Q: C++ function with multiple defaulted parameters used as a polynomial: compute two outputs for x = 2.3. #include double CuriousTabFunction(double, double, double = 0, double = 0, double = 0); int main() { double d = 2.3; cout << Cur…

Introduction / Context: The function evaluates a cubic-like expression using Horner-style nesting. Default arguments make most coefficients zero unless explicitly provided. You must track which parameters receive values and which default to zero.Given Data / Assumptions:x = 2.3.First call: p = 7, q = r = s = 0.Second call: p = 7, q = 6, r = s = 0.Concept / Approach: For the first call, every nested term beyond p cancels to zero, leaving p. For the second call, only q contributes beyond p, scaled by x once. Evaluating carefully yields numeric results.Step-by-Step Solution:Call 1: p + (q + (r + sx)x)x with q=r=s=0 → 7 + (0 + (0 + 0)x)x = 7.Call 2: p=7, q=6, r=s=0 → 7 + (6 + (0 + 0)x)x = 7 + 6x = 7 + 62.3 = 7 + 13.8 = 20.8.Verification / Alternative check: Writing the general form as p + qx + rx^2 + sx^3 confirms that only qx contributes in the second call.Why Other Options Are Wrong:7 20: Misses the 0.8 contribution of 60.133..; numerically incorrect.7 19.8: Off by 1.0; likely mistaken sign.7 Garbage / None of the above: All arguments are defined and the math is straightforward.Common Pitfalls: Misplacing parentheses in nested expressions, or forgetting that only q survives when r and s are zero.Final Answer: 7 20.8

Q: C++ overloading with a competing overload that has a default argument: which function is called and what is printed? #include int CuriousTabTest(int x, int y); int CuriousTabTest(int x, int y, int z = 5); int main() { cout << Curiou…

Introduction / Context: This item examines overload resolution when one overload exactly matches the argument list and another is also viable by supplying a default argument. The C++ rules select the best viable function, favoring an exact arity match over one that relies on defaults.Given Data / Assumptions:Two overloads exist: f(int,int) and f(int,int,int=5).Call site: CuriousTabTest(2, 4).Both overloads are viable, but one requires using a default argument for the third parameter.Concept / Approach: Overload resolution prefers the function that requires no default arguments and exactly matches the number and types of the provided parameters. Therefore CuriousTabTest(int,int) is selected, computing 2 * 4 = 8.Step-by-Step Solution:Evaluate candidates: (int,int) and (int,int,int=5).Both match types; the three-parameter overload needs its default for z.Best viable function is the exact two-parameter overload.Return 2 * 4 → 8; cout prints 8.Verification / Alternative check: If the call were CuriousTabTest(2,4,5), the three-parameter version would be chosen and print 40. Removing the two-parameter overload would also yield 40 due to the default.Why Other Options Are Wrong:5: Not computed by either overload.40: Would require calling the three-parameter function explicitly (or removing the two-parameter overload).Compile time error / None of the above: The overload set is resolvable.Common Pitfalls: Thinking the presence of a defaulted parameter forces selection of that overload; in reality, defaults only make a function viable, not necessarily the best match.Final Answer: The program will print the output 8.

Question 1

C++ function with multiple defaulted parameters used as a polynomial: compute two outputs for x = 2.3.  #include <iostream.h>  double CuriousTabFunction(double, double, double = 0, double = 0, double = 0);  int main() {  double d = 2.3;  cout << Cur…

