Q: C++ lvalue references cannot bind to rvalues: detect the error with post-increment operands. #include int main() { int m = 2, n = 6; int &x = m++; int &y = n++; m = x++; x = m++; n = y++; y = n++; cout << m << " " << n; return…

Introduction / Context: This question targets reference binding rules in C++. Non-const lvalue references (int&) cannot bind to temporaries (rvalues). Using a post-increment expression as a reference initializer creates such a temporary, triggering a compilation error. Given Data / Assumptions: int &x = m++; and int &y = n++; Post-increment yields rvalues (the old values), not lvalues. Classic iostream.h headers do not affect this rule. Concept / Approach: The initializer for x is m++ which is an rvalue; a non-const lvalue reference cannot bind to it. The same applies to y. Modern C++ permits binding a const reference to a temporary, but not a non-const reference. Therefore compilation fails before any runtime output is possible. Step-by-Step Solution: Recognize m++ / n++ are rvalues.int& requires an lvalue; binding fails.Compilation stops at the first invalid binding. Verification / Alternative check: Replacing with const int &x = m++; and const int &y = n++; would compile (binding to temporaries allowed for const refs), though subsequent ++ uses would fail due to constness. Why Other Options Are Wrong: All numeric outputs assume successful compilation and execution. Common Pitfalls: Assuming references can bind to any expression; confusing lvalues and rvalues; forgetting the special rule for const references to temporaries. Final Answer: It will result in a compile time error.

Q: C++ constructor parameters assigned via this-> and no Display call: what prints? #include class Tab { int x, y; public: Tab(int x, int y) { this->x = x; this->y = y; } void Display() { cout << x << " " << y; } }; int main() { int x…

Introduction / Context: Here, the goal is to check understanding that simply constructing an object does not print anything unless output is explicitly performed (e.g., by calling Display or writing from the constructor). The code initializes members using this-> but never outputs them. Given Data / Assumptions: Tab::Display exists but is never invoked. Constructor sets internal x and y to 50 and 50 respectively. Main returns immediately. Concept / Approach: C++ programs only print when code writes to std::cout. Since Display is not called and the constructor contains no output, nothing is written, and the program produces no console output. Step-by-Step Solution: Create x = 50 and reference y = x.Construct Tab b(y, x) → members set to 50 and 50.No cout call occurs; program ends. Verification / Alternative check: Adding b.Display(); before return 0 would print 50 50. Likewise, adding output inside the constructor would print during construction. Why Other Options Are Wrong: 50 50: would require Display() or constructor output. Garbage values: members are initialized deterministically. Compile error: code is valid with old headers. Common Pitfalls: Assuming the presence of a Display method implies it is called; expecting output from mere object construction. Final Answer: The program will print nothing.

Question 1

C++ lvalue references cannot bind to rvalues: detect the error with post-increment operands.  #include <iostream.h> int main() {  int m = 2, n = 6;  int &x = m++;  int &y = n++;  m = x++;  x = m++;  n = y++;  y = n++;  cout << m << " " << n;  return…

Accepted Answer

Introduction / Context: This question targets reference binding rules in C++. Non-const lvalue references (int&) cannot bind to temporaries (rvalues). Using a post-increment expression as a reference initializer creates such a temporary, triggering a compilation error. Given Data / Assumptions: int &x = m++; and int &y = n++; Post-increment yields rvalues (the old values), not lvalues. Classic iostream.h headers do not affect this rule. Concept / Approach: The initializer for x is m++ which is an rvalue; a non-const lvalue reference cannot bind to it. The same applies to y. Modern C++ permits binding a const reference to a temporary, but not a non-const reference. Therefore compilation fails before any runtime output is possible. Step-by-Step Solution: Recognize m++ / n++ are rvalues.int& requires an lvalue; binding fails.Compilation stops at the first invalid binding. Verification / Alternative check: Replacing with const int &x = m++; and const int &y = n++; would compile (binding to temporaries allowed for const refs), though subsequent ++ uses would fail due to constness. Why Other Options Are Wrong: All numeric outputs assume successful compilation and execution. Common Pitfalls: Assuming references can bind to any expression; confusing lvalues and rvalues; forgetting the special rule for const references to temporaries. Final Answer: It will result in a compile time error.

Question 2

C++ constructor parameters assigned via this-> and no Display call: what prints? #include class Tab { int x, y; public: Tab(int x, int y) { this->x = x; this->y = y; } void Display() { cout << x << " " << y; } }; int main() { int x…

Accepted Answer

Introduction / Context: Here, the goal is to check understanding that simply constructing an object does not print anything unless output is explicitly performed (e.g., by calling Display or writing from the constructor). The code initializes members using this-> but never outputs them. Given Data / Assumptions: Tab::Display exists but is never invoked. Constructor sets internal x and y to 50 and 50 respectively. Main returns immediately. Concept / Approach: C++ programs only print when code writes to std::cout. Since Display is not called and the constructor contains no output, nothing is written, and the program produces no console output. Step-by-Step Solution: Create x = 50 and reference y = x.Construct Tab b(y, x) → members set to 50 and 50.No cout call occurs; program ends. Verification / Alternative check: Adding b.Display(); before return 0 would print 50 50. Likewise, adding output inside the constructor would print during construction. Why Other Options Are Wrong: 50 50: would require Display() or constructor output. Garbage values: members are initialized deterministically. Compile error: code is valid with old headers. Common Pitfalls: Assuming the presence of a Display method implies it is called; expecting output from mere object construction. Final Answer: The program will print nothing.

Question 3

In C++ programming, consider the following two statements about references:  A reference is not a constant pointer, but rather an alias for an existing object.  A reference is automatically dereferenced when it is used in an expression, so you acces…

Accepted Answer

Introduction: This question tests your understanding of C++ references and how they differ from pointers. Many beginners confuse references with constant pointers, but at the language level they behave differently, even if compilers may implement them similarly under the hood. Given Data / Assumptions: Statement 1: A reference is not a constant pointer. Statement 2: A reference is automatically dereferenced when it is used. We must decide which combination of these statements is correct in C++. Concept / Approach: In C++, a reference is an alias for an existing variable. Once a reference is bound to a variable, it cannot be reseated to refer to another variable. A pointer, in contrast, is an object that stores a memory address. Even though references may be implemented using constant pointers internally, the C++ language rules treat them as a separate construct with simpler syntax. Step-by-Step Solution: Consider the declaration: int x = 10; int &ref = x; Here ref is a reference, not a pointer. You use ref exactly like x. There is no need to write *ref to access the value; ref itself gives the value 10. At the C++ language level, ref is not a pointer type. It is a reference type and cannot be null or reseated after initialization. Therefore, statement 1 is correct: a reference is not a constant pointer in the type system. Statement 2 is also correct because when you write ref in an expression, the compiler automatically treats it as the underlying object, just as if you had dereferenced a pointer. Verification / Alternative check: If you try to take sizeof on a reference, the result is the same as sizeof of the referred type, not the size of a pointer. This further shows that references are not exposed as pointers in the language model, although they behave pointer like in terms of indirect access. Why Other Options Are Wrong: Saying only one of the statements is correct ignores that both accurately describe references. Saying both are incorrect contradicts both the C++ standard description and how code using references is written and compiled. Common Pitfalls: A frequent misconception is to believe that because many compilers implement references as constant pointers internally, they should be treated as such in code. In fact, the programmer should think of references as safer, always valid aliases, not as pointers that require explicit dereferencing and address handling. Final Answer: The correct description is that both statements 1 and 2 are correct.

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