对于std::set像std::set<std::string> set={ "aaa", "bbb", "ccc", "ddd", "eee", "fff" };,如何在这个集合的特定范围a,b上迭代,而不跟踪额外的索引?
就像这样:
for(auto it = set.begin(); it!=set.begin()+b; ++it)
std::cout << *it << " ";或者像这样:
for(auto it = set.begin()+a; it!=set.begin()+b; ++it)
std::cout << *it << " ";a<=b和b<=set.size()
发布于 2021-03-03 08:12:09
你不能做set.begin()+a,但是你可以做std::advance(it, a)
#include <iostream>
#include <set>
#include <string>
int main()
{
std::set<std::string> set={ "aaa", "bbb", "ccc", "ddd", "eee", "fff" };
size_t a = 1, b = 4;
auto it = set.begin(), it_end = set.begin();
std::advance(it, a);
std::advance(it_end, b);
for(; it!= it_end; ++it)
std::cout << *it << " ";
}不幸的是,您不能执行类似auto it = std::advance(set.begin(), a);的操作,因为advance通过引用获取迭代器并将其更改。
更好的解决方案使用std::next,这要感谢@Evg:
#include <iostream>
#include <set>
#include <string>
int main()
{
std::set<std::string> set={ "aaa", "bbb", "ccc", "ddd", "eee", "fff" };
size_t a = 1, b = 4;
for(auto it = std::next(set.begin(), a), it_end = std::next(it, b-a); it != it_end; ++it)
std::cout << *it << " ";
}https://stackoverflow.com/questions/66452879
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