Application is causing exception c000001d on a certain computer in 64-bit mode
Hi,
my testing application, available as 32/64 for XP and newer, is working almost everywhere. But I had just a second report that on a certain i7 system with Windows 7 an exception c000001d occurs and the app doesn't even start. Here's the log:
Problem signature:
Problem Event Name: APPCRASH
Application Name: mwindowstest64.exe
Application Version: 0.0.0.0
Application Timestamp: 53eb8d37
Fault Module Name: mwindowstest64.exe
Fault Module Version: 0.0.0.0
Fault Module Timestamp: 53eb8d37
Exception Code: c000001d
Exception Offset: 0000000000caeb53
OS Version: 6.1.7600.2.0.0.256.1
Locale ID: 1043
Additional Information 1: 9a73
Additional Information 2: 9a735485f25f4708cf6362595f6a2ca1
Additional Information 3: 3833
Additional Information 4: 38330f6cd6677b6d586238fd85fa379b
What could cause this? How can I check the potential problem? Is there a way to view what code is on the address 0000000000caeb53 in Visual
Studio Express 2013?
Thanks in advance.
Hi,
Just to clarify this issue, is it the time that project throw exception when debugging that project with your VS2013?
If not, and if you have the source code, I would recommend you debug it with VS2013 to get which line throws this exception.
In addition, since this is a development issue, I would recommend you consider posting this issue in develop forum depends on the language you are using.
C# forum
Vb.net Forum
C++ Forum
Regards.
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try {
finally {
}pattern is used, without the catch block, purelly for flow control purpose, a return statement in the finally block could be problematic. If a finally block ends with a return statement, any exception that might be thrown in the try block would be ignored. The unhandled exception will not propergate, it simply lost in the midst. Because any code could throw unchecked exception, if the exception cannot afford to be lost, then one should be careful not to use return in the finally block. See the following test code:
class Finally {
void test() {
int result;
System.out.println("test()");
try {
result = f1();
System.out.println("result = " + result);
} catch(Exception e) {
System.out.println("Caught an Exception: " + e);
try {
result = f2();
System.out.println("result = " + result);
} catch(Exception e) {
System.out.println("Caught an Exception: " + e);
try {
f3();
} catch(Exception e) {
System.out.println("Caught an Exception: " + e);
try {
f4();
} catch(Exception e) {
System.out.println("Caught an Exception: " + e);
int f1() {
System.out.println("f1()");
int a = 0;
int b = 0;
try {
a = 1/0; // this generate an unchecked exception
b = 1;
} finally {
return b; // this return clobber the unchecked exception
int f2() {
System.out.println("f2()");
int a = 0;
int b = 0;
a = 1/0; // this generate an unchecked exception
b = 1;
return b;
void f3() {
System.out.println("f3()");
int a = 0;
int b = 0;
try {
a = 1/0; // this generate an unchecked exception
b = 1;
} finally {
return; // this return clobber the unchecked exception
void f4() {
System.out.println("f4()");
int a = 0;
int b = 0;
try {
a = 1/0; // this generate an unchecked exception
b = 1;
} finally {
public static void main(String[] args) {
Finally f = new Finally();
f.test();
/code]
Here are the results:test()
f1()
result = 0
f2()
Caught an Exception: java.lang.ArithmeticException: / by zero
f3()
f4()
Caught an Exception: java.lang.ArithmeticException: / by zero
We see that the devide by zero exception throw by f1() simply vanished, the unintended result 0, instead of the intended result 1, is returned. The test() method did not detect any abnormality in f1(), which is scary. To drive the point home, we try f3() and f4(), which has void return type. A simple return statement in f3() causes exception be lost. The stack frame seems intact, for otherwise, the test() method would exit abruptly before f2() even get a chance to be called. -
Code causing exceptions at its own sweet time and place
import java.lang.*;
class newthread implements Runnable
Thread t;
int click=0;
public volatile boolean running=true;
newthread(int p)
t=new Thread(this);
t.setPriority(p);
void start()
t.start();
public void run()
while(running)
click++;
void start1()
running=true;
void stop()
running=false;
class threadpriority2
public static void main(String argv[])
