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SAP C_ABAPD_2309 Exam Questions
- SAP Certified Associate Certifications
- SAP Back-End Developer - ABAP Cloud Certifications
- Topic 1: Core ABAP programming: This topic covers ABAP data types, the ABAP dictionary, modularization, exceptions SAP HANA database tables, and logical expressions, operator precedence.
- Topic 2: ABAP core data services and data modeling: It focuses on Core Data Services (CDS) views, SAP HANA database tables, foreign key relationships, and annotations.
- Topic 3: ABAP RESTful Application Programming Model: This topic explains the ABAP Restful Application Programming model, ABAP development, and the architecture of the ABAP Restful Application Programming model.
- Topic 4: Object-oriented design: It measures your knowledge about encapsulation, upcast, inheritance, polymorphism, and interfaces. Moreover, the topic evaluates your knowledge about constructor calls, Exception classes, and singleton pattern.
- Topic 5: SAP clean core extensibility and ABAP cloud: The topic explains extension pattern, extension rules, ABAP cloud development, and ABAP cloud rules.
- Topic 6: ABAP SQL and code pushdown: It discusses ABAP SQL, arithmetic expressions, manage dates, and create joins.
Free SAP C_ABAPD_2309 Exam Actual Questions
Note: Premium Questions for C_ABAPD_2309 were last updated On Apr. 10, 2025 (see below)
In an Access Control Object, which clauses are used? Note: There are 3 correct answers to this question.
You cannot do any of the following:
Which ABAP SQL clause allows the use of inline declarations?
The following code snippet uses the INTO clause with an inline declaration to declare a local variable itab and store the result of the SELECT query into it:
SELECT * FROM scarr INTO TABLE @DATA (itab).
The following code snippet uses the INTO clause with an inline declaration to declare a field symbol <fs> and store the result of the SELECT query into it:
SELECT SINGLE * FROM scarr INTO @<fs>.
You cannot do any of the following:
Refer to the exhibit.
with which predicate condition can you ensure that the CAST will work?
The following code snippet uses the IS INSTANCE OF predicate condition to check whether the variable g_super is an instance of the class lcl_super. If it is, the CAST will work and the variable g_sub1 will be assigned the value of g_super.
DATA: g_super TYPE REF TO lcl_super, g_sub1 TYPE REF TO lcl_sub1. IF g_super IS INSTANCE OF lcl_super. g_sub1 = CAST #( g_super ). g_sub1->method( ... ). ENDIF.
You cannot do any of the following:
Using ABAP SQL, which select statement selects the mat field on line #17?
Using ABAP SQL, the select statement that selects the mat field on line #17 is:
SELECT mat FROM demo_sales_cds_so_i_ve...
The other options are not valid because:
A . SELECT mat FROM Material... is not valid because Material is not a valid data source in the given code. There is no CDS view or database table named Material.
C . SELECT mat FROM demo_sales_so_i... is not valid because demo_sales_so_i is not a valid data source in the given code. There is no CDS view named demo_sales_so_i, only a database table. To access a database table, the keyword TABLE must be used, such as SELECT mat FROM TABLE demo_sales_so_i...
D . SELECT mat FROM demo sales cds material ve... is not valid because demo sales cds material ve is not a valid data source in the given code. There is no CDS view or database table named demo sales cds material ve. The correct name of the CDS view is demo_sales_cds_material_ve, with underscores instead of spaces.
In this nested join below in which way is the join evaluated?
The nested join is evaluated from the top to the bottom in the order of the ON conditions. This means that the join expression is formed by assigning each ON condition to the directly preceding JOIN from left to right. The join expression can be parenthesized implicitly or explicitly to show the order of evaluation. In this case, the implicit parentheses are as follows:
SELECT * FROM (a INNER JOIN (b INNER JOIN c ON b~c = c~c) ON a~b = b~b)
This means that the first join expression is b INNER JOIN c ON b~c = c~c, which joins the columns of tables b and c based on the condition that b~c equals c~c. The second join expression is a INNER JOIN (b INNER JOIN c ON b~c = c~c) ON a~b = b~b, which joins the columns of table a and the result of the first join expression based on the condition that a~b equals b~b. The final result set contains all combinations of rows from tables a, b, and c that satisfy both join conditions.
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