Q: C++ non-virtual method call through a base pointer: which class name is printed when the static type is A*? #include class A { public: void CuriousTabFunction(void) { cout<< "Class A" << endl; } }; class B: public A { public: voi…

Introduction / Context: This program tests dynamic dispatch vs. static binding. Member functions are not declared virtual, so calls through a base pointer use static type information, not the dynamic type of the object. Given Data / Assumptions: ptr has static type A*. It ultimately points to an object of type C (via new C()). Each class defines a method with the same name and signature, but none are virtual. Concept / Approach: Without virtual, the call is statically bound to A::CuriousTabFunction because the expression type is A*. Dynamic dispatch would require virtual on the base declaration to select the most-derived override at runtime. Step-by-Step Solution: 1) ptr becomes an A* pointing at a C object. 2) ptr->CuriousTabFunction() is compiled to call A::CuriousTabFunction due to static binding. 3) The output is “Class A”. Verification / Alternative check: Mark CuriousTabFunction as virtual in A and rerun; the output becomes “Class C”. Why Other Options Are Wrong: “Class B” or “Class C” would require virtual dispatch. There is no compile-time error since method lookup succeeds and signatures match. Common Pitfalls: Assuming polymorphism without declaring methods virtual; forgetting that overriding without virtual still hides but does not polymorphically override. Final Answer: Class A.

Question 1

C++ deleting this inside a member: what happens if Function() calls delete this and then the pointer is used again?  #include<iostream.h> class CuriousTab {  int x;  float y;  public:  void Function()  {  x = 4;  y = 2.50;  delete this;  }  void Dis…

Accepted Answer

Introduction / Context: This program demonstrates a dangerous anti-pattern: calling delete this; inside a member function and then continuing to use the same pointer. Accessing a deleted object is undefined behavior and often leads to a crash or other runtime failure. Given Data / Assumptions: Function() writes members, then immediately destroys the current object using delete this. After the first deletion, the code calls Function() and Display() again on the dangling pointer. No custom allocator or safety checks exist. Concept / Approach: Once delete this executes, the pointer becomes dangling. Any subsequent call through it invokes undefined behavior. Typical outcomes include segmentation faults, heap corruption detection, or seemingly random output depending on memory reuse, but none of these are well-defined or reliable. Step-by-Step Solution: 1) Allocate object with new. 2) First Function(): sets x and y, then frees *this. 3) Second Function(): calls a method on a freed object → undefined behavior. 4) Display() also dereferences a dangling pointer → undefined behavior; commonly triggers a runtime error. Verification / Alternative check: If you remove the second and third calls (or set the pointer to nullptr after deletion), the program would terminate safely but still exhibit questionable design. Why Other Options Are Wrong: Any deterministic printed output assumes defined behavior; there is none after deletion. Compile-time error does not apply because the code is syntactically valid. Common Pitfalls: Using delete this without a strict ownership protocol; forgetting to set the pointer to nullptr and guard all subsequent uses. Final Answer: Runtime error / undefined behavior due to use-after-free.

Question 2

C++ string scan: count characters immediately following spaces in "Welcome to CuriousTab.com!" and return the count length. #include #include class CuriousTab { char str[50]; char tmp[50]; public: CuriousTab(char s) { st…

Accepted Answer

Introduction / Context: The routine scans a C-string and copies into tmp each character that immediately follows a space in str. It then returns the length of tmp. This tests pointer arithmetic and pre/post-increment order in a loop. Given Data / Assumptions: Input text: “Welcome to CuriousTab.com!”. Loop condition: while current character is non-zero. On seeing a space, post-increment advances i before reading the next character. Concept / Approach: We need each character after a space. In the sample text there are two spaces: between “Welcome” and “to”, and between “to” and “CuriousTab.com!”. Therefore, the characters copied are the first letters of the following words, namely 't' and 'C'. Step-by-Step Solution: 1) Initialize i=0, j=0. 2) Traverse until null terminator. 3) When str[i] equals space, post-increment makes i point to the next character; that character is assigned to tmp[j++]. 4) For the two spaces, copied characters are 't' and 'C', so tmp becomes “tC”. 5) strlen(tmp) is 2, which is printed. Verification / Alternative check: Add another space in the source string; the count increases accordingly, confirming the algorithm. Why Other Options Are Wrong: 1 undercounts spaces; 24/25 misinterpret the function as returning the whole string length; 0 would require no spaces or an empty string. Common Pitfalls: Misunderstanding post-increment usage; off-by-one when reading the character after the space; forgetting to null-terminate tmp. Final Answer: 2

Question 3

C++ composition with explicit member initialization: what product does Show() print for objBase(yy, yy)?  #include<iostream.h> class CuriousTabBase {  int x, y;  public:  CuriousTabBase(int xx = 10, int yy = 10)  {  x = xx;  y = yy;  }  void Show() …

Accepted Answer

Introduction / Context: Here we compare base construction with a composed member that is explicitly initialized using constructor arguments. Only the composed member objBase is printed, not the base subobject. Given Data / Assumptions: Derived constructor: base part gets (10,20); member objBase gets (yy,yy) i.e., (20,20). objBase.Show() prints x*y of the member object. Concept / Approach: The initializer objBase(yy, yy) sets x=20 and y=20 for the composed object. Therefore, Show() prints 20*20 = 400. The base subobject's state is irrelevant to the printed result. Step-by-Step Solution: 1) Construct base subobject with (10,20) → not printed. 2) Construct member objBase with (20,20). 3) objBase.Show() prints 400. Verification / Alternative check: Change the member initialization to objBase(xx, yy); the printout would become 200, proving the dependency. Why Other Options Are Wrong: 100 and 200 correspond to different initializations; “Garbage” and compile errors do not apply because all members are correctly constructed. Common Pitfalls: Assuming Show() refers to the base subobject; overlooking that member initializers run before the constructor body. Final Answer: 400

Question 4

C++ non-virtual method call through a base pointer: which class name is printed when the static type is A*?  #include<iostream.h> class A {  public:  void CuriousTabFunction(void)  {  cout<< "Class A" << endl;  } }; class B: public A {  public:  voi…

Accepted Answer

Introduction / Context: This program tests dynamic dispatch vs. static binding. Member functions are not declared virtual, so calls through a base pointer use static type information, not the dynamic type of the object. Given Data / Assumptions: ptr has static type A*. It ultimately points to an object of type C (via new C()). Each class defines a method with the same name and signature, but none are virtual. Concept / Approach: Without virtual, the call is statically bound to A::CuriousTabFunction because the expression type is A*. Dynamic dispatch would require virtual on the base declaration to select the most-derived override at runtime. Step-by-Step Solution: 1) ptr becomes an A* pointing at a C object. 2) ptr->CuriousTabFunction() is compiled to call A::CuriousTabFunction due to static binding. 3) The output is “Class A”. Verification / Alternative check: Mark CuriousTabFunction as virtual in A and rerun; the output becomes “Class C”. Why Other Options Are Wrong: “Class B” or “Class C” would require virtual dispatch. There is no compile-time error since method lookup succeeds and signatures match. Common Pitfalls: Assuming polymorphism without declaring methods virtual; forgetting that overriding without virtual still hides but does not polymorphically override. Final Answer: Class A.

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