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| Vendor: | SAP |
|---|---|
| Exam Code: | C_ABAPD_2309 |
| Exam Name: | SAP Certified Associate - Back-End Developer - ABAP Cloud Exam |
| Exam Questions: | 81 |
| Last Updated: | October 5, 2026 |
| Related Certifications: | SAP Certified Associate, Back-End Developer - ABAP Cloud |
| Exam Tags: | Associate Level SAP Back-end DevelopersSAP consultants |
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When does SAP recommend to use a sorted or a hashed table respectively? Note: There are 2 correct answers to this question.
Which function call returns 0?
The function find_any_not_of returns the position of the first character in the string val that is not contained in the string sub. If no such character is found, the function returns 0. In this case, the string val contains only the characters A, B, and a, which are all contained in the string sub, so the function returns 0. The other functions return positive values, as follows:
Count_any_of returns the number of occurrences of any character in the string sub within the string val. In this case, it returns 8, since there are 8 A's and B's in val.
Count returns the number of occurrences of the string sub within the string val. In this case, it returns 2, since there are 2 AB's in val.
Exhibit:
With Icl_super being superclass for Icl_subl and Icl_sub2 and with methods subl_methl and sub2_methl being subclass-specific methods of Id_subl or Icl_sub2, respectivel. What will happen when executing these casts? Note:
There are 2 correct answers to this question
The following are the explanations for each statement:
Which type of legacy code does SAP recommend you eliminate when you review modifications as part of an SAP S/4HANA system conversion? Note: There are 2 correct answers to this question.
SAP recommends that you eliminate the following types of legacy code when you review modifications as part of an SAP S/4HANA system conversion:
Code that now is identical to a standard SAP object. This type of code is redundant and unnecessary, as it does not provide any additional functionality or customization. It can also cause conflicts or errors during the system conversion, as the standard SAP object may have changed or been replaced in SAP S/4HANA. Therefore, you should delete this type of code and use the standard SAP object instead.
Code that can be redesigned as a key user extension. This type of code is usually related to UI or business logic adaptations that can be achieved using the in-app tools provided by SAP S/4HANA. By redesigning this type of code as a key user extension, you can simplify and standardize your code base, reduce maintenance efforts, and avoid compatibility issues during the system conversion. Therefore, you should migrate this type of code to the key user extensibility framework and delete the original code.
The other types of legacy code are not recommended to be eliminated, as they may still be relevant or necessary for your business processes. However, you should still review and adjust them according to the SAP S/4HANA simplification items and best practices. These types of code are:
Code that supports a critical business process. This type of code is essential for your business operations and cannot be easily replaced or removed. However, you should check if this type of code is compatible with SAP S/4HANA, and if not, you should adapt it accordingly. You should also consider if this type of code can be optimized or enhanced using the new features and capabilities of SAP S/4HANA.
Code that has less than 10% usage according to usage statistics. This type of code is rarely used and may not be worth maintaining or converting. However, you should not delete this type of code without verifying its relevance and impact on your business processes. You should also consider if this type of code can be replaced or consolidated with other code that has higher usage or better performance.
Given the following Core Data Service View Entity Data Definition:
1 @AccessControl.authorizationCheck: #NOT_REQUIRED
2 DEFINE VIEW ENTITY demo_flight_info_join
3 AS SELECT
4 FROM scarr AS a
5 LEFT OUTER JOIN scounter AS c
6 LEFT OUTER JOIN sairport AS p
7 ON p.id = c.airport
8 ON a.carrid = c.carrid
9 {
10 a.carrid AS carrier_id,
11 p.id AS airport_id,
12 c.countnum AS counter_number
13 }
In what order will the join statements be executed?
The order in which the join statements will be executed is:
scarr will be joined with scounter first and the result will be joined with sairport.
Therefore, the join statements will be executed as follows:
First, scarr AS a will be joined with scounter AS c using the join condition a.carrid = c.carrid. This means that all the rows from scarr will be included in the result, and only the rows from scounter that have the same value for the carrid field will be included. If there is no matching row from scounter, the countnum field will be filled with an initial value.
Second, the result of the first join will be joined with sairport AS p using the join condition p.id = c.airport. This means that all the rows from the first join will be included in the result, and only the rows from sairport that have the same value for the id field as the airport field from the first join will be included. If there is no matching row from sairport, the id field will be filled with an initial value.
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