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C_ABAPD_2309 Exam Dumps and Practice Questions

Preview page 1 of 16 with real questions 1 to 5 from the full 81-question SAP set.

Most Recent SAP C_ABAPD_2309 Exam Dumps

 

Prepare for the SAP Certified Associate - Back-End Developer - ABAP Cloud Exam exam with our extensive collection of questions and answers. These practice Q&A are updated according to the latest syllabus, providing you with the tools needed to review and test your knowledge.

QA4Exam focus on the latest syllabus and exam objectives, our practice Q&A are designed to help you identify key topics and solidify your understanding. By focusing on the core curriculum, These Questions & Answers helps you cover all the essential topics, ensuring you're well-prepared for every section of the exam. Each question comes with a detailed explanation, offering valuable insights and helping you to learn from your mistakes. Whether you're looking to assess your progress or dive deeper into complex topics, our updated Q&A will provide the support you need to confidently approach the SAP C_ABAPD_2309 exam and achieve success.

Actual SAP C_ABAPD_2309 Exam Questions

Question 1

What RESTful Application Programming object contains only the fields required for a particular app?

Correct answer: C

A projection view is a RESTful Application Programming object that contains only the fields required for a particular app. A projection view is a CDS view entity that defines a projection on an existing CDS view entity or CDS DDIC-based view. A projection view exposes a subset of the elements of the projected entity, which are relevant for a specific business service. A projection view can also define aliases, virtual elements, and annotations for the projected elements. A projection view is the top-most layer of a CDS data model and prepares data for a particular use case. A projection view can have different provider contracts depending on the type of service it supports, such as transactional query, analytical query, or transactional interface.

A database view is a CDS DDIC-based view that defines a join or union of database tables. A database view has an SQL view attached and can be accessed by Open SQL or native SQL. A database view can be used as a projected entity for a projection view, but it does not contain only the fields required for a particular app.

A metadata extension is a RESTful Application Programming object that defines additional annotations for a CDS view entity or a projection view. A metadata extension can be used to enhance the metadata of a CDS data model without changing the original definition. A metadata extension does not contain any fields, but only annotations.

A data model view is a CDS view entity that defines a data model based on database tables or other CDS view entities. A data model view can have associations, aggregations, filters, parameters, and annotations. A data model view can be used as a projected entity for a projection view, but it does not contain only the fields required for a particular app.


Question 2

What are some characteristics of secondary keys for internal tables? Note: There are 3 correct answers to this question.

Correct answer: A, B, D

Secondary keys are additional keys that can be defined for internal tables to optimize the access to the table using fields that are not part of the primary key. Secondary keys can be either sorted or hashed, depending on the table type and the uniqueness of the key.Secondary keys have the following characteristics1:

A . Secondary keys must be chosen explicitly when you actually read from an internal table. This means that when you use a READ TABLE or a LOOP AT statement to access an internal table, you have to specify the secondary key that you want to use with the USING KEY addition. For example, the following statement reads an internal table itab using a secondary key sec_key:

READ TABLE itab USING KEY sec_key INTO DATA(wa).

If you do not specify the secondary key, the system will use the primary key by default2.

B . Multiple secondary keys are allowed for any kind of internal table. This means that you can define more than one secondary key for an internal table, regardless of the table type. For example, the following statement defines an internal table itab with two secondary keys sec_key_1 and sec_key_2:

DATA itab TYPE SORTED TABLE OF ty_itab WITH NON-UNIQUE KEY sec_key_1 COMPONENTS field1 field2 sec_key_2 COMPONENTS field3 field4.

You can then choose which secondary key to use when you access the internal table1.

D . Sorted secondary keys do NOT have to be unique. This means that you can define a sorted secondary key for an internal table that allows duplicate values for the key fields. A sorted secondary key maintains a predefined sorting order for the internal table, which is defined by the key fields in the order in which they are specified. For example, the following statement defines a sorted secondary key sec_key for an internal table itab that sorts the table by field1 in ascending order and field2 in descending order:

DATA itab TYPE STANDARD TABLE OF ty_itab WITH NON-UNIQUE SORTED KEY sec_key COMPONENTS field1 ASCENDING field2 DESCENDING.

You can then access the internal table using the sorted secondary key with a binary search algorithm, which is faster than a linear search3.

The following are not characteristics of secondary keys for internal tables, because:

C . Hashed secondary keys do NOT have to be unique. This is false because hashed secondary keys must be unique. This means that you can only define a hashed secondary key for an internal table that does not allow duplicate values for the key fields. A hashed secondary key does not have a predefined sorting order for the internal table, but uses a hash algorithm to store and access the table rows. For example, the following statement defines a hashed secondary key sec_key for an internal table itab that hashes the table by field1 and field2:

DATA itab TYPE STANDARD TABLE OF ty_itab WITH UNIQUE HASHED KEY sec_key COMPONENTS field1 field2.

You can then access the internal table using the hashed secondary key with a direct access algorithm, which is very fast.

E . Secondary keys can only be created for standard tables. This is false because secondary keys can be created for any kind of internal table, such as standard tables, sorted tables, and hashed tables. However, the type of the secondary key depends on the type of the internal table.For example, a standard table can have sorted or hashed secondary keys, a sorted table can have sorted secondary keys, and a hashed table can have hashed secondary keys1.


Question 3
Question 4

In a test method you call method cl_abap_unit_assert=>assert_equals( .. ) in the following way:

CLASS Itcl1 DEFINITION FOR TESTING RISK LEVEL HARMLESS DURATION SHORT.

PRIVATE SECTION.

METHODS m1 FOR TESTING.

ENDCLASS.

CLASS Itcl1 IMPLEMENTATION.

METHOD m1.

DATA: go_test_object TYPE REF TO zcl_to_be_tested.

CONSTANTS: Ico_exp TYPE string VALUE 'test2'.

CREATE OBJECT go_test_object.

cl_abap_unit_assert=>assert_equals(

EXPORTING

act = go_class->mv_attribute

exp = lco_exp

msg = 'assert equals failed ' && go_test_object->mv_attribute && ' ' && lco_exp

ENDMETHOD.

ENDCLASS.

What will happen if method parameters act and exp are not equal?

Correct answer: D

Question 5

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