c++11 - 如何设置C++string迭代器循环终止条件？

Question

大家好，我刚开始学习C++，学习教材是C++ Primer 第五版。今天在完成书上的一个练习时（324页，练习9.43），总是发生out_of_range错误，经过调试发现，函数find_and_replace中的for循环没有终止。

代码如下：

#include <iostream>
#include <string>

int find_and_replace(std::string& org_str, const std::string& str4find, const std::string& str4replace);

int main(int argc, char const *argv[])
{

    std::string str("I am a very loooooong string to be process!");
    int find_times;
    find_times = find_and_replace(str, "a", "###");
    std::cout << find_times << std::endl;
    std::cout << str << std::endl;

    return 0;
}

int find_and_replace(std::string& org_str, const std::string& str4find, const std::string& str4replace)
{
    int find_times = 0;
    // 被检查字符串长度小于等于匹配字符串长度的情况
    if (org_str.size() < str4find.size())
    {
        std::cout << "Error: The original string is too short!" << std::endl;
        return find_times;
    }
    else if (org_str == str4find)
    {
        org_str.assign(str4replace);
        ++find_times;
        return find_times;
    }

    // 其他情况
    for (auto i = org_str.begin(), j = i + str4find.size(); j != org_str.end(); )
    {
        std::string temp(i, j);
        std::cout << temp << std::endl;
        if (temp == str4find)
        {
            j = org_str.erase(i, j);
            org_str.insert(i, str4replace.begin(), str4replace.end());
            std::cout << org_str << std::endl;
            std::cout << *i << std::endl;
            j = i + str4find.size();
            std::cout << *j << std::endl;
            ++find_times;
        }
        else
        {
            ++i;
            ++j;
        }
    }

    // 跳出循环时，迭代器i和org_str.end()中间的字符串还没有检查
    if (org_str.substr(org_str.size() - str4find.size()) == str4find)
    {
        org_str.erase(org_str.size() - str4find.size(), str4find.size());
        org_str.insert(org_str.end(), str4replace.begin(), str4replace.end());
        ++find_times;
    }

    return find_times;
}

我在擦除和插入操作后都及时更新了迭代器j而且能够保证i始终在j之前，请问为什么我的循环终止条件j != org_str.end()没有起作用呢？

Answer 1

Since you know that the iterator is provided by the return value after erase, then insert may also invalidate the iterator, and there is a problem with the logic of your substitution method. If "a" is replaced with "abc", you This is infinitely replaceable. . .

Answer 2

If you change this string when traversing, it is likely that the iterator has expired. Check to see if this is the problem

Answer 3

Now that I understand this question, it is a self-question and answer. As the two friends above said, the problem lies in the fact that i and j are invalid iterators after erasing and inserting data, although std::cout outputs It looks correct, but it's essentially invalid. Since my compiler does not support i = org_str.insert(i, str4replace.begin(), str4replace.end());, I had to write a loop. The new version of the corrected code is as follows:

#include <iostream>
#include <string>

int find_and_replace(std::string& org_str, const std::string& str4find, const std::string& str4replace);

int main(int argc, char const *argv[])
{
    std::string str("I am a very loooooong string to be process!");
    int find_times;
    find_times = find_and_replace(str, "o", "!@#");
    std::cout << find_times << std::endl;
    std::cout << str << std::endl;

    return 0;
}

int find_and_replace(std::string& org_str, const std::string& str4find, const std::string& str4replace)
{
    int find_times = 0;
    // 被检查字符串长度小于等于匹配字符串长度的情况
    if (org_str.size() < str4find.size())
    {
        std::cout << "Error: The original string is too short!" << std::endl;
        return find_times;
    }
    else if (org_str == str4find)
    {
        org_str.assign(str4replace);
        return ++find_times;
    }

    // 其他情况
    // 终止条件是为了保证最后一段能够处理完毕，如果按照之前的j != org_str.end()，则还需要在循环外写一段代码处理最后一段字符串
    for (auto i = org_str.begin(), j = i + str4find.size(); (i + (str4find.size() - 1)) != org_str.end(); )
    {
        std::string temp(i, j);
        // std::cout << temp << std::endl;
        if (temp == str4find)
        {
            // 更新i
            i = org_str.erase(i, j);
            for (auto k = str4replace.begin(); k != str4replace.end(); ++k, ++i)
            {
                // 每次都更新i
                i = org_str.insert(i, *k);
            }
            // 用i更新j
            j = i + str4find.size();
            ++find_times;
        }
        else
        {
            ++i;
            ++j;
        }
    }

    return find_times;
}