使用模板链接错误 [英] Link error using templates

问题描述

我将函数转换为模板，并开始遇到此错误。我一定不了解模板的局限性。有人可以告诉我为什么会这样吗？

I converted a function to a template, and started getting this error. I must not be understanding a limitation of templates. Can someone tell me why this is broken?

我收到此错误：

Undefined symbols:
  "bool foo<int>(int const&, int const&)", referenced from:
      _main in file1.o
ld: symbol(s) not found

当我链接以下代码时。该代码已简化，但仍然失败。第一个文件包含：

When I link the following code. The code is simplified, but still fails. The first file contains:

#include <iostream>
template <class T> bool foo (const T&, const T&);

int main ()
{
  int left = 1;
  int right = 2;

  if (foo <int> (left, right))
    std::cout << "foo!" << std::endl;

  return 0;
}

第二个文件包含：

template <class T> bool foo (const T& left, const T& right)
{
  return true;
}

推荐答案

由于Uri给的原因，模板方法通常在头文件中定义。因为您的函数是函数而不是类的方法，所以将其（在可能包含多个CPP文件的头文件中）显式定义为静态或内联。

For the reason Uri gave, template methods are usually defined in the header file. Because yours is a function and not a method of a class, explicitly define it (in the header file which may be included by more than one CPP file) as static or inline.

将其放入您的foo.h

template<class T> inline bool foo (const T& left, const T& right)
{
  return true;
}

将此内容放入main.cpp

#include <iostream>
#include "foo.h"

int main ()
{
  int left = 1;
  int right = 2;

  if (foo <int> (left, right))
    std::cout << "foo!" << std::endl;

  return 0;
}

cpp代码现在可以看到模板函数的整个声明。

The cpp code now sees the whole declaration of the template function.

此处列出了其他解决方案：如何避免模板函数出现链接器错误？

Other solutions are listed here: How can I avoid linker errors with my template functions?

这篇关于使用模板链接错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！