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Salesforce Plat-Arch-205 - Salesforce Certified Platform Sharing and Visibility Architect Exam Braindumps

Salesforce Plat-Arch-205 - Salesforce Certified Platform Sharing and Visibility Architect Exam Braindumps

  • Certification Provider:Salesforce
  • Exam Code:Plat-Arch-205
  • Exam Name:Salesforce Certified Platform Sharing and Visibility Architect Exam Exam
  • Total Questions:276 Questions and Answers
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NEW QUESTION: 1
What happens when you attempt to compile and run the following code?
#include <vector>
#include <set>
#include <iostream>
#include <algorithm>
using namespace std;
class B {
int val;
public:
B(int v):val(v){}
operator int() { return val;}
};
template<class T>struct Out {
ostream & out;
Out(ostream & o): out(o){}
void operator() (const T & val ) { out<<val<<" "; } };
struct Sequence { int start;
Sequence(int start):start(start){}
int operator()() { return start++; } };
bool predicate(int v) { return v%2==0; }
int main() {
vector<int> v1(10);
generate_n(v1.begin(), 10, Sequence(1));
for_each(v1.begin(), remove_if(v1.begin(), v1.end(), predicate), Out<int>(cout));cout<<endl; return 0;} Program outputs:
A. 1 3 5 7 9 6 7 8 9 10
B. compilation error
C. no output
D. 1 3 5 7 9
E. 2 4 6 8 10
Answer: D

NEW QUESTION: 2
Sie sind Datenbankentwickler einer Microsoft SQL Server 2014-Datenbank. Sie entwerfen eine Tabelle, in der Kundendaten aus verschiedenen Quellen gespeichert werden. Die Tabelle enthält eine Spalte mit der CustomerID aus dem Quellsystem und eine Spalte mit der SourceID. Ein Beispiel für diese Daten ist in der folgenden Tabelle aufgeführt.

Sie müssen sicherstellen, dass die Tabelle keine doppelte CustomerID innerhalb einer SourceID enthält. Sie müssen auch sicherstellen, dass die Daten in der Tabelle in der Reihenfolge SourceID und CustomerID vorliegen. Welche Transact-SQL-Anweisung sollten Sie verwenden?
A. CREATE TABLE Kunde (SourceID int NOT NULL, CustomerID int NOT NULL PRIMARY KEY CLUSTERED, Kundenname varchar (255) NOT NULL);
B. CREATE TABLE Kunde (SourceID int NOT NULL PRIMARY KEY CLUSTERED, CustomerID int NOT NULL UNIQUE, Kundenname varchar (255) NOT NULL);
C. CREATE TABLE Customer (Quell-ID: NICHT NULL IDENTITY, Kundennummer: NICHT NULL IDENTITY, Kundenname: varchar (255) NICHT NULL);
D. TABELLE ERSTELLEN Kunde (SourceID int NICHT NULL, CustomerID int NICHT NULL, CustomerName varchar (255) NICHT NULL, CONSTRAINT PK_Customer PRIMARY KEY CLUSTERED (SourceID, CustomerID));
Answer: D

NEW QUESTION: 3
Which action is required for EIGRP to auto-summarize external routes?
A. external routes that stop at a classful boundary
B. an internal route that is part of the same major network
C. EIGRP does not auto-summarize external routes.
D. a class map that defines the external routes
Answer: B

NEW QUESTION: 4
攻撃軽減技術を左側から右側に軽減する攻撃のタイプにドラッグアンドドロップします。

Answer:
Explanation:


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