Re: [Boost-users] Using boost::format with Lambda expressions

On Jan 29, 2004, at 3:47 PM, Samuel Krempp wrote:

On Thu, 2004-01-29 at 18:10, Jaakko Jarvi wrote:

...

format f("%-10s|") ; for_each(Foos.begin(), Foos.end(), cout << ret(var(f) % bind(&foo::name, _1))); // format defines operator% as a member, which takes precedence over // % defined by lambda. Therefore one must make the format object to // be a lambda functor. var does that. // Var, however, cannot take a temporary object (it holds a reference to // the wrapped object). That's why the variable f. // ret informs lambda about the return type of formats % operator.

ah I hadn't thought about the interaction of the operator% with lambda.

What would the situation be if the operator% were out of the class ? If it improves the usability with lambda, that would seem a reason enough, for me. In fact I had made it a member at first, because it needs access to the private data, but I had to put the real code in out-of-class functions anyhow because of msvc6 problems with member function templates, so I might as well make the whole operator% declared out of class.

You couldn't do that I guess. The prototype would be template<class T> operator%(format& , const T& ) now a call: format("...") % 1 would fail, as format("...") is an rvalue, and rvalues cannot be bind to non-const references (except as the this argument). The reason why the compiler prefers format % over lambda % is that in the match to lambda %, const is added to the format argument, whereas in format % it is not. Probably leaving things as they are is the best solution, there's doesn't seem to be much one can do for this. Best, Jaakko

-- Samuel

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