如何正确使用此迭代器？

Question

I'm trying to use a container to store the iterator of string s , but I got segment fault when I run it. As you will see in the following, the error seems to come with char temp = **itbegin; , it might means I can't assign any value via the previous iterator.

Why is it? Did I misuse the iterator?How to use the iterator properly?

#include <iostream>
#include <vector>
using namespace std;
string reverseVowels(string s);

int main()
{
    string s ="hello";
    cout << reverseVowels(s);
}
string reverseVowels(string s) {
    string::iterator iter = s.begin();
    string::iterator iterend = s.end();
    vector<string::iterator> iteratorbox;
    for(;iter != iterend ; ++iter){
        if((*iter) == 'a' &&
           (*iter) == 'e' &&
           (*iter) == 'i' &&
           (*iter) == 'o' &&
           (*iter) == 'u'){
               iteratorbox.push_back(iter);
           }

    }
    auto itbegin = iteratorbox.begin();
    auto itend =   iteratorbox.end() ;
    --itend;
    //for(;itbegin < itend ; ++itbegin, --itend)
    {
        char temp = **itbegin;
    //    *(*itbegin) = *(*itend); 
    //    *(*itend) = temp;
    }
    return s;
}

Answer 1

Your problem comes from the condition upon which you insert iterators in your iteratorbox vector.

You used the && operator, implying that every letter of the string has to be equal to all vowels at the same time . This means no iterator will be inserted, and you are then trying to dereference the begin() iterator of the vector, which happens to be its past-the-end iterator. This causes undefined behavior, which in your case manifests itself as a crash.

You probably meant to use

((*iter) == 'a' ||
(*iter) == 'e' ||
(*iter) == 'i' ||
(*iter) == 'o' ||
(*iter) == 'u')

as a condition.

Answer 2

The answer by @rems4e is perfectly fine, but I find such code easier to read and less error-prone if you put the vowels into an array

char const vowels[] = { 'a', 'e', 'i', 'o', 'u' };

so that you can encapsulate the matching logic inside your reverseVowels into the Standard algorithm any_of

if (std::is_any_of(std::begin(vowels), std::end(vowels), [](auto const& v) {
   return *iter == v;
}) {
    iteratorbox.push_back(iter);
}

This avoids much of the repetitive testing (which can get out-of-sync quickly should you ever use a different alphabet (German eg))