Question 1

C varargs diagnostics: point out the specific error(s) in this program using . Rewrite focuses on why ellipsis functions need a named parameter and a correct va_start anchor. #include #include fun(...); int main() { …

Accepted Answer

Introduction / Context: This question examines correct use of C variable-argument functions (varargs) with . Specifically, it tests whether the function with ellipsis is declared and defined properly and whether va_start is anchored to a valid, named parameter that precedes the ellipsis. Understanding these rules prevents undefined behavior and compilation errors in real-world code such as logging or formatting routines. Given Data / Assumptions: A function is declared as fun(...); with no named fixed parameter. The definition fun(...) also uses only an ellipsis. va_start(ptr, n) uses an identifier n that is not a parameter. Old-style (K&R) return type is omitted. Concept / Approach: In standard C, a variadic function must have at least one fixed, named parameter before .... The macro va_start requires that the second argument be that last named parameter. Using only ... makes it impossible to position the va_list at the first variable argument. Therefore, such a declaration/definition is invalid. Missing an explicit return type is also an error in modern C, but the most fundamental issue is the invalid variadic signature. Step-by-Step Solution: Identify function signature: fun(...) → no named parameter → invalid for varargs.Check va_start(ptr, n) → n is not a parameter → misuse of va_start.Conclusion → declaration/definition of fun(...) is invalid. Verification / Alternative check: Correct pattern: void fun(int count, ...)va_list ap; va_start(ap, count);This compiles and allows safe extraction of arguments with va_arg. Why Other Options Are Wrong: Error: fun() needs return type — true in modern C, but not the primary varargs error being tested. Error: ptr Lvalue required — ptr is a proper lvalue; the issue is not lvalueness. No error — incorrect; there is a clear signature misuse. Common Pitfalls: Forgetting the required named parameter, anchoring va_start to an identifier that is not a parameter, and assuming legacy implicit-int rules still apply. Final Answer: Error: Invalid declaration of fun(...)

Question 2

C varargs with default promotions: identify the error when passing floating values through an ellipsis and retrieving them with va_arg. #include #include int main() { void display(int num, ...); display(4, 12.5, 13.5, 14.5, …

Accepted Answer

Introduction / Context: This problem targets the rule of default argument promotions in C varargs. When arguments are passed through an ellipsis, float values are promoted to double, and integer types narrower than int are promoted to int. Failing to read them with the correctly promoted type leads to undefined behavior and typically wrong outputs. Given Data / Assumptions: Function prototype: void display(int num, ...). Call site passes four floating constants (type double by default). Loop retrieves with va_arg(ptr, float). Concept / Approach: Because of the default promotions, any float passed via ... becomes double. The receiving code must therefore use va_arg(ptr, double). Storing into a float variable is fine after that conversion, but the type specified to va_arg must match the promoted type exactly. Step-by-Step Solution: At call: literals 12.5, 13.5, 14.5, 44.3 → type double.Default promotions in varargs → remain double.Correct retrieval → c = (float)va_arg(ptr, double);Given code uses va_arg(ptr, float) → mismatch → undefined behavior. Verification / Alternative check: Compile with warnings (e.g., -Wformat=2 -Wvarargs) or instrumented sanitizers to observe misreads. Correcting to double yields stable output. Why Other Options Are Wrong: Invalid va_list declaration — declaration is fine. No error options ignore the promotion rule and would print garbage on many platforms. Common Pitfalls: Retrieving float instead of double from varargs, and mixing integer and floating types without a controlling count or format string. Final Answer: Error: var c data type mismatch

Question 3

C varargs correctness: spot the error involving parameter naming, shadowing, and va_start anchor in this display function. #include #include int main() { void display(char *s, int num1, int num2, ...); display("Hello", 4, 2,…

