C++從兩個迭代器創建一個列表對

Question

我有兩個迭代器和一個列表對元素字符串和size_t 我想要做的是將我的student_id和成績添加到結果列表中。

Answer 1

經過您的解釋和添加一些東西，我認爲您的錯誤是使用std::string::compare。我測試過下面的代碼（避免使用addRegister因爲Transcript沒有定義）：

#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; 
} 

請注意我如何使用std::string::compare功能。這段代碼產生所需的輸出：

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