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operation on ‘numsels’ may be undefined

14 messages · Romain Francois, Kasper Daniel Hansen, Kevin Ushey +2 more

Original
cstrato
# 
Dear all,

Since many years the following C++ code does compile on ALL Bioconductor 
servers (Linux, Windows, Mac) without any warnings:

    Int_t numsels = 0;  //number of selected entries
    ...
    for (Int_t i=0; i<size; i++) {
       numsels = (arrMask[i] == 1) ? ++numsels : numsels;
    }//for_i

Even on the recently added release server 'zin2' Linux (Ubuntu 12.04.4 
LTS) the above code compiles w/o warnings.

However, on the new development server 'zin1' Linux (Ubuntu 12.04.4 LTS) 
I get suddenly the following warning message:

Found the following significant warnings:
   XPSPreProcessing.cxx:3026:56: warning: operation on ?numsels? may be 
undefined [-Wsequence-point]

Interestingly, both servers do not only run the same version of Ubuntu, 
but also the same version of the C++ compiler, i.e. g++ (Ubuntu/Linaro 
4.6.3-1ubuntu5) 4.6.3, and use the same flags, see:
http://bioconductor.org/checkResults/2.14/bioc-LATEST/zin2-NodeInfo.html
http://bioconductor.org/checkResults/devel/bioc-LATEST/zin1-NodeInfo.html

My question is now, why do I suddenly get the compiler warning?

The reason why I ask at R-devel and not Bioc-devel is that it may not 
only be a Bioc question, since I found the following links:
http://c-faq.com/expr/seqpoints.html
http://stackoverflow.com/questions/16838884/why-i-got-operation-may-be-undefined-in-statement-expression-in-c

I am not sure if I understand the meaning, but until now I have never 
got any warning from any compiler the I have used (including MS Visual 
C++).

Do I really have to replace '++numsels' with 'numsels+1'?

Best regards,
Christian
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C.h.r.i.s.t.i.a.n   S.t.r.a.t.o.w.a
V.i.e.n.n.a           A.u.s.t.r.i.a
e.m.a.i.l:        cstrato at aon.at
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Romain Francois
# 
Le 23 juin 2014 ? 15:20, cstrato <cstrato at aon.at> a ?crit :
This is confusing. I would write the loop body like this: 

numsels += (arrMask[i] == 1) ;


or preferably using the STL: 

Int_t numsels = std::count( begin(arrMask), end(arrMask), 1 ) ;

or some other variation of this, i.e. perhaps you don?t have a C++11 compiler, so perhaps one of these depending on what is arrMask:

Int_t numsels = std::count( arrMask.begin(), arrMask.end(), 1 ) ;
Int_t numsels = std::count( arrMask, arrMask + size, 1 ) ;

Romain
cstrato
# 
Dear Romain,

Thank you for your suggestions, I like especially the first one.

However, you did not explain why I have never got this warning message 
on any compiler, and why only one of the two identical Ubuntu compilers 
did give this warning message?

Best regards,
Christian
On 6/23/14 3:45 PM, Romain Fran?ois wrote:
Romain Francois
# 
Le 23 juin 2014 ? 18:28, cstrato <cstrato at aon.at> a ?crit :
I don?t know, but this:

numsels = ++numsels ; 

seems fishy to me, and so it keeps feeling weird with the addition of the ternary operator. 

There is obviously a difference of setup between these two machines, but I don?t have time to sherlock that for you. One of the compilers is getting more careful than the other. Getting warnings you did not get before is a good thing, as it helps you update the code with that new insight. 

Welcome to my world, I?m sometimes thrown all kinds of new warnings from esoteric compilers, all of them have value .

Romain
cstrato
# 
Dear Romain,

I do not know enough about compilers, but as far as I remember, they 
'work' from right to left, so numsels = ++numsels should not confuse the 
compiler. Anyhow I will change my code to your first suggestion since it 
is more elegant.

Best regards,
Christian
On 6/23/14 7:13 PM, Romain Fran?ois wrote:
cstrato
# 
Dear Kasper,

What do you mean with 'undefined aspect of ++'?