Accepted Answer

Introduction / Context: This question probes two classic varargs mistakes: (1) va_start must be given the last named parameter in the function's parameter list, and (2) accidental shadowing of parameters with local variables. Both lead to broken argument retrieval and undefined behavior. Given Data / Assumptions: Function signature: void display(char *s, int num1, int num2, ...). Inside the function, a local char s; shadows the pointer parameter char *s. va_start(ptr, s) is used incorrectly. Concept / Approach: The second argument of va_start must be exactly the identifier of the last named parameter before the ellipsis. In this prototype, that is num2. Using s (and worse, a local variable that shadows the parameter) violates the contract; the runtime will compute the wrong starting address for the variable arguments. Shadowing also confuses readers and compilers, but the key semantic error here is the bad va_start anchor. Step-by-Step Solution: Parameter list ends with int num2 before ....Correct anchor: va_start(ptr, num2);Remove the conflicting char s; or rename it.Retrieve floating arguments as double due to default promotions. Verification / Alternative check: Fixing to va_start(ptr, num2) and removing the shadowed s produces stable output. Tools like sanitizers can help detect UB caused by wrong va_start. Why Other Options Are Wrong: Invalid arguments and too many parameters are generic and do not pinpoint the varargs rule being broken. No error is clearly incorrect. Common Pitfalls: Anchoring va_start to the wrong parameter, shadowing names, and forgetting default promotions. Final Answer: Error: in va_start(ptr, s);

Question 4

C varargs under Turbo C note: does this program compile and what does it print when characters are passed through an ellipsis? #include #include void display(int num, ...); int main() { display(4, 'A', 'a', 'b', 'c'); return…

Accepted Answer

Introduction / Context: This item references legacy Turbo C behavior with varargs. In standard C, arguments passed through an ellipsis undergo default promotions, meaning char and short are promoted to int. Strictly, one should therefore retrieve such values using va_arg(ptr, int). However, many old compilers (including Turbo C in common configurations) accepted retrieving as char and still produced the expected output in simple cases. Given Data / Assumptions: Compiler context is explicitly “Turbo C”. Arguments are the character constants 'A', 'a', 'b', 'c'. Loop prints exactly four characters, no delimiters. Concept / Approach: With Turbo C, retrieving the arguments as char from va_list often works because of implementation-defined calling conventions and conversions that end up truncating the promoted int back to char. While this is undefined behavior per the C standard, the question asks specifically for Turbo C outcome, which historically prints the characters as expected. Step-by-Step Solution: Call display(4, 'A', 'a', 'b', 'c').Loop iterates j = 1..4 and reads each character-like value.Under Turbo C, va_arg(ptr, char) yields the low byte of the promoted int, matching the intended character.Printed result → Aabc (contiguously). Verification / Alternative check: Portably correct code would be int x = va_arg(ptr, int); then cast to char for printing. That works across compilers and standards-compliant toolchains. Why Other Options Are Wrong: Unknown variable ptr and Lvalue required do not match the code. Printing a leading 4 would require explicitly printing num, which the loop does not do. Common Pitfalls: Assuming Turbo C behavior implies standard-conforming code. For modern, portable C, always retrieve types exactly as they are after default promotions. Final Answer: No error and print A a b c

Question 5

C varargs API usage: identify the error when va_list is used without calling va_start in a variable-argument function. #include #include void varfun(int n, ...); int main() { varfun(3, 7, -11, 0); return 0; } void varfun(int…

Accepted Answer

Introduction / Context: Correct sequencing of varargs macros is crucial. The lifetime of a va_list begins with va_start, continues through one or more va_arg calls, and ends with va_end. Skipping va_start results in undefined behavior when attempting to read arguments. Given Data / Assumptions: Function signature: void varfun(int n, ...). va_list ptr; declared but not initialized. va_arg(ptr, int) is called immediately. Concept / Approach: va_start(ptr, n) initializes ptr so it points just past the last fixed parameter n. Only after this initialization may the code legally use va_arg to read each successive argument. Without initialization, ptr has an indeterminate value, and dereferencing it via va_arg is undefined. Step-by-Step Solution: Add va_start(ptr, n); before the first va_arg.Use num = va_arg(ptr, int); to fetch the first promoted integer.After finishing, call va_end(ptr); to clean up. Verification / Alternative check: Compilers with warnings enabled often flag missing va_start. Runtime sanitizers may catch crashes or misreads that arise from uninitialized va_list. Why Other Options Are Wrong: ptr must be type of va_list — it already is. invalid access to list member is vague; the concrete root cause is the missing initialization. No error is incorrect, as behavior is undefined. Common Pitfalls: Forgetting va_end after finishing, and mixing types without a controlling count or format descriptor. Final Answer: Error: ptr has to be set at begining

