The bug is in the documentation.
All of the compilers are right. The complier can choose to evaluate
i=2 before _1+i which results in the lambda functor
_1 + 2
or _1+i before i=2 which gives
_1 + 1
Anyway, the point of the documentation was to say that the lambda functor
stores the value of i at the time of creating the lambda functor, and
the actual argument to later replace _1 will be passed by reference.
So no side effects via the stored arguments.
The example in the docs is just badly chosen.
Jaakko
On Fri, 30 May 2003, firingme wrote:
> consider the following code :
>
> test.cpp
> *************************************************************
> #include <boost/lambda/lambda.hpp>
> #include <boost/lambda/bind.hpp>
>
> #include <list>
> #include <algorithm>
> #include <iostream>
> #include <vector>
>
> using namespace std ;
> using namespace boost ;
> using namespace boost::lambda;
>
> int main(){
> int i = 1 ;
> cout << ( _1+i)(i=2) << endl;
> }
> *************************************************************
>
> Result Table:
> *************************************************************
> Compiler Result
>
> VC7.1 4
> GCC3.2.3 ( MingW versiong ) 4
> GCC3.3 (i686-pc-cygwin) 3
>
> as in lambda's doc , the result should be 3 .
>
> Any help is welcome
>
>
> thx
>
>
>
>
>
>
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