cannot convert parameter 1 from 'char' to

2019-01-10 23:02发布

I keep getting this error: cannot convert parameter 1 from 'char' to 'LPCWSTR'

int main(int argc, char argv[])    
{

   // open port for I/O
   HANDLE h = CreateFile(argv[1],GENERIC_READ|GENERIC_WRITE,0,NULL,OPEN_EXISTING,0,NULL);

   if(h == INVALID_HANDLE_VALUE) {
       PrintError("E012_Failed to open port");

can someone help?

5条回答
放荡不羁爱自由
2楼-- · 2019-01-10 23:25

It should be

int main(int argc, char* argv[]) 

And

HANDLE h = CreateFileA(argv[1],GENERIC_READ|GENERIC_WRITE,0,NULL,OPEN_EXISTING,0,NULL);
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虎瘦雄心在
3楼-- · 2019-01-10 23:29

This is the main function that Visual Studio creates by default:

int _tmain(int argc, _TCHAR* argv[])

Where _TCHAR is defined to be char or wchar_t depending if _UNICODE is defined or not. The same thing happens with API functions. I would advise you against using explicit CreateFileA. Change your main and use CreateFile.

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何必那么认真
4楼-- · 2019-01-10 23:41

Go to the Properties for your Project and under Configuration Properties/General, change the Character Set to "Not Set". This way, the compiler will not assume that you want Unicode characters, which are selected by default:

alt text

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神经病院院长
5楼-- · 2019-01-10 23:41

I guess you're compiling with Unicode enabled. Then with char argv[], argv is a char array, so argv[1] is a char, and CreateFile wants a const wchar_t* as first parameter, not a char.

That said, your main definition is also broken, it should have char* argv[]. With that change, you can call CreateFileA.

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神经病院院长
6楼-- · 2019-01-10 23:43

Depending on your compiler setting for CharacterSet, you may need to perform a multibyte / widechar conversion, or change the CharacterSet if you don't care what it is.

For converting with MultiByteToWideChar, see the following...

http://www.codeguru.com/forum/showthread.php?t=231165

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