C#.net code Exception

Question

1.00/5 (1 vote)

See more:

C#

public void Opens()
{
    if (this.serialPort.IsOpen == false)
    {
        this.serialPort.Open();
    }
}

Showing Exception on this.serialPort.Open();
AccessDenied and IOException occured

Posted 15-Oct-13 18:36pm

Member 10314517

Updated 15-Oct-13 19:11pm

StM0n

v3

Add a Solution

Comments

Sergey Alexandrovich Kryukov 16-Oct-13 0:38am

To start with, if (this.serialPort.IsOpen == false) is silly; this.serialPort.IsOpen is already a Boolean...
—SA

bryce 16-Oct-13 0:50am

i'm guessing you're using the .net serial port class...
and when you try and open the serial port its throwing an exception and yo don't know why?
How about you tell us some more about your problem so we have data to work on

however access denied would suggest something (antherprogram) has already got the serial port in question open.

bryce

StM0n 16-Oct-13 1:13am

Same thought...

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Answer 1 · 2013-10-15T21:18:00

Just because you can create a SerialPort object and set the port to the appropriate value does not mean that the port is available: physical ports cannot be shared by applications since that would bean the received data being "shared" by two applications. This would beam that either all the data would have to be buffered to both - which is unrealistic since Windows would have to buffer an unknown number of applications, which takes time, and serial port processing is supposed to be as quick as possible to prevent overrun - or which application got which data would be random.

As has been suggested, another application got there first - so you cannot open the port. Use a try...catch block to detect this and report the problem to your user - you can't "share" ports!