问：

当我尝试编译代码时，我遇到了链接器错误。

当我使用编译时，出现以下错误：g++ *.cpp -std=c++11 -o run

main.cpp:(.text+0x355): undefined reference to `Actions(mBoard&, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::set<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::less<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > > >)'
collect2: error: ld returned 1 exit status

我尝试单独编译和链接所有文件，但无济于事。它不是成员函数，所以不是标签问题。我还试图用它来确保它实际上是在编译和链接这两个文件，但没有运气。g++ -std=c++11 main.cpp mAI.cpp -o run

真正让我抓狂的是，它并没有抱怨以与 Actions 函数相同的方式声明和定义的函数。Translate

任何帮助都非常感谢。

mAI.hpp（英语：mAI.hpp）

#ifndef MAI_HPP
#define MAI_HPP

#include <iostream>
#include <set>
#include <tuple>
#include "mBoard.hpp"
using namespace std;

void Actions(mBoard const &state, string const playerColor, set<string> moveList);
tuple<int, int> Translate(string index);

#endif

mAI.cpp

#include "mAI.hpp"

std::set<char> blackPieces = {'r', 'n', 'b', 'q', 'k', 'p'};
std::set<char> whitePieces = {'R', 'N', 'B', 'Q', 'K', 'P'};

void Actions(mBoard const &state, string const playerColor, set<string> moveList)
{
    //Do some stuff
}

tuple<int, int> Translate(string index)
{
    //Do different stuff
}

主要

#include "mAI.hpp"
#include "mBoard.hpp"
#include <set>
#include <tuple>
#include <string>
#include <iostream>

int main()
{

  mBoard dasBoard;
  set<string> moveList;
  string color = "White";

  cout<<"TEST - Translate: f6 to file, rank: ";
  tuple<int, int> test;
  test = Translate("f6");
  cout << get<0>(test) << ", " << get<1>(test) << endl;

  Actions(dasBoard, color, moveList);
  return 0;
}