Re: Failure to compile with std::set, compiles OK with std::vector,
why?
Hi Jason,
Thanks for the answer.
jason.cipriani@gmail.com wrote:
There is no set<Integer> constructor that takes two integers as
parameters. See:
http://www.sgi.com/tech/stl/set.html
You are right. The code should have been:
int main(int argc, char* argv[])
{
// std::vector<Integer> collection(5,2);
std::set<Integer> collection;
collection.insert(0);
std::for_each(collection.begin(),
collection.end(),
std::mem_fun_ref(&Integer::print));
return 0;
}
But that is not the error I am referring to. I am referring to this error:
c++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"Main.d" -MT"Main.d"
-o"Main.o" "../Main.cpp"
/usr/include/c++/4.3/bits/stl_algo.h: In function ???_Funct
std::for_each(_IIter, _IIter, _Funct) [with _IIter =
std::_Rb_tree_const_iterator<Integer>, _Funct = std::mem_fun_ref_t<void,
Integer>]???:
.../Main.cpp:27: instantiated from here
/usr/include/c++/4.3/bits/stl_algo.h:3791: error: no match for call to
???(std::mem_fun_ref_t<void, Integer>) (const Integer&)???
/usr/include/c++/4.3/bits/stl_function.h:560: note: candidates are: _Ret
std::mem_fun_ref_t<_Ret, _Tp>::operator()(_Tp&) const [with _Ret = void,
_Tp = Integer]
Unrelated, consider declaring Integer::print as const up in the class
declaration, if Integer::print() modifies the Integer, it can break
order invariants in the set. I may be misunderstanding something, but
it seems like some information about const is lost in mem_fun_ref /
for_each somewhere, I'd expect a warning to be generated.
Actually, an error is issued (see above). And indeed, changing the print
function to "void print() const" makes it compile both with the vector
and the set.
I still do not understand why there is a difference between the vector
and the set version. As far as I know, std::for_each allows for mutating
(non-const) operations and there is nothing in either
collection.begin()/end() or std::mem_fun_ref that suggests const-ness
somewhere.
Joost