联盟黑客为端测试和字节交换 [英] Union hack for endian testing and byte swapping

问题描述

有关工会，写入一个成员和阅读其他成员（除字符数组）是UB。

For a union, writing to one member and reading from other member (except for char array) is UB.

//snippet 1(testing for endianess): 

union
{
    int  i;
    char c[sizeof(int)];
} x;

x.i = 1;                     // writing to i
if(x.c[0] == 1)              // reading from c[0]
{   printf("little-endian\n");
}
else
{   printf("big-endian\n");
}

//snippet 2(swap bytes using union):

int swapbytes()
{
    union                   // assuming 32bit, sizeof(int)==4
    {        
        int  i;
        char c[sizeof(int)];
    } x;
    x.i = 0x12345678;       // writing to member i
    SWAP(x.ch[0],x.ch[3]);  // writing to char array elements
    SWAP(x.ch[1],x.ch[2]);  // writing to char array elements
    return x.i;             // reading from x.i 
}   

1片段是合法的C或C ++而不是片段2.我是否正确？有人点标准的部分，在那里它说，它可以OK写信给工会会员，并从另一个成员是一个字符数组阅读。

Snippet 1 is legal C or C++ but not snippet 2. Am I correct? Can some one point to the section of standard where it says its OK to write to a member of union and read from another member which is a char array.

推荐答案

我相信它（片段1）在技术上的不可以允许的，但大多数编译器允许也无妨，因为人们使用这种$ C的$ C。 GCC即使它支持的文件。

I believe it (snippet 1) is technically not allowed, but most compilers allow it anyway because people use this kind of code. GCC even documents that it is supported.

您的会有一些机器的问题是sizeof（int）的== 1，可能在某些既不是大尾数也不小尾数。

You will have problems on some machines where sizeof(int) == 1, and possibly on some that are neither big endian nor little endian.

要么使用会改变的话正确的顺序，或设置这个具有配置宏可用的功能。你可能需要认识到编译器和OS无妨。

Either use available functions that change words to the proper order, or set this with a configuration macro. You probably need to recognize compiler and OS anyway.

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