Accepted Answer

Introduction / Context: The function evaluates a cubic-like expression using Horner-style nesting. Default arguments make most coefficients zero unless explicitly provided. You must track which parameters receive values and which default to zero.Given Data / Assumptions:x = 2.3.First call: p = 7, q = r = s = 0.Second call: p = 7, q = 6, r = s = 0.Concept / Approach: For the first call, every nested term beyond p cancels to zero, leaving p. For the second call, only q contributes beyond p, scaled by x once. Evaluating carefully yields numeric results.Step-by-Step Solution:Call 1: p + (q + (r + sx)x)x with q=r=s=0 → 7 + (0 + (0 + 0)x)x = 7.Call 2: p=7, q=6, r=s=0 → 7 + (6 + (0 + 0)x)x = 7 + 6x = 7 + 62.3 = 7 + 13.8 = 20.8.Verification / Alternative check: Writing the general form as p + qx + rx^2 + sx^3 confirms that only qx contributes in the second call.Why Other Options Are Wrong:7 20: Misses the 0.8 contribution of 60.133..; numerically incorrect.7 19.8: Off by 1.0; likely mistaken sign.7 Garbage / None of the above: All arguments are defined and the math is straightforward.Common Pitfalls: Misplacing parentheses in nested expressions, or forgetting that only q survives when r and s are zero.Final Answer: 7 20.8

Question 2

C++ defaults, compound assignments, and parameter passing by value: trace member updates and print x and y.  #include <iostream.h>  class CuriousTab {  int x, y, z; public:  void Apply(int xx = 12, int yy = 21, int zz = 9)  {  x = xx; // x = 12  y =…

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Introduction / Context: This exercise tests understanding of default arguments, pass-by-value, and chained assignments with compound operators. It asks you to follow exact state changes inside Apply and then print two members using Display.Given Data / Assumptions:SetValue calls Apply(12, 0, 20).Parameters are passed by value; modifications to yy do not escape Apply except through explicit assignments to members.Operators used: yy += 10 and x -= 2, with their results assigned to y and z respectively.Concept / Approach: Carefully evaluate compound assignments which both change the left-hand operand and yield a value. Keep track of x and y after each statement to determine what Display prints.Step-by-Step Solution:Entry to Apply: xx = 12, yy = 0, zz = 20.x = xx → x = 12.y = yy += 10 → yy becomes 10 (locally) and y receives 10.z = x -= 2 → x becomes 10 and z receives 10.Display prints x and y → "10 10".Verification / Alternative check: Rewriting as yy = yy + 10; y = yy; x = x - 2; z = x; confirms the same final member values.Why Other Options Are Wrong:12 10: Ignores the later x -= 2.12 21 / 12 31: Misapply defaults or assume pass-by-reference, which is not used here.The program will report compilation error.: The code compiles as intended.Common Pitfalls: Thinking yy += 10 returns the old value, or assuming parameters are references; both lead to incorrect y values.Final Answer: 10 10

Question 3

C++ modulo is not defined for floating-point operands: identify the outcome when using % with float values in a method.  #include <iostream.h>  void Tester(float xx, float yy = 5.0); // unrelated free function  class CuriousTab {  float x;  float y;…

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Introduction / Context: In C++, the % (remainder) operator is defined only for integral types. Attempting to apply it to floating-point operands is ill-formed. This snippet intentionally uses % with float values to provoke a compiler diagnostic.Given Data / Assumptions:Method Tester takes float parameters and assigns to float members.Expression: ++x % --y with x and y of type float.Classic headers (iostream.h) do not affect the core language rule about %.Concept / Approach: Both operands to % must be integral. Since ++x and --y are still floats, there is no valid operator% overload, and the code fails to compile with an error similar to “invalid operands to binary %”.Step-by-Step Solution:Set x = 5.0 and y = 5.0.Attempt to compute ++x % --y with float operands.Compiler rejects the expression at compile time; no program output is produced.Verification / Alternative check: Changing the members to int (and passing integer literals) would allow % and yield a value. Alternatively, use fmod from for floating-point remainders: fmod(++x, --y).Why Other Options Are Wrong:Any printed numeric output assumes successful compilation and execution, which does not occur.“Garbage value” implies runtime behavior; the failure is compile-time.Common Pitfalls: Assuming % behaves like mathematical modulus for floats, or forgetting to use fmod for double/float types.Final Answer: The program will report compile time error.

