使用C89可变参数的函数而不传递参数的数目或最终的说法？ [英] Use variadic functions in C89 without passing number of arguments or a final argument?
问题描述
让我们说我有一个可变参数函数美孚（INT TMP，...），调用foo函数时，我需要知道有多少争论有。我所知道的找出两种方法有多少参数有：
Let's say I have a variadic function foo(int tmp, ...), when calling foo function I need to know how many arguments there are. I'm aware of two ways of finding out how many arguments there are:
-
调用foo时，像-1使用最后一个参数，所以你的函数调用将是这样的：
美孚（TMP，1，2，9，-1），当你在里面foo和一个在va_arg调用返回-1你知道你已经阅读所有的函数参数
Use a final argument when calling foo, like -1, so your function call will be like this:
foo(tmp, 1, 2, 9, -1)and when you are inside foo and a va_arg call returns -1 you know you have read all the function arguments
添加的富多了一个参数，其中程序员将有参数的总数，所以你将不得不调用到foo这样的：美孚（TMP，5，1，2，3 ，4，5）或美孚（TMP，2，7，8）
Add one more argument in foo where the programmer will have the total number of arguments, so you will have calls to foo like this: foo(tmp, 5, 1, 2, 3, 4, 5) or foo(tmp, 2, 7, 8)
我曾经跟随第一种方式，一旦有以下错误。随着code：
I used to follow the first way and once had the following bug. With the code:
expr_of_type(expr, boolexpr_e, newtable_e, nil_e, -1)
在这里expr_of_type是一个可变参数函数，是检查是否EXPR（第一个参数）是下列类型之一（boolexpr_e或new_table_e或nil_e拥有所有类型枚举类型）。
我合一无意中写道：
where expr_of_type was a variadic function and was checking if expr(the first argument) was one of the following types (boolexpr_e or new_table_e or nil_e had all type of an enumerated type). I one accidently wrote:
expr_of_type(expr, boolexpr_e, newtable_e, nil_e -1)
我忘了nil_e和-1之间的逗号，因为nil_e有一个枚举类型，nil_e - 1是一个有效的前pression因为nil_e不是0试图让expr_of_type参数没有当给定的可变参数函数找到-1作为最后一个参数，并继续寻找创造这我花了一些时间去找出一个bug。
I forgot the comma between nil_e and -1, because nil_e had an enumerated type, nil_e - 1 was a valid expression and because nil_e was not 0 the given variadic function when trying to get expr_of_type arguments didn't find -1 as last argument and continued searching creating a bug which took me some time to find out.
我没有找到很好的第二种方式要么，因为添加或从可变参数函数删除多一个说法，当您需要更改包含总参数的数量参数。
I don't find the second way nice either, because when adding or removing one more argument from a variadic function you need to change the parameter that contains the number of total arguments.
在寻找一种更好的方式来使用/创建可变参数的功能，我发现复杂的宏从而可以解决我用第一种方式，当有错误。但是复杂的宏可用于C99的标准。我一直在寻找更好的方法来使用/ C89中创建可变参数的函数。任何想法？
In searching for a better way to use/create variadic functions I found variadic macros which can solve the bug I had when using the first way. But variadic macros are available to C99 standard. I was looking for a better way to use/create variadic functions in C89. Any ideas?
推荐答案
在一般情况下，你仍然必须沿着参数传递计数不知何故，无论是通过的警戒值或通过明确的数量。
In general, you must still pass along the argument count somehow, whether via a sentinel value or via an explicit count.
您可以，但是，通过更好的定点解决您的定点问题。这就是为什么扩大到负常量preprocessor宏应该在括号包围的原因之一：
You could, however, fix your sentinel problem by making a better sentinel. This is one of the reasons why preprocessor macros that expand to negative constants should be surrounded in parentheses:
#define VARARG_SENTINEL (-1)
然后 nil_e VARARG_SENTINEL 将产生一个编译错误。
使用枚举或 const int的将工作太：
enum { VARARG_SENTINEL = -1 };
使用符号常量的标记值将是其他原因也更好（更自我记录，便于以后更改基础值）。
Using a symbolic constant for the sentinel value would be better for other reasons too (more self-documenting, easier to change the underlying value later).
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