Curs 6

Question 1

class A { public: int a1, a2, a3; };
class B { public: int b1, b2; };
class C : private A, private B { public: int c1, c2; };
void main(void)
{
C c;
A* a = &c;
}

Answer 1

11
no
derivarea e privata
error C2243: ‘type cast’: conversion from
‘C *’ to ‘A *’ exists, but is inaccessible

Question 2

class A { public: int a1, a2, a3; };
class B { public: int b1, b2; };
class C : private A, private B { public: int c1, c2; };
void main(void)
{
C c;
A* a = reinterpret_cast<A*> (&c);
}

Answer 2

25
yes
This code will compile. It is important to understand the reinterpret_cast is
not bounded by class type or access modifier type.

Question 3

int x = reinterpret_cast <int>(1000);</int>

Answer 3

26
eroare

Question 4

int y = 100;
int x = reinterpret_cast <int>(y);</int>

Answer 4

26
eroare

Question 5

int x = reinterpret_cast<int>(“test”);</int>

Answer 5

26
yes

Question 6

char x = reinterpret_cast<char>(“test”);</char>

Answer 6

26
yes cu warning trunchiere date

Question 7

float x = reinterpret_cast<float>(1.2f);</float>

Answer 7

27
eroare

Question 8

double x = reinterpret_cast<double>(1.2);</double>

Answer 8

27
eroare

Question 9

int number = 10;
reinterpret_cast<int &>(number) = 20;

Answer 9

27
yes

Question 10

int number = 10;
reinterpret_cast<char &>(number) = 20;

Answer 10

27
yes

Question 11

int x = static_cast<int>(1000);</int>

Answer 11

30
yes

Question 12

char x = static_cast<char>(1000);</char>

Answer 12

30
yes
232=1000%256

Question 13

char x = static_cast<char>(3.75);</char>

Answer 13

30
yes
3=int(3.75)

Question 14

char x = static_cast<char>("test");</char>

Answer 14

30
eroare

Question 15

const char * x = static_cast<const char *>(9);

Answer 15

30
eroare

Question 16

class A { public: int a1, a2, a3; };
class B { public: int b1, b2; };
void main(void)
{
B b;
A* a = static_cast<A *>(&b);
}

Answer 16

31
eroare

Question 17

class A { public: int a1, a2, a3; };
class B
{
public:
int b1, b2;
operator A* () { return new A(); }
};
void main(void)
{
B b;
A* a = static_cast<A *>(&b);
}

Answer 17

32
eroare
A* a = static_cast<A *>(&b); are parametru &b in loc de b

Question 18

class A { public: int a1, a2, a3; };
class B
{
public:
int b1, b2;
operator A* () { return new A(); }
};
void main(void)
{
B b;
A* a = static_cast<A *>(b);
}

Answer 18

33
yes

Question 19

int Add(int x, char y)
{
return x + y;
}
int Add(char x, int y)
{
return x + y;
}
void main(void)
{
int suma = static_cast<int(*) (int, char)> (Add)(100, 200);
}

Answer 19

35
yes

Question 20

class A { public: int a1, a2, a3; };
class B
{
public:
int b1, b2;
};
class C : public A, public B { public: int c1, c2; };
void main(void)
{
B b;
C c = dynamic_cast<C>(&b);
}

Answer 20

38
eroare
‘dynamic_cast’: ‘B’ is not a polymorphic type
note: see declaration of ‘B’

Question 21

class A { public: int a1, a2, a3; };
class B
{
public:
int b1, b2;
virtual void f() {};
};
class C : public A, public B { public: int c1, c2; };
void main(void)
{
B b;
C c = dynamic_cast<C>(&b);
}

Answer 21

39
yes
c este nullptr pt ca b este obiect de sine statator

Question 22

class A { public: int a1, a2, a3; };
class B
{
public:
int b1, b2;
virtual void f() {};
};
class C : public A, public B { public: int c1, c2; };
void main(void)
{
C c;
B b = (B)&c;
C c2 = dynamic_cast<C>(b);
}

Answer 22

40
yes

Question 23

class A {
public: int a1, a2, a3;
virtual void f2() {};
};
class B {
public: int b1, b2;
virtual void f() {};
};
class C : public A, public B { public: int c1, c2; };
void main(void) {
C c;
A a = (A)&c;
B b = dynamic_cast<B>(a);
}

Answer 23

41
This code also compiles.
 While there is no relationship between A and B , since local variable “a”
points to an object of type C , that contains a “B” component, this conversion
is possible.