Question 4

C++ overloaded constructors and default arguments in a base class: predict what is printed when the derived object chooses the 1-arg base constructor.  #include <iostream.h>  class Base { public:  char S, A, M;  Base(char x, char y)  {  S = y - y;  …

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Introduction / Context: This example probes constructor selection with default arguments and the order of initialization in inheritance. The derived class calls Base(x) with one argument, forcing selection of the Base overload that has defaults for the remaining parameters, and then chooses which Display to invoke at runtime.Given Data / Assumptions:Derived’s initializer list uses Base(x) with a single argument.Base has an overload Base(char, char y = 'A', char z = 'B') that can be called with one argument.Display(0 - 1) passes a non-zero integer, so Base::Display() runs.Concept / Approach: With Base(x) and defaults, y becomes 'A' and z becomes 'B'. The chosen Base constructor sets S = y, A = y + 1 - 1 (thus also 'A'), and M = z - 1 ('B' - 1 → 'A'). Therefore all three printed characters are 'A'.Step-by-Step Solution:Call Base(x) → selects Base(char, char y = 'A', char z = 'B').Assign S = 'A'.Compute A = 'A' + 1 - 1 → 'A'.Compute M = 'B' - 1 → 'A'.Base::Display prints: A A A.Verification / Alternative check: If Derived called Base(x, y) or Base(x, y, z), the printed values could differ. The code as written uses the 1-argument form with defaults.Why Other Options Are Wrong:Any option not equal to A A A contradicts the explicit arithmetic on characters in the selected constructor.“Garbage” would suggest uninitialized output, which is not the case for Base members.Compile-time error does not occur under the assumed environment.Common Pitfalls: Worrying about using member x before it is assigned; while x is indeterminate when passed, the selected Base constructor ignores its value for S, A, and M because they depend only on defaulted y and z, leading to deterministic output here.Final Answer: A A A

Question 5

C++ overloading with a competing overload that has a default argument: which function is called and what is printed?  #include <iostream.h>  int CuriousTabTest(int x, int y); int CuriousTabTest(int x, int y, int z = 5);  int main() {  cout << Curiou…

Accepted Answer

Introduction / Context: This item examines overload resolution when one overload exactly matches the argument list and another is also viable by supplying a default argument. The C++ rules select the best viable function, favoring an exact arity match over one that relies on defaults.Given Data / Assumptions:Two overloads exist: f(int,int) and f(int,int,int=5).Call site: CuriousTabTest(2, 4).Both overloads are viable, but one requires using a default argument for the third parameter.Concept / Approach: Overload resolution prefers the function that requires no default arguments and exactly matches the number and types of the provided parameters. Therefore CuriousTabTest(int,int) is selected, computing 2 * 4 = 8.Step-by-Step Solution:Evaluate candidates: (int,int) and (int,int,int=5).Both match types; the three-parameter overload needs its default for z.Best viable function is the exact two-parameter overload.Return 2 * 4 → 8; cout prints 8.Verification / Alternative check: If the call were CuriousTabTest(2,4,5), the three-parameter version would be chosen and print 40. Removing the two-parameter overload would also yield 40 due to the default.Why Other Options Are Wrong:5: Not computed by either overload.40: Would require calling the three-parameter function explicitly (or removing the two-parameter overload).Compile time error / None of the above: The overload set is resolvable.Common Pitfalls: Thinking the presence of a defaulted parameter forces selection of that overload; in reality, defaults only make a function viable, not necessarily the best match.Final Answer: The program will print the output 8.

Question 6

C++ operator overloading with a default argument on operator+ (member): compute the result of objA + 5 and print it.  #include <iostream.h>  class Addition {  int x; public:  Addition() { x = 0; }  Addition(int xx) { x = xx; }   Addition operator+(i…

