CPA-21-02 無料問題集「C++ Institute CPA - C++ Certified Associate Programmer」
What happens when you attempt to compile and run the following code?
#include <iostream>
#include <string>
using namespace std;
class A {
protected:
int y;
public:
int x;
int z;
A() { x=2; y=2; z=3; }
A(int a, int b) : x(a), y(b) { z = x ? y;}
void Print() {
cout << z;
}
};
int main () {
A a(2,5);
a.Print();
return 0;
}
#include <iostream>
#include <string>
using namespace std;
class A {
protected:
int y;
public:
int x;
int z;
A() { x=2; y=2; z=3; }
A(int a, int b) : x(a), y(b) { z = x ? y;}
void Print() {
cout << z;
}
};
int main () {
A a(2,5);
a.Print();
return 0;
}
正解:B
解答を投票する
What happens when you attempt to compile and run the following code?
#include <iostream>
#include <string>
using namespace std;
class complex{
double re;
double im;
public:
complex() : re(1),im(0.4) {}
bool operator==(complex &t);
};
bool complex::operator == (complex &t){
if((this?>re == t.re) && (this?>im == t.im))
return true;
else
return false;
}
int main(){
complex c1,c2;
if (c1==c2)
cout << "OK";
else {
cout << "ERROR";
}
}
#include <iostream>
#include <string>
using namespace std;
class complex{
double re;
double im;
public:
complex() : re(1),im(0.4) {}
bool operator==(complex &t);
};
bool complex::operator == (complex &t){
if((this?>re == t.re) && (this?>im == t.im))
return true;
else
return false;
}
int main(){
complex c1,c2;
if (c1==c2)
cout << "OK";
else {
cout << "ERROR";
}
}
正解:B
解答を投票する
What happens when you attempt to compile and run the following code?
#include <iostream>
#include <string>
using namespace std;
class A {
int x;
protected:
int y;
public:
int z;
};
class B : private A {
string name;
public:
void set() {
x = 1;
}
void Print() {
cout << x;
}
};
int main () {
B b;
b.set();
b.Print();
return 0;
}
#include <iostream>
#include <string>
using namespace std;
class A {
int x;
protected:
int y;
public:
int z;
};
class B : private A {
string name;
public:
void set() {
x = 1;
}
void Print() {
cout << x;
}
};
int main () {
B b;
b.set();
b.Print();
return 0;
}
正解:D
解答を投票する
What happens when you attempt to compile and run the following code?
#include <iostream>
using namespace std;
class A {
public:
int x;
A() { x=0;}
};
class B : protected A {
public:
int y;
using A::x;
B(int y) {this?>y = y;}
void Print() { cout << x << y; }
};
int main () {
B b(5);
b.Print();
return 0;
}
#include <iostream>
using namespace std;
class A {
public:
int x;
A() { x=0;}
};
class B : protected A {
public:
int y;
using A::x;
B(int y) {this?>y = y;}
void Print() { cout << x << y; }
};
int main () {
B b(5);
b.Print();
return 0;
}
正解:C
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What happens when you attempt to compile and run the following code?
#include <iostream>
using namespace std;
class First
{
public:
void Print(){ cout<<"from First";}
};
class Second:public First
{
public:
void Print(){ cout<< "from Second";}
};
void fun(First *obj);
int main()
{
First FirstObject;
fun(&FirstObject);
Second SecondObject;
fun(&SecondObject);
}
void fun(First *obj)
{
obj?>Print();
}
#include <iostream>
using namespace std;
class First
{
public:
void Print(){ cout<<"from First";}
};
class Second:public First
{
public:
void Print(){ cout<< "from Second";}
};
void fun(First *obj);
int main()
{
First FirstObject;
fun(&FirstObject);
Second SecondObject;
fun(&SecondObject);
}
void fun(First *obj)
{
obj?>Print();
}
正解:C
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