Every compiler has to evaluate first the expression on the right side 
and then apply the result to the variable on the left side, as in:
    i = i + 1;
I understand that the expression:
    i = i++;
may be confusing, but the expression:
    i = ++i;
should work. What about:
    i = (++i); or i = {++i}
Would this also result in a warning message?

Best regards,
Christian
On 6/23/14 10:05 PM, Kasper Daniel Hansen wrote:
Hervé Pagès
# 
Hi Christian,
On 06/23/2014 11:54 AM, cstrato wrote:
Not necessarily. So you should not rely on that. This is what the
(somewhat obscure) warning you see on zin1 is trying to tell you.
Personally I would just do:

     if (arrMask[i] == 1) numsels++;

which is the standard way to implement the "if (some condition)
then increment counter" idiom.

As Kasper mentioned, a similar issue was recently discussed here:

   https://stat.ethz.ch/pipermail/bioc-devel/2014-June/005858.html

Cheers,
H.

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Hervé Pagès
    

    

      
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On 06/23/2014 03:18 PM, Herv? Pag?s wrote:

            

      
      
      
    

        
Actually, I don't see an ambiguity in your code:

   numsels = (arrMask[i] == 1) ? ++numsels : numsels;

Yes it's confusing and unnecessarily complicated but I don't see that
it relies on some undefined behavior. It's not like in the thread on
Bioc-devel where the expression was:

   *p++ = tolower(*p);

In that case the left-value of the assignment is itself an expression
that needs to be evaluated and the outcome of the assignment depends
on whether the left-value is evaluated before the right expression or
not. But in your case the left-value is a variable name so there is
nothing to evaluate.

So I don't know. Looks like a false positive from the gcc compiler to
me. Someone on the list might have a better insight.

Cheers,
H.

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        
      

    

    

      
      

        


  

        

      Kevin Ushey
    

    

      
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I don't see what's so surprising here.

That statement is identical to writing:

    if (arrMask[i] == 1) {
        numsels = ++numsels;
    } else {
        numsels = numsels;
    }

and

    numsels = ++numsels;

has two statements modifying the value of numsels (= and prefix-++) in
a single sequence point. (Do we increment then assign, or assign then
increment? The C / C++ standards leave this undefined.)

Imagine writing the operations out as functions: we have the `=`
function, and the `prefix-++` function -- both of these 'modify' (one
of) their arguments. Do we evaluate it as `=`(a, `prefix-++`(a)) or
`prefix-++`(`=`(a, a))? The C standard leaves this undefined, so
compilers are free to do what they wish (and the nice ones warn you
when there is such an ambiguity). I guess the net result of the
operation is the same in each case _here_, but this is of course not
the case for other functions that modify the value of their
operand(s). And, in truth, this is _undefined behaviour_ and so the
compiler could still rightly make demons fly out of your nose if it
wanted to upon program execution.

I highly recommend reading the slides at
http://www.slideshare.net/olvemaudal/deep-c, especially the bit on
sequence points.

Cheers,
Kevin

On Mon, Jun 23, 2014 at 9:22 PM, Kasper Daniel Hansen

        
<kasperdanielhansen at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      cstrato
    

    

      
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Dear all,

Thank you very much for this interesting discussion, I appreciate it.

I think that Kevin gave a very good explanation why my code is 
ambiguous, although it is not clear to me why the C/C++ standard leaves 
this case undefined. One last question:
If I would write
    numsels = {++numsels;}
Is this also undefined or is this then allowed?

Best regards,
Christian

        
On 6/24/14 7:28 AM, Kevin Ushey wrote:

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

        

      
      
    

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Hervé Pagès
    

    

      
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Hi Kevin,

        
On 06/23/2014 10:28 PM, Kevin Ushey wrote:

            

      
      
      
    

        
As the former? Isn't it that `prefix-++` has precedence over `=`? Also
according to the standard, `prefix-++` must perform the increment
*before* returning the incremented value. So I'm still a little bit
confused about this warning. Although I would understand it if
`suffix-++` was used instead of `prefix-++`:

     numsels = numsels++;

In that case the incrementation happens *after* the right value
of the assignment is evaluated but it's undefined whether it will
happen before or after the assignment itself (on my machine it
happens after).

            

      
      
      
    

        
Nice slides. Pretty entertaining! The numsels = numsels++ case is
actually discussed at slide 142. Thanks for the link!

H.