Accepted Answer

Introduction / Context: This question validates understanding of member operator overloading and default arguments. A member operator+(int) computes a new Addition whose x is the receiver’s x plus the provided integer. Default arguments on operators are allowed in C++ and are not the cause of any compilation issue here.Given Data / Assumptions:objA is constructed with x = 15.Expression objA + 5 invokes Addition::operator+(int).The operator returns a new Addition with x = 15 + 5 = 20, then Display prints that value.Concept / Approach: Member operator+(int) is analogous to a method call objA.operator+(5). It builds a temporary result object with the sum and returns it by value. No undefined behavior occurs; the temporary is copied into objB.Step-by-Step Solution:Construct objA with x = 15; objB with default x = 0.Invoke objA + 5 → creates objTemp with x = 20.Assign objTemp to objB (copy elision may apply) → objB.x = 20.objB.Display() prints 20.Verification / Alternative check: If you omit the argument (objA + ), the default argument 0 would be used and the result would be 15. The code as written provides 5 explicitly.Why Other Options Are Wrong:Run time error / garbage: nothing invalid or uninitialized occurs.Compilation fails due to operator + cannot have default arguments.: False; operators may have default parameters like any function, subject to signature rules.Common Pitfalls: Confusing member versus non-member operator overload constraints, or assuming default arguments are disallowed for overloaded operators.Final Answer: The program will print the output 20.

Question 7

C++ overload resolution on (float, char, char): compute and print K from character arithmetic.  #include <iostream.h>  class CuriousTab {  int K; public:  void CuriousTabFunction(float, int, char);  void CuriousTabFunction(float, char, char); };  in…

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Introduction / Context: The call selects the overload CuriousTabFunction(float, char, char) because both the second and third arguments are chars, which is a better match than promoting the second argument to int. The body then performs character arithmetic, relying on integer promotion rules, which can expose the signedness of char on the target platform.Given Data / Assumptions:Invocation: (15.09f, 'A', char('A' + 'A')).'A' has code 65.char('A' + 'A') computes a char from 65 + 65 = 130. On implementations where plain char is signed (common on x86), char(130) becomes -126.K is assigned y + z after integer promotions.Concept / Approach: Overload selection chooses (float, char, char). Then y = 65 and z is the char resulting from 130, which on signed-char targets is -126. Adding with integer promotion gives 65 + (-126) = -61. The program prints this integer value.Step-by-Step Solution:Overload resolution: (float, char, char) is a better match than (float, int, char).Map y = 65.Compute z = char(130) → typically -126 when char is signed.Compute K = y + z = 65 + (-126) = -61.cout prints "K = -61".Verification / Alternative check: On systems where plain char is unsigned, char(130) stays 130 and K becomes 65 + 130 = 195. The provided options reflect both possibilities; the conventional assumption for such questions is signed char, yielding -61.Why Other Options Are Wrong:K = 130: Would be printed only if the function printed the intermediate assignment K = int(z) instead of the final sum.K = 195: Occurs on platforms with unsigned char; not the typical assumption here.K = -21: Does not match any consistent interpretation of char sums here.Compile time error: Both overloads are valid; one is selected unambiguously.Common Pitfalls: Forgetting that plain char may be signed or unsigned, expecting cout with char to print a glyph rather than an integer (here it prints an int), or assuming the (float, int, char) overload is chosen.Final Answer: The program will print the output K = -61.

Question 8

C++ (member access across objects) — predict the output printed by show() when it is invoked on an object whose fields were set just before the call.  #include<iostream.h> class Tab {  int x, y; public:  void show(void);  void main(void); }; void Ta…

Accepted Answer

Introduction / Context:This program tests understanding of member access across different instances of the same class in C++. The function show() constructs its own local object b but then prints the data members x and y of the current object (the one on which show() was called), not of the temporary b created inside show().Given Data / Assumptions:Class Tab has two private ints: x and y.Tab::main creates a local object b, assigns b.x = 6 and b.y = 8, then calls b.show().Tab::show creates its own local Tab b, sets b.x = 2 and b.y = 4, but prints x and y of the current object.Assume an older environment where is accepted (as used by many legacy MCQs).Concept / Approach:Within a non-static member function, an unqualified member name (x, y) refers to this->x and this->y. Creating another instance inside the member function does not change which object this points to. As a member, show() can assign to private members of any Tab instance it has a reference to, but the cout line is explicitly referencing the current object’s fields.Step-by-Step Solution:Construct local b in Tab::main and set b.x = 6, b.y = 8.Call b.show(). Inside show(), a new local Tab b is created and set to (2, 4).The cout statement prints this->x and this->y from the caller object, which are 6 and 8.Therefore the output is the two numbers: 6 8.Verification / Alternative check:Rename the inner object to b2 and print b2.x, b2.y to see 2 4; but the given code does not do that, it prints the caller’s fields.Why Other Options Are Wrong:2 4: Would be printed only if the code printed the inner object’s members.Compilation/Link/Run-time errors: The code is valid in a legacy header setting and runs normally.Common Pitfalls:Assuming that the most recently declared variable b is what gets printed; forgetting that x and y refer to this->x and this->y.Final Answer:6 8

