c++ - 将 short 转换为 char 是否合法？

Question

unsigned char HexLookup[513] = {
    "000102030405060708090a0b0c0d0e0f"
    "101112131415161718191a1b1c1d1e1f"
    "202122232425262728292a2b2c2d2e2f"
    "303132333435363738393a3b3c3d3e3f"
    "404142434445464748494a4b4c4d4e4f"
    "505152535455565758595a5b5c5d5e5f"
    "606162636465666768696a6b6c6d6e6f"
    "707172737475767778797a7b7c7d7e7f"
    "808182838485868788898a8b8c8d8e8f"
    "909192939495969798999a9b9c9d9e9f"
    "a0a1a2a3a4a5a6a7a8a9aaabacadaeaf"
    "b0b1b2b3b4b5b6b7b8b9babbbcbdbebf"
    "c0c1c2c3c4c5c6c7c8c9cacbcccdcecf"
    "d0d1d2d3d4d5d6d7d8d9dadbdcdddedf"
    "e0e1e2e3e4e5e6e7e8e9eaebecedeeef"
    "f0f1f2f3f4f5f6f7f8f9fafbfcfdfeff"
};

void to_hex(char* in_data, unsigned int in_data_size, char* out_data)
{
    unsigned short *pwHex =  (unsigned short*)HexLookup;

    for (unsigned int i = 0; i < in_data_size; i++)
    {
        *out_data = pwHex[*in_data];
        in_data++;
        out_data++;
    }
}

在上面的代码中，什么是行为:

unsigned short *pwHex = (unsigned short*)HexLookup;
*out_data = pwHex[*in_data];

Answer 1

根据C++03 - 不，它不合法，因为它违反了严格的别名规则。一些像 gcc 这样的编译器接受这一点，即使没有警告。您将必须使用移位将字符转换为短裤，一个能够处理此问题而不是字符缓冲区的类，或者甚至是一个 union 来命名几个选项。