结构内部的结构："对非静态成员的非法引用"错误

Question

我需要在struct内部创建一个结构。如何正确编码构造函数？我是否需要为A和B分别创建两个构造函数，或者我只能在下面的示例中使用一个"外部"构造函数？我的尝试导致C2597错误。

struct A
{
    struct B
    {
        unsigned n;
        int* d;
    };
    A(unsigned c)
    {
        B::n = c; // C2597
        B::d = new int[c]; // C2597
    }
};

Answer 1

您需要使B的成员静态(但是在这种情况下，在Windows的情况下，整个程序或DLL都有一个值(：

struct A
{
    struct B
    {
        static unsigned n;
        static int* d;
    };
    A(unsigned c)
    {
        B::n = c;
        B::d = new int[c];
    }
};

或创建B的实例：

struct A
{
    struct B
    {
        unsigned n;
        int* d;
    };
    A(unsigned c)
    {
        B b;
        b.n = c;
        b.d = new int[c];
    }
};

如果您的意思是A包含B-这就是您需要做的：

struct A
{
    struct B
    {
        unsigned n;
        int* d;
    } b;
    A(unsigned c)
    {
        b.n = c;
        b.d = new int[c];
    }
};

Answer 2

您需要B的实例才能访问成员n和d。

struct B
{
    unsigned n;
    int* d;
} b; // Create an instance of `B`
A(unsigned c)
{
    b.n = c;
    b.d = new int[c]; // ToDo - you need to release this memory, perhaps in ~B()
}

要么使它们 static。

Answer 3

您没有声明成员，在您的代码B中是nested a。

您必须声明该课程的成员。

struct A
{
    struct B
    {
        unsigned n;
        int* d;
    }; // this is a class-type declaration
    B data; // this is member of class A having type of class B
    A(unsigned c) 
    {
        data.n = c; 
        data.d = new int[c]; 
    }
};