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Putting ints into CStrings

Posted on 2003-11-04
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Last Modified: 2010-04-02
How can i get my

int age

into this CString correctly. When the method is called like this cout << myClient.toString();
it prints funny characters instead of the the number stored in the int. How do i get it to "go in" correctly

CODE EXTRACT--------------------

CString Client::toString()
{
      CString retVal = "Name:\t"  + name + "\nAge\t:" + age + "\nSex:\t" + sex + "\nTelephone:\t" + telephone + '\n';
      return retVal;
}

Surely this is obvious to someone!!!

Thanks in advance!

Phill!
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Question by:Philluminati
2 Comments
 
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Expert Comment

by:grg99
ID: 9682836
You're probably getting the right answer :)

You're saying: Pls Mr Compiler, take the address of where you're going to put the string "Name:\t", then add that to my variable "name",
then add a bunch more addresses and integers together too.

In classsic C, you'd probably do something like:  
retVal = malloc(200);
sprintf( retVal, "Name:\t:%d\nAge:\t:%d ..... ",   name, age, ... );

In C++ you probably wan t to wrap all the integer variables with some function that converts them to CStrings, then the "+" operator will be interpreted as the overloaded CString concat operator, and things will work a bit better.    Except you're returning a local variable, which is unlikely to work very well in the long run.  











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LVL 15

Accepted Solution

by:
efn earned 50 total points
ID: 9683486
You may be able to use CString::Format.  It formats the string contents according to a format string like sprintf.  For example:

CString retval;
retval.Format( "Name:\t%s\nAge\t:%d\nSex:\t%c\nTelephone:\t%s\n", name, age, sex, telephone);

This only works if your data has types that sprintf knows.  You can put out a CString with a format specification of %s if you convert the corresponding argument.  For example, if name is a CString, you would match the %s with name.operator LPCTSTR().

Maybe I'm missing something, but I don't see any problem with returning a local variable by value.

--efn
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