why exception in not caught in below code snippet?

When exception is thrown in a function and not handled there, it is passed to calling function where it can be handled...exceptoin occurs in A() which shd get handled by catch block in main which doesnt happen ....Why?

Code:

#include <iostream>
using namespace std;

class A
{
int i;int j;
public: 
A(int ii,int jj)
{
  i=ii;
  j=jj;
  if(i<0)
    throw exception(); // It should be passed in main function 

}


};

int main()
{
  A a1(-2,1);
  try
  {
  }
  catch(...) //It should actually catch the exception thrown by constructor 
  {             // but it doesn't happen...Pls clarify
    cout<<"Exception caught in main";
  }



}

You are creating the object outside of the try block. It should be:

Code:

try
{
    A a1(-2,1);
}

no it doesnt work

Originally Posted by forumuser

no it doesnt work

What didn't work? Did it not compile? Did the exception not get thrown?

Post your modified code.

didnt compile

This looks very similiar to another thread that you currently have open.

Originally Posted by forumuser

didnt compile

You're telling me that if you move the construction of the a1 variable into the try block that it won't compile? Sounds like you have a broken compiler, or you're not telling me the whole story.

I suspect that further down in the function (in code that is not shown here) a1 is being used, and since it was declared inside the try block, it's not visible outside it.

I could come to some conlusion after experimenting this.

When exception is thrown in a method and not immediately caught, terminate method is called.
But, if the exceptoin is caught and rethrown from CATCH BLOCK, it is propagated to the called method i.e. main method in this case...See belo code.....

If you have other views, pls do reply to this thread.

Code:

#include <iostream>
#include <string>
using namespace std;

class A
{
int i;int j;
public: 
A(int ii,int jj)
{
  i=ii;
  j=jj;
  if(i<=0)
	 throw exception();
    try
	{}
	catch(...)
	{
	 throw;
	}

};

int main()
{
  try
  {
	  A a1(-3,1); 
  
  }
  catch(...)
  {
    cout<<"Exception caught in main";
  }



}

You can safely remove

Code:

    try
	{}
	catch(...)
	{
	 throw;
	}

since that does absolutely nothing. A try block means that you try to execute some code (between the braces), and catch exceptions thrown by that code. An empty block cannot throw any exceptions.

Originally Posted by forumuser

When exception is thrown in a method and not immediately caught, terminate method is called.

If that is so, then you must always have a try/catch whenever you want to throw an exception, which is a rather silly requirement. Rather, this is what the C++ Standard says:

Originally Posted by C++03 Section 15.1 Paragraph 8

If no exception is presently being handled, executing a throw-expression with no operand calls terminate().

This is precisely what anon pointed out in post #10 as problematic in your example.

EDIT:
Oh wait, that is not quite right: since that throw is in a catch block, that should not be the problem. Rather, anon's analysis that the code is a no-op should be correct.