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.
An example of this is this program that on my system outputs:
00416808 00416808
one time I run it
#include <iostream>
int main()
{
const char* foo = "Hello";
const char* bar = "Hello";
std::cout << reinterpret_cast<const int*>( foo ) << " " <<
reinterpret_cast<const int*>( bar ) << "\n";
}
The compiler doesn't bother to create two constant arrays for "Hello",
since
it's constant it just creates it once and uses the same address for both
foo
and bar.
--
Jim Langston
tazmaster@[EMAIL PROTECTED]
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