SAP C_ABAPD SAP Certified - Backend Developer - ABAP Cloud (Beta Version) (C_ABAPD_2601) v4.1 DEMO QUESTIONS Sample Q&A Preview Preview content before purchase Get Full Version & Premium Features https://examsempire.com/exam/c_abapd Premium Benefits Included Free Updates 90 days of exam updates Money Back 30-day guarantee policy Instant Access Download immediately 24/7 Support Expert assistance anytime Page 1 of 6 https://examsempire.com/exams/c_abapd Question 1. (Single Select) <strong>CHALLENGE 1 — Calculating Days to Expiry from Stored Timestamps</strong> Branch staff report that a few batches appear in the near-expiry list one day before their printed use-by date, while others drop off a day late. The days-remaining value is computed by subtracting today's date from the date part of the stored use-by timestamp. What is the most likely cause of the one-day discrepancy? A: The batch master already holds the use-by value as a plain calendar date, so the extra conversion step shifts every computed result by one full day during the read. B: The near-expiry threshold is configured too tightly, so borderline batches move in and out of the list as the day-count rounding changes between runs. C: The use-by moment is stored in UTC, so reading its date part without a local-date conversion makes batches near midnight shift by a day. D: The list query orders batches by their text description instead of by the use-by value, so near-expiry rows surface in an unstable position on each run. Answer: C Explanation: The use-by value is held as a UTC timestamp, so reading only its date part ignores the offset to the branch's local zone. For batches expiring close to midnight the UTC date and the local date differ by one day, which is exactly why some rows appear a day early and others a day late. Converting to the local date before extracting the day removes the shift. Question 2. (Single Select) <strong>CHALLENGE 3 — Reading Dispensing Records Through a Driver Table</strong> A branch manager opens the dispensing summary on a quiet morning and unexpectedly sees dispensing rows from several other branches. The summary reads dispensing records with a key-driven set-based read, driven from an internal table of that branch's batch keys. What is the most likely cause? Page 2 of 6 https://examsempire.com/exams/c_abapd A: The dispensing table lacks a secondary index on the branch field, so the database widens the read to every branch in order to satisfy the original query during opening hours. B: The branch filter sits in the display layer rather than in the query itself, so all rows are fetched first and only some of them are hidden afterwards. C: The internal table of keys holds duplicate entries, so the read multiplies its result set and drags in records that belong to other branches entirely. D: The driver table of batch keys was empty for that quiet branch, and a key-driven set-based read over an empty table returns all rows, not none. Answer: D Explanation: A key-driven set-based read returns all rows when its driver internal table is empty. On a quiet morning the branch had no batch keys, so the driver table was empty and the read pulled every branch's dispensing rows instead of none. The empty-driver case is the second-order cause behind the unexpected foreign rows. Question 3. (Single Select) <strong>CHALLENGE 3 — Reading Dispensing Records Through a Driver Table</strong> You must make the dispensing summary reliable for branches with little or no recent activity. Which correction is complete? A: Sort the driver table of keys and strip out duplicates before the read, so the result set stays clean and no foreign-branch rows slip in. B: Check that the driver table of keys is not empty before issuing the read, and return nothing when there are no keys to drive it. C: Add the branch as an extra condition inside the read so the database also filters on branch, while leaving the driver-table handling exactly as it is now. D: Raise the app's result limit so the larger row set loads fully, then total only those rows whose branch matches the current user's branch afterwards. Answer: B Page 3 of 6 https://examsempire.com/exams/c_abapd Explanation: The reliable fix is to confirm the driver table of keys is not empty before issuing the read and to return nothing when there are no keys, which removes the empty-driver case that exposes every branch. It addresses the root cause rather than masking the symptom. Question 4. (Single Select) <strong>CHALLENGE 3 — Reading Dispensing Records Through a Driver Table</strong> The corrected read must protect both the branch data boundary and response time for low-activity branches. Which test best confirms this? A: Run the summary for a branch with no recent dispensing and confirm it returns zero rows quickly, revealing no other branch's records. B: Run the summary for the busiest branch and confirm its totals match the figures produced by the previous version of the same report. C: Run the summary repeatedly for a single active branch and confirm the response time stays within the agreed limit even under sustained, repeated load. D: Run the summary across every branch at once and confirm the combined total equals the chain-wide dispensing figure for the whole period. Answer: A Explanation: The defect surfaces specifically for low-activity branches, so running the summary for a branch with no recent dispensing and confirming it returns zero rows quickly — with no other branch's records — proves both the boundary and the response time are protected. This directly exercises the empty-driver path. Question 5. (Single Select) <strong>CHALLENGE 1 — Validating the Collection-Site Code Format Reliably</strong> Clerks at newer sites report that some correctly formed collection-site codes are rejected at entry, while a Page 4 of 6 https://examsempire.com/exams/c_abapd few malformed codes are accepted. What best explains this behavior? A: The released site source orders newer regional prefixes under a different collation, so the lookup no longer matches them, and aligning the source's sort sequence would let those codes validate again at entry. B: The check reads the prefix from a fixed character position assuming a three-letter prefix, so codes with a different prefix length are misread, and describing the format as a pattern accepts every valid code. C: The newer codes lose their leading zeros to an automatic numeric conversion before the length test runs, so they fail it, and restoring those zeros ahead of the check would stop the rejections. D: The collection-site code field is defined too short to hold the newer values, so the input is truncated before validation runs, and widening that field would allow the complete code through. Answer: B Explanation: Reading a fixed number of characters from a fixed position only works while every prefix is the same length; once newer sites use prefixes of a different length, the fixed read lands on the wrong characters, rejecting valid codes and sometimes accepting malformed ones. Describing the whole permitted structure as a pattern removes the dependence on a single assumed width. Page 5 of 6 https://examsempire.com/exams/c_abapd Ready for Success? Get the complete exam package today https://examsempire.com/exam/c_abapd Thank You for Choosing ExamCertify! Your Success is Our Mission Special Discount Code 15OFFTODAY Contact Us Sales: sales@examsempire.com Support: support@examsempire.com Page 6 of 6 https://examsempire.com/exams/c_abapd