Question

I am trying to learn a little c++ and I have a silly question. Consider this code:

TCHAR tempPath[255];
GetTempPath(255, tempPath);

Why does windows need the size of the var tempPath? I see that the GetTempPath is declared something like:

GetTempPath(dword size, buf LPTSTR);

How can windows change the buf value without the & operator? Should not the function be like that?

GetTempPath(buf &LPTSTR);

Can somebody provide a simple GetTempPath implementation sample so I can see how size is used?

EDIT:

Thanks for all your answers, they are all correct and I gave you all +1. But what I meant by “Can somebody provide a simple GetTempPath implementation) is that i have tried to code a function similar to the one windows uses, as follow:

void MyGetTempPath(int size, char* buf) 
{
 buf = "C:\\test\\";
}

int main(int argc, char *argv[])
{
    char* tempPath = new TCHAR[255];
    GetTempPathA(255, tempPath);
    MessageBoxA(0, tempPath, "test", MB_OK);
    return EXIT_SUCCESS;
}

But it does not work. MessageBox displays a “##$’ string. How should MyGetTempPath be coded to work properly?

Answer 1

Windows needs the size as a safety precaution. It could crash the application if it copies characters past the end of the buffer. When you supply the length, it can prevent that.

Array variables work like pointers. They point to the data in the array. So there is no need for the & operator.

Not sure what kind of example you are looking for. Like I said, it just needs to verify it doesn’t write more characters than there’s room for.