newthread ob1=new newthread(Thread.NORM_PRIORITY+2);
newthread ob2=new newthread(Thread.NORM_PRIORITY-2);
ob1.start();
ob2.start();
try
Thread.sleep(5000);
catch(Exception e)
System.out.println("error detected:"+e);
System.out.println("first one:"+ob1.click);
System.out.println("first one:"+ob2.click);
System.out.println("the priority of ob1 is:"+ob1.t.getPriority());
System.out.println("the priority of ob2 is:"+ob2.t.getPriority());
ob1.stop();
ob2.stop();
ob2.t.setPriority(Thread.NORM_PRIORITY+3);
ob1.t.setPriority(Thread.NORM_PRIORITY-3);
ob1.start1();
ob2.start1();
ob1.start();
ob2.start();
System.out.println("the priority of ob1 is:"+ob1.t.getPriority());
System.out.println("the priority of ob2 is:"+ob2.t.getPriority());
try
Thread.sleep(5000);
catch(Exception e)
System.out.println("error detected:"+e);
ob1.stop();
ob2.stop();
System.out.println("first one:"+ob1.click);
System.out.println("first one:"+ob2.click);
errors during runtime:
Exception in thread "main" java.lang.IllegalThreadStateException
at java.lang.Thread.start(Unknown Source)
at newthread.start(threadpriority2.java:17)
at threadpriority2.main(threadpriority2.java:77)
when i run this code it causes exceptions at its own sweet locations when i run it separately.if once it displays some valid input it would then be followed by the exception
next time if i run it it will throw exception as soon as i run it at the very start without displaying even an iota of valid data like the previous case
is ti because the os is placing restrictions on the threads growth or something else.incase there is an error in my code it should atleast pop up the exceptions at one particular loaction in the output and not give different outputs1) When you post code, please use and tags as described in Formatting tips on the message entry page. It makes it much easier to read.
wasnt aware of it.will keep it in mind i post nexttime.sorry for the inconvenience[\b]
3) You have multiple threads, right? The scheduling of which thread gets how much CPU time when is not subject to your control or prediction, so of course you'll see different timings in subsequent runnings of the same code.[b]actually the problem is not with multiple threads showing different times.whats happening is that exceptions are being thrown randomly at runtime.once when i ran it the exception was displayed at the very beginning of the output and the program snapped.next when i ran it first some output which was slated to be displayed got displayed and then the exceptions appeared in the output.now if the exceptions are thrown at the very beginning its obvious theres some problem at the background coz in the second case that never happened[\b]
i guess the formatting tips have taken effect[:D] -
Will adding a second ADFS Web Application Proxy cause service disruption
Today I have attempted to add a second ADFS WAP server to an existing (working) ADFS solution based on 2012 R2.
I am able to install and configure the required role/services successfully but then I'm presented with the Remote Access Management console. This shows the two WAP servers but not the existing published application from the original WAP server and only seems
to let me Publish a new application.
I'm not sure if I should go ahead and run the Publish Application wizard again in case it impacts on the existing application and causes disruption to the service/users.
Any suggestions would be much appreciated.
Cheers for now
Russellthe config for the Web Application Proxy is stored in the ADFS v3 configuration database.
As soon as you add a new WAP to the farm it will get its config from the database
WAP can be domain joined or not. The reason for having it be domain joined is if you need to manage the system centrally and you need to leverage Kerberos Constrained Delegation for Windows based apps
If you have more than one WAP, you should use some kind of load balancing mechanism such as either Windows NLB or a hardware loadbalancer
adding a new WAP should not impact, you just need to make sure it is actually used
Cheers,
Jorge de Almeida Pinto
Principal Consultant | MVP Directory Services | IAM Technologies
COMMUNITY...:
DISCLAIMER: This post is provided "AS IS" with no warranties of any kind, either expressed or implied, and confers no rights! Always evaluate/test yourself before using/implementing this!
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