On May 6, 5:28 pm, James Kanze <james.ka...@[EMAIL PROTECTED]
> wrote:
> On May 6, 11:58 am, Ian Collins <ian-n...@[EMAIL PROTECTED]
> wrote:
>
>
>
> > S S wrote:
> > > 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 =3D 6;
> > > const char* a[N] =3D {"isomer", "ephemeral", "prosaic",
> > > "nugatory", "artichoke", "serif"};
> > > set<const char*, ltstr> A(a, a + N);
> > > if (A.find("ephemeral") !=3D 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) =3D=3D 0 (in case of match)
> > > it will return false. So how set/map are able to find the const
char*
> > > value?
> > Given an ordered set, A < B < C and a search item X,
> > if (X < C) and !(B < X) then X is equivalent to B.
>
> The issue in std::set is a bit more complicated, because the
> domain may include elements which are not members of std::set.
> Basically, set<>::find first locates an element i such that for
> all previous elements j, j < x, and i for i, !(i < x) (which
> implies that the same holds for all following elements, of
> course). Having found this i, if !(x < i) determines whether
> the elements are equivalent or not: if !(i < x) && !(x < i), the
> elements are considered to be equivalent.
>
> Note the "considered to be". You can easily define an operator<
> which is inconsistent with =3D=3D. (He has, in fact: his ltstr is
> inconsistent with =3D=3D over char const*. Of course, if he were
> also interested in =3D=3D, he'd define an eqstr, which returned
> strcmp(s1, s2) =3D=3D 0.)
>
> --
> James Kanze (GABI Software) email:james.ka...@[EMAIL PROTECTED]
> Conseils en informatique orient=E9e objet/
> Beratung in objektorientierter Datenverarbeitung
> 9 place S=E9mard, 78210 St.-Cyr-l'=C9cole, France, +33 (0)1 30 23 00 34
Thanks James. It was a really nice explanation, I was linking my ltstr
operator with the find function. find has it's own searching algorithm
as you mentioned, no matter what operator I provide. My operator is
used for arranging the tree while insertion.
There is no meaning providing eqstr operator for set, it defies it's
purpose. I won't be able to add any other element (which differs from
first) to the set except the first one.
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