Thank you so much Kumar for such a good explanation. It is very useful for interview purpose. You explained with mainframe practical in very less time. Keep doing more videos and big thanks from our end.
Hi Kumar, Nice video this helps a lot for all of us to refresh our knowledge….I request you to please start some series on refreshment so that we can keep it up 👍
Thanks for the Detailed explanation. I have one doubt... PIC S9(13)V(2) USAGE COMP-3... For this it will take 8 bytes and each digit will take 1/2 a byte? Please correct me if I am wrong. And also this shows 13 digits.. In that 13 it will take 2 for decimal point? Or 13 digits for the whole number and extra 2 digits for the decimal point( total-15)?
Hi Kumar. I have a question, if comp saves space and is more fast and efficient then why normal numeric like 9(5) etc is used in cobol code. Why we are not using comp for all numeric values?
Hi kumar sir,Sir I want to get comp-1 variable as input from user through jcl . In working storage section I given 77 ws-num1 comp1 77 ws- num2 comp1 In jcl I pass 77.77 and 33.33 through sysin Jcl sir BUT am getting IGYPA3018-S as the Identifer(floating pt) was used in accept should be discarded. What I need to change sir?
Try with 01 Level number COMP1 should be COMP-1 Ex: 01 WS-NUM1 COMP-1. 01 WS-NUM2 COMP-1. Ex: Number 1.234 can be represented as +0.1234 x 101. In the above example :+ is the sign of the number, 1 in 101 is the exponent and 1234 is the mantissa which is represented using 3 bytes.
Excellent
Thank you very much, your videos are very useful
Thanks for such a clear explanation.. I am sharing the videos with my friends too..
Thank you so much Kumar for such a good explanation. It is very useful for interview purpose. You explained with mainframe practical in very less time. Keep doing more videos and big thanks from our end.
Excellent sir🙏🏻.....thank you so much....waiting for this for a long time.....
THANK YOU MR. KUMAR FRO TLV
Thanks for such a wonderful explanation Kumar.. Stay safe and keep teaching us
That was a brilliant explanation.
Please make complete mainframe in-depth course with simple to complex maninframe prgms.
Thank you so much for upload the vdeo☺️☺️
Hi Kumar, Nice video this helps a lot for all of us to refresh our knowledge….I request you to please start some series on refreshment so that we can keep it up 👍
Thanks for the Detailed explanation. I have one doubt... PIC S9(13)V(2) USAGE COMP-3... For this it will take 8 bytes and each digit will take 1/2 a byte? Please correct me if I am wrong.
And also this shows 13 digits.. In that 13 it will take 2 for decimal point? Or 13 digits for the whole number and extra 2 digits for the decimal point( total-15)?
Hi Kumar. I have a question, if comp saves space and is more fast and efficient then why normal numeric like 9(5) etc is used in cobol code. Why we are not using comp for all numeric values?
Thanks man!
Happy to help!
Sounds like this is Hitesh here from lco
PIC S9(9) USAGE COMP-5 ,can you tell me how to declare in Easytrive
Great concept explained...Honest work man... Keep it up👍🏻
Hi kumar sir,Sir I want to get comp-1 variable as input from user through jcl . In working storage section I given
77 ws-num1 comp1
77 ws- num2 comp1
In jcl
I pass 77.77 and 33.33 through sysin Jcl sir
BUT am getting IGYPA3018-S as the Identifer(floating pt) was used in accept should be discarded.
What I need to change sir?
Try with 01 Level number
COMP1 should be COMP-1
Ex:
01 WS-NUM1 COMP-1.
01 WS-NUM2 COMP-1.
Ex:
Number 1.234 can be represented as +0.1234 x 101.
In the above example :+ is the sign of the number, 1 in 101 is the exponent and 1234 is the mantissa which is represented using 3 bytes.