Question 6

C varargs across functions: identify the incorrect usage when forwarding varargs and misusing va_start. #include #include void display(char *s, ...); void show(char *t, ...); int main() { display("Hello", 4, 12, 13, 14, 44); …

Accepted Answer

Introduction / Context: Forwarding variable arguments between functions requires care. You cannot “pass” ... directly; instead you must rebuild the call with explicit arguments or use va_list/va_copy. Moreover, va_start must reference the last named parameter in the definition's parameter list. Given Data / Assumptions: display attempts show(s, ...), which is not valid C syntax for forwarding varargs. show calls va_start(ptr, s) even though its last named parameter is t, not s. printf("%f", a) prints an int with a %f specifier, which is also mismatched. Concept / Approach: The error most centrally tested is in show: va_start must be invoked as va_start(ptr, t), because t is the last named parameter. Additionally, forwarding from display requires building a va_list inside display and then either consuming it there or using va_copy to pass to show. Writing show(s, ...) is invalid. Step-by-Step Solution: Fix display to accept va_list forwarding: e.g., have show accept va_list or reconstruct arguments.In show, replace va_start(ptr, s) with va_start(ptr, t).Match printf format with type: use %d for int. Verification / Alternative check: Correct patterns include: void vshow(char *t, va_list ap); and calling it from display with va_start/va_copy. Why Other Options Are Wrong: Invalid function display() call is also problematic but the more concrete, immediate error flagged is within show (wrong va_start anchor and format mismatch). No error and other distractors ignore these standard violations. Common Pitfalls: Trying to forward ... literally, using the wrong identifier in va_start, and mismatching printf formats. Final Answer: Error: invalid function show() call

Question 7

C varargs typing: identify the primary error when the va_list variable is declared with the wrong type and used with va_start. #include #include void varfun(int n, ...); int main() { varfun(3, 7, -11.2, 0.66); return 0; } vo…

Accepted Answer

Introduction / Context: Using variable arguments in C requires a dedicated opaque type va_list. You cannot substitute other pointer types for va_list. The macros va_start, va_arg, and va_end are defined to operate only on objects of type va_list. Given Data / Assumptions: ptr is declared as float *. va_start(ptr, n) is invoked with ptr. Subsequent va_arg is called using the wrong ptr type. Concept / Approach: va_list is an implementation-defined type that stores the state necessary to iterate over unnamed arguments. Treating it as an ordinary pointer (e.g., float *) is invalid and will not compile or will result in undefined behavior. The correct declaration is va_list ptr; followed by va_start(ptr, n);. Step-by-Step Solution: Replace float *ptr; with va_list ptr;Invoke va_start(ptr, n); to initialize list traversal.Use va_arg(ptr, int) to retrieve the first promoted integer.Finish with va_end(ptr); Verification / Alternative check: Compiling with a conforming compiler will error on applying va_start to a non-va_list object. Static analyzers flag this immediately. Why Other Options Are Wrong: Too many parameters is unrelated. Invalid access to list member is vague and not the fundamental issue. No error is incorrect because the type is wrong. Common Pitfalls: Assuming va_list is just a pointer; it may be a struct or array type internally. Always use the standard macros and types. Final Answer: Error: ptr must be type of va_list

Question 8

C varargs theory check: the macro va_start initializes a pointer to the beginning of the list of fixed (named) arguments — True or False?

