
    two = 2,
    three = 3
};

int main()
{
    EnumStuff arr[10][10]{};
    for (auto const&x : arr)
    {
        for (auto y : x)
            std::cout << static_cast<int>(y) << ' ';
        std::cout << '\n';
    }
}

There is no 0-value in EnumStuff. Just something to keep in mind. A switch looking for only those three values, for example, will be sorely-disappointed (and shortly thereafter, likely you with it).

EnumStuff中没有0值。请记住一些事情。例如,仅仅寻找这三个值的开关将会非常失望(此后不久,很可能会使用它)。

Best of luck

祝你好运

#2

Value initialization (i.e. with {}) for multidimensional arrays value-initializes all elements, so all 10000 elements will have value 0.

多维数组的值初始化(即使用{})值初始化所有元素,因此所有10000个元素的值都为0。

#1

#include <iostream>

enum EnumStuff
{
    one = 1,
    two = 2,
    three = 3
};

int main()
{
    EnumStuff arr[10][10]{};
    for (auto const&x : arr)
    {
        for (auto y : x)
            std::cout << static_cast<int>(y) << ' ';
        std::cout << '\n';
    }
}