C++ 从两个迭代器创建一个列表对

Question

我有两个迭代器和一个包含对元素字符串和 size_t 的列表我想做的是将我的 student_id 和成绩添加到结果列表中。

Answer 1

在你的解释和添加一些东西之后，我认为你的错误在于使用 std::string::compare。我测试了以下代码（避免使用 addRegister 因为 Transcript 未定义）：

#include <iostream>
#include <list>
#include <algorithm>

typedef struct Records_r {
        std::string name; // Name of the records
        std::string student_id;
        std::list<std::pair<std::string, size_t>> grades; // List of (course, grade) pairs
} Records;

typedef std::list<std::pair<std::string, size_t>>::const_iterator Personaliteraattori;

std::list<std::pair<std::string, size_t>> findCourseResults(const std::list<Records>& registry, const std::string& course) {

    std::list<std::pair<std::string, size_t>> results;
    for (std::list<Records>::const_iterator it = registry.begin(); it != registry.end(); ++it) {
        for (Personaliteraattori itt = it->grades.begin(); itt != it->grades.end(); ++itt) {
            if (itt->first.compare(course) == 0) {
                results.push_back(std::make_pair(it->student_id, itt->second));
            }
        }
    }
    return results;
}

int main()
{
 std::string s = "Record of student_123456";
    std::string id = "123456";
    std::list<std::pair<std::string, size_t>> grades;
    grades.emplace_back("Math", 2);
    grades.emplace_back("Basic Math", 4);
    grades.emplace_back("Advanced Math", 5);
    grades.emplace_back("Math 1", 3);
    Records t;
    t.name = s;
    t.student_id = id;
    t.grades = grades;
    std::list<Records> reg;
    reg.push_back(t);

    t.name = "Record of student_34567";
    t.student_id = "34567";
    t.grades.clear();
    t.grades.push_back(std::pair<std::string,size_t>("Math",1));
    t.grades.push_back(std::pair<std::string,size_t>("Catalan",5));
    t.grades.push_back(std::pair<std::string,size_t>("Basic Math",9));
    reg.push_back(t);

    for (auto& tr : reg ) {
        std::cout << "Name: " << tr.name << std::endl;
        std::cout << "Student id: " << tr.student_id << std::endl;
        std::cout << "Grades:" << std::endl;
        std::for_each(tr.grades.begin(), tr.grades.end(), [](std::pair<std::string, size_t> e)
        {
            std::cout << e.first << " : " << e.second << std::endl; 
        });
    }

    while(true){
        std::cout<<"Enter a lecture: ";
        std::string lecture;
        std::getline(std::cin, lecture);
        std::list<std::pair<std::string, size_t>> results = findCourseResults(reg,lecture);
        for (auto& tr: results)
        {
            std::cout<<"StudentId: "<<tr.first<<" grades: "<<tr.second<<std::endl;
        }
    }

    return 0;
}

请注意我是如何使用 std::string::compare 函数的。此代码产生所需的输出：

Name: Record of student_123456
Student id: 123456
Grades:
Math : 2
Basic Math : 4
Advanced Math : 5
Math 1 : 3
Name: Record of student_34567
Student id: 34567
Grades:
Math : 1
Catalan : 5
Basic Math : 9
Enter a lecture: Math
StudentId: 123456 grades: 2
StudentId: 34567 grades: 1
Enter a lecture: Catalan
StudentId: 34567 grades: 5

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