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我需要一个 Java 应用程序来调用非托管 C++。我将 MSVCR90.dll 从 Visual Studio 2008 redist 路径手动复制到 vmware 的 Windows Server Datacenter。

这是我得到的错误：

Java 运行时环境检测到一个致命错误：

  EXCEPTION_ACCESS_VIOLATION (0xc0000005) at pc=0x73b4ae7a, pid=1108, tid=2272

 JRE version: 6.0_38-b05
 Java VM: Java HotSpot(TM) Client VM (20.13-b02 mixed mode, sharing windows-x86 )
 Problematic frame:
 C  [MSVCR90.dll+0x3ae7a]

 An error report file with more information is saved as:
 ...\hs_err_pid1108.log

 If you would like to submit a bug report, please visit:
 http://java.sun.com/webapps/bugreport/crash.jsp
 The crash happened outside the Java Virtual Machine in native code.
 See problematic frame for where to report the bug.

这是 C++ 代码：

 #include "stdafx.h"

 #include <stdio.h>

 #include "CCCheckString.h"

 #include <vector>

 #include <String>

 using namespace std;
 #include "jobHandler.h"

JNIEXPORT jbolean JNICALL Java_CCCheckString_Login
  (JNIEnv *env, jobject object, jstring host, jstring UserName, jstring Domain, jstring Password)
{
    bool result;

    jobHandler *handler = new jobHandler(); 

     const char *hostStr = (env)->GetStringUTFChars(host, NULL);
     string hostS(hostStr);
     const char *UserNameStr = (env)->GetStringUTFChars(UserName, NULL);
     string UserNameS(UserNameStr);
     const char *DomainStr = (env)->GetStringUTFChars(Domain, NULL);
     string DomainS(DomainStr);
     const char *PasswordStr = (env)->GetStringUTFChars(Password, NULL);
     string PasswordS(PasswordStr); 

         //if comment this line everthing is okey 
     **result = handler->Login(hostS,UserNameS,DomainS,PasswordS);**

    (env)->ReleaseStringUTFChars(host, NULL);
    (env)->ReleaseStringUTFChars(UserName, NULL);
    (env)->ReleaseStringUTFChars(Domain, NULL);
    (env)->ReleaseStringUTFChars(Password, NULL);


    delete handler;

    return result;

}

下面是Java中的处理代码：

CCCheckString ccCheckString = new CCCheckString();

 result=ccCheckString.Login("xxx", "xxx", "xx", "xxx");

我该如何解决这个错误？

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1 回答 1

1

我解决了。

问题是 Login() 接受 std::string 但我们不能发送它

我们重新组织我们的 Login() 并且它必须采用 const char []

前任：

bool ClassName::Login(std::string host,std::string userName,std::string userDomain,std::string userPassword)
{ //Orginal Code Here ....}

bool ClassName::Login(const char host[],const char userName[],const char userDomain[],const char userPassword[])
{
std::string strHost(host);
std::string strUserName(userName);
std::string strUserDomain(userDomain);
std::string strUserPassword(userPassword);
return Login(strHost,strUserName,strUserDomain,strUserPassword);

}

一切都很好。

于 2013-02-05T14:03:21.533 回答