Multiple choice technology mainframe

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

  1. 9

  2. 4

  3. 8

  4. 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

COMP (binary computational) format in COBOL stores numbers in binary representation. S9(8) COMP uses 4 bytes (32 bits) because binary storage uses the minimum number of bytes needed to represent the precision - 8 digits fits in 32 bits (4 bytes). COMP values are typically 2, 4, or 8 bytes depending on precision.

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.

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