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我是“Qt for Android”的初学者，现在我正在使用它来开发基于移动的通信软件。我开发了java函数，将android的api作为.java文档中的一个类来调用。为了简化UI开发，UI是基于Qt Widget程序。然后我使用Qt-JNI类“QAndroidJniObject”根据Qt5.2 API文档调用这些java函数。Android资源文件存放在目录：./android/src/ com/comm/sipcall/SipCallSend.java。由于这方面的资料比较少，所以我根据API文档开发了java和c++程序。但是遇到了以下问题，也希望得到答案： 1）Qt程序是基于Qt Widget。java程序需要获取当前应用程序对象Context才能初始化java对象。我开发的c++代码如下：

//c++:
    QPlatformNativeInterface *interface = QApplication::platformNativeInterface();
    jobject activity = (jobject)interface->nativeResourceForIntegration("QtActivity");
    QAndroidJniObject* at = new QAndroidJniObject(activity);
    QAndroidJniObject appctx = at->callObjectMethod("getApplicationContext","()Landroid/content/Context;");

//.pro：
    QT += core gui gui-private

这是正确的吗？

2）java类包含一个构造函数和三个公共函数：

java:
package com.comm.sipcall;
improt XXXX
....
....
public class SipCallSend extends Activity {
    private Context context; // 接收QT的context
    public String sipToAddress = "";
    public String sipDomain = "";
    public String user_name = "";
    public String pass_word = "";

        public SipCallSend(){

        Log.i("ddd","init");
        sipToAddress = "";
        sipDomain = "";
        user_name = "";
        pass_word = "";
        }
        public void SetContext(Context cnt)
        {
            this.context = cnt;
            //Log.i("ccc",user_name);
        }
        public int Login(String domain,String username,String password){
            ....
            sipDomain = domain;
            user_name = username;
            pass_word = password;
            ...
            return 0;
        }
        public int Call(String addrNum) {
            ...
            return 0;
        }
}

C++：
    QPlatformNativeInterface *interface = QApplication::platformNativeInterface();
    jobject activity = (jobject)interface->nativeResourceForIntegration("QtActivity");
    QAndroidJniObject* at = new QAndroidJniObject(activity);
    QAndroidJniObject appctx = at->callObjectMethod("getApplicationContext","()Landroid/content/Context;");
    QAndroidJniObject* m_sipcall = new QAndroidJniObject("com/comm/sipcall/SipCallSend");
    if (!m_sipcall->isValid())
        return;
    m_sipcall->callMethod<void>("SetContext","(Landroid/content/Context;)V",
                                            appctx.object<jobject>());

    QAndroidJniObject domain = QAndroidJniObject::fromString("10.3.56.54");
    QAndroidJniObject username = QAndroidJniObject::fromString("1006");
    QAndroidJniObject password = QAndroidJniObject::fromString("1234");
    jint res = m_sipcall->callMethod<jint>("Login","(Ljava/lang/String;Ljava/lang/String;Ljava/lang/String;)I",
                                            domain.object<jstring>(), username.object<jstring>(),password.object<jstring>());

    if (res!=0)
        return;

    QAndroidJniObject addrNum = QAndroidJniObject::fromString("1018");
    res = m_sipcall->callMethod<jint>("Call","(Ljava/lang/String;)I",addrNum.object<jstring>());

“new QAndroidJniObject”返回非空，但是下面的函数没有执行，为什么？

3) 我曾尝试将 c++ 中的 Context 作为构造函数的参数，但我发现代码没有运行，为什么？

java:
        public SipCallSend(Context cnt){

        this.context = cnt;
        }
C++:
    QAndroidJniObject m_sipcall("com/comm/sipcall/SipCallSend","(Landroid/content/Context;)V",appctx.object<jobject>());

4）根据Qt5.2 API文档，QAndroidJniObject提供了一个名为“callObjectMethod”的方法：

QAndroidJniObject myJavaString; ==> "Hello, Java"
QAndroidJniObject mySubstring = myJavaString.callObjectMethod("substring", "(II)Ljava/lang/String;" 7, 10);

但是当我如下使用它时，IDE提示我参数不正确，为什么？

jint res = m_sipcall->callMethod<jint>("Login","(Ljava/lang/String;Ljava/lang/String;Ljava/lang/String;)I",
                                        domain.object<jstring>(), username.object<jstring>(),password.object<jstring>());

谢谢...

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