Multiple choice technology mainframe

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

  1. 7 bytes

  2. 6 bytes

  3. 5 bytes

  4. 4 bytes
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

COMP-3 (packed decimal) format stores each digit in 1/2 byte (nibble). For S9(7), we have 7 digits plus 1/2 byte for the sign, totaling 8 half-bytes = 4 bytes. The formula is (number of digits + 1) / 2 bytes.

AI explanation

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

A COMP-3 field, also known as a packed decimal field, is a type of numeric data storage format in COBOL. It is used to store decimal numbers with a fixed number of digits.

In a COMP-3 field, each digit is represented by a half-byte (4 bits). The sign (positive or negative) is stored in the last half-byte.

The S9(7) notation indicates a signed numeric field with a total of 7 digits.

To calculate the number of bytes occupied by a COMP-3 field, we divide the total number of digits by 2 (since each digit occupies half a byte) and round it up to the nearest whole number.

In this case, the S9(7) field has a total of 7 digits. Dividing 7 by 2 gives us 3.5, which we round up to 4.

Therefore, a S9(7) COMP-3 field occupies 4 bytes.

Let's go through each option to understand why it is correct or incorrect:

Option A) 7 bytes - This option is incorrect because the S9(7) COMP-3 field occupies 4 bytes, not 7. Option B) 6 bytes - This option is incorrect because the S9(7) COMP-3 field occupies 4 bytes, not 6. Option C) 5 bytes - This option is incorrect because the S9(7) COMP-3 field occupies 4 bytes, not 5. Option D) 4 bytes - This option is correct because the S9(7) COMP-3 field occupies 4 bytes.

The correct answer is D.

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