Computer Science : Compile Time Errors
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Example Questions
Example Question #1 : Compile Time Errors
class Base{};
class Derived : public Base{
public:
void method(){ cout<< "method1\n"; }
};
class Derived2 : public Base{
public:
void method() { cout<< "method2\n"; }
};
int main(){
Base* bp = new Derived();
Derived2* d2p = bp;
d2p -> method();
}
What is the result of compiling and running the program in C++?
The program compiles and runs, printing "method1"
The program compiles and runs to completion without printing anything
The program compiles and crashes when it runs.
The program compiles and runs, printing "method2"
The program does not compile.
The program does not compile.
In this problem, Derived1 and Derived2 are children of the Base class. If we take a look at this line:
Base* bp = new Derived();
We are assigning a new Derived class to a base pointer. This will compile. Think of the Base as a larger object because it is the parent, so copying a smaller object into a larger one is acceptable.
Now let's look at this one:
Derived2* d2p = bp;
This line will cause the program to not compile. Since the Base class is considered the "bigger" object, copying a bigger object into a "smaller" one will result in a failure to copy everything over, this is known as a Slicing Problem.
We don't even have to look at the next line because we know that the program wil crash.
Example Question #1 : Debugging
Given:
const int x = 10;
Which of the following will compile?
int * const q = &x
const int * r = &x
None of the above
All of the above
int * p = &x
const int * r = &x
First we take a look at the given statement:
const int x = 10;
the const in front of "int" means that x will always hold the value of 10 and it will not change.
Let's observe all the choices.
int *p =&x
This line says to assign the address of x (in memory) to the pointer p. This however, will not compile because int * p is not marked as const. x is marked as a const so this forces int * p to be a const as well.
int * const q = &x
There is a const in this case but it is in the wrong place
const int * r = &x
The const is in the correct place and this is the correct answer
Example Question #1 : Compile Time Errors
class Base{
protected:
void method();
};
class Derived : public Base{
};
int main(){
Base b;
b.method(); //Line A
Derived d;
d.method(); //Line B
}
Which of the following is true?
None of these
Line A will compile
Line B will not compile
Line A will not compile
Line B will not compile
Line A will compile
Line B will compile
Line A will not compile
Line B will compile
Line A will not compile
Line B will not compile
To understand this question, we have to understand what protected method means. A protected method is a method that is accessible to methods inside it's own class as well as it's children. This means that a protected method can be called in the child class.
We can see that method() is called inside main. This should already raise a red flag. A protected class is being called outside of the child class so it will not compile. Even those it's being called on the Base and Derived objects, the calls are not made inside Base and Derived class so neither line will compile.
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