other/9693: sequence point consistency

[lxg8906@yahoo.com]

>Number:         9693
>Category:       other
>Synopsis:       sequence point consistency
>Confidential:   no
>Severity:       non-critical
>Priority:       medium
>Responsible:    unassigned
>State:          open
>Class:          sw-bug
>Submitter-Id:   net
>Arrival-Date:   Thu Feb 13 17:06:01 UTC 2003
>Closed-Date:
>Last-Modified:
>Originator:     Frank G
>Release:        g++ 3.2
>Organization:
>Environment:
Solaris 2.8 Sparc9
>Description:

>How-To-Repeat:

>Fix:

>Release-Note:
>Audit-Trail:
>Unformatted:
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Dear Sir/Madam,

The version I am using is g++ 3.2, platform is Solaris 2.8.
<program 1>
int main()
{
  int i=2;
  int j=(++i)+(++i);
  cout<<j<<endl;
  return 0;
}

The output is 7.  It seems that (++i) is evaluated first, yields 3, increments, then 3+4=7.  I am a little confused here. C++ std 5.1/4 reads: "Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expresion..."  Should we have i modified twice here?   Or this statement is just undefined? Solaris CC gives j=8 which I think is more reasonable.


Consider the following <program 2> 

int& foo(int& i)
{
  return ++i;
}

int main()
{
  int i=2;
  int j=foo(i)+foo(i);
  cout<<j<<endl;
  return 0;
}

foo() is just an in-place increment function, g++ gives j=8.  Is it that 
j=foo(i)+foo(i) is the same as j=(++i)+(++i) ? why they give two different results?  


If I wrap the integer in a class, as follows, 
<program 3>

#include <iostream>
using namespace std;

class T
{
public:
  int m_b;
  T(int i=0): m_b(i) {}
  T& operator ++(void) {  ++m_b; return *this;}
  T operator +(cosnt T& rhs) { m_b+=rhs.m_b; return *this;}
  void foo(void) {cout<<m_b<<endl;}
};
  
int main()
{
  T t1(2);
  T t2=(++t1)+(++t1);
  t2.foo();
  return 0;
}

The result is j=8.  

In the program 2 and 3, if we change the return type in foo() and operator++ to T&, j=7.

Is it reasonble to have program 1-3 have different results when they in fact are doing the same thing?

Thank you.


Frank G.