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Mainframe Interview Questions and Answers

Mainframe Interview Questions and Answers

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Best Mainframe Interview Questions and Answers

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Q1) Name the units in the COBOL project.

Answer: Identification Division, Environmental Division, Data Division, Practical Division.

Q2) What are the various data types in COBOL?

Answer: Alpha-numeric (fx), alphabet (a) and number (9).

Q3) What is verb �

Answer: Alphabetical, alphanumeric fields and alphaned edited items are set to SPACES.
Numeric, edited edited items are set to ZERO.
Removing FILLER and OCCURS items has not been resumed.

Q4) What is 77 level?

Answer: Startup item. Do not have subdivisions of other objects (can not qualify), or they can not separate themselves.

Q5) What is the status of 88?

Answer: Conditional names.

Q6) What level 66 is used?

Answer: For RENAMES section.

Q7) What is the ISER NUMERIC rule?

Answer: NUMERICs can be used in alphanumeric items, countless & packed decimal materials can be entered using digital & packed decimal items. Turns NUMERIC TRUE to the item only 0-9. However, if the tested item is a signed item, it can have 0-9, + and -.

Q8) Are you defining a table / row in COBOL?

Answer: 01Arace.
05 ARRAY1 PIC X (9) OCCURS 10 TIMES.
05 ARRAY2 PIC X (6) OCCURS 20 TIMES WS-INDEX.

Q9) Can OCCURS be at the level 01?

Answer: No.

Q10) What is the difference between index and subscription?

Answer: The symbol refers to the array arrangement when the displacement (in the absence of bytes) from the beginning of the start. Only one code can be changed using PERFORM, SEARCH & SET.
SEARCH must index a table to use SEARCH ALL.

Q11) What are the difference between SEARCH and SEARCH ALL?

Answer: Search – a series search.
Search ALL – SEARCH ALL A SEX BUTTON AND TABLE BEFORE BEFORE USING ALL (SELECTING / ERROR DATA).

Q12) What is sort ordering for search all

Answer: It can be done by Ashoka or Designed. ASCENDING default. If you want to be sorted in a sequence defined in the order, you must provide key terms when defining the sequence. (You have to load the table in the specified order).

Q13) What is binary search?

Answer: Search on the sorted sequence. Compare the item to search with the item in the center. If it does not fit, the left half or the right half will restart the process depending on where else the item is.

Q14) My plan has a range of 10 items. Due to an error, if the program accesses the 11th item in this line, the program will not reject it. What’s wrong with that?

Answer: If you want to check the range limits, you need to use the client option SSRANGE. The default is NOSSRANGE.

Q15) How can you sort in a COBOL scheme? Define type file type, line statement and syntax.

Answer: Syntax:
SORT file-1 key key animation / formatting ….
Using file-2
Providing file-3
INPUT can be used instead of the PAR-1 THRU para-2
PIV-1 THRU Para-2 through the OUTPUT process.
file-1 is to be defined using the SD entry in a sort workfile and the FILE SECTION.
file-2 is a SORT input file and should be defined using the FD entry in the FILE section and SELECT section of FILE CONTROL.
file-3 must be defined using SORT from the FTP entry in the outfile and FILE section and the SELECT section in FILE CONTROL.
file-1, file-2 & file-3 can not be opened openly.
The INPUT process must be pre-activated and the registry will be RELEASEd the task file from the input process.
The output process is implemented after all records are sorted. Records from a type of work file should be reset once to the output process.

Q16) How is a type of file defined in the JCL that executes the COBOL program?

Answer: SORTWK01, SORTWK02, …… dd Use the names in the step. The number of sorted data depends on the size of the sorted data, but at least 3 is required.

Q17) What are the two ways to sort a COBOL scheme? Give shapes.

Answer: See Question 16.

Q18) Provide the design and use of the SORT report. What are the restrictions with it?

Answer: Question 16. Controls – can not massage the records.

Q19) What is the difference between SECTION and PARAGRAPH?

Answer: SECTION implementation causes all the columns that are part of the area.
PARAGRAPH will only do that paragraph.

Q20) What is the use of the EVALUATE statement?

Answer: An assessment is a case statement and can be used to replace local IFS. The difference between EVALUATE and the trial is that EVALATEATE requires any ‘breakdown’ from EVALATEATE, with a competition taking place.

Q21) What are the different forms of the EVALUATE report?

Answer: Evaluate EVALUATE SQLCODE file level
A = B and C = D When 100 ALOO ’00’
Forced to force
When else
END-EVALUATE END-EVALUATE
EVALUATE SQLCODE ALSO A = B Rating SQLCODE is also true
100 ALOO A = B When is 100 true?
WHEN -305 FALSE WHEN -305 ALSO (A / C = 4)
Forced to force
END-EVALUATE END-EVALUATE

Q22) How do you come out with a valuation report?

Answer: Whenever the debates take place, the control will automatically be sent to the next penalty after the EVALUATE report. No additional code needed.

Q23) In a valuation statement, when can I give a complicated situation?

Answer: Yes.

Q24) What is the purpose terminator? Give examples.

Answer: The terminal is used to indicate the end of a verb, e.g. EVALUATE, END-EVALUATE; IF, END-IF.

Q25) How do you act?

Answer: PERFORM … …
END PERFORM

Q26) When should you use inline program?

Answer: The body of the action will not be used in other columns. If the body of the event is a common type of code (used from many places in the program), it is best to place the code in a custom package and use the PERFORM paragraph than the Internet program.

Q27) What is the difference between the series and the next week?