Accepted Answer

Introduction / Context: This conceptual question clarifies what va_start actually does in the C varargs mechanism. Misunderstandings about whether it references fixed (named) arguments or variable (unnamed) arguments can lead to incorrect code and undefined behavior. Given Data / Assumptions: We have a typical prototype like void f(int count, ...); va_start(ap, count) is used to begin traversal. Concept / Approach: va_start initializes a va_list object to allow access to the unnamed (variable) arguments that appear after the last fixed parameter. Its second argument must be exactly the last named parameter in the function’s parameter list. It does not “point to the fixed arguments”; it uses the location of the last fixed argument as an anchor to locate the first variable argument. Step-by-Step Solution: Define va_list ap;Call va_start(ap, count); → positions ap to retrieve the first variable argument with va_arg.Retrieve each successive unnamed argument → va_arg(ap, type) repeatedly.End with va_end(ap); Verification / Alternative check: Reading the standard or any reputable C reference confirms va_start enables access to the variable argument list, not the fixed ones. Why Other Options Are Wrong: True reverses the direction of the relationship: fixed argument merely identifies the boundary; iteration is over the variable list. Other distractors are irrelevant (C++ has its own facilities, but the question is about C). Common Pitfalls: Mis-anchoring va_start to the wrong parameter and assuming it provides access to named parameters, which it does not. Final Answer: False

Question 9

C varargs usage check: should a variadic function use va_arg() specifically to extract the last argument in the variable list — True or False?

Accepted Answer

Introduction / Context: This statement probes understanding of how va_arg works. The macro does not “jump” to the last argument; it sequentially retrieves the next argument of a specified type from the variable argument list represented by a va_list. Given Data / Assumptions: Variadic function signature such as void logf(int n, ...); Caller may pass an arbitrary number of arguments. Concept / Approach: va_arg(ap, T) returns the next argument of type T and advances the va_list state. It has no knowledge of how many arguments were passed; determining the count is up to the function (via an explicit count parameter, a sentinel value, or a format string). Thus, the claim that va_arg() is for “the last argument” is false. Step-by-Step Solution: Initialize with va_start(ap, last_named_param);Iterate: for (...){ value = va_arg(ap, expected_type); }Stop based on a known count or a sentinel condition.Finish with va_end(ap); Verification / Alternative check: Review any formatted print implementation strategy: it repeatedly calls va_arg for each conversion specifier, never “seeking to the last”. Why Other Options Are Wrong: Architecture width (32-bit) and parity of count are irrelevant. The mechanism is the same across platforms that conform to the C standard. Common Pitfalls: Misconstruing va_arg as random-access; it is strictly sequential. Passing mismatched types against the expected types is undefined behavior. Final Answer: False

Question 10

C varargs fundamentals: is va_list an array type that “holds information needed by va_arg and va_end” — True or False?

Accepted Answer

Introduction / Context: This concept check clarifies what va_list actually is. Many learners assume it is a simple array or pointer. In reality, va_list is an implementation-defined type that may be a pointer, a struct, an array type, or something else depending on the platform and ABI. Given Data / Assumptions: Use of with the macros va_start, va_arg, and va_end. Portability across compilers and architectures is desired. Concept / Approach: The C standard defines va_list abstractly. Code must treat it as opaque and manipulate it only via the standard macros. Assuming it is an “array” is incorrect and leads to non-portable code. On some platforms it may be a single pointer; on others, a small structure holding register saves and stack pointers, or even an array type. The only portable way to use it is through the macros and, when needed, va_copy to duplicate it. Step-by-Step Solution: Declare: va_list ap;Initialize: va_start(ap, last_named_param);Iterate: value = va_arg(ap, type);Finish: va_end(ap); Verification / Alternative check: Inspect headers for different compilers; you will observe varying typedefs for va_list. This confirms it is not universally an array. Why Other Options Are Wrong: “True only on GCC/MSVC” is misleading; each implementation may choose a different representation, and user code should not depend on any particular one. Common Pitfalls: Taking the address of a va_list and assuming pointer arithmetic works portably, or copying it by assignment without va_copy where required. Final Answer: False

