On May 6, 2:45 pm, Ian Collins <ian-n...@[EMAIL PROTECTED]
> wrote:
> Jim Langston wrote:
> > S S wrote:
> >> Hi
>
> >> I have a very basic question, but it's a good one. Below is the code
> >> fragment.
>
> >> struct ltstr
> >> {
> >> bool operator()(const char* s1, const char* s2) const
> >> {
> >> return strcmp(s1, s2) < 0;
> >> }
> >> };
>
> >> int main()
> >> {
> >> const int N = 6;
> >> const char* a[N] = {"isomer", "ephemeral", "prosaic",
> >> "nugatory", "artichoke", "serif"};
> >> set<const char*, ltstr> A(a, a + N);
> >> if (A.find("ephemeral") != A.end())
> >> cout << "Found";
> >> else
> >> cout << "Not found";
> >> return 0;
> >> }
>
> >> Output will be -> "Found"
> >> My question is , Why?
>
> >> How it is able to compare a const char* with another const char* to
> >> find that value? I did not specify any equality operator? I just
> >> mentioned strcmp(s1, s2) < 0
> >> which means, when strcmp(s1, s2) == 0 (in case of match)
> >> it will return false. So how set/map are able to find the const char*
> >> value?
>
> > I believe it's undefined behavior. You have the constant "ephemeral"
in
> > your program twice. Once for the char array, once for the parameter
of the
> > find. The compiler probably sees that they are exactly the same, so
instead
> > of making 2 it just makes one and uses the same pointer for both.
>
> That does not matter, note the use of the second template argument of
> std::set.
>
> --
> Ian Collins.
Ian,
You wanted to explain something? I did not get you, sorry. Could you
elaborate please?
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