C varargs API usage: identify the error when va_list is used without calling va_start in a variable-argument function.\n\n#include <stdio.h>\n#include <stdarg.h>\nvoid varfun(int n, ...);\n\nint main()\n{\n varfun(3, 7, -11, 0);\n return 0;\n}\nvoid varfun(int n, ...)\n{\n va_list ptr;\n int num;\n num = va_arg(ptr, int); /* va_start not called */\n printf("%d", num);\n}

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