Re: [Boost-users] [Phoenix] Simple if_ issue

On 7/2/2012 12:10 PM, Nathan Crookston wrote:

Aaron,

On Mon, Jul 2, 2012 at 1:05 PM, Nathan Crookston wrote:

...

On Mon, Jul 2, 2012 at 12:51 PM, wrote: <snip code>

...

What do I need to do in the second test to get it to compile? Can I tell if_ that its return type is not void?

Sadly no -- this was a point of confusion for me too. The correct way to do this is with if_else

Sorry to reply twice. The advice is correct, but the link is wrong (I'm not sure where if_else is documented. . .). <snip>

http://boost-sandbox.sourceforge.net/libs/phoenix/doc/html/phoenix/modules/o... And let me fill in the rationale. The if_else function models the ternary conditional operator (cond ? <true-case> : <false-case>). This is an expression, which has a type and a value. When you use if_(x)[y].else_[z], you're emulating if(x){y;}else{z;}, which is a statement. Statements don't have types or values in C++. Why does it matter? Because y and z might have incompatible types, and that's OK in an if/else statement, but not in a conditional expression. HTH, -- Eric Niebler BoostPro Computing http://www.boostpro.com