Re: [Boost-users] Bind operator== in standard algorithms

On 8/18/05, Alan M. Carroll <amc@network-geographics.com> wrote:

It's all about types. In the stand alone expression, the bind() result is evaluated by passing it x, yielding an int which is compared to the int y, so operator== is called on two ints. In the find_if call, the bind() result is *not* evaluated so operator== is called on the bind type and an int.

I suspected I wasn't using 1.33 (which indeed allows use of operators in bind), but this apparent partial operator functionality had me doubting. But this obviously explains it. And I've checked it now, I am using 1.32, time to go badger the sysadmin to upgrade :-) Thanks for the help, all. -- Alex Borghgraef