Thread: [Mingw-users] Question about increment..

Hi,

I recently to try some toy things in DevC++ (version I am using is 4.9.9.2)

Hi,

I recently to try some toy things in DevC++ (version I am using is 4.9.9.2)

The code I had was this:

    int a = 1, b = 2;
    int c = a++ + b++ + a + b;

    cout << "a = " << a << endl;
    cout << "b = " << b << endl;
    cout << "c = " << c << endl;

I expected the c value to be 8 (by operator precedence, ++ has higher
precedence than +, so do in order:
a++, then b++
then do (a++ + b++)
then add the previous result and a
then add the previous result and b

but the answer I got was 6..

what might be the reason for this..??

thanks, murali

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On 18.09.2012 20:28, Mani wrote:
> Hi,
> 
> I recently to try some toy things in DevC++ (version I am using is
> 4.9.9.2)
> 
> The code I had was this:
> 
> int a = 1, b = 2; int c = a++ + b++ + a + b;
> 
> cout << "a = " << a << endl; cout << "b = " << b << endl; cout <<
> "c = " << c << endl;
> 
> I expected the c value to be 8 (by operator precedence, ++ has
> higher precedence than +, so do in order: a++, then b++ then do
> (a++ + b++) then add the previous result and a then add the
> previous result and b
> 
> but the answer I got was 6..
> 
> what might be the reason for this..??
> 
It's called "undefined behaviour" [1].
There's a good article [2] about this. Sadly, it's in Russian, but
there's an image [3] in that article that neatly illustrates the point.

[1] http://en.wikipedia.org/wiki/Undefined_behavior
[2] http://lurkmore.to/%2B%2Bi_%2B_%2B%2Bi
[3] http://lurkmore.so/images/9/95/Cpp.png

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Hi,

The answer is that your code describes undefined-behavior for C (C++)
language.  If you take a closer look to C specification, you will
learn that the pre/post-increment|decrement operator has to be
executed before next sequence-point.  Within your calculation, the
sequence-point is the assignment operator, so order of execution of
your code is undefined.

Regards,
Kai

On 18/09/12 17:48, Kai Tietz wrote:
> The answer is that your code describes undefined-behavior for C (C++)
> language.  If you take a closer look to C specification, you will
> learn that the pre/post-increment|decrement operator has to be
> executed before next sequence-point.

This is correct.

> Within your calculation, the sequence-point is the assignment operator,

This is incorrect.  The OP's original expression:

     int c = a++ + b++ + a + b;

has only one sequence point -- the terminal semicolon.

> so order of execution of your code is undefined.

And again, correct.  The sequence is complete when the assignment to 
variable c and the two side effects of post-incrementing variables a and 
b have all been completed.  However, each of the three effects is 
independent of the other two, and the order in which they must be 
completed is undefined.

-- 
Regards,
Keith.

2012/9/18 Keith Marshall <kei...@us...>:
> On 18/09/12 17:48, Kai Tietz wrote:
>> The answer is that your code describes undefined-behavior for C (C++)
>> language.  If you take a closer look to C specification, you will
>> learn that the pre/post-increment|decrement operator has to be
>> executed before next sequence-point.
>
> This is correct.
>
>> Within your calculation, the sequence-point is the assignment operator,
>
> This is incorrect.  The OP's original expression:
>

This is correct.  Please see in Annex C the following description.
 ... "— The end of a full expression: an initializer (6.7.8); the
expression in an expression
statement (6.8.3); the controlling expression of a selection statement
(if or switch)
(6.8.4); the controlling expression of a while or do statement
(6.8.5); each of the
expressions of a for statement (6.8.5.3); the expression in a return statement
(6.8.6.4)."

>      int c = a++ + b++ + a + b;

We have here an initializer (6.7.8).

> has only one sequence point -- the terminal semicolon.

Well, the terminal semicolon itself isn't a sequence-point, but of
course it describes the end of the expression.

Kai

Forget about initializers ... misread it.

Correct, the sequence-point here is the expression itself, which is
terminated by the semicolon.

Kai

Hi,

I was wondering if you had any comments as to how we can explain the
below.. thanks, murali.


---------- Forwarded message ----------

Hi,

I recently to try some toy things in DevC++ (version I am using is 4.9.9.2)

The code I had was this:

    int a = 1, b = 2;
    int c = a++ + b++ + a + b;

    cout << "a = " << a << endl;
    cout << "b = " << b << endl;
    cout << "c = " << c << endl;

I expected the c value to be 8 (by operator precedence, ++ has higher
precedence than +, so do in order:
a++, then b++
then do (a++ + b++)
then add the previous result and a
then add the previous result and b

but the answer I got was 6..

what might be the reason for this..??

thanks, murali

On Thu, Sep 20, 2012 at 12:53 PM, Mani wrote:
> Hi,
>
> I was wondering if you had any comments as to how we can explain the
> below.. thanks, murali.
>

This was responded to by many, are you receiving list mail?

-- 
Earnie
-- https://sites.google.com/site/earnieboyd

thanks for pointing out.. I had not received the original responses..
Probably I changed my settings..
In any case, thanks for all the responses.. they were very helpful..

best, murali.

On Thu, Sep 20, 2012 at 2:06 PM, Earnie Boyd
<ea...@us...> wrote:
> On Thu, Sep 20, 2012 at 12:53 PM, Mani wrote:
>> Hi,
>>
>> I was wondering if you had any comments as to how we can explain the
>> below.. thanks, murali.
>>
>
> This was responded to by many, are you receiving list mail?
>
> --
> Earnie
> -- https://sites.google.com/site/earnieboyd