Solved

char to CString

Posted on 2002-03-05
8
4,694 Views
Last Modified: 2012-05-04
How can I convert a array of chars to a CString?
for example:
char* buf[50]
CString myString;

myString = myString + charTostr(buf);
0
Comment
Question by:el_rooky
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
8 Comments
 
LVL 4

Expert Comment

by:pagladasu
ID: 6842611
myString = buff; //converts char* to CString;

0
 

Author Comment

by:el_rooky
ID: 6842708
When I do that I get the following error:

error C2679: binary '=' : no operator defined which takes a right-hand operand of type 'char *[2052]' (or there is no acceptable conversion)

I am using VC++6 and MFC
thanks
0
 
LVL 3

Expert Comment

by:jcgd
ID: 6842818
myString.Format("%s",buf);
0
Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 6

Expert Comment

by:thienpnguyen
ID: 6842830


    char buf[50]
    CString myString;
    ...

    myString = buf;

 
or

    char *buf[50] // array of pointer
    CString myString;
    ...

    myString = *buf;

0
 

Expert Comment

by:dchan_4544
ID: 6842906
First of all, are you trying to convert an array of chars or an array of "strings" ? (according to your example, you have char* buf[50], which is an array of strings (buf[50] is a string or array of characters)).  So, if you are just trying to convert an array of characters, then maybe you can do this:

char buf[50];
CString mystring;

strncpy(buf, "Bla Bla", sizeof(buf));
mystring = buf;

If you have an array of "strings" (array of array of characters), then maybe you can do this:

char* buf[50];
CString mystring;

buf[0] = new char[50];
strncpy(buf[0], "bla Bla", 50);
mystring = buf[0];
delete buf[0];

Hope that helps.
0
 
LVL 30

Expert Comment

by:Axter
ID: 6842917
You should always initialize your variables when you declare them.

For example:
const char* buf = "Hello World";
CString myString = "Hey";

myString = myString + buf;

AfxMessageBox(myString);
0
 
LVL 4

Accepted Solution

by:
pagladasu earned 50 total points
ID: 6842981
Change the declaration of
char *buf[50];
to
char buf[50];
0
 

Author Comment

by:el_rooky
ID: 6844041
Thanks, I wasn't thinking yesterday.
0

Featured Post

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Errors will happen. It is a fact of life for the programmer. How and when errors are detected have a great impact on quality and cost of a product. It is better to detect errors at compile time, when possible and practical. Errors that make their wa…
Basic understanding on "OO- Object Orientation" is needed for designing a logical solution to solve a problem. Basic OOAD is a prerequisite for a coder to ensure that they follow the basic design of OO. This would help developers to understand the b…
The goal of the tutorial is to teach the user how to use functions in C++. The video will cover how to define functions, how to call functions and how to create functions prototypes. Microsoft Visual C++ 2010 Express will be used as a text editor an…
The viewer will learn how to use the return statement in functions in C++. The video will also teach the user how to pass data to a function and have the function return data back for further processing.

738 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question