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Which code, inserted at line 19, generates the output "23"?
#include
#include
using namespace std;
class A {
int x;
protected:
int y;
public:
int z;
A() { x=1; y=2; z=3; }
};
class B : public A {
string z;
public:
int y;
void set() { y = 4; z = "John"; }
void Print() {
//insert code here
}
};
int main () {
B b;
b.set();
b.Print();
return 0;
}
Suggested Answer: C
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22 comments · newest first
SH
Shala
5 days ago
Why B?
FA
Fanny
10 days ago
I think option B is the right choice.
BI
Billye
15 days ago
Exactly! B gives "23" because of A::z.
HU
Hui
20 days ago
D? But it prints B's z, not A's.
GW
Gwen
25 days ago
I feel like D could work too.
ST
Stephania
1 month ago
It accesses the base class's z.
HU
Hui
1 month ago
Why B?
ST
Stephania
1 month ago
I think option B is the right choice.
VI
Vicki
2 months ago
I’m not so sure about that, seems tricky.
CH
Chara
2 months ago
No way, it's definitely A!
JA
Janessa
2 months ago
Wait, does that mean A::z is 3?
MA
Marjory
2 months ago
I think it's actually D.
MA
Marion
2 months ago
Option B is the right choice!
GI
Gilma
2 months ago
Not sure about that, seems tricky.
ZO
Zona
3 months ago
Totally agree with B!
KI
Kimi
3 months ago
Wait, does that mean A::z is different from B::z?
AL
Alona
3 months ago
I think it's actually option D.
MA
Malcolm
3 months ago
Option B is the right choice!
SI
Silvana
3 months ago
I'm a bit confused about the protected members. Does option A even work? I thought protected members couldn't be accessed like that.
QU
Quinn
3 months ago
I practiced a similar question, and I feel like option D could work since it prints the values from class B directly.
BE
Beata
4 months ago
I'm not sure, but I remember something about scope resolution. Maybe option C is the right choice since it uses A::z.
JA
Jacklyn
5 months ago
I think option B might be correct because it accesses the public variable from class A.