Question 11

In C programming (stdarg.h), what does the macro va_arg do when iterating through a variable argument list? Clarify whether it both extracts the current argument value of the specified promoted type and advances the internal va_list position to the …

Accepted Answer

Introduction / Context: Functions that accept a variable number of arguments (variadic functions) in C use facilities from stdarg.h: va_list, va_start, va_arg, and va_end. A common point of confusion is the exact role of va_arg in stepping through the unnamed arguments. This item checks whether you understand that va_arg both fetches the current argument and moves the internal pointer forward to the next one. Given Data / Assumptions: The function is declared with at least one fixed (named) parameter followed by an ellipsis ... . A va_list object has been initialized with va_start using the last fixed parameter. Each variadic argument undergoes default argument promotions (e.g., float → double, types smaller than int → int). Concept / Approach: After va_start initializes traversal, va_arg(list, type) is invoked repeatedly. For each call, C expects you to pass the promoted type that matches how the caller provided the argument. va_arg returns the value of the current argument and then advances the internal position in the va_list to point to the next argument. When you finish, you must call va_end to allow the implementation to clean up. Step-by-Step Solution: Declare va_list ap; then call va_start(ap, last_named_param).Call v1 = va_arg(ap, int); // extracts current promoted int and advances.Call v2 = va_arg(ap, double); // extracts next promoted double and advances again.After all arguments are consumed, call va_end(ap). Verification / Alternative check: Examine a typical printf-like loop that reads arguments until a format string is exhausted. Each va_arg corresponds to one conversion and automatically steps to the next parameter. If va_arg did not advance, the loop would return the same value repeatedly and never progress. Why Other Options Are Wrong: “Incorrect” ignores the specification that va_arg advances. “Only for the first variadic argument” is wrong because va_arg is used for each argument. “Works only with stdio.h” confuses library layers; stdarg.h is independent. “Not applicable in C” is false because stdarg.h is a core C feature. Common Pitfalls: Using the wrong type with va_arg (mismatch with promotions), forgetting va_end, or assuming va_arg peeks without advancing. Final Answer: Correct

Question 12

For a C function that receives a variable number of arguments (declared with an ellipsis ...), should the function use va_arg() to sequentially extract each unnamed argument from the va_list after calling va_start()?

Accepted Answer

Introduction / Context: Processing variadic parameters in C requires an explicit iteration mechanism. This question verifies that you know the standard workflow: va_start to initialize traversal, va_arg to read each value, and va_end to finish. Given Data / Assumptions: The function prototype includes at least one named parameter and then ... . stdarg.h macros are available: va_list, va_start, va_arg, va_end. Default argument promotions apply to the unnamed arguments. Concept / Approach: Calling va_start prepares a va_list to access the unnamed arguments. The only standard way to fetch each value is va_arg(list, type). There is no automatic extraction, and no reflection in C to discover argument types; the callee must know what types to request, often guided by a format string or a count parameter. Step-by-Step Solution: Initialize: va_list ap; va_start(ap, last_named_param).Loop: value = va_arg(ap, expected_promoted_type); process value.Repeat until your stopping condition (e.g., count reached or format parsed).Finish: va_end(ap). Verification / Alternative check: Compare printf and vprintf. The latter accepts a va_list created by another variadic function. Both ultimately use va_arg internally to traverse unnamed parameters. Why Other Options Are Wrong: “Incorrect” denies standard practice. Requiring stdio.h or 64-bit is irrelevant; stdarg.h is portable. “Extraction happens automatically” is false; va_arg is mandatory for reading values. Common Pitfalls: Requesting the wrong type from va_arg, forgetting default promotions (float → double), or skipping va_end. Final Answer: Correct

Question 13

In C variadic functions, must there be at least one fixed (named) parameter before the ellipsis so that va_start() can be given a valid last named argument?

