David Greene wrote:

All right, I'm still trying to figure out the MPL lambda stuff.

I had expected this to work:

template<typename Arg> some func { typedef ... type; };

typedef transform< sequence, some_func<_>

...

::type result;

But I get a compiler error about a missing apply<> in some_func.

Because some_func isn't defined anywhere? What compiler, BTW?

This works:

typedef transform< sequence, quote1

...

::type result;

I don't understand why the placeholder doesn't convert some_func to an appropriate metafunction class.

I don't understand why/how you could expect it to. It's not as though the MPL can retroactively generate a specialization of some_func that has a nested ::apply<> metafunction. Assuming you meant to write some_func at the top and not func, some_func<_> is a placeholder expression and therefore a lambda expression. Unless your compiler is broken, that should be sufficient to allow it to be passed as the 2nd argument to transform.

Similarly, I tried to use unpack args:

template some_func2 { typedef ... type; };

typedef transform< zipped_sequence, unpack_args

...

::type result;

The compiler complained about a missing apply<> in some_func2.

Looks like it should have been a syntax error. Your <>s don't nest.

Again, this worked:

typedef transform< zipped_sequence, unpack_args >

...

::type result;

Is it because I'm using _ rather than _1 and _2?

Shouldn't be. Examples from http://www.boost-consulting.com/metaprogramming-book that use _ in that way work just fine, and were tested with at least 2 compilers.

The examples in the MPL manual using is_same, etc. work fine for me.

I suggest you post the precise examples that are failing for you rather than summaries, and tell us what compiler you're testing on. -- Dave Abrahams Boost Consulting http://www.boost-consulting.com