Re: Guarantee of side-effect free assignment

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.std.c++
Date:
Sun, 7 Oct 2007 12:57:53 CST
Message-ID:
<1191752016.820149.141930@w3g2000hsg.googlegroups.com>
On Oct 7, 2:21 am, jdenn...@acm.org (James Dennett) wrote:

Alf P. Steinbach wrote:

From discussions in [comp.lang.c++] and [comp.lang.c++.moderated], as
well as articles on the net about concurrency in C++, I'm reasonably
sure that given

  #include <iostream>
  #include <ostream>

  struct S { S(){ throw 123; } int foo(){ return 666; } };

  int main()
  {
      S* p = 0;

      try
      {
          p = new S();
      }
      catch( ... )
      {}

      if( p ) { std::cout << p->foo() << std::endl; }
  }

there is no guarantee that this code will not end up in a call to
p->foo() with an invalid pointer p, i.e., that might well happen.

Surely that couldn't have been the committee's intention?


I wouldn't imagine so.

Why isn't assignment treated as a function call?


It doesn't need to be. The assignment cannot occur until the
new value is known, which means that the "new" operator
has returned its result, which means that the object has been
constructed.


The construction of the object is a side effect. The compiler
can use the new value as soon as it knows it. All that the
standard requires is that side effects occur before the next
sequence point.

If the constructor throws, there's no value from "new" above,
and the assignment cannot occur; p will remain null.


How is this any different from the compiler generating the
actual assignment in ++i after it uses the value?

--
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