在c++中以不同方式处理多个输入命令的正确方法是什么？
在c++中以不同方式处理多个输入命令的正确方法是什么？
提问于 2013-07-21 09:06:48
我有一个程序，接受来自用户的命令，它将处理不同的命令不同。例如：
ADD_STUDENT ALEX 5.11 175
ADD_TEACHER MERY 5.4 120 70000
PRINT MERY
REMOVE ALEX
PRINT TEACHER SALARY
PRINTALL 因此，我需要检查每一行，看看输入是由什么组成的。
以下是我的代码，但我想我误解了iss<<的工作方式。有人能给我一个建议吗？告诉我为什么我的代码不能像我预期的那样工作？
string line;
while(getline(cin, line))
{
//some initialization of string, float variable
std::istringstream iss(line);
if(iss >> command >> name >> height >> weight)
..examine the command is correct(ADD_STUDENT) and then do something..
else if(iss >> command >> name >> height >> weight >> salary)
..examine the command is correct(ADD_TEACHER) and then do something...
else if(iss >> command >> name)
..examine the command is correct(REMOVE) and then do somethin...
}我的想法是，如果所有参数都已填充，iss>> first >>second >> third将返回true，如果参数不足，则返回false。但显然我错了。
回答 5
Stack Overflow用户
回答已采纳
发布于 2013-07-21 09:15:38
如下所示：
iss >> command;
if (!iss)
cout << "error: can not read command\n";
else if (command == "ADD_STUDENT")
iss >> name >> height >> weight;
else if (command == "ADD_TEACHER")
iss >> name >> height >> weight >> salary;
else if ...Stack Overflow用户
发布于 2013-07-22 04:32:26
你的问题被巧妙地低估了。这总是促使我使用Boost Spirit提供一个夸张的示例实现。
备注：请不要把这个作为你的家庭作业交上去。
使用以下示例输入查看：
ADD_STUDENT ALEX 5.11 175
ADD_STUDENT PUFF 6 7
ADD_STUDENT MAGIC 7 8
ADD_STUDENT DRAGON 8 9
ADD_TEACHER MERY 5.4 120 70000
PRINT MERY
ADD_TEACHER DUPLO 5.4 120 140000
PRINTALL 10
REMOVE ALEX
PRINT TEACHER SALARY
PRINT MERY PUFF MAGIC DRAGON
REMOVE MERY PUFF MAGIC DRAGON
PRINT TEACHER SALARY完整代码：
更新当包含如图所示的make_visitor.hpp时，您可以更优雅地编写访问者代码：
auto print_salary = [&] ()
{
for(auto& p : names)
boost::apply_visitor(make_visitor(
[](Teacher const& v) { std::cout << "Teacher salary: " << v.salary << "\n"; },
[](Student const& v) {}),
p.second);
};请参阅改编的示例
#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
namespace qi = boost::spirit::qi;
namespace phx= boost::phoenix;
struct Person
{
std::string name;
double height, weight;
friend std::ostream& operator<<(std::ostream& os, Person const& s) {
return os << "Person { name:" << s.name << ", height:" << s.height << ", weight:" << s.weight << " }";
}
};
struct Student : Person
{
Student() = default;
Student(std::string n, double h, double w) : Person {n,h,w} {}
};
struct Teacher : Person
{
Teacher() = default;
Teacher(std::string n, double h, double w, double s) : Person {n,h,w}, salary(s) {}
double salary;
};
int main()
{
std::stringstream ss;
ss << std::cin.rdbuf();
std::map<std::string, boost::variant<Student, Teacher> > names;
using namespace qi;
auto add_student = phx::ref(names)[_1] = phx::construct<Student>(_1, _2, _3);
auto add_teacher = phx::ref(names)[_1] = phx::construct<Teacher>(_1, _2, _3, _4);
auto remove = phx::erase(phx::ref(names), _1);
auto print_all = [&] (int i) { for(auto& p : names) { std::cout << p.second << "\n"; if (--i==0) break; } };
auto print_salary = [&] ()
{
struct _ : boost::static_visitor<> {
void operator()(Teacher const& v) const { std::cout << "Teacher salary: " << v.salary << "\n"; }
void operator()(Student const& v) const { }
} v_;
for(auto& p : names) boost::apply_visitor(v_, p.second);
};
auto name_ = as_string[lexeme[+graph]];
if (phrase_parse(begin(ss.str()), end(ss.str()),
(
("ADD_STUDENT" >> name_ >> double_ >> double_) [ add_student ]
| ("ADD_TEACHER" >> name_ >> double_ >> double_ >> double_) [ add_teacher ]
| (eps >> "PRINT" >> "TEACHER" >> "SALARY") [ print_salary ]
| ("PRINTALL" >> int_) [ phx::bind(print_all, _1) ]
| ("PRINT" >> +name_ [ std::cout << phx::ref(names)[_1] << std::endl ])
| ("REMOVE" >> +name_ [ remove ])
) % +eol,
qi::blank))
{
std::cout << "Success";
}
else
{
std::cout << "Parse failure";
}
}输出：
Person { name:MERY, height:5.4, weight:120 }
Person { name:ALEX, height:5.11, weight:175 }
Person { name:DRAGON, height:8, weight:9 }
Person { name:DUPLO, height:5.4, weight:120 }
Person { name:MAGIC, height:7, weight:8 }
Person { name:MERY, height:5.4, weight:120 }
Person { name:PUFF, height:6, weight:7 }
Teacher salary: 140000
Teacher salary: 70000
Person { name:MERY, height:5.4, weight:120 }
Person { name:PUFF, height:6, weight:7 }
Person { name:MAGIC, height:7, weight:8 }
Person { name:DRAGON, height:8, weight:9 }
Teacher salary: 140000
SuccessStack Overflow用户
发布于 2013-07-21 09:27:28
您的问题是使用>>操作符从流中读取和清除令牌。
if(iss >> command >> name >> height >> weight)这个(上图)尝试从流中读取4个令牌，每次成功读取时，它都会从流中清除读取的数据。
else if(iss >> command >> name >> height >> weight >> salary)当你达到这一点(上面)时，这意味着一些令牌无法被读取并转换为适当的类型，但是很可能至少命令令牌已经从流中剥离。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/17768152
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