Java – JNI pass parameters to method of c++

cjavajava-native-interface

I have a c++ file myCppTest.cpp which has method

int myFunction(int argv, char **argc) {
}

and
a Java native method in myClass.java

public native int myFunction (int argv, char[][] argc);

After generate the header file using javah -jni myClass, i have the header

JNIEXPORT jint JNICALL Java_JPTokenizer_init
  (JNIEnv *, jobject, jint, jobjectArray);

In my myClass.cpp, I defined

JNIEXPORT jint JNICALL Java_JPTokenizer_init
  (JNIEnv *env, jobject obj, jint argv, jobjectArray argc) {
        //need to call int myFunction(int argv, char **argc) in myCppTest.cpp 
}

How could I pass the arguments "jint argv, jobjectArray argc" to "int argv, char **argc", thanks.

EDIT:

I THINK I MADE A MISTAKE

The Java native method in myClass.java should be

public native int init (int argv, char[][] argc);

So there is

JNIEXPORT jint JNICALL Java_myClass_init (JNIEnv *, jobject, jint, jobjectArray);

generated after javah.
And in myClass.cpp, i have

JNIEXPORT jint JNICALL Java_myClass_init (JNIEnv *env, jobject obj, jint argv, jobjectArray argc) { //need to call int myFunction(int argv, char **argc) in myCppTest.cpp }

Best Answer

There is no direct mapping between Java objects and C++ primitives, so you will have to convert the arguments that are passed by the Java runtime environment, and then call your function.

Java will call Java_JPTokenizer_init -- this is where you perform your conversion and invoke your "plain old" C++ function.

To convert the array of strings, you will first need to access the array, then the individual strings.

For array access, see http://java.sun.com/docs/books/jni/html/objtypes.html#5279.
For string access, see http://java.sun.com/docs/books/jni/html/objtypes.html#4001.

Related Solutions

Java – Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.

Java – Failed to load the JNI shared Library (JDK)

You need a 64-bit trio:

64-bit OS
64-bit Java
64-bit Eclipse

Best Answer

Related Solutions

Java – Is Java “pass-by-reference” or “pass-by-value”

Java – Failed to load the JNI shared Library (JDK)

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