INTERVIEW QUESTIONS MAINFRAME COBOL DETAILS

Question: 77 I PI 9.
PERFORM VARYING I FROM 1 BY 1 UNTIL>10
DISPLAY 'OK'
END-PERFORM.

What output/msg is likely when this program is executed thru JCL?

Answer: This loop will give compilation error becoz value of I is declared for only one digit , when loop reaches to 10 it will find mismatching b/w I & the current value i.e. 10.

This loop will give compilation error becoz value of I is declared for only one digit , when loop reaches to 10 it will find mismatching b/w I & the current value i.e. 10. Source: CoolInterview.com

It Won't give the compilation error.

At Run time when Varibale I reached to 10 then in working storage variable which is declared as 9(01) can hold 0 not 10
So Program goes into loop result in Timeout Error S322.
Source: CoolInterview.com

I think it will lead to infinite loop. As I will never reach value of 10. After 9 if you add 1 again it will overflow and start from 0... Source: CoolInterview.com

infinite loop will occur.becoz when the value reaches 10,since it is a sigle digit it will take only 0. so the loop goes indefinitely Source: CoolInterview.com

The above code will give COMPILATION ERROR since the syntax is wrong.

But when you correct the syntax i.e changing it to UNTIL I > 10 ..it will go to infinite loop as stated correctly by others. Source: CoolInterview.com

syntax is wrongg Source: CoolInterview.com

it will throw subscript error at the time of execution.
s.koteeswaran Source: CoolInterview.com

it will give the syntax error Source: CoolInterview.com

Infinite loop as everytime when the value of i reaches 10,it will take zero as input and the loop will keep continuing. Source: CoolInterview.com

it displays only once,
sys will take first digit bcoz I length is 9(1).
for second iteration perform condition is thrue Source: CoolInterview.com

syntax error,it should be UNTIL I>10 Source: CoolInterview.com

77 I PI 9.
PERFORM VARYING I FROM 1 BY 1 UNTIL I >10
DISPLAY 'OK'
END-PERFORM.

--> Hi.. Nothing is there in the above program..we have one loop in the program it is will run until I > 10

--> that means it will run until I value becomes 11. Now observe the I values when the loop is running
--> When the first time loop runs I value is I=1.

--> When the second time loop runs I value is I=2.

--> When the third time loop runs I value is I=3.
........
........
--> When the tenth time loop runs I value is I=10. But here comes the trick..

--> I was declared as 77 I PIC 9. it is same as 77 I PIC 9(1). when the loop runs 10th time I value becomes 10.. 10 is a two digit number. But i can hold only one digit number..So here data truncation will happen.

--> when at the time of running loop 10th time instead of 10 I will have the value 0 this is because of data truncation..So the condition UNTIL 0 >10 fails and continue to the loop. then I value started from 0... It will go On and get into infinite Loop..Then we will get one ABEND Named s322..this because for maximum execution time reached.. Source: CoolInterview.com

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