Answer: CONTINUE is a zero statement (do not do anything), while the next article next to the next sentence (!!) (a sentence for the rest of the period)

Q28) What does EXIT do?

Answer: Nothing done! If used, it should be a sentence in one sentence.

Q29) Can I retrieve a (X) (100) field with X (200) field?

Answer: Yes. Again and again, both fields begin at one place. For example:
01 WS-TOP PIC X (1)
01 WS-TOP-RED REDIFINES WS-TOP PIC X (2).
If you are ’12’ for WS-TOP-RED
The display will show WS-TOP 1 times
The display will show WS-TOP-RED 12.

Q30) Can I retrieve X (200) domain with X (100) field?

Answer: Yes.

Q31) What should you do to resolve the SOC-7 error?

Answer: Basically you have to fix the harmful data.
Many times the reason for SOC7 is an unreleased item. Explore that opportunity first.
Many installs provide you with a padding for the run time (which is created by calling some subroutines or OS server as an assembly language). This is the last step that has been overcome to dumps Offline Offline. Get the serial number of the source code in this corner to test the output of the XREF list and output it. You can see the source code to find the error. To capture the run time, you need to define some data (such as SYSABOUT) at JCL.
If none of these are helpful, use judgment and display to change the source of the error.
Some installation packages may contain program diagnostic tools. Use them.

Q32) How does the pack be stored in the Descell Fields and Zoned Decimal Fields?

Answer: Bucket Demiic fields: The last numbers of storage (4 bits) are encoded as a hex value.
Zoned decim fields: By default, the last signature is signed with the value of the saved number.

Q33) How is the identity stored in a comp-3 domain?

Answer: This is stored in the last nibble. For instance if your number is 100, it is the last byte, hex 1c in your number 101, hex 2c is your number 102, number 1D if number 101, number 2D number, 102 etc …

Q34) How is the identity of a COMP in the field stored?

Answer: Very significant bit. If bit, if i + v.

Q35) What else is there between COMP & COMP-3?

Answer: COMP-3 is a binary storage form, but COMP-3 fills the decimal format.

Q36) What is COMP-1? COMP-2?

Answer: COMP-1 – single precise floating point. Uses 4 bytes.
COMP-2 – double precision floating point. Uses 8 bytes.

Q37) How do you define the variable of COMP-1? COMP-2?

Answer: You have to give any picture. Example 01 WS-VAR Application COMP-1.

Q38) How many bytes are occupied by an S9 (7) COMP-3 field?

Answer: 4 bytes will take. The hex value is stored as a symbol in the last nibble.
General form INT ((n / 2) + 1)), where n = 7 in this example.

Q39) How many bytes a S9 (7) SIGN TRAILING SEPARATE Field Occupied?

Answer: 8 bytes (one extra byte sign) will occupy.

Q40) How many bytes a S9 (8) COMP field invades?

Answer: 4 bytes.

Q41) What is the maximum value that can be saved in S9 (8) COMP?

Answer: 99999999

Q42) What is COMP SYNC?

Answer: Reasons for the item to change the natural boundaries. LEFT or RIGHT can be SYNCHRONIZED.
For binary data objects, if they are in word boundaries in memory, the address clarity will be faster. For example, the memory key is 4 bytes in the main frame. That is, each word starts at an address divided by 4. If my first variable is x (3) and the next one
S9 (4) comp, if you do not specify the SYNC section, S9 (4) will start from COMPB 3 (this starts from 0). If you specify SYNC, the binary data item will start from address 4. You can find some waste of memory, but access to it
The computational field is fast.

Q43) What is the maximum level of 01 level item in COBOL I? In COBOL II?

Answer: In COBOL II: 16777215

Q44) How you can follow the file format from COBOL programs:

Answer: Fixed block file – application ORGANIZATION is parallel. Use the Recording MODE F, BLOCK CONTAINS 0.
Unblocked – Use ORGANIZATION. Do not use the Recording MOD F, BLOCK CONTAINS
DARK BLOCK FILE – Use ORGANIZATION. Recording uses MODE V, BLOCK CONTAINS 0. FD Do not tag 4 bytes of registration length, ie the JCL record length pgm + 4
Unsubscribed – use ORGANIZATION. Use V for recording mode, do not use BLOCK CONTAINS. Do not encrypt 4 bytes for log length in FD, which means that the JCL record length pgm + 4 has the maximum length of the wreck.
Use ESDS VSAM file – ORGANIZATION.
KSDS VSAM file – ORGANIZATION is INDEXED, KEY IS, ALTERNATE RECORD KEY IS
RRDS file – use ORGANIZATION related, relevant keyword IS
Printer file – ORGANIZATION. Use Recording MODE F, BLOCK CONTAINS 0. (Use RECFM = FBA in JCL DCB).

Q45) What are the open modes available in another file gopal?

Answer: INPUT, OUTPUT, I-O, EXTEND.

Q46) What is the method of opening a file that you write?

Answer: OUTPUT, EXTEND

Q47) GLC How do you define the files mentioned in a Ubuntu?

Answer: Provide DD cards for files mentioned in the main scheme.

Q48) Can you get a record in an ESDS file? Can you delete a record from it?

Answer: Can not write (record length should be), but can not be deleted.

Q49) What is file level 92?

Answer: Logic error. E.g., a file is opened to the input and an attempt to write it.

Q50) What is file 39?

Answer: Your COBOL is not applicable to either LRECL or BLOCKSIZE or RECFM between pgm & JCL (or database label). The file 39 is available in OPEN.

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