Question 9

C++ (constructor overload selection and copy construction) — determine the output printed after constructing Derived via two different constructors.  #include<iostream.h> class Base {  int x, y; public:  Base() { x = y = 0; }  Base(int xx) { x = xx;…

Accepted Answer

Introduction / Context:This exercise probes C++ constructor resolution and object slicing versus copy construction. The class Derived has two user-declared constructors: one forwarding arguments to the Base subobject, and a second that takes a Base by value with a default argument. When we pass a Derived to new Derived(objD), which constructor is invoked and what state does the Base subobject hold when Display() is called?Given Data / Assumptions:Derived(int,int) builds a Base subobject via Base(xx, xx + 1).Derived(Base) is a converting constructor taking Base by value and doing nothing in its body.An implicit copy constructor Derived(const Derived&) is still generated by the compiler.Display() prints the Base subobject of the object it is called on.Concept / Approach:Overload resolution chooses the best match. Passing a Derived to construct another Derived makes the implicit copy constructor an exact match, which is better than converting to Base and calling Derived(Base). Therefore, new Derived(objD) uses the copy constructor, not the Derived(Base) overload.Step-by-Step Solution:Construct objD(5,3): Base subobject uses Base(5,6) so x = 5 + 6 = 11, y = 6.Copy-construct *ptrD from objD: the Base subobject is copied, so it also has x = 11, y = 6.Call ptrD->Display(): prints 11 6.Verification / Alternative check:If you forcibly call the converting constructor (e.g., new Derived(static_cast(objD))) then the Base subobject in the new object would default-construct to 0 0 because the body does nothing and no base initializer is provided.Why Other Options Are Wrong:3 2, 8 3, 11 10: Do not match the actual Base state produced by the constructors shown.“Will not compile”: All declarations are valid in classic header context.Common Pitfalls:Assuming the presence of Derived(Base) suppresses the implicit copy constructor (it does not), or assuming the converting constructor will be chosen over the exact-match copy constructor.Final Answer:11 6

Question 10

C++ (overload resolution with defaults) — which overload is selected and what value of x is printed?  #include<iostream.h> class CuriousTab {  int x; float y; public:  void CuriousTabFunction(int = 0, float = 0.00f, char = 'A');  void CuriousTabFunc…

Accepted Answer

Introduction / Context:This question examines overload resolution when two viable overloads accept parameters that can be formed from the same call via standard conversions, and where defaults supply missing arguments. It highlights that exact matches and promotions out-rank other standard conversions.Given Data / Assumptions:Call site: CuriousTabFunction(10 * 1.0, int(56.0)) i.e., arguments are double and int.Two likely candidates: (int, float, char = 'A') and (float, int, char = 'Z').Third overload (char, char, char) is not viable for the given types without narrowing conversions to char.Concept / Approach:For the first parameter, both candidates require converting a double (to int or to float). For the second parameter, the (float, int, char) overload has an exact match for the provided int, while the (int, float, char) overload requires converting int to float. Therefore, the chosen overload is CuriousTabFunction(float, int, char) with zz defaulting to 'Z' (ASCII 90).Step-by-Step Solution:Select overload: (float, int, char).Compute x = zz + zz = 90 + 90 = 180.The function prints " x = 180".Verification / Alternative check:Force the other overload with an explicit cast: obj.CuriousTabFunction(int(10.0), float(56.0)) and the output will reflect x = 10 + 56.Why Other Options Are Wrong:65, 66, 130 correspond to other letter sums or different overloads; not chosen here.Compilation failure does not occur; all overloads are valid.Common Pitfalls:Overlooking that the second parameter is an exact match for int in the float-int-char overload, which tips the ranking.Final Answer:The program will print the output x = 180.

