I have the following code, which removes a pair (string, int) from the
set. I defined a predicate that returns true if the string value matches.
But I am getting compiler error on the remove_if function:
Here is the code:
typedef pair<string, int> my_pair;
typedef set<my_pair> my_set;
typedef my_set::iterato r my_set_p;
class my_pair_eq : public unary_function< my_pair, bool>
{
string s;
public:
explicit my_pair_eq (const string& ss) : s(ss) { }
bool operator() (const my_pair& mp) const {return mp.first == s; }
};
int main(void)
{
my_set mset;
mset.insert(my_ pair("one", 1));
mset.insert(my_ pair("one", 2));
mset.insert(my_ pair("two", 2));
mset.insert(my_ pair("one", 3));
mset.insert(my_ pair("three", 3));
remove_if(mset. begin(), mset.end(), my_pair_eq("one "));
return 0;
} 5 1849
On Sun, 07 Dec 2003 23:22:46 -0500, Wang Tong <wa******@seas. upenn.edu> wrote: I have the following code, which removes a pair (string, int) from the set. I defined a predicate that returns true if the string value matches.
But I am getting compiler error on the remove_if function:
You can't use remove_if on a set. It is a mutating algorithm (ie. it
makes changes to the underlying container), but set doesn't allow you
to do that.
It also makes no sense. remove_if, moves the items in the range that
satisfy the predicate to the end of the range (and returns a new end
iterator indicating where those "removed" elements start). A set is
ordered and hence you can't rearrange the elements (since it would then
no longer be ordered).
--
Sam Holden
While it was 8/12/03 4:56 am throughout the UK, Sam Holden sprinkled
little black dots on a white screen, and they fell thus:
<snip> It also makes no sense. remove_if, moves the items in the range that satisfy the predicate to the end of the range (and returns a new end iterator indicating where those "removed" elements start). A set is ordered and hence you can't rearrange the elements (since it would then no longer be ordered).
Correction:
A set is unordered, i.e. there is no such thing as the order of elements
within a set.
If, in normal mathematical notation, we write
S = { 3, 7, 12 }
then this is merely a way of writing it down - as much as 7 appears to
be between 3 and 12, it isn't. We could write the same set as
S = { 7, 12, 3 }
A list is ordered, i.e. the order of elements is defined. So <7, 3, 12>
and <12, 3, 7> are two distinct lists.
Of course, one might use a list in order to implement a set, and may
even automatically sort the underlying list giving a clear order in
which an iterator returns elements. But this order is not part of the
set - it is just provided for the purpose of being able to e.g. search a
set or operate on each element in turn.
Stewart.
--
My e-mail is valid but not my primary mailbox, aside from its being the
unfortunate victim of intensive mail-bombing at the moment. Please keep
replies on the 'group where everyone may benefit.
"Wang Tong" <wa******@seas. upenn.edu> wrote in message
news:br******** ***@netnews.upe nn.edu...
| I have the following code, which removes a pair (string, int) from the
| set. I defined a predicate that returns true if the string value matches.
|
| But I am getting compiler error on the remove_if function:
[snip]
| int main(void)
| {
| my_set mset;
|
| mset.insert(my_ pair("one", 1));
| mset.insert(my_ pair("one", 2));
| mset.insert(my_ pair("two", 2));
| mset.insert(my_ pair("one", 3));
| mset.insert(my_ pair("three", 3));
|
| remove_if(mset. begin(), mset.end(), my_pair_eq("one "));
mset.erase( std::remove_if( mset.begin(),
mset.end(), my_pair_eq("one ") ), mset.end() );
Cheers.
Chris Val
"Stewart Gordon" <sm*******@yaho o.com> wrote in message
news:br******** @sun-cc204.lut.ac.uk ... While it was 8/12/03 4:56 am throughout the UK, Sam Holden sprinkled little black dots on a white screen, and they fell thus: <snip> It also makes no sense. remove_if, moves the items in the range that satisfy the predicate to the end of the range (and returns a new end iterator indicating where those "removed" elements start). A set is ordered and hence you can't rearrange the elements (since it would then no longer be ordered). Correction:
A set is unordered, i.e. there is no such thing as the order of elements within a set.
If, in normal mathematical notation, we write
S = { 3, 7, 12 }
then this is merely a way of writing it down - as much as 7 appears to be between 3 and 12, it isn't. We could write the same set as
S = { 7, 12, 3 }
In mathematical terminology, yes. In the standard library, however,
std::set is ordered, std::list is not.
A list is ordered, i.e. the order of elements is defined. So <7, 3, 12> and <12, 3, 7> are two distinct lists.
Of course, one might use a list in order to implement a set, and may even automatically sort the underlying list giving a clear order in which an iterator returns elements. But this order is not part of the set - it is just provided for the purpose of being able to e.g. search a set or operate on each element in turn.
Stewart.
-- My e-mail is valid but not my primary mailbox, aside from its being the unfortunate victim of intensive mail-bombing at the moment. Please keep replies on the 'group where everyone may benefit.
While it was 8/12/03 2:52 pm throughout the UK, Thomas Wintschel
sprinkled little black dots on a white screen, and they fell thus:
<snip> In mathematical terminology, yes. In the standard library, however, std::set is ordered, std::list is not.
<snip>
You mean the STL bible uses the word "ordered" with a meaning contrary
to the mathematical one? That sense corresponds to the word "sorted" in
my vocabulary.
Stewart.
--
My e-mail is valid but not my primary mailbox, aside from its being the
unfortunate victim of intensive mail-bombing at the moment. Please keep
replies on the 'group where everyone may benefit. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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