0

mainframe Online Quiz - 145

Description: mainframe Online Quiz - 145
Number of Questions: 20
Created by:
Tags: mainframe
Attempted 0/20 Correct 0 Score 0

How can return codes be tested before execution of a job step?

  1. Through the RC keyword

  2. Through the CODE keyword

  3. Through the COND keyword

  4. Through the RETURNCD keyword


Correct Option: C

What is concatenating?

  1. Overriding or adding procedures

  2. Logically connecting input datasets

  3. Overriding a called PROC dsname

  4. All of the above


Correct Option: B

How can values be passed from the job stream to an executable program?

  1. Through the PARM keyword

  2. Through the VALUE parameter

  3. Through the PGM parameter

  4. All of the above


Correct Option: A

Must tape dataset definitions include VOL=SER specifications?

  1. Yes

  2. No

  3. Only for uncataloged datasets

  4. Only for cataloged datasets


Correct Option: C

What is the length of DFHCOMMAREA

  1. 16Kb

  2. 32Kb

  3. 64Kb

  4. 1 MB


Correct Option: B

By which commad we test the map amd mapset individually

  1. CEMT

  2. CECI

  3. CEDF

  4. CEBR


Correct Option: C

AI Explanation

To test the map and mapset individually, you can use the CECI command.

Option B) CECI - This option is correct because the CECI command is used to test individual maps and mapsets in IBM's mainframe operating system, z/OS. It allows you to verify the correctness and functionality of a specific map or mapset.

The correct answer is B) CECI.

by which command we switch amoung messages?

  1. CEMT

  2. CEBR

  3. CMSG

  4. CECKI


Correct Option: C

What error we get when we press enter with out entering any value in map

  1. MAPFAIL

  2. ASRA

  3. ASRB

  4. AEI9


Correct Option: A

how we position the cursor in a cics pogrem statically

  1. send map() cursor()

  2. cursor()

  3. send ()

  4. none of the above


Correct Option: A

How many bytes will a S9(8) COMP field occupy ?

  1. 9

  2. 4

  3. 8

  4. 5


Correct Option: B

AI Explanation

To answer this question, we need to understand the concept of the COMP field in COBOL.

In COBOL, the COMP field is used to represent numeric data. It is stored in binary format, which means that each digit is represented by a binary byte.

The S9(8) COMP field is defined as a signed numeric field with a length of 8 digits. The "S9" part indicates that it is a signed field, and the "(8)" part indicates the length.

Since each digit is represented by a binary byte, the S9(8) COMP field will occupy 8 bytes.

Therefore, the correct answer is B) 4.

How many bytes does a S9(7) COMP-3 field occupy

  1. 4

  2. 8

  3. 3

  4. 7


Correct Option: B

How many bytes does a S9(7) COMP-3 field occupy

  1. 4

  2. 8

  3. 3

  4. 7


Correct Option: A

AI Explanation

To answer this question, we need to understand the representation of a COMP-3 field in COBOL.

A S9(7) COMP-3 field is a signed numeric field that can hold up to 7 digits. It is stored in a packed decimal format, where each digit is represented by half a byte (nibble). The left-most nibble is used to store the sign of the number.

In a COMP-3 field, each digit occupies 4 bits (half a byte). Therefore, for a S9(7) COMP-3 field, we have 7 digits, which requires 7*4 = 28 bits.

Since 8 bits make up 1 byte, we need to divide 28 bits by 8 to get the number of bytes required.

28 bits / 8 = 3.5 bytes

Since we cannot have half a byte, we need to round up to the nearest whole number. Therefore, a S9(7) COMP-3 field occupies 4 bytes.

So, the correct answer is A) 4 bytes.

How many bytes does a S9(7) SIGN TRAILING SEPARATE field occupy ?

  1. 7

  2. 8

  3. 9

  4. 4


Correct Option: B

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

  1. 999999999

  2. 9999999

  3. 99999999

  4. 99999


Correct Option: C
Explanation:

To solve this question, the user needs to know the format and properties of a S9(8) COMP data type in mainframe systems.

The S9(8) COMP is a signed 4-byte (32-bit) binary integer data type in IBM mainframe systems. It can store values ranging from -2147483648 to 2147483647. Since S9(8) is a signed data type, the first bit of the binary representation is reserved for the sign of the number (0 for positive and 1 for negative).

To find the maximum value that can be stored in S9(8) COMP, we can use the formula:

Maximum value = (2^(number of bits - 1)) - 1

For S9(8) COMP, the number of bits is 32 (4 bytes x 8 bits per byte), so the formula becomes:

Maximum value = (2^(32 - 1)) - 1

Maximum value = 2,147,483,647

However, since S9(8) COMP is a signed data type, the maximum positive value that can be stored is half of the range, or 2,147,483,647 / 2 = 1,073,741,823.

Now, let's go through each option and explain why it is right or wrong:

A. 999999999: This option is incorrect because the maximum positive value that can be stored in S9(8) COMP is 1,073,741,823, which is less than 999,999,999.

B. 9999999: This option is incorrect because it is less than the maximum positive value that can be stored in S9(8) COMP, which is 1,073,741,823.

C. 99999999: This option is incorrect because it is greater than the maximum positive value that can be stored in S9(8) COMP, which is 1,073,741,823.

D. 99999: This option is incorrect because it is much less than the maximum positive value that can be stored in S9(8) COMP, which is 1,073,741,823.

Therefore, the correct answer is:

The Answer is: C. 99999999

  1. A READ statement is missing in the MAINLINE section after the PERFORM.

  2. MAINLINE must not contain any statements other than the PERFORM statement.

  3. S-EOF-CHECK is not properly initialized prior to the PERFORM statement

  4. The line "SET S-END-OF-FILE TO TRUE" is not a valid statement.

  5. In working storage, an 88 level must be defined for values of S-EOF-CHECK other than 'N' and 'Y'.


Correct Option: C
  1. There should not be a hyphen between PROGRAM and ID.

  2. Line 0100 should be DIVISION-ID.

  3. The compiler shortened the program name to "AM822P11".

  4. Line 0200 should be PROGRAM IDENTIFICATION. AM822P115.

  5. The program name must not contain numbers.


Correct Option: C
  1. 01 PRICE-TABLE. 05 PRICE-GROUP OCCURS 1 TO 100 TIMES ASCENDING KEY IS ITEM-CODE. 10 ITEM-CODE PIC X(7). 10 ITEM-PRICE PIC 99V99.

  2. 01 PRICE-TABLE. 05 PRICE-GROUP OCCURS UP TO 100 TIMES DEPENDING ON WS-ITEM-COUNT ASCENDING KEY IS ITEM-CODE. 10 ITEM-CODE PIC X(7). 10 ITEM-PRICE PIC 99V99.

  3. 01 PRICE-TABLE. 05 PRICE-GROUP OCCURS 1 TO 100 TIMES DEPENDING ON WS-ITEM-COUNT ASCENDING KEY IS ITEM-CODE. 10 ITEM-CODE PIC X(7). 10 ITEM-PRICE PIC 99V99.

  4. 01 PRICE-TABLE. 05 PRICE-GROUP OCCURS UP TO 100 TIMES DEPENDING ON WS-ITEM-COUNT AND KEY ITEM-CODE. 10 ITEM-CODE PIC X(7). 10 ITEM-PRICE PIC 99V99.


Correct Option: C
- Hide questions