内联函数链接器错误 [英] inline function linker error

问题描述

我想使用特定类的内联成员函数。例如，没有内联的函数声明和实现如下：

I am trying to use inline member functions of a particular class. For example the function declaration and implementation without inlining is as such:

在头文件中：

int GetTplLSize();

：

int NeedleUSsim::GetTplLSize()
{
    return sampleDim[1];
}

由于某种原因，如果我将inline实现和声明，以及在这两个地方，我得到如下所示的链接器错误：

For some reason if I put the "inline" keyword in either one of the implementation and declaration, as well as in both places, I get linker errors as shown:

 Creating library C:\DOCUME~1\STANLEY\LOCALS~1\TEMP\MEX_HN~1\templib.x and object C:\DOCUME~1\STANLEY\LOCALS~1\TEMP\MEX_HN~1\templib.exp 
mexfunction.obj : error LNK2019: unresolved external symbol "public: int __thiscall NeedleUSsim::GetTplLSize(void)" (?GetTplLSize@NeedleUSsim@@QAEHXZ) referenced in function _mexFunction 
mexfunction.mexw32 : fatal error LNK1120: 1 unresolved externals 

  C:\PROGRA~1\MATLAB\R2008B\BIN\MEX.PL: Error: Link of 'mexfunction.mexw32' failed. 

为了摆脱这个错误，需要做些什么（例如，在做这些内联成员函数时我做错了什么） ？

What needs to be in order to get rid of this error (i.e. what am I doing wrong in terms of making these inline member functions)?

推荐答案

您需要将函数定义放入标题。提示编译器内联的最简单的方法是在类声明中包含方法体：

You need to put function definition into the header then. The simplest way to hint the compiler to inline is to include method body in the class declaration like:

class NeedleUSsim
{
  // ...
  int GetTplLSize() const { return sampleDim[1]; }
  // ...
};

或者，如果您坚持单独声明和定义：

or, if you insist on separate declaration and definition:

class NeedleUSsim
{
  // ...
  int GetTplLSize() const;
  // ...
};

inline int NeedleUSsim::GetTplLSize() const
{ return sampleDim[1]; }

定义必须在使用该方法的每个翻译单元中可见。

The definition has to be visible in each translation unit that uses that method.

这篇关于内联函数链接器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！