Question

我需要在struct内部创建一个struct。如何正确编码构造函数？我需要分别为A和B创建两个构造函数，还是可以像下面的示例一样仅为A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的成员成为静态成员（但在这种情况下，对于Windows来说，整个程序或dll将只有一个值）：

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];
    }
};

如果您想让A包含B-那就是您需要做的：

struct A
{
    struct B
    {
        unsigned n;
        int* d;
    } b;

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

Answer 2

您需要一个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()
}

要么要么将它们设为static。

Answer 3

您没有声明成员，因为代码B中是A类的嵌套类。

您必须声明该类的成员。

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]; 
    }
};

结构内部结构：“非法引用非静态成员”错误

3 个答案: