using name space
时间: 2024-05-03 14:17:29 浏览: 125
命名空间(namespace)是一种将不同作用域中的标识符分隔开来的机制,它可以防止命名冲突,并允许程序中定义的标识符与其他库中定义的标识符重名而不会产生冲突。
在C++中,使用关键字“namespace”来定义命名空间,语法格式如下:
```
namespace namespace_name {
// 声明或定义变量、函数、类等标识符
}
```
在使用命名空间中的标识符时,需要在标识符前加上命名空间的名称和“::”运算符,例如:
```
namespace MyNamespace {
int my_variable;
void my_function() {
// 函数实现
}
}
int main() {
MyNamespace::my_variable = 10;
MyNamespace::my_function();
return 0;
}
```
相关问题
xiazai.py:10:0: C0301: Line too long (130/100) (line-too-long) xiazai.py:29:21: C0303: Trailing whitespace (trailing-whitespace) xiazai.py:30:0: W0311: Bad indentation. Found 10 spaces, expected 12 (bad-indentation) xiazai.py:40:0: C0301: Line too long (103/100) (line-too-long) xiazai.py:41:0: C0301: Line too long (153/100) (line-too-long) xiazai.py:53:0: C0305: Trailing newlines (trailing-newlines) xiazai.py:1:0: C0114: Missing module docstring (missing-module-docstring) xiazai.py:7:0: C0103: Constant name "url" doesn't conform to UPPER_CASE naming style (invalid-name) xiazai.py:13:13: W3101: Missing timeout argument for method 'requests.get' can cause your program to hang indefinitely (missing-timeout) xiazai.py:14:16: I1101: Module 'lxml.etree' has no 'HTML' member, but source is unavailable. Consider adding this module to extension-pkg-allow-list if you want to perform analysis based on run-time introspection of living objects. (c-extension-no-member) xiazai.py:19:0: C0103: Constant name "num" doesn't conform to UPPER_CASE naming style (invalid-name) xiazai.py:21:4: R1723: Unnecessary "elif" after "break", remove the leading "el" from "elif" (no-else-break) xiazai.py:24:17: W3101: Missing timeout argument for method 'requests.get' can cause your program to hang indefinitely (missing-timeout) xiazai.py:25:20: I1101: Module 'lxml.etree' has no 'HTML' member, but source is unavailable. Consider adding this module to extension-pkg-allow-list if you want to perform analysis based on run-time introspection of living objects. (c-extension-no-member) xiazai.py:28:8: C0103: Constant name "judge" doesn't conform to UPPER_CASE naming style (invalid-name) xiazai.py:28:31: C0209: Formatting a regular string which could be a f-string (consider-using-f-string) xiazai.py:30:22: C0209: Formatting a regular string which could be a f-string (consider-using-f-string) xiazai.py:31:14: C0209: Formatting a regular string which could be a f-string (consider-using-f-string) xiazai.py:34:8: C0103: Constant name "chapter_num" doesn't conform to UPPER_CASE naming style (invalid-name) xiazai.py:38:29: W3101: Missing timeout argument for method 'requests.get' can cause your program to hang indefinitely (missing-timeout) xiazai.py:39:32: I1101: Module 'lxml.etree' has no 'HTML' member, but source is unavailable. Consider adding this module to extension-pkg-allow-list if you want to perform analysis based on run-time introspection of living objects. (c-extension-no-member) xiazai.py:41:22: C0209: Formatting a regular string which could be a f-string (consider-using-f-string) xiazai.py:42:16: C0103: Constant name "all_content" doesn't conform to UPPER_CASE naming style (invalid-name) xiazai.py:44:20: R1713: Consider using str.join(sequence) for concatenating strings from an iterable (consider-using-join) ----------------------------------- Your code has been rated at 5.43/10
