Re: vc 6.0's bug?????

From:

"Greg Herlihy" <greghe@pacbell.net>

Newsgroups:

comp.std.c++

Date:

30 Apr 2006 18:40:05 GMT

Message-ID:

<1146414124.942160.78660@g10g2000cwb.googlegroups.com>

James Dennett wrote:

Greg Herlihy wrote:

Momchil Velikov wrote:

Alf P. Steinbach wrote:

* fisker0303@126.com:

In vc 6.0:

#include <iostream>
using namespace std;

int main()
{
        int a = 10;
        int b = 20;
        a = (a + b) - (b = a);
        cout << "a=" << a << ",b=" << b << endl;
        return 0;

}

Release output : a ,b=10
Debug output: a=10,b=10

If it were 'a' being modified, the behavior would not (IMO) be merely
unspecified, but undefined, which is much worse.

The behavior is *undefined*, because the expression "a = (a + b) - (b =
a);"
violates "5. Expressions" [#4]:

"... Furthermore, the prior value shall be accessed only to determine
      the value to be stored. The requirements of this paragraph shall
     be met for each allowable ordering of the subexpressions of a full
     expression; otherwise the behavior is undefined."

No, the behavior is merely unspecified - not undefined. In other words,
the variable "a" is certain to have either the value 20 or the value 10
after the expression is evaluated.

The expression would have to store a value in either "a" or "b" more
than once for it to have undefined behavior.

It writes to b, and also reads from it for a purpose other
than that to determine the value to be stored. By the quote
above, the behavior is undefined. It is *not* necessary for
there to be two writes to an object between sequence points
for there to be undefined behavior (though it is sufficient).

In that case, evaluating this expression:

a = b = 0;

must also lead to undefined behavior - because it too reads from b "for
a purpose other than that to determine the value to be stored". And
yet this kind of expression appears so frequently in C++ programs that
it is difficult to imagine that its behavior is, in fact, undefined.

Greg

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