使用bool_switch，我可以编写一个命令行选项来打开标志：
bool flag;
po::options_description options;
options.add_options()
("on", po::bool_switch(&flag)->default_value(false))
;
现在./a.out flag==false和./a.out --on将flag==true options.add_options()
("on", po::bool_switch(&flag)->default_value(false))
("off", po::anti_bool_switch(&flag)) // ????
;
。但是，为了明确，我还想添加一个命令行选项来关闭标志 off 。类似的东西：
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，还是我基本上必须编写代理bool引用？
答案 0 :(得分：3)
我能够提出的一件事（不确定这是否是最好的方法）是使用implicit_value()：
po::typed_value<bool>* store_bool(bool* flag, bool store_as)
{
return po::value(flag)->implicit_value(store_as)->zero_tokens();
}
value必须使用所需的默认值进行初始化，否则符合所需的功能：
bool value = false;
options.add_options()
("on", store_bool(&value, true))
("off", store_bool(&value, false))
;
答案 1 :(得分：1)
我不确定你的要求是否合理。如果用户输入“./a.out --on --off”，会发生什么？
否则，您只需确保在用户输入“./a.out --off”并且您将获得所需行为时，不会收到“无法识别的选项”消息。
bool flag;
bool flag_off
po::options_description options;
options.add_options()
("on", po::bool_switch(&flag)->default_value(false))
("off", po::bool_switch(&flag_off)->default_value(false)) ;
...
if( flag_on && flag_off )
{
cout << "nasty error" << endl; exit(1)
}