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How do I modify a C++ class to display which variable is uninitialized?

Posted on 2010-11-13
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Last Modified: 2012-05-10
How do I modify a C++ class to display which variable is unitialized?  My program accepts an expression and checks, among other things, that all variables are initialized.  Currently, it throws an error when there is an unitialized variable, but how to I get it to display that variable that is uninitialzed?

Example:

I enter: (x + y), x=2;

It throws an exception and displays:
Error: Uninitialzed variable.

I want it to display which variable is uninitialized:

Error: Uninitialized variable y
main.cpp
symboltable.cpp
variable.cpp
subexpression.cpp
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Question by:NSing9
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LVL 37

Expert Comment

by:TommySzalapski
ID: 34128525
Make your exception have a string constructor
Then do
	if (symbolTable.lookUp(name) == -1)		//
		throw UninitializedException(name);		// terminate function

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You didn't post see the code for UninitializedException, but I think you can figure it out from there.
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Author Comment

by:NSing9
ID: 34128622
I'm not sure how to do that.
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LVL 37

Expert Comment

by:TommySzalapski
ID: 34128676
Let me see the UninitializedException code.
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Author Comment

by:NSing9
ID: 34128737
See attached.
UninitializedException.h
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Accepted Solution

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TommySzalapski earned 1000 total points
ID: 34128793
Try this.
#include <stdexcept> // stdexcept header file contains runtime_error
#include <string>
using std::runtime_error; // standard C++ library class runtime_error
using std::string

class UninitializedException : public runtime_error
{
public:
	// constructor specifies default error message
	UninitializedException::UninitializedException()
		: runtime_error( "Uninitialized Variable" ) {}
	UninitializedException::UninitializedException(string name)
		: runtime_error( "Uninitialized Variable: " + name ) {}

}; // end class UninitializedException

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Expert Comment

by:TommySzalapski
ID: 34128797
Actually, the line probably needs to be

: runtime_error( ("Uninitialized Variable: " + name).c_str() ) {}

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LVL 37

Expert Comment

by:TommySzalapski
ID: 34128804
If name isn't a std::string type you'll need to adjust of course.
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Author Closing Comment

by:NSing9
ID: 34128875
Thank you!  It works after modifying the constructor.  Although I don't know C++ well, some of the basic programming concepts have escaped me after being away from it for so long.
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