在c ++中以不同方式处理多个输入命令的正确方法是什么？

Question

我有一个程序从用户接收命令,它将以不同的方式处理不同的命令.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.

这是我的代码,但我想我误解了"工作方式".有人可以给我一个建议吗？并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,如果所有参数都被填充,则iss >> first >> second >> third将返回true,如果参数不足则返回false.但显然我错了.

Answer 1

你的问题很糟糕.这总是促使我使用Boost Spirit提供夸大的示例实现.

注意:请不要把它作为你的家庭作业.

通过以下示例输入查看Live on Coliru:

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make_visitor.hpp如此处所示时,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Live on Coliru

#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
Success

Answer 2

这样做:

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 ...

Answer 3

您的问题是使用>>运算符从流中读取并清除令牌.

if(iss >> command >> name >> height >> weight)

这(上图)尝试从流中读取4个令牌,并且对于每次成功读取,它都会清除流中的读取数据.

else if(iss >> command >> name >> height >> weight >> salary)

当你到达这个(上面)时,这意味着无法读取某些令牌并将其转换为适当的类型,但是可能至少已经从流中剥离了命令令牌.