Re: Unpredictable result of Increment operation

On Thu, 8 May 2008 17:12:15 -0700 (PDT), bintom
<binoythomas1108@[EMAIL PROTECTED]
> wrote in comp.lang.c++:

> Is there any reason why the following C++ code behaves as it does ?
> 
> 
> 
> int i;
> 
> i=1;  cout << (++i)++;           Output: 2
> 
> i=1;  cout << ++(++i);           Output: 3
> 
> i=1;  cout << (i++)++;           Output:  Error. LValue required
> 
> i=1;  cout << ++(i++);           Output: Error. LValue required

For build-in types, such as int, the behavior of the first two is
undefined.  You are attempting to modify the value of 'i' more than
once without a sequence point in between.  The output could be
anything, as far as the language is concerned.

The last two are errors because the post increment operator on a
built-in type does not yield an object, instead it yields the value
that the object had before the post increment.  So it is the same as
if you had written:

  cout << 1++;
  cout << ++1;

You can't apply the increment operators to a value.

I'd suggest you consult your C++ book about sequence points and
undefined behavior.

-- 
Jack Klein
Home: http://JK-Technology.Com
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