Need help with Macros

[ngoyal1987]

Output of the following program for "J" should be ++y * ++y = 2 * 3 = 6. But when I execute this program, I get Output as 9. I tried on several compilers (GNU GCC).
Can someone please help me out with this

#include<iostream>
using namespace std;
#define DOUBLE(X) X*X

int main(){
    int y=1;
    int j = DOUBLE(++y);
    cout<<"J is -> "<<j;
    return 0;
}

[Ganado]

You don't want to do something like that, doing ++y * ++y is undefined behavior because you don't know in what order the program will modify the values.

In general, in order to be clear, you should increment y on its own line.

[Chervil]

The compiler may issue a warning, such as:
[Warning] operation on 'y' may be undefined [-Wsequence-point]

The behaviour is undefined. See for example:
http://www.cplusplus.com/forum/beginner/104484/

[ne555]

> ++y * ++y
that's undefined behaviour
http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points

You cannot fix the macro, but given that you are using c++ you may utilize a template instead

template <class T>
T square(T x){
   return x*x;
}

Your macro would also fail in the case DOUBLE( y+1 ) because of precedence
(remember that is simply text substitution)

[Jaybob66]

++ takes precedence over * so both sides get ++'ed before the *

I disagree this is a sequence point issue, it looks like simple operator precedence to me.


EDIT:

After reading ne555's link, I've changed my mind on the sequence point issue. Because it modifies y twice it is definitely a sequence point issue.

However, i stand by my answer to the question as to whats happening.

Take this example:
int y=1; int i = ++y + ++y * ++y;

gcc 4.3 and vs2010 equated to int i = 4 + 2 * 3; = 10

and JS and PHP equated to int i = 2 + 3 * 4; = 14

I think this is because interpreted languages manifest their side effects at parse time, while compiled languages manifest their side effects at run time.

[Chervil]

It really isn't related to interpreted versus compiled languages. Undefined behaviour is undefined behaviour, it's no use trying to deduce rules, since each compiler may give a different result.

(See for example the post I linked previously where different compilers give a variety of differing outputs).

[ngoyal1987]

Thanks a lot to all of you for such a quick responses. I will try to avoid this undefined behavior.

@jaybob66 - I agree to your point about operator precedence, which is true in the example which I posted. But after reading all the other replies, I think problem is not with macros, its with pre and post fix operation execution by compilers. See below example O/P for cout operator

<< i   << i++               => 2 1
<< i++ << i                 => 1 2
<< i++ << i++               => 2 1
<< i++ << i   << i          => 1 2 2
<< i++ << i++ << i          => 2 1 3
<< i++ << i++ << i++        => 3 2 1
<< i++ << i   << i   << i   => 1 2 2 2
<< i++ << i++ << i   << i   => 2 1 3 3
<< i++ << i++ << i++ << i   => 3 2 1 4
<< i++ << i++ << i++ << i++ => 4 3 2 1

after analyzing 1st line, I thought cout operator executes from right to left. Means first i++ is processed, then i which results in 2 1 as o/p.
But second o/p doesn't agree with the above logic.

[ne555]

> See below example O/P for cout operator
All those are undefined behaviour.

> I thought cout operator executes from right to left
the function arguments may be evaluated in any order

> I think problem is not with macros
macros are evil
http://www.parashift.com/c++-faq/inline-vs-macros.html