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From: "Andy Salnikov" <A.A.Salnikov@inp.nsk.su>
Newsgroups: comp.std.c++
Subject: Re: try block as main block of a function
Date: Tue, 31 Jul 2001 17:05:44 GMT
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"Francis Glassborow" <francis.glassborow@ntlworld.com> wrote in message
news:hSy9obBruRZ7EwIG@ntlworld.com...
> That is my point, they do not. If you wrap a function in a try block and
> an exception is thrown the function does not return after processing an
> exception but rethrows.
>
  This sounds strange and almost unreasonable. Where is the place in
standard which says so?

  The following simple program does what I would expect, i.e. prints both
lines to cout:

#include <iostream>
#include <stdexcept>

void f()
try {
  throw std::logic_error("no logic here") ;
} catch ( const std::exception& e ) {
  std::cout << "exception caught: " << e.what() << std::endl ;
}

int main(int,char**)
{
  f() ;
  std::cout << "returned from f()" << std::endl ;
}

Cheers,
Andy.


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