我正在尝试编写一个程序，该程序从文本文件中读取信息，并使用结构，联合和cstrings将其复制到数组中。我的文本文件看起来像这样。
F South Korea
Male Psy Park Jae Sang
31 - 12 - 1977
3 CSCI114 55 CSCI103 44 GangNam 100
S Female Super Junior
5 - 8 - 1978
2 CSCI114 60 CSCI103 80
F People Republic Of China
Unknown James Bond
11 - 12 - 1976
4 CSCI114 54 CSCI124 66 CSCI007 99 CSCI123 28
我的查询是，我编写了一个带有for循环的简单切换案例，该循环根据其条件读取信息。例如，每个段落前的第一个字母表示该学生是外国人，F还是新加坡人。根据他们的国籍，我选择将哪些结构/工会复制到其中。为了给你一个更好的主意，这是我从原始文本文件中获取信息并处理它之后我的最终文本输出文件的样子。
http://i.stack.imgur.com/Bv2YS.jpg
以下是我遇到问题的代码部分。看来char变量是否为“S”或“F”，我的switch语句只是将其读作“F”。
//file to array.
char dateJunk;
int numOfCourses;
int k = 0;
while (!afile.eof())
{
for (int k = 0; k < 3; k++)
{
afile >> locale;
switch (locale)
{
case 'F':
afile.getline(x[k].st.foreignStudent.nationality, MAX);
afile >> x[k].st.foreignStudent.gender;
afile.getline(x[k].st.foreignStudent.name, MAX);
afile >> x[k].st.foreignStudent.bd.day;
afile >> dateJunk;
afile >> x[k].st.foreignStudent.bd.month;
afile >> dateJunk;
afile >> x[k].st.foreignStudent.bd.year;
afile >> x[k].st.foreignStudent.numOfCourses;
for (int i = 0; i < x[k].st.foreignStudent.numOfCourses; i++)
{
afile >> x[i].st.foreignStudent.subjects[k];
afile >> x[i].st.foreignStudent.grades[k];
}
break;
case 'S':
afile >> x[k].st.localStudent.gender;
afile.getline(x[k].st.localStudent.name, MAX);
afile >> x[k].st.localStudent.bd.day;
afile >> dateJunk;
afile >> x[k].st.localStudent.bd.month;
afile >> dateJunk;
afile >> x[k].st.localStudent.bd.year;
afile >> x[k].st.localStudent.numOfCourses;;
for (int i = 0; i < x[k].st.localStudent.numOfCourses; i++)
{
afile >> x[i].st.localStudent.subjects[k];
afile >> x[i].st.localStudent.grades[k];
}
}
}
}
//Tests my cstring arrays to see everything is copied in correctly.
for (int k = 0; k < 3; k++)
{
cout << locale << " " << x[k].st.foreignStudent.nationality;
cout << endl;
cout << x[k].st.foreignStudent.gender;
cout << x[k].st.foreignStudent.name;
cout << endl;
cout << x[k].st.foreignStudent.bd.day << " - ";
cout << x[k].st.foreignStudent.bd.month << " - ";
cout << x[k].st.foreignStudent.bd.year;
cout << endl;
cout << x[k].st.foreignStudent.numOfCourses;
cout << endl;
for(int i = 0; i < x[k].st.foreignStudent.numOfCourses; i++)
{
cout << x[i].st.foreignStudent.subjects[k] << " ";
cout << x[i].st.foreignStudent.grades[k] << " ";
}
cout << endl;
}
return 0;
}
以下是完整的代码。
#include <iostream>
#include <fstream>
#include <cstdlib>
#include <ctime>
using namespace std;
const int MAX = 80;
struct Birthday
{
char day[MAX];
char month[MAX];
char year[MAX];
};
struct Local
{
char name[MAX];
char nationality[MAX];
char gender[MAX];
Birthday bd;
char subjects [MAX][MAX];
char grades [MAX][MAX];
int numOfCourses;
};
struct Foreigner
{
char name[MAX];
char nationality[MAX];
char gender[MAX];
Birthday bd;
char subjects [MAX][MAX];
char grades [MAX][MAX];
int numOfCourses;
};
union Student
{
Local localStudent;
Foreigner foreignStudent;
};
enum CountryType {S, F};
struct UowStudents
{
CountryType ct;
Student st;
};
int fileToArray (fstream& afile, char fileName [], UowStudents* x, char& locale);
int main ()
{
srand(time_t(NULL));
fstream afile;
UowStudents x [MAX];
char fileName[MAX];
char locale;
cout << "Enter filename: ";
cin >> fileName;
int size = fileToArray (afile, fileName, x, locale);
}
int fileToArray (fstream& afile, char fileName [], UowStudents* x, char& locale)
{
afile.open(fileName, ios::in);
if (!afile)
{
cout << fileName << "could not be opened for read" << endl;
exit (-1);
}
//file to array.
