问题描述
是以下代码标准？ （在g ++ - 4.0.2下干净地编译）：
struct Asc {bool operator（）（int a，int b）{return a< b;}};
struct Des {bool operator（）（int a，int b）{return b> a;}};
int main（）{
int arr [] = {1,2,3};
set< int ，ASC> asc（arr，arr + 3）;
set< int，Des> :: iterator beg = asc.begin（）; // [*]
set< int，Des> :: iterator end = asc.end（）; // [*]
copy（beg，end，ostream_iterator< int>（cout，""））; //打印123
返回0;
}
请注意作业双方的_sets_类型[*由于它们的比较器不同（Asc / Des），因此b $ b不同。尽管如此，[*]似乎是标准的，因为迭代器的模板参数
是''Key''和''Distance''（不是''Comparator''），所以迭代器
类型是平等的，对吗？ [我们使用用户定义的模板''比较器''
参数（Des / Asc），所以他们没有标准的专业化，
这意味着我们可以肯定两个迭代器的距离类型
是相同的; ''Key''类型（int）也明显相同]。
谢谢，
--dan
is the following code standard? (cleanly compiles under g++-4.0.2):
struct Asc { bool operator()(int a, int b) {return a < b;} };
struct Des { bool operator()(int a, int b) {return b > a;} };
int main() {
int arr[] = {1, 2, 3};
set<int,Asc> asc(arr, arr+3);
set<int,Des>::iterator beg = asc.begin(); //[*]
set<int,Des>::iterator end = asc.end(); //[*]
copy(beg, end, ostream_iterator<int>(cout, "")); // prints 123
return 0;
}
note that the types of the _sets_ in both sides of assignment[*] are
different due to their different comparators (Asc/Des). nevertheless,[*] seems to be standard as the template parameters of the iterator
are ''Key'' and ''Distance'' (not the ''Comparator''), and so the iterator
types are equal, right? [we use user defined template ''Comparator''
parameters (Des/Asc), so they don''t have standard specialization,
which means we can be sure the ''Distance'' type of the two iterators
is the same; the ''Key'' type (int) is also obviously the same].
thanks,
--dan
推荐答案
这段代码很好。 std :: copy不要求两个迭代器是相同类型的
。事实上，他们甚至不需要属于那些
保持相同类型的容器 - 只要被复制的类型可以转换为输出迭代器的类型。
所以在这个例子中，你可以复制一组< int>到一组< long>甚至是
设置< double>还是没问题。
Greg
This code is fine. std::copy does not require that the two iterators be
of the same type. In fact, they need not even belong to containers that
hold identical types - as long as the type being copied is convertible
to the output iterator''s type.
So in this example, you could copy a set<int> to a set<long> or even a
set<double> and still be OK.
Greg
注意1：您的Asc和Des仿函数具有相同的行为。
注2：Asc和Des中的operator（）应该是const方法，并且
Asc和Desc应该可以从std :: binary_function<>继承。
#include< functional>
struct Asc：public std :: binary_function< int， int，bool>
{
bool operator（）（int a，int b）const {return a< b; }
}
struct Des：public std :: binary_function< int，int，bool>
{
//假设你不想要相同的行为
// Asc和Des
bool operator（）（int a，int b）const {返回> b; }
}
Note 1: You Asc and Des functors have identical behavior.
Note 2: operator() in Asc and Des should be const methods, and
Asc and Desc should probably inherit from std::binary_function<>.
e.g.:
#include <functional>
struct Asc: public std::binary_function<int,int,bool>
{
bool operator()(int a, int b) const { return a < b; }
}
struct Des: public std::binary_function<int,int,bool>
{
// assumes you didn''t want identical behavior for
// Asc and Des
bool operator()(int a, int b) const { return a > b; }
}
此代码没问题。 std :: copy不要求两个迭代器是相同类型的。实际上，它们甚至不需要属于具有相同类型的容器 - 只要被复制的类型可以转换为输出迭代器的类型。
所以在此示例中，您可以复制一组< int>到一组< long>甚至是
设置< double>并且仍然可以。
This code is fine. std::copy does not require that the two iterators be
of the same type. In fact, they need not even belong to containers that
hold identical types - as long as the type being copied is convertible
to the output iterator''s type.
So in this example, you could copy a set<int> to a set<long> or even a
set<double> and still be OK.
问题不是关于copy（），而是关于[*]（副本（）就在那里
使程序有意义）。 [*]的问题是程序
为类型的对象设置类型的对象：set< int，asc> :: iterator
：set< ; int，Dec> :: iterator
谢谢，
--dan
the question is not about copy(), it''s about[*] (the copy() is just there
to make the program meaningful). the problem with[*] is that the program
assigns an object of the type: set<int,Asc>::iterator
to an object of the type : set<int,Dec>::iterator
thanks,
--dan
这篇关于具有不同比较器的集合具有相同的迭代器类型;标准？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！