Question

所有

我制作了一组c ++代码如下然后，我成功地用g ++将每个.cpp编译为.o 但是，我收到了链接错误

 Undefined symbols for architecture x86_64:
  "derive1<3>::derive1(int)", referenced from:
      derive2::derive2(int)in derive2.o
      derive2::derive2(int)in derive2.o
ld: symbol(s) not found for architecture x86_64
collect2: ld returned 1 exit status

你可以告诉我发生了什么事吗？如果我将所有代码片段绑定在一个文件中，我没有任何错误，并且它完全按照我的意图运行。我假设我应该在某处引入extern关键字，但我不知道该怎么做。

我的编译器是

i686-apple-darwin11-llvm-g ++ - 4.2（GCC）4.2.1（基于Apple Inc.） build 5658）（LLVM build 2336.11.00）

“base.hpp”

class base {
public:
  virtual int hoge( int num ) = 0;

};

“derive1.hpp”

#include "base.hpp"
#include <iostream>

template <int N>
class derive1 : public base {
public:
  explicit derive1( int n );
  virtual int hoge( int num ) {
    std::cout << "num = " << num << std::endl
          << "N = " << N << std::endl;
    return N;
  }    

};

“derive1.cpp”

#include "derive1.hpp"

template <int N>
derive1<N>::derive1 ( int n ) {
  std::cout << "This is the constructer of derive1" << std::endl
        << "N = " << N << ", n = " << n << std::endl;
}

“derive2.hpp”

#include "derive1.hpp"
#include <iostream>

class derive2 : public derive1<3> {
public:
  explicit derive2( int n );
};

“derive2.cpp”

#include "derive2.hpp"

derive2::derive2( int n ) : derive1<3>( n ) {
  std::cout << "This is the constructer of derive2" << std::endl
        << "n = " << n << std::endl;
}

“derive_test.cpp”

#include<iostream>
#include "derive2.hpp"

int main() {

  derive2 test(3);

  std::cout << "return value = "
        << test.hoge( 5 )
        << std::endl;
  return 0;
}

Answer 1

您不应该为模板类拆分.h和.cpp文件的定义和实现。

What is an undefined reference/unresolved external symbol error and how do I fix it?

Answer 2

不幸的是，大多数编译器都要求模板函数的实现（在这种情况下template <int N> derive1<N>::derive1 ( int n )可以使用它来编译单元。将它移动到“derive1.hpp”。

Answer 3

简单的答案是，您必须将所有模板化代码（derive1::derive1(int)等）放入头文件中，而不是放入自己的翻译单元中。编译器在尝试实例化模板时需要查看完整定义。当然，你可以在*.cpp文件中有模板的定义，但是你必须明确地规范它，并在头文件中声明这种特殊化。

非类型模板参数的链接器错误

3 个答案: