Question 1

In C++ source code, which characters begin a single-line comment so that the remainder of the line is ignored by the compiler?

Accepted Answer

Introduction / Context: Comments are essential for documentation and maintainability. C++ supports two primary comment styles: single-line comments and block comments. Knowing the correct introductory tokens prevents syntax errors and ensures that annotations do not accidentally become part of the program logic. Given Data / Assumptions: The code is compiled as C++ (not strictly C89). We focus on the most common, single-line comment style used in modern C++. No preprocessor tricks or unusual compiler extensions are considered. Concept / Approach: In C++, the token // begins a single-line comment. Everything from // to the end of that physical line is ignored by the compiler. C++ also supports block comments beginning with /* and ending with / for multi-line remarks. Symbols such as , &&, backslash, or at-sign have other syntactic meanings and do not introduce comments. Step-by-Step Solution: Identify the desired behavior: comment out the remainder of a single line.Recall that the correct token in C++ for single-line comments is //.Place // before the text to be ignored.Confirm that the compiler discards the trailing text and proceeds with parsing the next line. Verification / Alternative check: Create a minimal program that contains code, a // comment, and a statement on the next line. Observe that only the code outside comments is compiled. Replacing // with other tokens will either cause compilation errors or change program meaning (for example, && is a logical operator). Why Other Options Are Wrong: and && are operators or text without comment semantics. \ is an escape for a single backslash in string literals; it does not start comments. @ has no standard meaning for comments in C++. Common Pitfalls: Accidentally nesting block comments with / and */ can cause unexpected uncommenting. Using // avoids nesting issues for single lines. Final Answer: //.

Question 2

In C++ function declarations and definitions, what is the correct return type keyword used when a function is intended to return no value at all to its caller?

Accepted Answer

Introduction / Context: In C++ programming, every function has a declared return type that informs both the compiler and the caller what kind of result (if any) will be produced when the function completes. Many functions perform actions such as logging, printing, mutating objects via references, or orchestrating control flow without producing a value for the caller. For these cases, the language provides a dedicated keyword to indicate the complete absence of a return value. Given Data / Assumptions: The question asks for the correct C++ return type used when a function should not return a value. Standard C++ syntax and semantics are assumed (no compiler-specific extensions). We are not discussing templates, coroutines, or special member functions beyond how they relate to the idea of “no return value.” Concept / Approach: In C++, the keyword used to denote that a function returns no value is void. A function declared as void f(parameters) signals that control returns to the caller without any accompanying data. Such a function may use an empty return; to exit early, but it cannot return an expression. This is distinct from functions that return a type like int or std::string, where a value must be produced on all control paths. The term “void” is part of the core language and has precise, standardized meaning; alternatives such as “empty” or “barren” are not C++ keywords. Step-by-Step Solution: 1) Identify the requirement: the function should not yield any result value to the caller.2) Recall the C++ keyword that represents the absence of a return type: void.3) Form a declaration using that keyword, for example: void logEvent(const std::string& msg);4) Ensure the definition does not attempt to return an expression; using just return; is permitted for early exit. Verification / Alternative check: If a function is declared with a non-void return type but does not return a value on all paths, the compiler will typically issue an error or warning. This confirms that “no value” functions must be explicitly marked with void. Conversely, attempting to write return 42; inside a void function will produce a compile-time error, verifying that void truly indicates no return value. Why Other Options Are Wrong: not allowed in C++: Incorrect. Functions that return no value are fully supported via the void return type. type empty: Not a C++ keyword or type; has no meaning to the compiler. type barren: Also not a C++ keyword or recognized type. Common Pitfalls: Confusing a void function with one returning an “empty” object (e.g., an empty std::string). A void function returns nothing at all, not an “empty” value. Forgetting that constructors and destructors have no return type specified—this is a special rule; they are not written with void and still return no value. Using return expression; inside a void function, which is ill-formed. Final Answer: type void.

Question 3

Evaluate the Boolean expression in C++ operator precedence: 3 > 6 && 7 > 4 — is the overall result True or False?

