Please determine the output

led mike

Sushant Duggal wrote:

can you tell me the output for C++ and Java compiler and whats the reason for the answer.

Yes I can... but then I would be doing your class assignment for you. X|

led mike

Christian Graus

Sushant Duggal wrote:

cout< This code is making some assumptions that are now being shown for it to compile as C++. It's not going to compile for Java. Beyond that, I agree with the other reply, you need to do your own homework.

Christian Graus - Microsoft MVP - C++ Metal Musings - Rex and my new metal blog

Sushant Duggal

Thanks for your reply friends, But I just want to know the reason why the outputs differ. i In C++, The result is x=18.y=7 where as in Java, The result is x=19, and y=7 I want to know the reason for this output. Is there a difference of stack implementation in the two languages or there is some other reason.. Please let me know. Thanks,

Sushant Duggal.

Christian Graus

++x guarentees that the number is incremented before it's returned. x++ guarentees it's incremented after, but not how LONG after.

Christian Graus - Microsoft MVP - C++ Metal Musings - Rex and my new metal blog

Sushant Duggal

Yes, I agree on your comment. so do you want to say that there is no method to determine the value?

Sushant Duggal.

Sushant Duggal

OK now i got the point. its correct to get 19 in C++, but I still didnt understand why it comes 18 in java. Can you comment on this? Thanks

Sushant Duggal.

Christian Graus

I believe it goes like this: The ++ operator occurs in both languages AFTER x is given it's final value. The difference is that in Java, the values are by reference, so the value that x is assigned to, is the value that still has a ++ operator outstanding. so, x gets set to 18. y gets incremented, to 7, and x also gets incremented, but it's the same x that is 18, not a copy of x which is 16.

Christian Graus - Microsoft MVP - C++ Metal Musings - Rex and my new metal blog

Christian Graus

Oh - you also posted in the wrong forum, I didn't notice before.

Christian Graus - Microsoft MVP - C++ Metal Musings - Rex and my new metal blog

Sushant Duggal

Then where I would have posted this?

Sushant Duggal.

Christian Graus

Managed C++/CLI is the forum for discussion of .NET C++, as it clearly states. This[^] is the general C++ forum.

Christian Graus - Microsoft MVP - C++ Metal Musings - Rex and my new metal blog

Sushant Duggal

The ++ operator occurs in both languages AFTER x is given it's final
value. The difference is that in Java, the values are by reference, so
the value that x is assigned to, is the value that still has a ++ operator outstanding. so, x gets set to 18. y gets incremented, to 7,
and x also gets incremented, but it's the same x that is 18, not a
copy of x which is 16.

I can understand that in java, the values are by reference. that means X is incremented direectly where it is stored, it sets it to 12 + 6 = 18. Confusion ..... when the X is going to incremented? and why it is not reflected(X's value is still 18). Thanks For your Patience, Sushant Duggal. -- modified at 3:39 Thursday 28th September, 2006

Sushant Duggal.

Nish Nishant

The statement i = i++ has undefined behavior in C++ and different compilers can generate different code and still be standards compliant. The = operator does not define a separate sequence point, and C/C++ rules dictate that an expression can modify an object's value only once within a single sequence point - else the compiler's free to interpret the code in whatever way it wants to.

Regards, Nish

Nish’s thoughts on MFC, C++/CLI and .NET (my blog)
Currently working on C++/CLI in Action for Manning Publications. Also visit the Ultimate Toolbox blog