Multiple choice technology mainframe

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

  1. 2

  2. 4

  3. 8

  4. 16
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

S9(8) COMP is a binary field with 8 digits. In COBOL, COMP fields use binary storage where the size depends on the number of digits. For S9(1) through S9(4), it uses 2 bytes (halfword). For S9(5) through S9(9), it uses 4 bytes (fullword). For S9(10) through S9(18), it uses 8 bytes (doubleword). S9(8) falls in the 4-byte range.

AI explanation

To answer this question, you need to understand the concept of COMP fields in COBOL.

In COBOL, the COMP field is used to store numeric data. The size of a COMP field is determined by the number of bytes it occupies in memory.

In this case, the S9(8) represents a signed numeric field with a length of 8 digits. The "S" indicates that the field is signed, and the "9(8)" indicates that it can hold 8 digits.

In COBOL, the COMP field is typically represented using the binary format, which means that each digit occupies 4 bits. Therefore, to calculate the number of bytes required for the field, we divide the number of digits by 2.

In this case, the S9(8) COMP field has 8 digits, so it will occupy 8/2 = 4 bytes of memory.

Based on this explanation, the correct answer is B) 4.

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