删除first1.erase(std::next(first1.begin(), i));
后,进行第二个循环,这有点奇怪,因为first2.erase(first2.begin() + 4, first2.end());
可以正常工作
#include <iostream>
#include <vector>
int main ()
{
std::vector<int> first1 = {0,1,2,3,4,5};
std::vector<int> first2 = {0,1,2,3,4,5};
std::vector<int> second;
std::vector<int> third;
for(size_t i = 4; i < first1.size(); ++i){
auto child = first1[i];
second.push_back(child);
first1.erase(std::next(first1.begin(), i));
}
third.assign(first2.begin() + 4, first2.end());
first2.erase(first2.begin() + 4, first2.end());
std::cout << "Size of first: " << int (first1.size()) << '\n';
std::cout << "Size of second: " << int (second.size()) << '\n';
std::cout << "Size of first: " << int (first2.size()) << '\n';
std::cout << "Size of third: " << int (third.size()) << '\n';
return 0;
}
输出:
Size of first1: 5
Size of second: 1
Size of first2: 4
Size of third: 2
我期望
first1/second
与first2/third
相同你可以在这里测试http://cpp.sh/9ltkw
最佳答案
循环的第一次迭代之后
for(size_t i = 4; i < first1.size(); ++i){
auto child = first1[i];
second.push_back(child);
first1.erase(std::next(first1.begin(), i));
}
i
将等于5
,first1.size()也将等于5。因此,仅擦除了 vector 的一个元素。您可以像这样重写循环
for(size_t i = 4; i != first1.size(); ){
auto child = first1[i];
second.push_back(child);
first1.erase(std::next(first1.begin(), i));
}
获得预期的结果。
在这些陈述中
third.assign(first2.begin() + 4, first2.end());
first2.erase(first2.begin() + 4, first2.end());
分配并删除了2个元素。