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SAP C_ABAPD_2309 Exam Syllabus Topics:

Topic
Details

Topic 1

ABAP SQL and code pushdown: It discusses ABAP SQL, arithmetic expressions, manage dates, and create joins.

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.

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SAP Certified Associate - Back-End Developer - ABAP Cloud Sample Questions (Q42-Q47):

NEW QUESTION # 42
What are some characteristics of secondary keys for internal tables? Note: There are 3 correct answers to this question.

A. Sorted secondary keys do NOT have to be unique.
B. Multiple secondary keys are allowed for any kind of internal table.
C. Secondary keys can only be created for standard tables.
D. Hashed secondary keys do NOT have to be unique.
E. Secondary keys must be chosen explicitly when you actually read from an internal table.

Answer: A,B,E

Explanation:
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.
References: 1: Secondary Table Key - ABAP Keyword Documentation 2: READ TABLE - ABAP Keyword Documentation 3: Sorted Tables - ABAP Keyword Documentation : Hashed Tables - ABAP Keyword Documentation

NEW QUESTION # 43
In this nested join below in which way is the join evaluated?

A. From the bottom to the top in the order of the on conditions:
1.
a is joined with b
2.
b is joined with c
B. From the right to the left in the order of the tables:
1.
b is joined with c.
2.
b is joined with a.
C. From the top to the bottom in the order of the on conditions
1.
b is joined with c
2.
a is joined with b
D. From the left to the right in the order of the tables:
1.
a is joined with b
2.
b is joined with c

Answer: C

Explanation:
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.
References: 1: SELECT, FROM JOIN - ABAP Keyword Documentation - SAP Online Help

NEW QUESTION # 44
Which restrictions exist for ABAP SQL arithmetic expressions? Note: There are 2 correct answers to this question.

A. The operator is allowed only in floating point expressions.
B. Decimal types and integer types can NOT be used in the same expression.
C. The operator/is allowed only in floating point expressions.
D. Floating point types and integer types can NOT be used in the same expression.

Answer: A,C

Explanation:
ABAP SQL arithmetic expressions have different restrictions depending on the data type of the operands. The following are some of the restrictions:
* Floating point types and integer types can be used in the same expression, as long as the integer types are cast to floating point types using the cast function. For example, CAST ( num1 AS FLTP ) / CAST ( num2 AS FLTP ) is a valid expression, where num1 and num2 are integer types.
* The operator / is allowed only in floating point expressions, where both operands have the type FLTP or f. For example, num1 / num2 is a valid expression, where num1 and num2 are floating point types. If the operator / is used in an integer expression or a decimal expression, a syntax error occurs.
* Decimal types and integer types can be used in the same expression, as long as the expression is a decimal expression. A decimal expression has at least one operand with the type DEC, CURR, or QUAN or p with decimal places. For example, num1 + num2 is a valid expression, where num1 is a decimal type and num2 is an integer type.
* The operator ** is allowed only in floating point expressions, where both operands have the type FLTP or f. For example, num1 ** num2 is a valid expression, where num1 and num2 are floating point types.
If the operator ** is used in an integer expression or a decimal expression, a syntax error occurs.
References: sql_exp - sql_arith - ABAP Keyword Documentation, SQL Expressions, Arithmetic Calculations
- ABAP Keyword Documentation

NEW QUESTION # 45
Refer to the Exhibit.
Image:

In the following ABAP SQL code, what are valid case distinctions? Note: There are 2 correct answers to this question.

Answer: B,C

NEW QUESTION # 46
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?

A. scounter will be joined to sairport first and the result will be joined with scarr.
B. scarr will be joined with scounter first and the result will be joined with sairport.
C. sairport will be joined to scounter first and the result will be joined with scarr.
D. scarr will be joined with sairport first and the result will be joined with scounter.

Answer: B

Explanation:
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.

NEW QUESTION # 47
......

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