根据你提供的代码检查结果,有一些问题需要注意和修正:
1. 第10行代码的长度超过了100个字符,建议将其分割成多行或缩短长度。
2. 第29行代码存在尾部空格,建议删除。
3. 第30行代码的缩进不正确,应该是12个空格而不是10个空格。
4. 第40行和第41行的代码长度超过了100个字符,建议进行分割或缩短长度。
5. 第53行代码存在多余的换行符,建议删除。
6. 第1行代码缺少模块文档字符串,建议添加以提供模块的说明。
7. 第7行和第19行的常量命名不符合大写命名规范,建议改为大写命名。
8. 第13行和第24行的`requests.get`方法缺少超时参数,可能导致程序无限期挂起,建议添加超时参数。
9. 第14行和第25行的`lxml.etree`模块没有`HTML`成员,可能需要将其添加到`extension-pkg-allow-list`中。
10. 第21行的`elif`语句在`break`之后是不必要的,建议删除`elif`关键字。
11. 第28行代码中的字符串格式化可以改为使用f-string进行简化。
12. 第30行和第31行的字符串格式化也可以改为使用f-string进行简化。
13. 第34行和第44行的常量命名不符合大写命名规范,建议改为大写命名。
14. 第38行和第39行的`requests.get`方法缺少超时参数,可能导致程序无限期挂起,建议添加超时参数。
15. 第39行的`lxml.etree`模块没有`HTML`成员,可能需要将其添加到`extension-pkg-allow-list`中。
16. 第41行的字符串格式化可以改为使用f-string进行简化。
17. 第42行的常量命名不符合大写命名规范,建议改为大写命名。
18. 第44行的字符串连接可以考虑使用`str.join(sequence)`来简化。
以上是对你提供的代码进行的静态分析和建议,希望对你有所帮助。
7-3 Score Processing 分数 10 作者 翁恺 单位 浙江大学 Write a program to process students score data. The input of your program has lines of text, in one of the two formats: Student's name and student id, as <student id>, <name>, and Score for one student of one course, as <student id>, <course name>, <marks>. Example of the two formats are: 3190101234, Zhang San 3190101111, Linear Algebra, 89.5 Comma is used as the seperator of each field, and will never be in any of the fields. Notice that there are more than one word for name of the person and name of the course. To make your code easier, the score can be treated as double. The number of the students and the number of the courses are not known at the beginning. The number of lines are not known at the beginning either. The lines of different format appear in no order. One student may not get enrolled in every course. Your program should read every line in and print out a table of summary in .csv format. The first line of the output is the table head, consists fields like this: student id, name, <course name 1>, <course name 2>, ..., average where the course names are all the courses read, in alphabet order. There should be one space after each comma. Then each line of the output is data for one student, in the ascended order of their student id, with score of each course, like: 3190101234, Zhang San, 85.0, , 89.5, , , 87.3 For the course that hasn't been enrolled, leave a blank before the comma, and should not get included in the average. The average has one decimal place. There should be one space after each comma. And the last line of the output is a summary line for average score of every course, like: , , 76.2, 87.4, , , 76.8 All the number output, including the averages have one decimal place. Input Format As described in the text above. Output Format As described in the text above. The standard output is generated by a program compiled by gcc, that the round of the first decimal place is in the "gcc way". Sample Input 3180111435, Operating System, 34.5 3180111430, Linear Algebra, 80 3180111435, Jessie Zhao 3180111430, Zhiwen Yang 3180111430, Computer Architecture, 46.5 3180111434, Linear Algebra, 61.5 3180111434, Anna Teng Sample Output student id, name, Computer Architecture, Linear Algebra, Operating System, average 3180111430, Zhiwen Yang, 46.5, 80.0, , 63.2 3180111434, Anna Teng, , 61.5, , 61.5 3180111435, Jessie Zhao, , , 34.5, 34.5 , , 46.5, 70.8, 34.
