Hello,
In previous discussions on rounding of double on x86 I wanted
to find an example that points to the absurdity of current
gcc behavior.
At last I found such an example:
---- t.cpp start ---
#include <tr1/unordered_set>
#include <iostream>
double x=3.0;
int main()
{
std::tr1::unordered_set<double> myset;
myset.insert(2/x);
myset.insert(2/x);
std::cout << myset.size() << " elements\n";
if (myset.size() > 1)
std::cout << "Are last and first equal? "
<< std::boolalpha
<< ( *myset.begin() == *++myset.begin())
<< "\n";
return 0;
}
--- t.cpp end ---
Here is what I get (Pentium 4):
--- trace begin ---
test$ /opt/gcc-4.0-20050602/bin/g++ t.cpp
test$ ./a.out
1 elements
test$ /opt/gcc-4.0-20050602/bin/g++ -O3 -finline-limit=1000000 t.cpp
test$ ./a.out
2 elements
Are last and first equal? true
--- trace end ---
The behavior of the second run does not look right. What does it mean?
1. Is it forbidden by tr1 to define unordered_set<double> ?
2. Is it a bug in the tr1 document (which should have forbidden this).
3. Is it OK to have repetitions in unordered_set?
4. Is it a bug in gcc, for handling double the way it does?
5. Is it a bug in the implementation of tr1 in libstdc++ ?
Maybe handling of double should move to a different
translation unit, to avoid aggressive inlining. Or maybe
there should be a specialization for equal_to<double>,
where error bands will be used.
Using error bands will work fine for unordered_set<doble> insertion.
It may lead to the "loss" of close entries, but in case of double it sounds
reasonable.
P.S.
std::tr1::hash<dobule> is implemented in a very bad way.
it casts double to size_t, which of course does a very poor job on big
values (is the result of 1.0e100 cast to size_t defined ?).
Michael
