SAP C_ABAPD_2309 Exam Dumps

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C_ABAPD_2309 Pack
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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Free SAP C_ABAPD_2309 Exam Actual Questions

Question No. 1

When does SAP recommend to use a sorted or a hashed table respectively? Note: There are 2 correct answers to this question.

Show Answer Hide Answer
Correct Answer: A, B

Question No. 2

Which function call returns 0?

Show Answer Hide Answer
Correct Answer: D

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.

find_any_of returns the position of the first character in the string val that is contained in the string sub. In this case, it returns 1, since the first character A is in sub.Reference:String Functions - ABAP Keyword Documentation,Examples of String Functions - ABAP Keyword Documentation


Question No. 3

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

Show Answer Hide Answer
Correct Answer: A, D

The following are the explanations for each statement:

A: This statement is correct. go_subl = CAST #(go_super) will not work. This is because go_subl is a data object of type REF TO cl_subl, which is a reference to the subclass cl_subl. go_super is a data object of type REF TO cl_super, which is a reference to the superclass cl_super. The CAST operator is used to perform a downcast or an upcast of a reference variable to another reference variable of a compatible type. A downcast is a conversion from a more general type to a more specific type, while an upcast is a conversion from a more specific type to a more general type. In this case, the CAST operator is trying to perform a downcast from go_super to go_subl, but this is not possible, as go_super is not pointing to an instance of cl_subl, but to an instance of cl_super.Therefore, the CAST operator will raise an exception CX_SY_MOVE_CAST_ERROR at runtime12

B: This statement is incorrect. go_sub2 = CAST #(go_super) will work. go_subl = CAST #(go_super) will not work. This is because go_sub2 is a data object of type REF TO cl_sub2, which is a reference to the subclass cl_sub2. go_super is a data object of type REF TO cl_super, which is a reference to the superclass cl_super. The CAST operator is used to perform a downcast or an upcast of a reference variable to another reference variable of a compatible type. A downcast is a conversion from a more general type to a more specific type, while an upcast is a conversion from a more specific type to a more general type. In this case, the CAST operator is trying to perform a downcast from go_super to go_sub2, and this is possible, as go_super is pointing to an instance of cl_sub2, which is a subclass of cl_super. Therefore, the CAST operator will assign the reference of go_super to go_sub2 without raising an exception.However, the CAST operator will not work for go_subl, as explained in statement A12

C: This statement is incorrect. go_sub2 = CAST #(go_super) will work. go_sub2->sub2_meth1(...) will not work. This is because go_sub2 is a data object of type REF TO cl_sub2, which is a reference to the subclass cl_sub2. go_super is a data object of type REF TO cl_super, which is a reference to the superclass cl_super. The CAST operator is used to perform a downcast or an upcast of a reference variable to another reference variable of a compatible type. A downcast is a conversion from a more general type to a more specific type, while an upcast is a conversion from a more specific type to a more general type. In this case, the CAST operator is trying to perform a downcast from go_super to go_sub2, and this is possible, as go_super is pointing to an instance of cl_sub2, which is a subclass of cl_super. Therefore, the CAST operator will assign the reference of go_super to go_sub2 without raising an exception. However, the method call go_sub2->sub2_meth1(...) will not work, as sub2_meth1 is a subclass-specific method of cl_sub2, which is not inherited by cl_super.Therefore, the method call will raise an exception CX_SY_DYN_CALL_ILLEGAL_METHOD at runtime123

D: This statement is correct. go_subl->subl_meth1(...) will work. This is because go_subl is a data object of type REF TO cl_subl, which is a reference to the subclass cl_subl. subl_meth1 is a subclass-specific method of cl_subl, which is not inherited by cl_super.Therefore, the method call go_subl->subl_meth1(...) will work, as go_subl is pointing to an instance of cl_subl, which has the method subl_meth1123


Question No. 4

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.

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Correct Answer: B, D

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.


Question No. 5

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?

Show Answer Hide Answer
Correct Answer: A

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.

This is because the join statements are nested from left to right, meaning that the leftmost data source is joined with the next data source, and the result is joined with the next data source, and so on. The join condition for each pair of data sources is specified by the ON clause that follows the data source name. The join type for each pair of data sources is specified by the join operator that precedes the data source name. In this case, the join operator is LEFT OUTER JOIN, which means that all the rows from the left data source are included in the result, and only the matching rows from the right data source are included. If there is no matching row from the right data source, the corresponding fields are filled with initial values1.

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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