c++ - “MessageBoxA”：无法将参数 2 从“std::vector<_Ty>”转换为“LPCSTR”

Question

以下代码有效：

void CMyPlugin8::myMessageBox(std::string& myString)
{
    myString = "Received the following string\n" + myString;
    char * writable = new char[myString.size() + 1];
    std::copy(myString.begin(), myString.end(), writable);
    writable[myString.size()] = '\0'; // don't forget the terminating 0 "delete[] writable;"

    int msgboxID = MessageBox(
        NULL,
        writable,
        "Notice",
        MB_OK
    );
    delete[] writable;
}

为了自动清理，我使用了以下信息：如何将 std::string 转换为 const char* 或 char*？.

以下代码引发错误：

void CMyPlugin8::myMessageBox(std::string& myString)
{
    myString = "Received the following string\n" + myString;
    std::vector<char> writable(myString.begin(), myString.end());
    writable.push_back('\0');

    int msgboxID = MessageBox(
        NULL,
        writable,
        "Notice",
        MB_OK
    );
}

我收到此错误： “MessageBoxA”：无法将参数 2 从“std::vector<_Ty>”转换为“LPCSTR”

Answer 1

你不能将向量作为 LPCSTR 传递，你必须。采用：

&writable[0]

或者：

writable.data()

反而。或者干脆使用myString.c_str()

Answer 2

MessageBox需要一个const char*. 为此，您无需先复制字符串。只需使用c_str：

void CMyPlugin8::myMessageBox(std::string& myString)
{
    myString = "Received the following string\n" + myString;
    int msgboxID = MessageBox(
        NULL,
        myString.c_str(),
        "Notice",
        MB_OK
    );
}

请注意，我认为您的 API 很差：您正在修改传入的字符串的值。通常调用者不会期望这样。我认为你的函数应该是这样的：

void CMyPlugin8::myMessageBox(const std::string& myString)
{
    std::string message = "Received the following string\n" + myString;
    int msgboxID = MessageBox(
        NULL,
        message.c_str(),
        "Notice",
        MB_OK
    );
}

Answer 3

void CMyPlugin8::myMessageBox(const std::string& myString)
{
    std::string message = "Received the following string\n" + myString;
    int msgboxID = MessageBox(
        NULL,
        message.c_str(),
        "Notice",
        MB_OK
    );
}

谢谢大家和@Falcon