[Phoenix] Simple if_ issue - Boost-users - lists.preview.boost.org

newer
FW: Does the boost library free...

[Phoenix] Simple if_ issue

older
[exception] Passing existing...

Aaron_Wright＠selinc.com

2 Jul 2012 2 Jul '12

6:51 p.m.

I'm trying to do a simple if/else statement using boost::phoenix and I seem to mis-understanding something. namespace bp = boost::phoenix; namespace bpa = bp::arg_names; boost::function< std::string () > test1 = bp::val("true"); boost::function< std::string (bool) > test2 = bp::if_(bpa::arg1) [ bp::val("true") ] .else_ [ bp::val("false") ]; The first one works, but the second one doesn't. I get an error about not being able to convert void to std::string: cannot convert parameter 1 from 'void' to 'const std::basic_string<_Elem,_Traits,_Ax> & 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? --- Aaron Wright

Reply

Sign in to reply online Use email software

Show replies by date

Nathan Crookston

2 Jul 2 Jul

7:05 p.m.

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 http://www.boost.org/doc/libs/1_50_0/libs/spirit/phoenix/doc/html/phoenix/co.... HTH, Nate

Reply

Sign in to reply online Use email software

Nathan Crookston

7:10 p.m.

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. . .). The following shows usage: #include #include #include <iostream> namespace px = boost::phoenix; using px::arg_names::arg1; int main() { boost::functionstd::string(bool) f = px::if_else(arg1, "true", "false"); std::cout << f(true) << std::endl; std::cout << f(false) << std::endl; return 0; } HTH, Nate

Reply

Sign in to reply online Use email software

Eric Niebler

7:54 p.m.

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

Reply

Sign in to reply online Use email software

4463

Age (days ago)

4463

Last active (days ago)

Download

3 comments

3 participants

tags

participants (3)

Aaron_Wright＠selinc.com
Eric Niebler
Nathan Crookston