JavaCPP 错误：无法将 `char*` 转换为 `Abc*` 以将参数 '1' 转换为 `void testMyObject(Abc*)`

Posted 2023-02-17

【中文标题】JavaCPP 错误：无法将 `char*` 转换为 `Abc*` 以将参数 \'1\' 转换为 `void testMyObject(Abc*)`

【英文标题】：JavaCPP error: cannot convert `char*` to `Abc*` for argument ‘1’ to `void testMyObject(Abc*)`JavaCPP 错误：无法将 `char*` 转换为 `Abc*` 以将参数 '1' 转换为 `void testMyObject(Abc*)`

【发布时间】：2020-06-22 07:15:37

【问题描述】：

我有一个带有外部函数的简单 cpp 文件。一个函数返回类 Abc 的引用，另一个接收它并调用某个方法：

extern "C" 

void* getMyObject(int arg1, int arg2)
     return new Abc();


void testMyObject(Abc* p) 
     p->test()

还有一个调用外部函数的 Java 代码：

@Platform(include="export.cpp", link="mylib")
class MyIterface 
  static  
     Loader.load()
  

public static native Pointer getMyObject(int a, int b);
public static native void testMyObject(Pointer op);

我这样称呼他们：

MyInterface lib = new MyInterface();
Pointer op = lib.getMyObject(111, 222); //get the reference to Abc object
lib.testMyObject(op); //Pass the Abc pointer to invoke its method

但我收到以下错误："

/home/Project/target/classes/com/proj/dl4j/jniMyInterface.cpp: In function ‘void Java_com_proj_dl4j_MyInterface_testMyPointer(JNIEnv*, jclass, jobject)’:
/home/Project/target/classes/com/proj/dl4j/jniMyInterface.cpp:629:27: error: cannot convert ‘char*’ to ‘Abc*’ for argument ‘1’ to ‘void testMyPointer(Abc*)’
         testMyPointer(ptr0);
                           ^
[INFO] ------------------------------------------------------------------------
[INFO] BUILD FAILURE
[INFO] ------------------------------------------------------------------------
[INFO] Total time:  5.464 s
[INFO] Finished at: 2020-06-22T12:22:47+05:30
[INFO] ------------------------------------------------------------------------
[ERROR] Failed to execute goal org.bytedeco:javacpp:1.5.3:build (process-classes) on project MyProject: Execution process-classes of goal org.bytedeco:javacpp:1.5.3:build failed: Process exited with an error: 1 -> [Help 1]

我在void* 上进行了投射，即：Abc* ob = (Abc*)p 有效。但我仅限于修改这些外部函数（设计问题！！）。有什么解决办法吗？？

【参考方案1】：

将方法修改为public static native void testMyObject(@Cast("Abc *") Pointer op); 成功了！