C++ wchar_t 指针
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wchar_t pointer
提问by Tony The Lion
What's wrong with this:
这有什么问题：
wchar_t * t = new wchar_t;
t = "Tony";
I thought I could use a wchar_t pointer as a string...
我以为我可以使用 wchar_t 指针作为字符串...
回答by Michael
Your code has two issues.
您的代码有两个问题。
First, "Tony"is a pointer to a string of char's. L"Tony"is the appropriate wide string.
首先，"Tony"是一个指向char's字符串的指针。 L"Tony"是合适的宽字符串。
Second, you allocate a single wchar_tvia new, then immediately lose track of it by reassigning the pointer to Tony. This results in a memory leak.
其次，您wchar_t通过 new分配了一个单曲，然后通过将指针重新分配给 Tony 立即失去对它的跟踪。这会导致内存泄漏。
回答by GManNickG
A pointer just points to a single value. This is important.
一个指针只指向一个值。这个很重要。
All you've done is allocated room for a single wchar_t, and point at it. Then you try to set the pointer to point at a string (remember, just at the first character), but the string type is incorrect.
您所做的只是为单个 分配了空间wchar_t，然后指向它。然后您尝试将指针设置为指向一个字符串（记住，只是在第一个字符处），但字符串类型不正确。
What you have is a string of char, it "should" be L"Tony". But all you're doing here is leaking your previous memory allocation because the pointer holds a new value.
你所拥有的是一串char，它“应该”是L"Tony". 但是你在这里所做的只是泄漏你以前的内存分配，因为指针保存了一个新值。
Rather you want to allocate enough room to hold the entire string, then copythe string into that allocated memory. This is terrible practice, though; never do anything that makes you need to explicitly free memory.
相反，您希望分配足够的空间来容纳整个字符串，然后将字符串复制到分配的内存中。不过，这是一种可怕的做法；永远不要做任何需要明确释放内存的事情。
Just use std::wstringand move on. std::wstring t = L"Tony";. It handles all the details, and you don't need to worry about cleaning anything up.
只需使用std::wstring并继续。std::wstring t = L"Tony";. 它处理所有细节，您无需担心清理任何东西。
回答by josefx
Since you are a C# developer I will point out a few things c++ does different.
由于您是 C# 开发人员，因此我将指出 C++ 的一些不同之处。
This allocates a new wchar_t and assigns it to t
这将分配一个新的 wchar_t 并将其分配给 t
wchar_t* t = new wchar_t
This is an array of constant char
这是一个常量字符数组
"Tony"
To get a constant wchar_t array prefix it with L
获得一个常量 wchar_t 数组前缀为 L
L"Tony"
This reasigns t to point to the constant L"Tony" instead of your old wchar_t and causes a memory leak since your wchar_t will never be released.
这将 t 重新指定为指向常量 L"Tony" 而不是您的旧 wchar_t 并导致内存泄漏，因为您的 wchar_t 永远不会被释放。
t = L"Tony"
This creates a string of wide chars (wchar_t) to hold a copy of L"Tony"
这将创建一串宽字符 (wchar_t) 来保存 L"Tony" 的副本
std::wstring t = L"Tony"
I think the last line is what you want. If you need access to the wchar_t pointer use t.c_str(). Note that c++ strings are mutable and are copied on each assignment.
我认为最后一行是你想要的。如果您需要访问 wchar_t 指针，请使用 t.c_str()。请注意，c++ 字符串是可变的，并且在每次赋值时都会被复制。
The c way to do this would be
这样做的c方法是
const wchar_t* t = L"Tony"
This does not create a copy and only assigns the pointer to point to the const wchar array
这不会创建副本，只会分配指向 const wchar 数组的指针
回答by Mark B
What this does is first assign a pointer to a newly allocated wchar_tinto t, and then try to assign a non-wide string into t.
它的作用是首先分配一个指向新分配的wchar_tinto的指针t，然后尝试分配一个非宽字符串 into t。
Can you use std::wstringinstead? That will handle all your memory management needs for you.
你可以用std::wstring吗？这将为您处理所有内存管理需求。
回答by gbjbaanb
you can, its just that "Tony" is a hardcoded string, and they're ANSI by default in most editors/compilers. If you want to tell the editor you're typing in a Unicode string, then prefix it with L, e.g. t = L"Tony".
你可以，只是“Tony”是一个硬编码的字符串，在大多数编辑器/编译器中它们默认是ANSI。如果你想告诉编辑器你正在输入一个 Unicode 字符串，那么在它前面加上 L 前缀，例如t = L"Tony".
You have other problems with your code, your allocation is allocating a single Unicode character (2 bytes), then you're trying to point the original variable to the constant string, thus leaking those 2 bytes.
您的代码还有其他问题，您的分配分配了一个 Unicode 字符（2 个字节），然后您试图将原始变量指向常量字符串，从而泄漏了这 2 个字节。
If you want to create a buffer of Unicode data and place data into it, you want to do:
如果要创建 Unicode 数据缓冲区并将数据放入其中，则需要执行以下操作：
wchar_t* t = new wchar_t[500];
wcscpy(t, "Tony");
回答by Pavel Radzivilovsky
this is completely wrong. There's no need to allocate two bytes, make t to point to them, and then overwrite the pointer t leaking the lost memory forever.
这是完全错误的。没有必要分配两个字节，让 t 指向它们，然后覆盖指针 t 永远泄漏丢失的内存。
Also, "Tony" has a different type. Use:
此外，“托尼”有不同的类型。用：
wchar_t *t = L"Tony";
wchar_t *t = L"Tony";
IMHO better don't use wchars at all - See https://stackoverflow.com/questions/1049947/should-utf-16-be-considered-harmful
恕我直言，最好根本不要使用 wchars - 请参阅https://stackoverflow.com/questions/1049947/should-utf-16-be-thinked-harmful
