On 22/04/12 02:42, Mostafa wrote:
On Fri, 20 Apr 2012 05:15:50 -0700, Olaf van der Spek <ml@vdspek.org> wrote:
Hi,
What do you expect this code to do? Is b true or false? And why? Is this expected behaviour?
[1] #include <boost/range/iterator_range.hpp> [2] #include <string> [3][4] int main() [5] { [6] std::string s = "Olaf"; [7] boost::iterator_range<std::string::iterator> r(s); [8] bool a = r == s; [9] bool b = r == "Olaf"; [10] assert(a); [11] assert(b); [12] return 0; [14] }
I would expect both lines 8 and 9 to cause compilation failure.
There is absolutely no reason to, this code makes perfect sense. iterator_range has an overloaded operator== that allows it to compare it to any other range.
To someone unfamiliar with the library, it's not /immediately/ clear what the intent of either line is.
It is immediately clear once you know what types are ranges and what operator== on iterator_range does.
If one wants to treat "s" as a range
s is a std::string. It is already a range. There is no need to "treat it" as such.
Another note on implicit conversions, if line 8 means implicitly adapt "s" to "r"'s type and then does a comparison, then the following:
[*] bool c = (s == r);
by symmetry should mean adapt "r" to "s"'s type and then do a comparison, which clearly isn't the current behaviour. Even if it was, then [*] and [8] may then give different results when unicode comes into play. Which is another reason in favor of explicit conversions for both [8] and [9].
I have no idea what you're talking about. There is no conversion involved. s == r just does std::equal(begin(s), end(s), begin(r))