Question

好吧所以我知道人们已经开始讨论这个话题了，但是我一遍又一遍地搜索问题并反复编辑我的代码，我似乎仍然遇到了这个问题。这段代码基本上用于运行向量并比较/组合其内容。为了做到这一点，我使用＆＃39;它＆＃39;和＆＃39; in＆＃39;浏览向量和.erase以删除已合并到向量的新部分的内容。知道我基本上需要.erase函数和迭代器，我在StackOverflow上的其他问题中使用了一些似乎在这种情况下工作的代码（auto it stuff）。由于我不熟悉该代码，我可能在这种情况下使用不正确，但我对此并不确定。

无论如何，正如标题所示，我得到了一个不可递增的矢量迭代器＆＃39;错误。经过第二次for循环几次后，程序似乎遇到了这个错误。我已经尝试了很多东西，但我似乎无法弄清楚我的代码的哪一部分是真正的问题。

非常感谢任何帮助！

\if (!strvector.empty()) {
        for (auto in = strvector.begin(); in != strvector.end() - 1;) {//in order to process if there is more than one final piece

            str = ' ' + *in + ' ';//adds spaces so it only compares beginnings & ends
            if (strvector.size() < 2) {
                break;
            }//doesn't do anything if there are too few objects to process

            for (auto it = strvector.begin(); it != strvector.end() - 1;) { 
                if (strvector.size() < 2) {
                    break;
                }//doesn't continue if there are too few objects to process
                str2 = ' ' + *it + ' '; //adds spaces so it only compares beginnings & ends
                substr = str; //sets substring of string to be chopped up
                while (substr.length() >= 6) { //only processes substrings >= 6 characters bc smaller bits are useless
                    size_t found = str2.find(substr); //searches str2 for substr
                    if (found != string::npos) {
                        str = str.substr(0, substr.length()) + ' '; //cuts substr off of str
                        str.append(str2); //adds str and str2
                        it = strvector.erase(it);
                        substr = 'a'; //shortens substr to get out of while
                        test=1;
                    }//end if
                    else {
                        substr.erase(substr.size() - 1, 1); //if substr was not found, decrement substr and compare again
                    }
                }//end while

                substr = str; //resets substr
                while (substr.length() >= 6) { //only processes substrings >= 6 characters bc smaller bits are useless
                    size_t found = str2.find(substr); //searches str2 for substr
                    if (found != string::npos) {
                        str = str.substr(substr.length()) + ' '; //cuts substr from beginning of string
                        str = str2 + str; //adds str2 and str, with str2 at the beginning
                        it = strvector.erase(it++);
                        substr = 'a'; //shortens substr to get out of while 
                        test=1;

                    }//end if
                    else {
                        substr.erase(0, 1); //erases the first character of the string
                    }
                    if (test < 1) {
                        it++; //increments if the erase function did not already do that
                    }

                }//end while
                if (test != 1) {
                    it++; //increments if the erase function did not already do that
                }
                if (test < 2) {
                    strvector.push_back(str); //adds new str to the vector
                    test = 0;
                }
            }//end ptr2 for
            if (strvector.size() < 2) {
                in = strvector.erase(in - 1);
            }
            else {
                in++;
            }
            cout << "str1 is " << str << endl; //prints out str
        }//end ptr for
    }//end if vector is not empty

Answer 1

在erase上呼叫std::vector后，您使用的迭代器将变为无效。相反，对erase的调用会返回一个你应该继续使用的全新迭代器。

在您的情况下，您正在使用postfix ++运算符，该运算符将在erase方法使用后尝试递增迭代器。此时迭代器无效，因此您会收到错误。

你可能想要的是it = strvector.erase(it);删除迭代器中的元素然后返回一个新的迭代器，该迭代器位于你擦除的元素之后。您不需要额外的++因为erase有效地为您做了这一点。

矢量迭代器不可增量for for循环

1 个答案: