Combine 3 hexvalues by using left shift operator. Examples: take decimal values as 2, 129025, 127 .combine the all hex values and need to come as 2, 1F801, 7F

Question

1.00/5 (2 votes)

See more:

combine 3 hexvalues by using left shift operator.

examples:

take decimal values as 2,129025,127 .combine the all hex values and need to come as 2,1F801,7F

i need to write as 2,1F801,7F IN THE kvaser CANKING output.

What I have tried:

int h1=2;
int h2=129025;
int h3=127;
int combinedID = h1 + (h2<<8) + (h3<<25);

//for writing on kvaser canking
canlib.canwrite(handle,combinedID,data,data.length,0);

Posted 21-Jan-24 20:46pm

Naveen Kumar Amujuru

Add a Solution

Comments

CPallini 22-Jan-24 2:53am

And what's wrong with your code?

jeron1 yesterday

Are there range restrictions on h1,h2,and h3?

:edit:
:sigh: I didn't realize how old this was, sorry.

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)

RainHat · Answer 1 · 2024-05-10T04:13:00

You need to shift by 28, not 25.

When shifting by hexadecimal digits your shifts will always need to be multiples of 4, as that is the size of a hexadecimal digit.

When you shift by 25 bits, you are trying to shift the 2 in the first number into the high bits of the 1 in the second number, resulting in 5F8017F instead of the expected 21F8017F