架构x86_64(curlpp)的未定义符号

时间:2017-06-16 06:14:25

标签: c++ gcc macos-sierra clang++ curlpp

在我正在开发的项目中,我正在尝试使用 curlpp 库来创建一个简单的html GET请求。我使用以下选项将cpp文件传递给clang ++:

clang++ -std=c++11 -stdlib=libc++ -I /usr/local/Cellar url_test.cpp

然后我得到了这些错误:

Undefined symbols for architecture x86_64:
  "curlpp::OptionBase::OptionBase(CURLoption)", referenced from:
      curlpp::Option<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > >::Option(CURLoption, std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > const&) in url_test-541b93.o
  "curlpp::OptionBase::~OptionBase()", referenced from:
      curlpp::Option<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > >::Option(CURLoption, std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > const&) in url_test-541b93.o
      curlpp::Option<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > >::~Option() in url_test-541b93.o
  "curlpp::UnsetOption::UnsetOption(char const*)", referenced from:
      curlpp::Option<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > >::updateMeToOption(curlpp::OptionBase const&) in url_test-541b93.o
      curlpp::OptionTrait<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >, (CURLoption)10002>::updateHandleToMe(curlpp::internal::CurlHandle*) const in url_test-541b93.o
      curlpp::Option<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > >::getValue() const in url_test-541b93.o
  "curlpp::RuntimeError::~RuntimeError()", referenced from:
      curlpp::UnsetOption::~UnsetOption() in url_test-541b93.o
  "curlpp::libcurlRuntimeAssert(char const*, CURLcode)", referenced from:
      void curlpp::internal::CurlHandle::option<void*>(CURLoption, void*) in url_test-541b93.o
  "curlpp::Easy::perform()", referenced from:
      _main in url_test-541b93.o
  "curlpp::Easy::Easy()", referenced from:
      _main in url_test-541b93.o
  "curlpp::Easy::~Easy()", referenced from:
      _main in url_test-541b93.o
  "curlpp::Cleanup::Cleanup()", referenced from:
      _main in url_test-541b93.o
  "curlpp::Cleanup::~Cleanup()", referenced from:
      _main in url_test-541b93.o
  "curlpp::OptionBase::operator<(curlpp::OptionBase const&) const", referenced from:
      vtable for curlpp::OptionTrait<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >, (CURLoption)10002> in url_test-541b93.o
      vtable for curlpp::Option<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > > in url_test-541b93.o
  "typeinfo for curlpp::LogicError", referenced from:
      GCC_except_table0 in url_test-541b93.o
  "typeinfo for curlpp::OptionBase", referenced from:
      curlpp::Option<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > >::updateMeToOption(curlpp::OptionBase const&) in url_test-541b93.o
      typeinfo for curlpp::Option<std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> > > in url_test-541b93.o
  "typeinfo for curlpp::RuntimeError", referenced from:
      GCC_except_table0 in url_test-541b93.o
      typeinfo for curlpp::UnsetOption in url_test-541b93.o
  "_curl_easy_setopt", referenced from:
      void curlpp::internal::CurlHandle::option<void*>(CURLoption, void*) in url_test-541b93.o
ld: symbol(s) not found for architecture x86_64

我认为这意味着编译器找不到任何curlpp库。

这是我正在尝试运行的代码:

  1 #include <string>
  2 #include <sstream>
  3 #include <iostream>
  4 #include <curlpp/cURLpp.hpp>
  5 #include <curlpp/Easy.hpp>
  6 #include <curlpp/Options.hpp>
  7 #include <fstream>
  8 
  9 using namespace curlpp::options;
 10 
 11 int main(int, char **)
 12 {
 13     try
 14     {
 15         curlpp::Cleanup testCleanup;
 16         curlpp::Easy miRequest;
 17         miRequest.setOpt<Url>("http://www.wikipedia.org");
 18         miRequest.perform();
 19     }
 20     catch(curlpp::RuntimeError & e)
 21     {
 22         std::cout << e.what() << std::endl;
 23     }
 24     catch(curlpp::LogicError & e)
 25     {
 26         std::cout << e.what() << std::endl;
 27     }
 28 
 29     return 0;
 30 }

我正在使用Xcode命令行工具和自制软件安装运行macOS Sierra 10.12.4。我自制了curlpp库。我知道其他人能够使用gcc在Ubuntu 16.04上编译这个项目,所以我认为这个问题与macOS有关。

我对cpp很新,所以任何帮助都会非常感激!

2 个答案:

答案 0 :(得分:1)

您必须告诉编译器哪些库与您的可执行文件链接。在您的情况下,通过添加选项-lcurlpp -lcurl来解决问题。

顺便说一句,您提供的路径不正确,因为标题实际上位于 / usr / local / Cellar / curlpp / [在此处插入版本] / include 。无论如何,这可以很好地编译,而无需为通过自制软件安装的库添加包含搜索路径,因为它会自动在 / usr / local / include 中创建符号链接,这是编译器搜索的默认位置之一。

答案 1 :(得分:0)

您的问题是您没有链接您使用的库。添加库,你应该很好。