C++, increment operator

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C++

int k, n = 2;
k = (k = n + 5)++;
cout << k << '\n';
k = ++(k = n + 5);
cout << k << '\n';

I thought the 1st one is 7 and the 2nd is 8. But the output is surprisingly both 8.
Anyone can tell me about this? Thx in advance.

Posted 20-Apr-12 1:36am

Krauze

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Krauze 20-Apr-12 7:48am

Thanks, P. Salini

P.Salini 20-Apr-12 7:53am

welcome Krauze

2 solutions

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Philippe Mori · Accepted Answer · 2012-04-21T03:02:00

Solution 2

In C++, it is undefined as is i = i++ since i is modified twice in the same expression and the order is not defined as operator= is not a sequence point.

See http://en.wikipedia.org/wiki/Sequence_point[^]

Posted 21-Apr-12 3:02am

Philippe Mori

Comments

Lakamraju Raghuram 21-Apr-12 9:13am

I agree with this.
Statements like k = (k = n + 5)++; looks fancy but no serious developer should not ever use them as many ambitiousness are involved

[no name] 21-Apr-12 9:27am

Are you saying that the result of k = (k = n + 5)++; is undefined? Don't the enclosing brackets define the order of evaluation?

Philippe Mori 21-Apr-12 9:57am

Yes, check the Wikipedia link I gave above. Essentially, the compiler will not take into account that it is the same variable k that is modified twice in the same expression. Thus the compiler might store the result of k=n+5 in a temporary variable. Increment k and do the assigment or assign the result twice to k before doing the increment for example. It will give the expected result except in the case where the same variable is modified multiple time (either directly or through an alias) where the result is undefined (but in practice, it will probably be either 7 or 8).

[no name] 21-Apr-12 10:06am

I agree with another poster that it is not nice to see this kind of expression. But if it is dangerous why doesn't the compiler issue a warning? It is very easy to pick up this condition.

Philippe Mori 21-Apr-12 10:51am

Well, simple cases would be easy for the compiler to give a warning... but the you just have to suppose you have a function say f(int &i, int &j, int &k) { i = ++j + k++; }. It would normally not be a problem except if the caller call it with the same variable for multiple argument in which case you will still have the problem.

By the way, expression like k = i++ + ++i are also undefined as it is not defined which subexpression is evaluated first (and when intermediate results are loaded and stored).

[no name] 21-Apr-12 11:27am

Your answers are very interesting. Can't vote for them but have 5'd your solution. It is always good not just to know something is bad practice but why.

Philippe Mori 21-Apr-12 10:04am

Parenthesis only tell the compiler that k=n+5 is done first and that the increment will apply to that expression as a whole.

By the way, k = k++ or k = ++k are undefined as any variation that modify the same variable more than once in the same expression (without defined sequence point).

[no name] 21-Apr-12 10:16am

Thanks - I see it now.

P.Salini · Accepted Answer · 2012-04-20T01:43:00

Solution 1

The 1st statement also returning 8 because
when the execution of this statement is completed

k = (k = n + 5)++;

then the postincrement also takes place and you are getting result as 8.

if you print (k = n + 5)++ then the output will be 7.

Posted 20-Apr-12 1:43am

P.Salini

Comments

sravani.v 20-Apr-12 7:53am

my 5!

P.Salini 20-Apr-12 7:54am

Thank you sravani

Prasad_Kulkarni 20-Apr-12 8:04am

Good one +5!

P.Salini 20-Apr-12 23:46pm

Thank you Prasad

Philippe Mori 21-Apr-12 9:03am

This explanation is not correct. See my solution.

P.Salini 23-Apr-12 0:41am

Hi Mori, I had gone through the link provided by you, but i am not able to understand the exact concept. can you provide other link which is more descriptive.Is there any similar concept in .net?