error C2664: 'MessageBoxA' : cannot convert parameter 2 from 'std::string' to 'LPCSTR'

Question

I tried this in c++:

std::string teststring = "hello";
MessageBox(NULL,teststring,NULL, NULL);

The last parameter isn't even a pointer, yet you pass `NULL`. — chris, Sep 19 '13 at 12:06
@chris Well, NULL actually works. I don't have a problem with it if it works. — user10056, Sep 19 '13 at 16:44

score 4 · Accepted Answer · answered Sep 19 '13 at 12:02

4

First, it looks like Visual C++ so tag it properly.

You can get the inside buffer using c_str() method on a std::string, so your code becomes:

std::string teststring = "hello";
MessageBox(NULL,teststring.c_str(),NULL, NULL);

answered Sep 19 '13 at 12:02

Aamir

why visual C++? you could use any C++ compiler here. However answer is correct ) – Andrey Sep 19 '13 at 12:04
1

Yes, but normally in Windows environment when you are using a Windows API (MessageBox), most probably you are on Visual C++. That's why I said, 'it looks like' – Aamir Sep 19 '13 at 12:05

user1233963 · Answer 2 · 2013-09-19T12:23:30.360

1

MessageBox's second and third parameter expect a C string.

To get a C string from a std::string you call c_str(), therefor the correct way to call it is:

std::string teststring = "hello";
MessageBox(NULL, teststring.c_str(), NULL, NULL);

edited Sep 19 '13 at 12:23

answered Sep 19 '13 at 12:03

user1233963

score 0 · Answer 3 · answered Sep 19 '13 at 12:03

0

what about try this?

std::string teststring = "hello";
LPCSTR tmp = teststring .c_str()
MessageBox(NULL,tmp ,NULL, NULL);

answered Sep 19 '13 at 12:03

dhein

3 Answers3