Question 11

C++ (constructor selection, shadowing, and references) — determine the output after calling SetValue and Display.  #include<iostream.h> class CuriousTab {  int x; float y; public:  CuriousTab(int x) { x = x; }  CuriousTab(int p = 0, int q = 10) { x …

Accepted Answer

Introduction / Context:This question explores constructor overload choice when defaults are present, name shadowing inside constructors, the most-vexing parse, and effects of passing by reference. It asks what value is ultimately printed by Display().Given Data / Assumptions:Two constructors: CuriousTab(int) and CuriousTab(int,int) (the latter has defaults).Call CuriousTab objCuriousTab(val); with val = 12.CuriousTab objTmp(); is a function declaration (most-vexing parse), not an object.SetValue(int& y, float z) assigns to the member and mutates the caller’s variable.Concept / Approach:Overload resolution prefers the exact single-argument match (CuriousTab(int)) over using a default parameter in the two-argument constructor. Inside that constructor, the parameter name shadows the member, so x = x; assigns the parameter to itself and leaves the member x uninitialized (though it will be assigned later). SetValue(val, 3.14f) sets the member x to 12 and then casts 3.14f to 3 and stores it back into val via the reference.Step-by-Step Solution:Select one-arg ctor: member x remains uninitialized but will be set next.Call SetValue: assign x = y → x = 12; then y = (int)z → caller’s val becomes 3.Display prints the current member x, which is 12.Verification / Alternative check:Had the two-arg ctor been chosen, x would become p += 2. But default-argument overload is not preferred over the exact one-parameter constructor here.Why Other Options Are Wrong:2: Misreads the second constructor as selected.Errors: The program compiles and runs (with legacy headers).Common Pitfalls:Name shadowing in constructors (x = x;) and misinterpreting CuriousTab objTmp(); as an object instead of a function declaration.Final Answer:The program will print the output 12.

Question 12

C++ strings (C-style) — predict the concatenated message built by the constructor with default parameters and print via Display(). #include #include class CuriousTabString { char x[50]; char y[50]; char z[50]; public: Curiou…

Accepted Answer

Introduction / Context:This item checks understanding of C-style string operations (strcpy, strcat) and how default parameters work when some but not all constructor arguments are provided. The Display() method prints the buffer z built by the three-argument constructor.Given Data / Assumptions:Constructor signature: (char* xx, char* yy = " C++", char* zz = " Programming!").Call: CuriousTabString objStr("Learn", " Java"); so xx = "Learn", yy = " Java", zz takes the default " Programming!".Buffers are large enough for the concatenation in this example.Concept / Approach:The constructor copies xx into z, then concatenates yy, then concatenates zz. Because yy and zz include leading spaces, they create word spacing naturally in the result.Step-by-Step Solution:Start with z = "Learn".Concatenate yy: z = "Learn Java".Concatenate default zz: z = "Learn Java Programming!".Display() prints the final string.Verification / Alternative check:Changing the call to ("Learn") would use the one-parameter constructor and not initialize z; Display() would then print an uninitialized buffer (undefined). The two-parameter constructor is the correct usage here.Why Other Options Are Wrong:Other choices either omit the Learn prefix or use the wrong defaults ordering.Common Pitfalls:Forgetting that default arguments keep their specified leading spaces; omitting them causes words to touch.Final Answer:Learn Java Programming!