Accepted Answer

Introduction / Context: va_start requires the name of the last fixed parameter to locate the start of the unnamed argument area. Without at least one named parameter, standard C cannot correctly initialize traversal of the variadic arguments. Given Data / Assumptions: We are following the C standard’s stdarg.h facilities. Some nonstandard extensions may exist, but the question targets portable, standard C. Concept / Approach: The macro va_start(ap, parmN) expands in a way that depends on the address and calling convention of parmN (the last named parameter). Therefore, a zero-named-parameter signature like func(...) is not valid in standard C for using va_start. Portable variadic functions always declare at least one fixed parameter (e.g., int count, ... or const char *fmt, ...). Step-by-Step Solution: Design the API with at least one named parameter (e.g., fmt).Call va_start(ap, fmt); // requires fmt to exist.Use va_arg to retrieve each value in order.Call va_end when done. Verification / Alternative check: Look at the standard library’s printf family: printf(const char *fmt, ...). The existence of fmt enables correct va_start behavior. Families like execl also include at least one fixed parameter before ... . Why Other Options Are Wrong: “No” conflicts with the stdarg.h contract. References to compiler era or stdio.h are irrelevant. Floating-point is not special here; the rule applies to all types. Common Pitfalls: Trying to declare func(...) without a fixed arg; assuming the implementation can somehow infer the boundary automatically. Final Answer: Yes

Question 14

In a C function that accepts a variable number of arguments, may the fixed (named) parameters appear after the ellipsis, or must the ellipsis be last in the parameter list?

Accepted Answer

Introduction / Context: The grammar and calling convention for variadic functions in C require that the ellipsis ... terminates the parameter list. This determines where the unnamed arguments begin in the call frame and allows va_start to compute their location based on the last named parameter. Given Data / Assumptions: We are using standard C (stdarg.h) with no compiler extensions. Function declarations follow the form: type f(type1 p1, type2 p2, ..., typeN pN, ...); Concept / Approach: Because va_start needs the last named parameter, the syntax and ABI expect the ellipsis to appear at the very end. Any attempt to place a named parameter after the ellipsis breaks that model and is not allowed by the language grammar. Compilers will reject such a declaration. Step-by-Step Solution: Declare the last fixed parameter (e.g., const char *fmt).Place the ellipsis immediately after fmt, making it the final token in the parameter list.Inside the body, call va_start(ap, fmt) and then va_arg repeatedly.End with va_end(ap). Verification / Alternative check: Try to compile a prototype like int f(..., int x); — it is ill-formed. In contrast, int f(int x, ...); is valid and reflects how printf and friends are defined. Why Other Options Are Wrong: Allowing fixed parameters after the ellipsis contradicts the C standard. The compiler does not reorder parameters. Pointer status or OS does not change the rule. Common Pitfalls: Believing that variadic arguments can be “inserted” in the middle; forgetting that the ellipsis must terminate the list. Final Answer: The ellipsis must be last; fixed arguments cannot follow it

Question 15

In a C variadic function, if va_start() has been called on a va_list, is it required to call va_end() before the function returns?

Accepted Answer

Introduction / Context: va_end is the counterpart to va_start in C’s stdarg.h. Although on some implementations va_end may appear to do nothing, the C standard requires its use to allow the implementation to perform any necessary cleanup or state reset for the va_list. Given Data / Assumptions: va_list ap has been initialized by va_start. The function will return normally or via a branch. Concept / Approach: Once you call va_start, the program has entered a region where the runtime may have reserved resources or needs to restore internal state. Calling va_end signals that iteration is finished. This is required even if you ended up reading zero variadic arguments after initialization. Step-by-Step Solution: va_list ap; va_start(ap, last_named_param);// zero or more calls to va_arg(ap, type)va_end(ap); // always before return Verification / Alternative check: Inspect the standard library’s v* functions or compiler documentation—each stresses the contract: every va_start must be paired with a va_end, similar to resource acquisition/cleanup patterns. Why Other Options Are Wrong: Conditioning va_end on platform, number of arguments, or types is incorrect. The rule is unconditional in portable C code. Common Pitfalls: Returning early without calling va_end on all paths; using va_list after va_end, which is undefined behavior. Final Answer: Yes

Question 16

Can you construct and pass a variable argument list to a variadic function at run time in standard C, or must you instead use a va_list and call a corresponding v* function (such as vprintf)?

