`std :: set`有什么问题? [英] What is wrong with `std::set`?
问题描述
在另一个主题,我试图解决这个问题。问题是从 std :: string
中删除重复的字符。
In the other topic I was trying to solve this problem. The problem was to remove duplicate characters from a std::string
.
std::string s= "saaangeetha";
因为顺序不重要,所以我排序 s
首先,然后使用 std :: unique
,最后调整大小以获得所需的结果:
Since the order was not important, so I sorted s
first, and then used std::unique
and finally resized it to get the desired result:
aeghnst
这是正确的!
现在我想做同样的事情,但同时我想要的字符顺序完好无损。方法,我想要这个输出:
Now I want to do the same, but at the same time I want the order of characters intact. Means, I want this output:
sangeth
所以我写了这:
template<typename T>
struct is_repeated
{
std::set<T> unique;
bool operator()(T c) { return !unique.insert(c).second; }
};
int main() {
std::string s= "saaangeetha";
s.erase(std::remove_if(s.begin(), s.end(), is_repeated<char>()), s.end());
std::cout << s ;
}
这将提供此输出:
saangeth
c> a 被重复,虽然其他重复也没有了。代码有什么问题?
That is, a
is repeated, though other repetitions gone. What is wrong with the code?
无论如何我更改我的代码有点: (见注释)
Anyway I change my code a bit: (see the comment)
template<typename T>
struct is_repeated
{
std::set<T> & unique; //made reference!
is_repeated(std::set<T> &s) : unique(s) {} //added line!
bool operator()(T c) { return !unique.insert(c).second; }
};
int main() {
std::string s= "saaangeetha";
std::set<char> set; //added line!
s.erase(std::remove_if(s.begin(),s.end(),is_repeated<char>(set)),s.end());
std::cout << s ;
}
输出:
sangeth
问题所在!
那么第一个解决方案有什么问题?
So what is wrong with the first solution?
此外,如果我不使成员变量
引用类型,然后问题不能解决。
Also, if I don't make the member variable unique
reference type, then the problem doesn't go.
std :: set
或 is_repeated
函子有什么问题?
What is wrong with std::set
or is_repeated
functor? Where exactly is the problem?
我还注意到如果 is_repeated
函子被复制到某处,它的每个成员也被复制。我在这里看不到问题!
I also note that if the is_repeated
functor is copied somewhere, then every member of it is also copied. I don't see the problem here!
推荐答案
在GCC(libstdc ++)中, remove_if
In GCC (libstdc++), remove_if
is implemented essentially as
template<typename It, typename Pred>
It remove_if(It first, It last, Pred predicate) {
first = std::find_if(first, last, predicate);
// ^^^^^^^^^
if (first == last)
return first;
else {
It result = first;
++ result;
for (; first != last; ++ first) {
if (!predicate(*first)) {
// ^^^^^^^^^
*result = std::move(*first);
++ result;
}
}
}
}
您的谓词被传递<-em> by-value 到 find_if
,所以结构体以及集合被修改在 find_if
不会传播回调用者。
Note that your predicate is passed by-value to find_if
, so the struct, and therefore the set, modified inside find_if
will not be propagated back to caller.
由于第一个副本出现在:
Since the first duplicate appears at:
saaangeetha
// ^
code> find_if 调用后将保留code>sa。同时,谓词
的集合为空( find_if
中的插入是局部的)。因此,循环之后将保持第三个 a 。
The initial "sa"
will be kept after the find_if
call. Meanwhile, the predicate
's set is empty (the insertions within find_if
are local). Therefore the loop afterwards will keep the 3rd a
.
sa | angeth
// ^^ ^^^^^^
// || kept by the loop in remove_if
// ||
// kept by find_if
这篇关于`std :: set`有什么问题?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IT屋!