What prints this program?

#include <iostream>

using namespace std;

int main(int argc, char ** argv)
{
    int a = 4;
    int b = --a + --a;

    cout << b << endl;

}

my compiler says  4, is it a standard behaviour?




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On Mon, 1 Dec 2003 01:11:56 +0000 (UTC), nafabbio@tiscali.it ("Fabio
Napoli") wrote in comp.std.c++:

> What prints this program?
>
> #include <iostream>
>
> using namespace std;
>
> int main(int argc, char ** argv)
> {
>     int a = 4;
>     int b = --a + --a;
>
>     cout << b << endl;
>
> }
>
> my compiler says  4, is it a standard behaviour?

There is no standard behavior, your program generates undefined
behavior.  You have modified the value twice without a sequence point
in between.

--
Jack Klein
Home: http://JK-Technology.Com
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In article <bq3fn0$ukn$1@lacerta.tiscalinet.it>, Fabio Napoli
<nafabbio@tiscali.it> writes
>What prints this program?
>
>#include <iostream>
>
>using namespace std;
>
>int main(int argc, char ** argv)
>{
>    int a = 4;
>    int b = --a + --a;
>
>    cout << b << endl;
>
>}
>
>my compiler says  4, is it a standard behaviour?

Yes, but so is outputting 27, 289, 3.14 or absolutely anything else
including sending an email to your employer informing him that you write
programs that exhibit undefined behaviour.


--
Francis Glassborow      ACCU
If you are not using up-to-date virus protection you should not be reading
this. Viruses do not just hurt the infected but the whole community.

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