3

The following code throws an exception

terminate called after throwing an instance of 'boost::exception_detail::clone_impl<boost::exception_detail::error_info_injector<boost::bad_function_call> >'
  what():  call to empty boost::function

at the line f() (while executing the block):

void foo(); // assume this is defined somewhere
boost::function<void()> f = boost::bind(&foo);

^(void) {
   f();
}();

However, according to the documentation on blocks,

In general you can use C++ objects within a block. Within a member function, references to member variables and functions are via an implicitly imported this pointer and thus appear mutable. There are two considerations that apply if a block is copied:

  • If you have a __block storage class for what would have been a stack-based C++ object, then the usual copy constructor is used.

  • If you use any other C++ stack-based object from within a block, it must have a const copy constructor. The C++ object is then copied using that constructor.

This seems to be true normally; if I replace f above with an instance of a simple class with an operator()(), the above code runs as expected.

Why doesn't the version with boost::function work?

Jesse Beder
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    I'm no boost expert, but I'd hazard a guess that it relates to the *const copy constructor* for the `boost::function<>` type (per the quoted documentation). – Richard Mar 28 '11 at 16:24
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    Do you get the same error without the bind call (not needed for free functions, see [here](http://www.boost.org/doc/libs/1_46_1/doc/html/function/tutorial.html#id1274976))? What about using the portable syntax, function0<> instead of function<>? – academicRobot Apr 03 '11 at 17:01
  • I'm curious about the `bind()` call as well, unneeded and could be responsible for the error. – lefticus Apr 06 '11 at 04:47
  • @academicRobot, @lefticus - there's no difference without the `bind()` call (and in my actual situation, I need to bind some parameters) – Jesse Beder Apr 06 '11 at 04:58

2 Answers2

1

It appears that if I modify the declaration with __block, it works correctly:

__block boost::function<void()> f = boost::bind(&foo);

I'm still not sure why this is - as @Richard mentions in a comment above, it must have to do with the "const copy constructor" as opposed to the "usual copy constructor". I don't know how to check this difference, though; the following works fine:

const boost::function<void()> f = boost::bind(&foo);
const boost::function<void()> g(f);
g();

and if that doesn't call a "const copy constructor", I'm not sure what will.

Jesse Beder
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0

The example in your answer,

const boost::function<void()> f = boost::bind(&foo);
const boost::function<void()> g(f);
g();

works because you're calling operator() in the same scope in which its callee was created. f() would work outside of block scope as well, and calling foo() would work inside block scope because it is a classic function and is not considered to be "modified" when it is called. The referenced call to a const copy constructor is made when passing the object from containing scope to block scope. It would appear that whatever difference exists in the const copy constructor is failing to copy part of the object through to block scope.

matthias
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