As you already state in your question, the main difference between union and struct is that union members overlay the memory of each other so that the sizeof of a union is the one , while struct members are laid out one after each other (with optional padding in between). Also an union is large enough to contain all its members, and have an alignment that fits all its members. So let's say int can only be stored at 2 byte addresses and is 2 bytes wide, and long can only be stored at 4 byte addresses and is 4 bytes long. The following union

union test {
    int a;
    long b;
}; 

could have a sizeof of 4, and an alignment requirement of 4. Both an union and a struct can have padding at the end, but not at their beginning. Writing to a struct changes only the value of the member written to. Writing to a member of an union will render the value of all other members invalid. You cannot access them if you haven't written to them before, otherwise the behavior is undefined. GCC provides as an extension that you can actually read from members of an union, even though you haven't written to them most recently. For an Operation System, it doesn't have to matter whether a user program writes to an union or to a structure. This actually is only an issue of the compiler.

union和struct的另一个重要属性是，它们允许指向它们的指针可以指向其任何成员的类型。因此，以下是有效的:

struct test {
    int a;
    double b;
} * some_test_pointer;

Some_test_pointer可以指向int*或double*。如果将一个test类型的地址转换为int*，它将指向它的第一个成员，实际上是a。工会也是如此。因此，因为联合将始终具有正确的对齐方式，您可以使用联合来使指向某些类型的指针有效:

union a {
    int a;
    double b;
};

这个联合实际上可以指向int型和double型:

union a * v = (union a*)some_int_pointer;
*some_int_pointer = 5;
v->a = 10;
return *some_int_pointer;    

实际上是有效的，正如C99标准所述:

对象的存储值只能由具有以下类型之一的左值表达式访问: 与对象的有效类型兼容的类型 .．. 在其成员中包含上述类型之一的聚合或联合类型

编译器不会优化出v->a = 10;因为它可能会影响*some_int_pointer的值(该函数将返回10而不是5)。

As you already state in your question, the main difference between union and struct is that union members overlay the memory of each other so that the sizeof of a union is the one , while struct members are laid out one after each other (with optional padding in between). Also an union is large enough to contain all its members, and have an alignment that fits all its members. So let's say int can only be stored at 2 byte addresses and is 2 bytes wide, and long can only be stored at 4 byte addresses and is 4 bytes long. The following union

union test {
    int a;
    long b;
}; 

could have a sizeof of 4, and an alignment requirement of 4. Both an union and a struct can have padding at the end, but not at their beginning. Writing to a struct changes only the value of the member written to. Writing to a member of an union will render the value of all other members invalid. You cannot access them if you haven't written to them before, otherwise the behavior is undefined. GCC provides as an extension that you can actually read from members of an union, even though you haven't written to them most recently. For an Operation System, it doesn't have to matter whether a user program writes to an union or to a structure. This actually is only an issue of the compiler.

union和struct的另一个重要属性是，它们允许指向它们的指针可以指向其任何成员的类型。因此，以下是有效的:

struct test {
    int a;
    double b;
} * some_test_pointer;

Some_test_pointer可以指向int*或double*。如果将一个test类型的地址转换为int*，它将指向它的第一个成员，实际上是a。工会也是如此。因此，因为联合将始终具有正确的对齐方式，您可以使用联合来使指向某些类型的指针有效:

union a {
    int a;
    double b;
};

这个联合实际上可以指向int型和double型:

union a * v = (union a*)some_int_pointer;
*some_int_pointer = 5;
v->a = 10;
return *some_int_pointer;    

实际上是有效的，正如C99标准所述:

对象的存储值只能由具有以下类型之一的左值表达式访问: 与对象的有效类型兼容的类型 .．. 在其成员中包含上述类型之一的聚合或联合类型

编译器不会优化出v->a = 10;因为它可能会影响*some_int_pointer的值(该函数将返回10而不是5)。

结构和联合的区别是什么?

简单的回答是:区别在于内存分配。 解释: 在结构中，将为结构内部的所有成员创建内存空间。 在联合内存空间将只为需要最大内存空间的成员创建。 考虑下面的代码:

struct s_tag
{
   int a; 
   long int b;
} x;

union u_tag
{
   int a; 
   long int b;
} y;

这里在struct和union内部有两个成员:int和long int。在32位操作系统中，int的内存空间为:4字节，long int的内存空间为:8。

因此，对于struct, 4+8=12个字节将被创建，而对于union，将创建8个字节

代码示例:

#include<stdio.h>
struct s_tag
{
  int a;
  long int b;
} x;
union u_tag
{
     int a;
     long int b;
} y;
int main()
{
    printf("Memory allocation for structure = %d", sizeof(x));
    printf("\nMemory allocation for union = %d", sizeof(y));
    return 0;
}

裁判:http://www.codingpractise.com/home/c-programming/structure-and-union/

“union”和“struct”是C语言的构造。谈论它们之间的“操作系统级别”差异是不合适的，因为如果您使用一个或另一个关键字，则是编译器产生不同的代码。

结构分配其中所有元素的总大小。

一个联合只分配它最大的成员所需要的内存。

在编写下面给出的字节排序函数时，联合非常方便。这在结构体中是不可能的。

int main(int argc, char **argv) {
    union {
        short   s;
        char    c[sizeof(short)];
    } un;

    un.s = 0x0102;

    if (sizeof(short) == 2) {
        if (un.c[0] == 1 && un.c[1] == 2)
            printf("big-endian\n");
        else if (un.c[0] == 2 && un.c[1] == 1)
            printf("little-endian\n");
        else
            printf("unknown\n");
    } else
        printf("sizeof(short) = %d\n", sizeof(short));

    exit(0);
}
// Program from Unix Network Programming Vol. 1 by Stevens.