c++英雄联盟

Unions are conceptually similar to structures. The syntax to declare/define a union is also similar to that of a structure. The only differences is in terms of storage. In structure each member has its own storage location, whereas all members of union uses a single shared memory location which is equal to the size of its largest data member.

工会在概念上类似于结构 。 声明/定义联合的语法也类似于结构的语法。 唯一的区别在于存储。 在结构上，每个成员都有其自己的存储位置，而联合的所有成员都使用一个共享存储位置，该位置等于其最大数据成员的大小。

This implies that although a union may contain many members of different types, it cannot handle all the members at the same time. A union is declared using the union keyword.

这意味着，尽管一个联合可以包含许多不同类型的成员， 但它不能同时处理所有成员 。 一名工会是使用声明的union关键字。

union item
{int m;float x;char c;
}It1;

This declares a variable It1 of type union item. This union contains three members each with a different data type. However only one of them can be used at a time. This is due to the fact that only one location is allocated for all the union variables, irrespective of their size. The compiler allocates the storage that is large enough to hold the largest variable type in the union.

这声明了一个union item类型的变量It1 。 该union包含三个成员，每个成员具有不同的数据类型。 但是，一次只能使用其中之一。 这是由于以下事实：所有union变量均仅分配一个位置，而不管其大小如何。 编译器会分配足够大的存储空间以容纳联合中最大的变量类型。

In the union declared above the member x requires 4 bytes which is largest amongst the members for a 16-bit machine. Other members of union will share the same memory address.

在上面声明的联合中，成员x要求4个字节 ，这是16位计算机中成员中最大的。 工会的其他成员将共享相同的内存地址。

在C中访问工会会员 (Accessing a Union Member in C)

Syntax for accessing any union member is similar to accessing structure members,

访问任何union成员的语法类似于访问结构成员的语法，

union test
{int a;float b;char c;
}t;t.a;    //to access members of union t
t.b;
t.c;

实例时间 (Time for an Example)

#include <stdio.h>union item
{int a;float b;char ch;
};int main( )
{union item it;it.a = 12;it.b = 20.2;it.ch = 'z';printf("%d\n", it.a);printf("%f\n", it.b);printf("%c\n", it.ch);return 0;
}

-26426 20.1999 z

-26426 20.1999 z

As you can see here, the values of a and b get corrupted and only variable c prints the expected result. This is because in union, the memory is shared among different data types. Hence, the only member whose value is currently stored will have the memory.

如您所见， a和b的值被破坏，只有变量c打印预期结果。 这是因为在联合中，内存在不同的数据类型之间共享。 因此，当前存储其值的唯一成员将具有内存。

In the above example, value of the variable c was stored at last, hence the value of other variables is lost.

在上面的示例中，变量c的值最后被存储了，因此其他变量的值丢失了。

翻译自: https://www.studytonight.com/c/unions-in-c.php

c++英雄联盟