很好,这是一道关于学生成绩数据处理的编程题。下面是我的解答:
```C++
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <iomanip>
using namespace std;
int main() {
map<string, string> name_map; // 学生姓名字典
map<string, vector<pair<string, double>>> score_map; // 学生成绩字典
map<string, double> sum_map; // 课程成绩和字典
map<string, int> cnt_map; // 课程成绩个数字典
string line; // 输入的一行数据
while (getline(cin, line)) {
int pos1 = line.find(", "); // 第一个逗号的位置
int pos2 = line.find(", ", pos1 + 2); // 第二个逗号的位置
string id = line.substr(0, pos1); // 学生 ID
string key = line.substr(pos2 + 2); // 课程名或学生姓名
if (key.find('.') != string::npos) { // 如果是学生成绩
double score = stod(key); // 成绩
string course = line.substr(pos1 + 2, pos2 - pos1 - 2); // 课程名
score_map[id].push_back(make_pair(course, score)); // 记录成绩
sum_map[course] += score; // 累加成绩和
cnt_map[course]++; // 记录个数
} else {
name_map[id] = key; // 记录学生姓名
}
}
vector<string> courses; // 课程列表
for (auto item : cnt_map) {
courses.push_back(item.first);
}
sort(courses.begin(), courses.end()); // 课程按字母顺序排序
cout << "student id, name, ";
for (string course : courses) {
cout << course << ", ";
}
cout << "average" << endl;
for (auto item : score_map) {
string id = item.first; // 学生 ID
string name = name_map[id]; // 学生姓名
cout << id << ", " << name << ", ";
double sum = 0; // 成绩总和
int cnt = 0; // 参加考试的课程数
for (string course : courses) {
bool found = false; // 是否参加考试
for (auto score : item.second) {
if (score.first == course) {
cout << fixed << setprecision(1) << score.second << ", ";
sum += score.second;
cnt++;
found = true;
break;
}
}
if (!found) {
cout << ", , ";
}
}
if (cnt > 0) {
cout << fixed << setprecision(1) << sum / cnt << endl;
} else {
cout << endl;
}
}
cout << ", , ";
for (string course : courses) {
cout << fixed << setprecision(1) << (cnt_map[course] > 0 ? sum_map[course] / cnt_map[course] : 0) << ", ";
}
cout << endl;
return 0;
}
```
这个程序的主要思路是使用四个字典来记录学生姓名、学生成绩、课程成绩和以及课程成绩个数,并且按照题目要求输出表格。
程序的详细注释如下:
1. 使用 `map<string, string> name_map` 记录每个学生的姓名,其中键为学生的 ID,值为学生的姓名。
2. 使用 `map<string, vector<pair<string, double>>> score_map` 记录每个学生的成绩,其中键为学生的 ID,值为包含课程名和成绩的一对 pair,使用 vector 来存储多个 pair。
3. 使用 `map<string, double> sum_map` 记录每门课程的成绩总和,其中键为课程名,值为成绩总和。
4. 使用 `map<string, int> cnt_map` 记录每门课程的成绩个数,其中键为课程名,值为成绩个数。
5. 使用 `getline(cin, line)` 读入每行输入,然后使用 `find()` 函数和 `substr()` 函数来解析每行输入。
6. 如果第二个逗号后面是数字,说明这是学生成绩,就把成绩和课程名记录在 `score_map`、`sum_map` 和 `cnt_map` 中。
7. 如果第二个逗号后面不是数字,说明这是学生姓名,就把学生姓名记录在 `name_map` 中。
8. 使用 `vector<string> courses` 记录所有参加考试的课程名,并且按字母顺序排序。
9. 使用 `cout` 输出表格头,先输出 `student id, name, `,然后按顺序输出所有课程名,最后输出 `average`。
10. 使用 `cout` 输出每个学生的成绩,先输出学生的 ID 和姓名,然后按顺序输出所有课程的成绩,如果某门课程没有参加考试,就留空,最后输出平均成绩。
11. 使用 `cout` 输出所有课程的平均成绩。
程序中使用了 `fixed` 和 `setprecision()` 函数来控制输出的小数点后一位。
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