Question 24

void main(void)
{
int x = 100;
const int * ptr = &x;
int * non_const_pointer = const_cast <int *>(ptr);
*non_const_pointer = 200;
printf(“%d”, x);
}

Answer 24

43
This code will compile and will convert ptr const pointer to a non-constant
pointer that can be used to change the value of “x”
 As a result, the program will print “200” on the screen

Question 25

void main(void)
{
const char * txt = “C++ exam”;
char * non_const_pointer = const_cast <char *>(txt);
non_const_pointer[0] = ‘c’;
printf(“%s”, non_const_pointer);
}

Answer 25

44
This code will compile but it will most likely produce a run-time error.

Question 26

void main(void)
{
const int x = 100;
(const_cast<int>(&x)) = 200;
}

Answer 26

Now , this code will compile (using the const_cast we can remove the const
attribute of “x”) allowing one to change “x” value.
 The behavior of the program is however , undefined → in this context
undefined means that the actual value of “x” may not be 200 !!!

Question 27

void main(void)
{
const int x = 100;
(const_cast<int>(&x)) = 200;
printf(“%d”, x);
}

Answer 27

47
nu stim ce se va afisa

Question 28

void main(void)
{
const int x = 100;
(const_cast<int>(&x)) = 200;
printf(“%d”, (const_cast<int>(&x))) ;
}

Answer 28

48
200

Question 29

class Test
{
public:
const int x;
Test(int value) : x(value) { }
void Set(int value) { (const_cast<int>(&x)) = value; }
};
void main(void)
{
Test t(100);
printf(“%d “, t.x);
t.Set(200);
printf(“%d “, t.x);
}

Answer 29

49
100 200

Question 30

define PRINT(format,…) \

{ \
printf(“\nPrint values:”); \
printf(format, __VA_ARGS__); \
}
void main(void)
{
int v1, v2;
v1 = 100;
v2 = 200;
PRINT(“%d,%d”, v1, v2);
}

Answer 30

57
yes
void main(void)
{
int v1, v2;
v1 = 100;
v2 = 200;
printf(“\nPrint values:”);
printf(“%d,%d”, v1, v2);
}

Question 31

define CHECK(condition) { \

if (!(condition)) { printf(“The condition ‘%s’ wasn’t evaluated correctly”, #condition);
};
void main(void)
{
int v1, v2;
v1 = 100;
v2 = 200;
CHECK(v1 > v2);
}

Answer 31

58
void main(void)
{
int v1, v2;
v1 = 100;
v2 = 200;
if (!(v1>v2)) { printf(“The condition ‘%s’ wasn’t evaluated correctly”, “v1 > v2”); };
}

Question 32

define SUM(a,b) a+b

#define SUM(a,b,c) a+b+c
void main(void)
{
int x = SUM(1, 2);
}

Answer 32

61
error C2059: syntax error: ‘;’

Question 33

void main(void)
{
#define SUM(a,b) a+b
int x = SUM(1, 2);
#define SUM(a,b,c) a+b+c
x = SUM(1, 2, 3);
}

Answer 33

61
yes

Question 34

define DIV(x,y) x/y

void main(void)
{
int x = DIV(10 + 10, 5 + 5);
}

Answer 34

62
incorect
void main(void)
{
int x = 10 + 10 / 5 + 5;
}

Question 35

define DIV(x,y) ((x)/(y))

void main(void)
{
int x = DIV(10 + 10, 5 + 5);
}

Answer 35

62
corect
void main(void)
{
int x = ((10 + 10) / (5 + 5));
}

Question 36

define PRINT(x,y) \

printf(“X=%d”,x);printf(“Y=%d”,y);
void main(void)
{
int x = 10, y = 20;
if (x > y)
PRINT (x, y);
}

Answer 36

63
incorect
void main(void)
{
int x = 10, y = 20;
if (x > y)
printf(“X=%d”,x);printf(“Y=%d”,y);
}

Question 37

define PRINT(x,y) \

{ printf(“X=%d”,x);printf(“Y=%d”,y); }
void main(void)
{
int x = 10, y = 20;
if (x > y)
PRINT (x, y);
}

Answer 37

63
corect
void main(void)
{
int x = 10, y = 20;
if (x > y)
{printf(“X=%d”,x);printf(“Y=%d”,y);}
}

Question 38

define set_min(result,x,y) \

result = ((x)>(y)?(x):(y));
void main(void)
{
int x = 2, y = 1;
int res;
set_min(res, x++, y);
}

Answer 38

64
incorect
res will be 3

Question 39

define set_min(result,x,y) {\

int t_1 = (x); \
int t_2 = (y); \
result = ((t_1)>(t_2)?(t_1):(t_2)); \
}
void main(void)
{
int x = 2, y = 1;
int res;
set_min(res, x++, y);
}

Answer 39

64
corect
void main(void)
{
int x = 2, y = 1;
int res;
{ int t_1 = (x); int t_2 = (y);
result = ((t_1)>(t_2)?(t_1):(t_2)); }
}

Question 40

void main(void)
{
int x = __COUNTER__;
int y = __COUNTER__;
int z = __COUNTER__;
int t = __COUNTER__;
}

Answer 40

66
void main(void)
{
int x = 0;
int y = 1;
int z = 2;
int t = 3;
}