C varargs typing: identify the primary error when the va_list variable is declared with the wrong type and used with va_start. #include <stdio.h> #include <stdarg.h> void varfun(int n, ...); int main() { varfun(3, 7, -11.2, 0.66); return 0; } void varfun(int n, ...) { float ptr; / WRONG: should be va_list / int num; va_start(ptr, n); / invalid: ptr is not a va_list */ num = va_arg(ptr, int); printf("%d", num); }

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