In legacy C++ (iostream.h), what happens when a private member function is called from main versus a public one? Carefully read the code and predict the program's behavior.
#include
class CuriousTabSample
{
private:
int AdditionOne(int x, int y = 1)
{
return x * y;
}
public:
int AdditionTwo(int x, int y = 1)
{
return x / y;
}
};
int main()
{
CuriousTabSample objCuriousTab;
cout << objCuriousTab.AdditionOne(4, 8) << " ";
cout << objCuriousTab.AdditionTwo(8, 8);
return 0;
}
-
AThe program will print the output 32 0.
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BThe program will print the output 32 garbage-value.
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CThe program will print the output 32 1.
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DThe program will report compile time error.
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EIt prints nothing because both functions are inaccessible.
Answer
Correct Answer: The program will report compile time error.
Explanation
Introduction / Context:This question checks C++ access control for class members and how it affects compilation. The class exposes two member functions: one is private and multiplies two integers, while the other is public and performs integer division. The main function attempts to call both, which leads to an access-control issue at compile time.
Given Data / Assumptions:
AdditionOneis declared in the private section of the class.AdditionTwois declared in the public section.maincallsobjCuriousTab.AdditionOne(4, 8)and thenobjCuriousTab.AdditionTwo(8, 8).- Legacy header
Concept / Approach:In C++, class members are private by default if not specified, and any member explicitly placed under private: can be accessed only by member functions, friends, or privileged code. A call from main to a private member function is not permitted and must be rejected by the compiler.
Step-by-Step Solution:Identify access: AdditionOne is in the private section.Attempted call: objCuriousTab.AdditionOne(4, 8) originates outside the class.C++ rule: calling a private member from outside the class is ill-formed.Therefore, compilation fails before any code runs or outputs are produced.
Verification / Alternative check:Move AdditionOne to the public section or create a public wrapper that calls it; compilation will then succeed and you can observe outputs.
Why Other Options Are Wrong:Option A/B/C assume successful execution. In reality, the program never links or runs due to the illegal private call.Option E is incorrect because AdditionTwo is public and would be callable if the first error did not exist.
Common Pitfalls:Forgetting that access control is enforced at compile time, not run time, and assuming the presence of default arguments changes accessibility (it does not).
Final Answer:The program will report compile time error.