char dateJunk;
int numOfCourses;
int k = 0;
while (!afile.eof())
{
for (int k = 0; k < 3; k++)
{
afile >> locale;
switch (locale)
{
case 'F':
afile.getline(x[k].st.foreignStudent.nationality, MAX);
afile >> x[k].st.foreignStudent.gender;
afile.getline(x[k].st.foreignStudent.name, MAX);
afile >> x[k].st.foreignStudent.bd.day;
afile >> dateJunk;
afile >> x[k].st.foreignStudent.bd.month;
afile >> dateJunk;
afile >> x[k].st.foreignStudent.bd.year;
afile >> x[k].st.foreignStudent.numOfCourses;
for (int i = 0; i < x[k].st.foreignStudent.numOfCourses; i++)
{
afile >> x[i].st.foreignStudent.subjects[k];
afile >> x[i].st.foreignStudent.grades[k];
}
break;
case 'S':
afile >> x[k].st.localStudent.gender;
afile.getline(x[k].st.localStudent.name, MAX);
afile >> x[k].st.localStudent.bd.day;
afile >> dateJunk;
afile >> x[k].st.localStudent.bd.month;
afile >> dateJunk;
afile >> x[k].st.localStudent.bd.year;
afile >> x[k].st.localStudent.numOfCourses;;
for (int i = 0; i < x[k].st.localStudent.numOfCourses; i++)
{
afile >> x[i].st.localStudent.subjects[k];
afile >> x[i].st.localStudent.grades[k];
}
}
}
}
//Tests my cstring arrays to see everything is copied in correctly.
//The print for foreign student cstrings also has information for the
//one Singaporean student "S" in the middle. Singaporean must go into
//the local cstrings.
for (int k = 0; k < 3; k++)
{
cout << locale << " " << x[k].st.foreignStudent.nationality;
cout << endl;
cout << x[k].st.foreignStudent.gender;
cout << x[k].st.foreignStudent.name;
cout << endl;
cout << x[k].st.foreignStudent.bd.day << " - ";
cout << x[k].st.foreignStudent.bd.month << " - ";
cout << x[k].st.foreignStudent.bd.year;
cout << endl;
cout << x[k].st.foreignStudent.numOfCourses;
cout << endl;
for(int i = 0; i < x[k].st.foreignStudent.numOfCourses; i++)
{
cout << x[i].st.foreignStudent.subjects[k] << " ";
cout << x[i].st.foreignStudent.grades[k] << " ";
}
cout << endl;
}
//as you can see, everything gets copied into localStudent struct as well.
for (int k = 0; k < 3; k++)
{
cout << x[k].st.localStudent.gender;
cout << x[k].st.localStudent.name;
cout << endl;
cout << x[k].st.localStudent.bd.day << " - ";
cout << x[k].st.localStudent.bd.month << " - ";
cout << x[k].st.localStudent.bd.year;
cout << endl;
cout << x[k].st.localStudent.numOfCourses;
cout << endl;
for(int i = 0; i < x[k].st.localStudent.numOfCourses; i++)
{
cout << x[i].st.localStudent.subjects[k] << " ";
cout << x[i].st.localStudent.grades[k] << " ";
}
cout << endl;
}
return 0;
}
答案 0 :(得分：0)
您正在看到在C ++中使用union的效果。
所有成员共享同一块内存的联合的主要特征。并且因为您的成员结构Local和Foreigner具有完全相同的布局，您可以使用它们来打印联合的内容。
如果更改其中一个结构（例如，从nationality删除Local），您会看到某些打印输出变得混乱（特别是那些打印外国学生的条目本地的，反之亦然）。这是因为用于学生的记忆块对本地和外国学生的解释不同。
注意：从技术上讲，这会导致未定义的行为，但在这种情况下，结果不太可能是灾难性的。在官方上，您只能读取您上次写入的联合的同一成员，或者（如果成员是结构）结构的初始成员与最后写入联合的结构中的成员相同。