C++ 如何将 char* 转换为 LPCWSTR？
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How to convert char* to LPCWSTR?
提问by sigil
I know this has already been discussed in several questions on SO, but none of those solutions have worked for me.
我知道这已经在关于 SO 的几个问题中讨论过，但是这些解决方案都没有对我有用。
I start with a char*because this is for a DLL that will be called from VBA, and char*is necessary for VBA to pass a string to the DLL.
我从 a 开始，char*因为这是一个将从 VBA 调用的 DLL，并且char*是 VBA 将字符串传递给 DLL 所必需的。
I need to return a LPCWSTRbecause that's the input parameter for the API function I'm trying to call, and I can't enable casting by switching from Unicode to multi-byte character set in the Properties window, because the API has this code:
我需要返回 aLPCWSTR因为这是我尝试调用的 API 函数的输入参数，并且我无法通过在属性窗口中从 Unicode 切换到多字节字符集来启用转换，因为 API 具有以下代码：
#if !defined(UNICODE) && !defined(NOUNICODE)
#error UNICODE is not defined. UNICODE must be defined for correct API arguments.
#endif
I tried this:
我试过这个：
LPCWSTR convertCharArrayToLPCWSTR(char* charArray)
{
const char* cs=charArray;
wchar_t filename[4096] = {0};
MultiByteToWideChar(0, 0, cs[1], strlen(cs[1]), filename, strlen(cs[1]));
}
which gave these errors:
这给出了这些错误：
error C2664: 'strlen' : cannot convert parameter 1 from 'const char' to 'const char *'
error C2664: 'MultiByteToWideChar' : cannot convert parameter 3 from 'const char' to 'LPCCH'
I tried this (same function header), loosely adapted from this post:
我试过这个（相同的函数头），松散地改编自这篇文章：
size_t retVal;
const char * cs = charArray;
size_t length=strlen(cs);
wchar_t * buf = new wchar_t[length](); // value-initialize to 0 (see below)
size_t wn = mbsrtowcs_s(&retVal,buf,20, &cs, length + 1, NULL);
return buf;
This compiled ok, but when I passed it an example string of "xyz.xlsx", mbsrtowcs_s()set bufto an empty string: L""
这编译没问题，但是当我传递给它一个示例字符串“xyz.xlsx”时，mbsrtowcs_s()设置buf为一个空字符串：L""
So, how do I make this conversion?
那么，我该如何进行这种转换？
采纳答案by cHao
Since csis a const char*, cs[1]is a const char. C++ won't convert it to a pointer for you, because in most cases that doesn't make sense.
由于cs是const char*，cs[1]是const char。C++ 不会为您将其转换为指针，因为在大多数情况下这没有意义。
You could instead say &cs[1]or cs+1if the intent is to skip the first char. (That's what you're doing when you pass a pointer to the 1th element; in C++, indexes start at 0.) If the intent is to pass the whole string, then just pass cs.
你可以说&cs[1]或者cs+1如果意图是跳过第一个字符。（这就是传递指向第 1 个元素的指针时所做的；在 C++ 中，索引从 0 开始。）如果意图传递整个字符串，则只需传递cs.
回答by sigil
Following Hans Passant's advice regarding pointers to local variables, I worked out this approach, which seems to work well:
遵循 Hans Passant 关于局部变量指针的建议，我制定了这种方法，它似乎工作得很好：
wchar_t *convertCharArrayToLPCWSTR(const char* charArray)
{
wchar_t* wString=new wchar_t[4096];
MultiByteToWideChar(CP_ACP, 0, charArray, -1, wString, 4096);
return wString;
}
I'm aware that the use of newrequires memory management, which I perform in the function that calls this one.
我知道使用new需要内存管理，我在调用这个的函数中执行。
