如果引用,我想将一些输入解析为public class MainApplication extends Application {
/**
* Dagger 2
*/
private AppComponent appComponent;
/**
* @param context
* @return
*/
public static AppComponent getAppComponent(Context context) {
return ((MainApplication) context.getApplicationContext()).appComponent;
}
@Override
public void onCreate() {
super.onCreate();
// Dagger 2
appComponent = DaggerAppComponent.builder()
.appModule(new AppModule(this))
.networkModule(new NetworkModule(getString(R.string.base_url)))
.build();
}
}
或long
。对此的合理解决方案是使用std::string
来存储数据。这是一个示例程序:
x3::variant<long, std::string>
我的预期结果是:
#include <iostream>
#include <string>
#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
namespace x3 = boost::spirit::x3;
const x3::rule<class number_tag, long> number = "number";
const auto number_def = x3::long_;
BOOST_SPIRIT_DEFINE(number);
const x3::rule<class string_tag, std::string> string = "string";
const auto string_def = x3::lexeme['"' >> *(x3::char_ - '"') >> '"'];
BOOST_SPIRIT_DEFINE(string);
using any_type = x3::variant<long, std::string>;
const x3::rule<class any_tag, any_type> any = "any";
const auto any_def = number | string;
BOOST_SPIRIT_DEFINE(any);
int main()
{
const std::string src = "18";
any_type result;
auto iter = src.begin();
bool success = x3::phrase_parse(iter, src.end(), any, x3::space, result);
if (!success || iter != src.end())
return 1;
else
std::cout << "Result: " << result << std::endl;
}
然而,实际结果很简单:
Result: 18
我做错了什么? Boost版本是1.61。
答案 0 :(得分:1)
您无法像这样打印变体。您必须将其传递给访客。例如(没有为转换完成错误检查):
struct Visitor
{
using result_type = long;
result_type operator()(long v) const { return v; }
result_type operator() (const std::string& v) { return std::atol(v.c_str()); }
};
应该从你的代码中调用,如:
if (!success || iter != src.end()) {
return 1;
} else {
Visitor v;
std::cout << "Result: " << boost::apply_visitor(v, result) << std::endl;
}