Union不同于struct，因为Union在其他的上面重复:它重新定义同一个内存，而struct一个接一个地定义，没有重叠或重定义。

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.

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。

简单的回答是:结构体是一个记录结构体:结构体中的每个元素分配新的空间。一个结构体

struct foobarbazquux_t {
    int foo;
    long bar;
    double baz; 
    long double quux;
}

为每个实例在内存中分配至少(sizeof(int)+sizeof(long)+sizeof(double)+sizeof(long double))字节。(“至少”是因为架构对齐约束可能迫使编译器填充结构。)

另一方面，

union foobarbazquux_u {
    int foo;
    long bar;
    double baz; 
    long double quux;
}

分配一个内存块并给它四个别名。因此sizeof(union foobarbazquux_u)≥max((sizeof(int)，sizeof(long)，sizeof(double)，sizeof(long double))，同样有可能添加一些对齐。

是的，struct和union之间的主要区别和你说的一样。 Struct使用其成员的所有内存，union使用最大的成员内存空间。

但所有的区别在于使用内存的需要。 在unix的进程中可以看到联合的最佳使用，在那里我们使用信号。 比如一个进程一次只能作用于一个信号。 所以一般声明是:

union SIGSELECT
{
  SIGNAL_1 signal1;
  SIGNAL_2 signal2;
  .....
};

在这种情况下，进程只使用所有信号中最高的内存。 但如果在这种情况下使用struct，内存使用量将是所有信号的总和。 有很大的不同。

总之，如果您知道一次访问任何一个成员，则应该选择Union。

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)。