Accepted Answer

Introduction / Context: The ellipsis “...” in C is a compile-time feature. Call sites fix the number and types of arguments syntactically; there is no portable facility to construct an argument pack at run time and pass it to an arbitrary variadic function. Instead, the standard pattern for forwarding is to use va_list with the v* versions of APIs. Given Data / Assumptions: Variadic functions like printf(const char fmt, ...) have companions like vprintf(const char fmt, va_list ap). va_list encapsulates the traversal state for an existing variadic call. Concept / Approach: To “pass along” a variadic sequence, write your function as variadic too, create a va_list with va_start, and then call a v function, e.g., vprintf(fmt, ap). You cannot, in portable C, generate a fresh “...” argument list dynamically; ABIs and calling conventions prevent standardizing such behavior. Step-by-Step Solution: Define wrapper: void logf(const char fmt, ...).Inside: va_list ap; va_start(ap, fmt);Forward: vprintf(fmt, ap);Cleanup: va_end(ap). Verification / Alternative check: Inspect standard library: every variadic family has a v counterpart that accepts va_list (vfprintf, vsnprintf, etc.). That is the endorsed mechanism for run-time forwarding. Why Other Options Are Wrong: “Yes, synthesize ... dynamically” is nonportable/incorrect. Hardware width, integer-only limitations, or C++ do not change the C rule; even in C++ the recommended approach uses va_list or variadic templates (a different language feature). Common Pitfalls: Trying to cast a pointer to emulate arguments, or assuming you can create a fake stack frame; this is undefined behavior. Final Answer: No, you cannot build “...” at run time; use va_list with a v function

Question 17

When defining a C function that accepts a variable number of arguments, should you include the ellipsis (...) in the prototype/declaration, or may you drop it?

Accepted Answer

Introduction / Context: The compiler needs explicit syntax to know a function is variadic. In C, that syntax is the ellipsis ... at the end of the parameter list. Omitting it declares a non-variadic function, and calling it with extra arguments produces undefined behavior. Given Data / Assumptions: Standard header stdarg.h is used inside the function body. The function will call va_start using the last named parameter. Concept / Approach: The ellipsis must appear in both the prototype (if one exists) and the definition to communicate the calling convention to both the compiler and linker. Without it, callers will pass no extra arguments and the callee cannot legally use va_start because there is no last named parameter before ... . Step-by-Step Solution: Declare: int sum(int count, ...);Define: int sum(int count, ...) { va_list ap; va_start(ap, count); /* use va_arg */ va_end(ap); }Do not omit the ellipsis in either place. Verification / Alternative check: Review standard library declarations (printf, execl, etc.); all include ... in their prototypes to signal variadic status. Why Other Options Are Wrong: The compiler cannot infer variadic behavior automatically. Attributes are nonportable. Floating point is not a condition for using ... . Using ... only in the definition mismatches the prototype and leads to undefined behavior. Common Pitfalls: Leaving the prototype non-variadic while defining a variadic function; forgetting that default promotions change the types to request via va_arg. Final Answer: Include the ellipsis (...); you must not drop it

Question 18

In C, can a function that accepts a variable argument list forward that same list to another function? (For example, wrap printf by calling vprintf with the captured va_list.)

