C++ 如何将 char* 转换为 wchar_t*？
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How to convert char* to wchar_t*?
提问by AutoBotAM
I've tried implementing a function like this, but unfortunately it doesn't work:
我试过实现这样的功能，但不幸的是它不起作用：
const wchar_t *GetWC(const char *c)
{
const size_t cSize = strlen(c)+1;
wchar_t wc[cSize];
mbstowcs (wc, c, cSize);
return wc;
}
My main goal here is to be able to integrate normal char strings in a Unicode application. Any advice you guys can offer is greatly appreciated.
我的主要目标是能够在 Unicode 应用程序中集成普通字符字符串。非常感谢你们提供的任何建议。
采纳答案by Cheers and hth. - Alf
Use a std::wstringinstead of a C99 variable length array. The current standard guarantees a contiguous buffer for std::basic_string. E.g.,
使用 astd::wstring而不是 C99 可变长度数组。当前标准保证 的连续缓冲区std::basic_string。例如，
std::wstring wc( cSize, L'#' );
mbstowcs( &wc[0], c, cSize );
C++ does not support C99 variable length arrays, and so if you compiled your code as pure C++, it would not even compile.
C++ 不支持 C99 可变长度数组，因此如果您将代码编译为纯 C++，它甚至无法编译。
With that change your function return type should also be std::wstring.
有了这个改变，你的函数返回类型也应该是std::wstring.
Remember to set relevant locale in main.
请记住在main.
E.g., setlocale( LC_ALL, "" ).
例如，setlocale( LC_ALL, "" )。
Cheers & hth.,
干杯 & hth.,
回答by Andrew Shepherd
In your example, wcis a local variable which will be deallocated when the function call ends. This puts you into undefined behavior territory.
在您的示例中，wc是一个局部变量，它将在函数调用结束时被释放。这会让你进入未定义的行为领域。
The simple fix is this:
简单的修复是这样的：
const wchar_t *GetWC(const char *c)
{
const size_t cSize = strlen(c)+1;
wchar_t* wc = new wchar_t[cSize];
mbstowcs (wc, c, cSize);
return wc;
}
Note that the calling code will then have to deallocate this memory, otherwise you will have a memory leak.
请注意，调用代码随后必须释放此内存，否则会出现内存泄漏。
回答by Maks
const char* text_char = "example of mbstowcs";
size_t length = strlen(text_char );
Example of usage "mbstowcs"
用法示例“mbstowcs”
std::wstring text_wchar(length, L'#');
//#pragma warning (disable : 4996)
// Or add to the preprocessor: _CRT_SECURE_NO_WARNINGS
mbstowcs(&text_wchar[0], text_char , length);
Example of usage "mbstowcs_s"
用法示例“mbstowcs_s”
Microsoft suggest to use "mbstowcs_s" instead of "mbstowcs".
Microsoft 建议使用“mbstowcs_s”而不是“mbstowcs”。
Links:
链接：
wchar_t text_wchar[30];
mbstowcs_s(&length, text_wchar, text_char, length);
回答by Greg Hewgill
You're returning the address of a local variable allocated on the stack. When your function returns, the storage for all local variables (such as wc) is deallocated and is subject to being immediately overwritten by something else.
您正在返回分配在堆栈上的局部变量的地址。当您的函数返回时，所有局部变量（例如wc）的存储空间将被释放，并且会立即被其他内容覆盖。
To fix this, you can pass the size of the buffer to GetWC, but then you've got pretty much the same interface as mbstowcsitself. Or, you could allocate a new buffer inside GetWCand return a pointer to that, leaving it up to the caller to deallocate the buffer.
要解决这个问题，您可以将缓冲区的大小传递给GetWC，但是您已经获得了与mbstowcs它几乎相同的接口。或者，您可以在内部分配一个新缓冲区GetWC并返回一个指向该缓冲区的指针，由调用者来释放缓冲区。
回答by Kerrek SB
Your problem has nothing to do with encodings, it's a simple matter of understanding basic C++. You are returning a pointer to a local variablefrom your function, which will have gone out of scope by the time anyone can use it, thus creating undefined behaviour(i.e. a programming error).
您的问题与编码无关，这是了解基本 C++ 的简单问题。您正在从您的函数返回一个指向局部变量的指针，当任何人都可以使用它时，该指针将超出范围，从而产生未定义的行为（即编程错误）。
Follow this Golden Rule: "If you are using naked char pointers, you're Doing It Wrong. (Except for when you aren't.)"
遵循这个黄金法则：“如果你使用裸字符指针，你就做错了。（除非你不是。）”
I've previously postedsome code to do the conversion and communicating the input and output in C++ std::stringand std::wstringobjects.
我之前已经发布了一些代码来进行转换并在 C++std::string和std::wstring对象中传达输入和输出。
回答by Gandalf458
I did something like this. The first 2 zeros are because I don't know what kind of ascii type things this command wants from me. The general feeling I had was to create a temp char array. pass in the wide char array. boom. it works. The +1 ensures that the null terminating character is in the right place.
我做了这样的事情。前 2 个零是因为我不知道这个命令想要我提供什么样的 ascii 类型的东西。我的总体感觉是创建一个临时字符数组。传入宽字符数组。繁荣。有用。+1 确保空终止字符在正确的位置。
char tempFilePath[MAX_PATH] = "I want to convert this to wide chars";
int len = strlen(tempFilePath);
// Converts the path to wide characters
int needed = MultiByteToWideChar(0, 0, tempFilePath, len + 1, strDestPath, len + 1);