Accepted Answer

Introduction / Context: Understanding relational and logical operators in C++ is fundamental to writing correct conditional statements. This item verifies knowledge of how comparison operators interact with the logical AND operator, including precedence and short-circuit evaluation, to determine a final Boolean value. Given Data / Assumptions: The expression is exactly: 3 > 6 && 7 > 4. Standard C++ operator precedence applies: relational comparisons are evaluated before logical AND. Values are integers, and there is no user-defined operator overload affecting semantics. Concept / Approach: Relational operators (such as >) produce Boolean results. The logical AND operator && yields true only if both operands are true. Therefore evaluate each comparison first, then compute the logical AND of the two truth values. Short-circuiting means that if the left operand of && is false, the right operand need not be evaluated to know the overall result; however, here we can still reason about both sides for clarity. Step-by-Step Solution: Evaluate 3 > 6, which is false because 3 is not greater than 6.Evaluate 7 > 4, which is true because 7 is greater than 4.Combine with logical AND: false && true is false because both operands must be true for AND to be true.Therefore the entire expression evaluates to false. Verification / Alternative check: A quick program printing the result of (3 > 6 && 7 > 4) will show 0 (false) using standard C++ stream formatting. Changing the first comparison to 8 > 6 would yield true && true and produce true, which demonstrates the rule. Why Other Options Are Wrong: True: Incorrect because one operand of the AND is false. Not applicable / Cannot be determined / None of the above: The expression is fully determined and evaluates to false. Common Pitfalls: Misreading operator precedence or forgetting short-circuiting. Always evaluate comparisons first, then apply logical operators according to precedence and associativity. Final Answer: False.

Question 4

In typical C/C++ projects, source header files frequently use which filename extension to indicate a header (choose the most common traditional extension)?

Accepted Answer

Introduction / Context: Header files declare interfaces, types, and function prototypes that can be shared across multiple translation units. While modern C and C++ standard headers are often referenced using angle brackets without a file extension (for example, ), many codebases and platforms still use file extensions to distinguish headers from implementation files. Given Data / Assumptions: The question targets conventional, legacy, or platform-neutral naming practices. We are not restricting to the C++ standard library naming scheme; we consider user headers as well. Case sensitivity may vary by operating system; .H and .h are conceptually the same family. Concept / Approach: The most traditional and widely recognized extension for header files is .h (lowercase), and on some systems .H (uppercase) appears equivalently. None of the other listed extensions are standard practice for C or C++ headers. Therefore, among the provided choices, .H is the correct selection standing for the common header extension family. Step-by-Step Solution: Identify the typical header extension used by C and C++ projects: .h.Map to the provided options, noting that .H represents the same convention in uppercase.Exclude nonstandard endings such as .HE, .HEA, and .HEAD.Conclude that .H is the best match among the available options. Verification / Alternative check: Examine open source projects across platforms; user headers consistently appear as .h. On case-sensitive filesystems, .H may be a stylistic or historical alternative, but remains recognizable as a header extension compared with the nonstandard alternatives listed. Why Other Options Are Wrong: .HE, .HEA, .HEAD are not recognized extensions in conventional C or C++ build systems. None of the above is incorrect because .H matches the common header extension convention. Common Pitfalls: Confusing modern standard headers like (no .h) with user headers. Even in modern C++, user headers often still use .h or platform variants such as .hpp for C++ specific headers. Final Answer: .H.

Question 5

In object-oriented C++, when a class requires special initialization steps for its objects, which special member function should you design to perform that initialization automatically at construction time?

Accepted Answer

Introduction / Context: Resource management and invariants are central to robust C++ class design. The language provides special member functions that are invoked automatically at key object lifecycle events. Correctly choosing and implementing the function that runs when an object is created is essential for ensuring valid state and exception safety from the very beginning of an object's lifetime. Given Data / Assumptions: Objects of the class need initialization beyond simple aggregate defaulting. Resources may include memory, file handles, mutexes, or network sockets. We are using standard C++ object model semantics without exotic extensions. Concept / Approach: The constructor is the special member function that initializes an object when it is created. It has the same name as the class (pre C++11 rule; in modern C++ it is spelled as ClassName(...) and has no return type). Constructors can be overloaded and can use member initializer lists to set base classes and data members efficiently. They are the cornerstone of RAII (Resource Acquisition Is Initialization), ensuring resources are tied to object lifetime and released in the destructor. Step-by-Step Solution: Identify the need for automatic initialization at object creation.Declare one or more constructors for the class, using parameters as needed.Use member initializer lists to initialize bases and data members efficiently and correctly.Verify that constructed objects satisfy all class invariants before any methods are invoked. Verification / Alternative check: Create a small class that opens a file in its constructor and closes it in the destructor. Instantiating the class automatically performs the initialization, demonstrating the constructor's role. Why Other Options Are Wrong: housekeeping routine and initializer are vague terms without special semantics in C++. compiler is the toolchain component, not a class member function. None of the above is incorrect because constructor is the precise C++ term. Common Pitfalls: Doing heavy work in the constructor without exception safety or leaving members uninitialized. Prefer member initializer lists and consistent invariants to avoid undefined behavior. Final Answer: constructor.

Question 6

In C++ assignment versus comparison operators, which of the following statements correctly assigns the numeric value 5 to the variable named area?

