从char *创建UTF-16字符串 [英] Create UTF-16 string from char*

问题描述

所以我有标准的C字符串：

So I have standard C string:

char* name = "Jakub";

我想将它转换为UTF-16。我发现，UTF-16将两倍长 - 一个字符需要两个字符。

所以我创建另一个字符串：

And I want to convert it to UTF-16. I figured out, that UTF-16 will be twice as long - one character takes two chars.
So I create another string:

char name_utf_16[10];  //"Jakub" is 5 characters

现在，我相信ASCII字符我只会使用低字节，所以对于所有这些，它将像 J 等等 74 00 。有了这个信念，我可以做这样的代码：

Now, I believe with ASCII characters I will only use lower bytes, so for all of them it will be like 74 00 for J and so on. With that belief, I can make such code:

void charToUtf16(char* input, char* output, int length) {
    /*Todo: how to check if output is long enough?*/
    for(int i=0; i<length; i+=2)  //Step over 2 bytes
    {
        //Lets use little-endian - smallest bytes first
        output[i] = input[i];
        output[i+1] = 0;  //We will never have any data for this field
    }
}

但是，通过这个过程，我以Jkb结束。我知道没有办法测试这个正确 - 我刚刚发送字符串到Minecraft Bukkit服务器。这是它断开连接时所说的：

But, with this process, I ended with "Jkb". I know no way to test this properly - I've just sent the string to Minecraft Bukkit Server. And this is what it said upon disconnecting:

13:34:19 [INFO]断开 jkb ？ [/127.0.0.1:53215]：过期的服务器！

13:34:19 [INFO] Disconnecting jkb?? [/127.0.0.1:53215]: Outdated server!

注意：我知道Minecraft使用big-endian。上面的代码只是一个例子，事实上，我已经在类中实现了转换。

Note: I'm aware that Minecraft uses big-endian. Code above is just an example, in fact, I have my conversion implemented in class.

推荐答案

output[i] = input[i];

这将分配输入的每个其他字节，因为增加 i 通过2.所以难怪你获得Jkb。
你可能想写：

This will assign every other byte of the input, because you increment i by 2. So no wonder that you obtain "Jkb". You probably wanted to write:

output[i] = input[i / 2];

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