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Top Mainframe Interview Questions and Answers - 2021 [UPDATED]

Are you looking for Mainframe Interview Questions and Answers? Than you are at the Right Place. Browse through Popular and Most Asked Interview Questions for Mainframe.  There is a Huge Demand for Mainframe Professionals in the Market. These Questions are suitable for both Freshers and Experienced Professionals and are based on Trending Topics and as per Current Industry Requirements.

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Mainframe Interview Questions and Answers

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Identification Division, Environmental Division, Data Division, Practical Division.

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

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.

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

Conditional names.

For RENAMES section.

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 -.


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.


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).

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.

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

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.

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.

Question 16. Controls – can not massage the records.

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

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.

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

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

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


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.

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)

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

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.

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.

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.

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 …

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

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

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

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

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.

8 bytes (one extra byte sign) will occupy.

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.

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).

Provide DD cards for files mentioned in the main scheme.

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

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

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.