Question 13

C++ (recursion with default arguments in a member function) — does the call terminate or loop forever, and what would be printed?  #include<iostream.h> class CuriousTab { public:  void CuriousTab(int x = 15) {  x = x / 2;  if (x > 0)  CuriousTab(); …

Accepted Answer

Introduction / Context:This snippet highlights a common trap: recursion with a default parameter that is re-applied at each call. Although the body halves a variable and seems to move toward a base case, the recursive call does not pass that new value along, so the base case is never reached.Given Data / Assumptions:Member function signature: void CuriousTab(int x = 15).Body halves x and recurses with CuriousTab() (no argument).Default argument value 15 is used whenever the call omits the parameter.Concept / Approach:Each recursive call with no explicit argument uses the default, setting x to 15 again, then halving to 7, then recursing again with the default, and so on. The halved value is never propagated into the next call, so the condition x > 0 remains true on every activation until the call stack overflows or the runtime aborts, i.e., it behaves like an infinite loop (non-terminating recursion).Step-by-Step Solution:Initial call: default x = 15 → x = 7.Recursive call with no argument applies default again: x = 15 → 7.This process repeats indefinitely; the else branch is never reached.Verification / Alternative check:Fix by passing the updated value: CuriousTab(x). Then the sequence 15 → 7 → 3 → 1 → 0 will terminate, printing 0 % 2 = 0.Why Other Options Are Wrong:Printing 1, 2, or 15 presumes termination; it does not terminate.Compilation error does not occur; the code is syntactically valid.Common Pitfalls:Confusing default arguments with persistent state, and forgetting that defaults are reapplied at each call when the argument is omitted.Final Answer:The program will go into an infinite loop.

Question 14

C++ (name qualification in multiple-base-like contexts) — evaluate the printed sum after calling the specifically qualified BaseTwo::Display.  #include<iostream.h> static double gDouble; static float gFloat; static double gChar; static double gSum =…

Accepted Answer

Introduction / Context:This question reinforces qualified calls to resolve ambiguity in multiply-inherited interfaces and shows how arguments flow into a function with defaults overridden at the call-site. The output is a straightforward arithmetic sum of three values stored in globals.Given Data / Assumptions:We explicitly call objDev.BaseTwo::Display(10, 20, 'Z').Inside BaseTwo::Display, the globals are set and gSum is computed as the sum of the three converted values.'Z' has integer code 90.Concept / Approach:The computation uses gDouble = 10, gFloat = 20.0f, gChar = 90, so the sum is 10 + 20 + 90. The function itself streams that sum via cout, so the visible output is the numeric result.Step-by-Step Solution:Assign inputs: 10, 20, 'Z'.Convert and store: 10.0, 20.0f, 90.0.Compute: gSum = 10 + 20 + 90 = 120.Print gSum.Verification / Alternative check:Calling BaseOne::Display(10.0, 20.0f, 'Z') would compute the same numeric sum since the same values are used after conversions.Why Other Options Are Wrong:0 / garbage: contradicts the explicit arithmetic.Errors: The qualified call resolves any ambiguity; the code compiles and runs.Common Pitfalls:Forgetting to qualify the call in multiple inheritance, which would make the call ambiguous and fail to compile.Final Answer:The program will print the output 120.

Question 15

C++ (recursion with static indices and post-recursion swapping) — determine the final array order printed by Display().  #include<iostream.h> struct CuriousTabArray {  int arr[5]; public:  void CuriousTabFunction();  void Display(); }; void CuriousT…

Accepted Answer

Introduction / Context:This problem checks your ability to trace a recursive function that uses static indices and performs its swap operations after unwinding recursion. The order of operations (pre-recursion index update, then post-recursion swap) is crucial to the final arrangement of the array elements.Given Data / Assumptions:Initial array: [5, 6, 3, 9, 0].Static indices start at i = 0, j = 4.Each call increments i, decrements j, recurses while j > 0, then swaps arr[i] and arr[j] while unwinding.Concept / Approach:Because the swap happens after recursion returns, the deepest frame swaps the ends first, then higher frames swap inner pairs, potentially swapping some positions back depending on indices restored by the static variables. Carefully tracking the evolving values of i and j is the key.Step-by-Step Solution:Deepest frame (after updates): swap indices 4 and 0 → [0, 6, 3, 9, 5]; i=3, j=1.Next: swap indices 3 and 1 → [0, 9, 3, 6, 5]; i=2, j=2.Next: swap indices 2 and 2 (no change) → remains [0, 9, 3, 6, 5]; i=1, j=3.Next: swap indices 1 and 3 → [0, 6, 3, 9, 5]; i=0, j=4.Verification / Alternative check:Compare with a version that performs the swap before recursion; it yields a different ordering (see next question for a variant).Why Other Options Are Wrong:Other sequences do not match the exact swap order based on post-recursion operations with static indices.Common Pitfalls:Forgetting that i and j are static and persist across frames; assuming each function activation has its own i, j.Final Answer:0 6 3 9 5

