Question

这是我想写的代码。

int opt;
po::options_description desc("Allowed options");
desc.add_options()
    ("help","produce help message")
    ("compression",po::value<int>(&opt)->default_value(10),"optimization level")
    ("include-path,I", po::value< std::vector<std::string> >(), "include path")
    ("input file", po::value< std::vector<std::string> >(),"input file")
    ;


po::variables_map vm;
po::store(po::parse_command_line(argc,argv,desc),vm);
po::notify(vm);

if (vm.count("help")){
    std::cout <<desc<<"\n";
    return 1;
}
if (vm.count("compression")){
    std::cout<<"Compression level was set to"<<vm["compression"].as<int>()<<".\n";
} else {
    std::cout << "compression level was not set.\n";
} 
if (vm.count("include-path")){
    std::cout << "Include paths are:     " << vm["include-path"].as< std::vector<std::string> > ()<< "\n";
    }

编译器为最终的if语句提供错误，我想打印 include-path 参数。它给出的错误是

rx_timed_samples.cpp：62：96：错误：'std :: operator＆lt;＆lt;'中的'operator＆lt;＆lt;'不匹配[with _Traits = std :: char_traits]（（*＆amp; std :: cout），（（const char *）“Include paths are：”））＆lt;＆lt; （＆amp; vm.boost :: program_options :: variables_map :: operator []（（*＆amp; std :: basic_string（（（const char *）“include-path”），（（const std :: allocator） ）（＆amp; std :: allocator（））））））） - ＆gt; boost :: program_options :: variable_value ::与T = std :: vector，std :: allocator＆gt; ＆GT;” 的

我不明白吗？请帮忙。

Answer 1

我认为这里的问题是没有＆lt;＆lt;运算符是为std::vector<std::string>定义的。这次电话需要哪个：

std::cout << "Include paths are:     " << vm["include-path"].as< std::vector<std::string> > ()<< "\n";

Answer 2

您需要可以处理向量的流运算符的专门化。

template<class T>
std::ostream& operator <<(std::ostream& os, const std::vector<T>& v)
{
    std::copy(v.begin(), v.end(), std::ostream_iterator<T>(std::cout, " ")); 
    return os;
}

带有boost库C ++的向量字符串给出了错误

2 个答案: