Assuming that public class B : A { public B(int i) :base(i) { } } compiles and runs correctly, what can we conclude about the constructors in the class A?
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One constructor takes an argument of type i
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There is only a default constructor
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One constructor takes an arguments of the type int
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a and b
The :base(i) syntax calls the parent constructor with an int parameter, so class A must have a constructor accepting int. Option A is wrong because i is the variable name, not the type. Option B is wrong because a parameterized constructor is being called. C correctly identifies an int-taking constructor exists.
To answer this question, let's analyze the given code:
public class B : A {
public B(int i) : base(i) {
}
}
This code snippet shows the class B inheriting from class A. The constructor in class B takes an argument of type int and calls the base constructor of class A with the same argument.
Based on this information, we can conclude that:
A. One constructor takes an argument of type i - This option is incorrect because the argument in the constructor of class B is named i, not i itself.
B. There is only a default constructor - This option is incorrect because the presence of the B(int i) constructor in class B indicates that there is a non-default constructor in class A.
C. One constructor takes an argument of the type int - This option is correct because the B(int i) constructor in class B calls the base constructor A(int i), indicating that there is a constructor in class A that takes an argument of type int.
D. a and b - This option is incorrect because option A is incorrect. Only option C is correct.
Therefore, the correct answer is option C.