I have a function foo(char *n, ...);
I need to get and use all of optional char parameters.
I had an idea to use
while(va_arg(argPtr, char) != NULL)
{
...
}
to understand when I reach the end of the list. So, will it work, if in function call I'll do foo(n, 't', 'm', '$', NULL); ?
Will NULL be read as a char by va_arg?
Or maybe there's a more common way to determine the end of list, without adding NULL as last parameter?
There is no direct way for a function that uses va_arg to determine the number or type(s) of the arguments passed by a given call.
Your proposed method in particular:
while(va_arg(argPtr, char) != NULL)
is incorrect. va_arg(argPtr, char) yields a value of type char, while NULL is a null pointer constant. (NULL is commonly defined as 0, which compares equal to the null character '', but you can't rely on that.)
Any variadic function must have a way for the caller to specify the number and types of arguments. The *printf functions, for example, do so via the (non-variadic) format string. The POSIX execl*() functions expect a sequence of char* arguments; the end of the argument list is marked by the caller with (char*)NULL. Other methods are possible, but they almost all depend on information given at run time in the arguments. (You could use some other method, such as a global variable. Please don't.)
This places a burden on the caller to ensure that the arguments passed to the function are consistent. The function itself has no way to confirm this. Incorrect calls, like printf("%d
", "hello") or execlp("name", "arg1") have undefined behavior.
One more thing: you can't use va_arg with an argument of type char. When you call a variadic function, arguments corresponding to the , ... are promoted. Integer arguments of types narrower than int are promoted to int or to unsigned int, and arguments of type float are promoted to double. If a caller passes an argument of type char, the function must invoke va_arg(argPtr, int).
(In very obscure circumstances that you're not likely to run into, char can be promoted to unsigned int. That can happen only if plain char is unsigned and sizeof (int) == 1, which implies that a byte is at least 16 bits.)