Accepted Answer

Introduction / Context: A common beginner error in C and C++ is confusing the assignment operator with the equality comparison operator. Clear understanding of operator spelling and semantics prevents logic bugs and compilation failures. This question focuses on the exact syntax required to store a literal value into a variable. Given Data / Assumptions: The variable area has been declared with a numeric type capable of storing the integer 5. We are performing a simple assignment, not a comparison or a pointer operation. The code is standard C++ with no operator overloading affecting these basics. Concept / Approach: The assignment operator in C++ is a single equals sign: =. It places the value on the right-hand side into the variable on the left-hand side. The equality comparison operator is a double equals: ==, which produces a Boolean result and does not assign. Other sequences shown in the options are not valid C++ operators for assignment and will not compile in this context. Step-by-Step Solution: Choose the assignment operator: =.Write the target variable first, then the operator, then the literal value.Form the statement: area = 5;Terminate with a semicolon in actual code to complete the statement. Verification / Alternative check: Compile a snippet with area = 5; and then print area to verify that the value has been stored. Replace = with == and observe the compiler either accept a meaningless comparison expression or produce warnings, depending on context, but it will not assign to area. Why Other Options Are Wrong: area 1 = 5 is invalid syntax and suggests a space-separated token. area == 5 performs a comparison, not an assignment. area --> 5 and area < > 5 are not C++ assignment operators. Common Pitfalls: Using == inside an if condition when you meant to assign, or using = in conditions when you meant to compare. Many compilers can warn about suspicious use of = inside conditionals; enable warnings. Final Answer: area = 5.

Question 7

In C and C++, what is the correct keyword used to define a user-defined structure type that groups related data members under a single name?

Accepted Answer

Introduction / Context: Structures provide a way to create composite types by grouping variables under one logical unit. This mechanism is central to modeling records, points, and other multi-field entities. Using the correct keyword ensures the compiler recognizes the declaration and generates the proper layout in memory. Given Data / Assumptions: The language is standard C or C++. We intend to define a new type with several named fields. No typedefs, classes, or unions are implied beyond the basic structure keyword. Concept / Approach: The reserved word struct initiates a structure definition. For example: struct Person { std::string name; int age; }; creates a type named Person with two members. In C++, struct and class are similar in capability, but the default access for struct members is public, whereas for class it is private. Tokens like stru, stt, or the word structure are not C or C++ keywords and will cause compile-time errors if used in place of struct. Step-by-Step Solution: Choose the keyword struct to begin the type definition.Provide a tag name and a brace-enclosed list of members.Optionally, create variables or use typedef/using to introduce aliases.Use the new type to declare variables that group related fields together. Verification / Alternative check: Compile a simple program with struct Point { int x; int y; }; Point p; p.x = 3; p.y = 4; and observe successful compilation and execution. Replacing struct with a nonkeyword token will fail to compile. Why Other Options Are Wrong: stru and stt are not language keywords. structure is English prose, not a reserved identifier in C or C++. None of the above is incorrect because struct is correct. Common Pitfalls: Confusing struct and class default access. In C++, either may be used to define rich types; choose based on intent and coding style. Final Answer: struct.

Question 8

C++ function overloading rules: given a function with header int function(double d, char c), which of the following additional function headers may legally coexist in the same program?

Accepted Answer

Introduction / Context: Function overloading in C++ allows multiple functions to share the same name as long as their parameter lists differ in number or types. The return type alone is not part of the function signature for overloading resolution. This question verifies your understanding of which changes create a distinct overload and which do not. Given Data / Assumptions: Existing declaration: int function(double d, char c). We are considering additional declarations with the same name but different signatures. Standard C++ overload resolution rules apply; namespaces and default parameters are not in play here. Concept / Approach: An overload must differ in its parameter types or count. Changing only the return type does not create a valid overload and leads to a redefinition error because the function name and parameter list still collide. Therefore a second function that keeps parameters (double, char) but changes the return type to char is illegal. However, changing a parameter type, such as using int in place of double for the first parameter, produces a distinct signature and is permitted. Step-by-Step Solution: Compare option (a): char function(double d, char c) — same parameters as the original, only the return type differs.Recognize that return type alone does not differentiate overloads; this is not allowed.Compare option (b): int function(int d, char c) — parameter list differs in the first type (int vs double), creating a distinct signature.Conclude that only option (b) is a valid additional function in the same program. Verification / Alternative check: Attempting to compile both int function(double, char) and char function(double, char) in the same scope will yield a redeclaration error. Replacing the first parameter type with int compiles successfully, demonstrating legal overloading. Why Other Options Are Wrong: Option (a) fails because only the return type is changed. Option (c) claims both are valid, which is incorrect. Option (d) claims neither, which ignores the valid distinction in (b). Common Pitfalls: Assuming that different return types are sufficient for overloads. Overload resolution relies on the argument types at the call site, not on the desired return type. Final Answer: int function(int d, char c).

Object Oriented Programming Using C++ Questions