Question 16

C++ (recursion with static indices and pre-recursion swapping) — determine the array order produced and printed.  #include<iostream.h> struct CuriousTab {  int arr[5]; public:  void CuriousTabFunction(void);  void Display(void); }; void CuriousTab::…

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Introduction / Context:This follow-up uses the same static-index idea but performs the swap before recursion. The base condition is j != i, so recursion stops when the two indices meet at the center. You must trace the swaps carefully with persistent i, j to get the final array ordering.Given Data / Assumptions:Initial array: [5, 6, 3, 9, 0].Static i = 0, j = 4.Each call swaps arr[i] with arr[j], then increments i and decrements j.Concept / Approach:Because the swap occurs immediately, we reverse pairs from the outermost inward, then stop when i == j. With static variables, each recursive frame sees updated indices rather than independent copies.Step-by-Step Solution:Call 1: swap 0↔4 → [0, 6, 3, 9, 5]; then i=1, j=3; recurse (1 != 3).Call 2: swap 1↔3 → [0, 9, 3, 6, 5]; then i=2, j=2; stop (2 == 2).Display prints: 0 9 3 6 5.Verification / Alternative check:Compare with the variant that swaps after recursion; you will notice the middle positions differ because of the different order of operations.Why Other Options Are Wrong:Other sequences do not match the exact swap-before-recursion trace.Common Pitfalls:Thinking that recursion will reverse the entire array including the center twice; the stopping condition prevents that.Final Answer:0 9 3 6 5

Question 17

C++ (method hiding vs base call) — what happens when a derived Display() calls Display() without qualification?  #include<iostream.h> class Base {  int x, y, z; public:  Base() { x = y = z = 0; }  Base(int xx, int yy = 'A', int zz = 'B') { x = xx; y…

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Introduction / Context:This question targets name lookup and method hiding in C++. The derived class defines its own Display() that writes two values, then calls Display() again without qualification. Does it call the base or itself?Given Data / Assumptions:Derived::Display() prints x and y, then calls Display() unqualified.There is a Base::Display(), but the derived function hides it for unqualified lookup.Constructors initialize Base subobject and derived members to known values (x = 66, y = 65) before the first call.Concept / Approach:Unqualified lookup finds the nearest match in the current scope first. Since Derived declares a function named Display, an unqualified call inside Derived::Display resolves to itself, causing infinite recursion. The compiler will not implicitly resolve to Base::Display()—you must qualify it (Base::Display()) to call the base version.Step-by-Step Solution:First invocation prints 66 65 (derived members).Unqualified call then re-enters Derived::Display().The cycle repeats indefinitely, printing 66 65 until stack overflow or program termination.Verification / Alternative check:Replace the recursive call with Base::Display() to get a single line beginning with 66 65 followed by the base’s computed triple values.Why Other Options Are Wrong:Compilation error does not occur; recursion is legal.Other outputs assume a base call or single execution; here the call is recursive.Common Pitfalls:Forgetting to qualify base calls when a derived function with the same name exists; assuming overload resolution will “find” the base automatically.Final Answer:The program will produce the output 66 65 infinite number of times (or till stack memory overflow).

Question 18

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<iostream.h> class CuriousTabSample {  private:  int AdditionOne(in…

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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:AdditionOne is declared in the private section of the class.AdditionTwo is declared in the public section.main calls objCuriousTab.AdditionOne(4, 8) and then objCuriousTab.AdditionTwo(8, 8).Legacy header is used but does not change access rules.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.

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