Multiple choice technology mainframe

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

  1. 7

  2. 8

  3. 9

  4. 4
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B Correct answer
Explanation

SIGN TRAILING SEPARATE stores the sign as a separate character byte at the end. For S9(7), 7 digit bytes plus 1 separate sign byte = 8 bytes total. This is different from SIGN TRAILING (included in last nibble) or SIGN LEADING SEPARATE (sign at beginning).

AI explanation

8 bytes is correct. S9(7) defines a signed numeric field with 7 digit positions. Normally (DISPLAY, no SEPARATE clause), the sign is embedded via zone overpunch on the last digit, so no extra byte is needed. But SIGN IS TRAILING SEPARATE forces the sign to occupy its own dedicated character position (a literal '+' or '-') appended after the digits, rather than being combined with a digit's zone bits. That adds one byte beyond the 7 digit bytes, giving 7 + 1 = 8 bytes total. 7 would be correct only without SEPARATE; 9 and 4 don't correspond to any standard COBOL storage rule for this picture clause.

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