未定义对作为lib .a生成并使用CMake的方法的引用。

huangapple
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英文:

undefined reference to method generated as lib .a with CMake

问题

你的函数AllToAll在示例中为什么未定义可以解释一下吗？我使用CMake生成了一个名为libNeuralNetwork.a的库，该库在示例中被调用。

LayerFactory.hpp

  1. #pragma once
  2. #include "LayerModel.hpp"
  3. #include "Layer.hpp"
  5. namespace nn
  6. {
  7.     extern internal::LayerModel AllToAll(int numberOfNeurons, activationFunction activation = sigmoid);
  8. }

LayerFactory.cpp

  1. #include "LayerFactory.hpp"
  2. #include "AllToAll.hpp"
  4. using namespace nn;
  5. using namespace internal;
  7. LayerModel AllToAll(int numberOfNeurons, activationFunction activation)
  8. {
  9.     LayerModel model
  10.     {
  11.         allToAll,
  12.         activation,
  13.         numberOfNeurons
  14.     };
  15.     return model;
  16. }

NeuralNetwork.hpp

  1. #pragma once
  2. #include "layer/LayerModel.hpp"
  3. #include "layer/LayerFactory.hpp"
  5. namespace nn
  6. {
  7. 	class NeuralNetwork
  8. 	{
  9.     public:
  10. 		NeuralNetwork(int numberOfInputs, std::vector<internal::LayerModel> models);
  11.         //...
  12.     };
  13. }

Example.cpp

  1. #include "../src/neural_network/NeuralNetwork.hpp"
  3. using namespace nn;
  5. int example1()
  6. {
  7.     NeuralNetwork neuralNetwork(3, {AllToAll(5), AllToAll(2)});
  8. }

错误消息：

  1. CMakeFiles/UnitTests.out.dir/ExamplesTest.cpp.o: In function `example1()':
  2. ExamplesTest.cpp:(.text+0x8b3): undefined reference to `nn::AllToAll(int, nn::activationFunction)'

(Note: I removed the HTML entity codes like &quot; in your code snippets to make the code more readable in the translation.)

英文:

Can you explain me why my function AllToAll is undefined in my example? I use CMake to generate a libNeuralNetwork.a which is called by the exemple.

LayerFactory.hpp

  1. #pragma once
  2. #include &quot;LayerModel.hpp&quot;
  3. #include &quot;Layer.hpp&quot;
  5. namespace nn
  6. {
  7.     extern internal::LayerModel AllToAll(int numberOfNeurons, activationFunction activation = sigmoid);
  8. }

LayerFactory.cpp

  1. #include &quot;LayerFactory.hpp&quot;
  2. #include &quot;AllToAll.hpp&quot;
  4. using namespace nn;
  5. using namespace internal;
  7. LayerModel AllToAll(int numberOfNeurons, activationFunction activation)
  8. {
  9.     LayerModel model
  10.     {
  11.         allToAll,
  12.         activation,
  13.         numberOfNeurons
  14.     };
  15.     return model;
  16. }

NeuralNetwork.hpp

  1. #pragma once
  2. #include &quot;layer/LayerModel.hpp&quot;
  3. #include &quot;layer/LayerFactory.hpp&quot;
  5. namespace nn
  6. {
  7. 	class NeuralNetwork
  8. 	{
  9.     public:
  10. 		NeuralNetwork(int numberOfInputs, std::vector&lt;internal::LayerModel&gt; models);
  11.         //...
  12.     };
  13. }

Example.cpp

  1. #include &quot;../src/neural_network/NeuralNetwork.hpp&quot;
  3. using namespace nn;
  5. int example1()
  6. {
  7. 	NeuralNetwork neuralNetwork(3, {AllToAll(5), AllToAll(2)});
  8. }

error message:

  1. CMakeFiles/UnitTests.out.dir/ExamplesTest.cpp.o: In function `example1()&#39;:
  2. ExamplesTest.cpp:(.text+0x8b3): undefined reference to `nn::AllToAll(int, nn::activationFunction)&#39;

答案1

得分: 2

你已经在顶级命名空间中声明了AllToAll，并在nn命名空间中定义了它。

以下内容不会在命名空间中声明函数：

  1. using namespace foo;
  3. extern void Bar();

你需要：

  1. namespace foo {
  2.   extern void Bar();
  3. }
英文:

You have declared AllToAll in the top-level namespace and defined it in the nn namespace.

The following will not declare the function in the namespace:

  1. using namespace foo;
  3. extern void Bar();

You need:

  1. namespace foo {
  2.   extern void Bar();
  3. }

huangapple
  • 本文由 发表于 2020年1月3日 21:55:10
  • 转载请务必保留本文链接：https://go.coder-hub.com/59579808.html
  • c++
  • cmake
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