Re: several newbie questions about stl
On Feb 9, 5:25 pm, Zeppe <ze...@remove.all.this.long.comment.yahoo.it>
wrote:
James Kanze wrote:
4. Why the first two arguments of std::find_first_of()
requires forward iterator, but not just input iterator?
Good question. (Typically, it's not very useful over an
input iterator, but there's no reason to forbid it.)
There actually is. An input iterator can read elements only
once. If you make a copy of an input iterator, the copy and
the original iterators will read different values when
incremented. There is no way to implement find_first_of with
input_iterators.
template< typename Iter1, typename Iter2 >
Iter1
find_first_of(
Iter1 begin1,
Iter1 end1,
Iter2 begin2,
Iter2 end2 )
{
while ( begin1 != end1
&& find( begin2, end2, *begin1 ) == end2 )
++ begin1 ;
}
return begin1 ;
}
Where's the problem is Iter1 is an input iterator?
7. The prototype of bsearch() is:
void *bsearch(const void *key, const void *base, size_t nmemb, size_t
size, int (*compar)(const void *, const void *));
The base argument is of type (void*) and the return value is of type
(void*). This seems to be of C-like style, which is for the sake of
ease of usage of the return value, such as the strchr() in C:
char *strchr(const char *s, int c);
But in C++, most of such functions have been changed to a pair of
overloaded functions:
const char* strchr(const char* s, int c);
char* strchr(char* s, int c);
So, why the same change is not done to bsearch()?
Probably oversight.
or because it was useless. You cannot assign a void* anyway,
you need to make a cast, and therefore there is not a big
difference to me between a void* and a const void*, as a
return type (as input it allows you to give const void* if
used in the code).
You need a static_cast, but not a const_cast.
8. As for std::basic_ostream and std::basic_istream, why the
<< and >> operation of bool, short, int, long are all defined
as member function, but only the equivalent of char is defined
as non-member function?
To confuse people.
Seriously, they were all members in the classical iostream. And
changing this broke a lot of code.
It looks like they tried to minimise the number of member
functions for basic_ostream. Since the output of characters
has several overloading, it was probably easier, and more
efficient at compilation time, to provide a simple template
put and implement all the others using it.
They didn't minimize very much. There's no real reason why <<
int, for example, has to be a member.
There are several advantages in making all of the overloaded
operators members. To do so means, however, that users must be
able to add members---in today's C++, this could be done by
making operator<< a member function template, and allowing users
to specialize it on their types, but this option wasn't
available when the classical iostream was designed.
If you decide to make as few of the overloads members as
possible, then none of them would be members; all of them are
easily defined in terms of the rest of the public interface.
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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