Accepted Answer

Introduction / Context: Wrappers around variadic interfaces are common. The correct pattern is to collect the arguments with va_start into a va_list and then call a companion function that accepts va_list directly (e.g., vprintf, vfprintf, vsnprintf). This is called variadic forwarding. Given Data / Assumptions: Your wrapper is itself declared with ... after at least one named parameter. You have access to the appropriate v* function for the target routine. va_list is valid only between va_start and va_end; if you need to traverse twice, use va_copy where available. Concept / Approach: Because you cannot synthesize a new “...” at run time, you forward by passing the va_list to a v* variant. Many standard families implement both forms for this reason: printf/vprintf, fprintf/vfprintf, snprintf/vsnprintf, etc. Step-by-Step Solution: Define: void logf(const char fmt, ...) { va_list ap; va_start(ap, fmt); vfprintf(stderr, fmt, ap); va_end(ap); }If you must forward the list to two consumers, use va_copy before the first one and call va_end on each instance. Verification / Alternative check: Check standard headers: prototypes for v functions accept va_list precisely to enable forwarding. Why Other Options Are Wrong: Prohibiting passing va_list is incorrect; it is explicitly supported. Platform or integer-only restrictions do not apply. Skipping va_end is never acceptable. Common Pitfalls: Reusing a va_list after it has been passed to a function that also iterates it; not using va_copy when multiple passes are needed. Final Answer: Yes, by capturing with va_start and forwarding to a v* function (e.g., vprintf)

Question 19

In a function that accepts a variable argument list, may the fixed (named) parameters appear at the end of the parameter list, or must they precede the ellipsis?

Accepted Answer

Introduction / Context: The language requires that the ellipsis ... be the final token in the parameter list. Therefore, any fixed (named) parameters must precede it. This guarantees that va_start can be given the correct last named parameter and that the compiler can generate the appropriate calling sequence. Given Data / Assumptions: We are using stdarg.h and portable, standard C. The function has the form: ret f(T1 a1, T2 a2, ..., TN aN, ...). Concept / Approach: Named parameters define the boundary between fixed and variadic arguments. Placing a named parameter after the ellipsis would make it impossible for the callee to find how many and what types of arguments were passed, and it is not allowed by the grammar. Step-by-Step Solution: Define all fixed parameters first (e.g., fmt).Append ... to indicate the start of unnamed arguments.Inside the function, use va_start(ap, fmt) to enter the unnamed region and process with va_arg. Verification / Alternative check: Trying to compile f(..., int x) yields a diagnostic; printf-style signatures demonstrate the correct order. Why Other Options Are Wrong: The compiler cannot infer parameter order. Constness and target ABI do not change the rule. Common Pitfalls: Believing you can add “optional trailing named parameters” after the ellipsis; that is not the model C uses. Final Answer: Fixed parameters must come before the ellipsis; they cannot occur at the end

Question 20

In standard C, is it required that a function which accepts a variable argument list has at least one fixed (named) parameter before the ellipsis so that va_start() can be used correctly?

Accepted Answer

Introduction / Context: This question reiterates a foundational requirement for variadic functions in C: a last fixed (named) parameter must exist so va_start can compute where the unnamed arguments begin. Without it, the behavior is not defined by the standard. Given Data / Assumptions: We rely on stdarg.h macros: va_list, va_start, va_arg, va_end. We target portable C, not compiler-specific extensions. Concept / Approach: va_start(ap, last) uses last to determine the starting address of the variadic arguments. Consequently, a function prototype must include at least one named parameter before ... . The classic pattern is int printf(const char *fmt, ...). Step-by-Step Solution: Declare f(T last, ...);Inside f: va_list ap; va_start(ap, last);Iterate with va_arg until a known condition (count or format) ends processing.Call va_end(ap) prior to return. Verification / Alternative check: All standard variadic functions follow this model. Attempting to write f(...){ va_list ap; va_start(ap, ???); } is impossible in portable C because there is no last named parameter to supply. Why Other Options Are Wrong: “No” conflicts with stdarg.h semantics. Numeric width, floating-point, or stdio.h helpers do not change the rule. Common Pitfalls: Assuming a compiler extension permits f(...); even if an extension exists, it is nonportable and should be avoided for educational and interview contexts. Final Answer: Yes

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