Cprogramming 简明教程

Variable Arguments in C

有时您可能会遇到一种情况，其中希望有一个可以接受可变数量的参数（而不是预定义数量的参数）的函数。C 编程语言为这种情况提供了解决方案。

Sometimes, you may come across a situation, when you want to have a function that can accept a variable number of arguments (parameters) instead of a predefined number of arguments. The C programming language provides a solution for this situation.

阅读本章以了解如何根据您的要求定义可接受可变数量的参数的函数。

Read this chapter to learn how you can define a function that can accept a variable number of parameters based on your requirement.

以下示例显示了这样的函数的 definition −

The following example shows the definition of such a function −

int func(int, ... ) {
   ...
   ...
}

int main() {
   func(1, 2, 3);
   func(1, 2, 3, 4);
}

应该指出，函数 func() 的最后一个参数是 ellipses ，即三个点（…），而紧挨在省略号之前的参数始终是一个 int ，它表示传入的总数量可变参数。

It should be noted that the function func() has its last argument as ellipses, i.e. three dotes (…) and the one just before the ellipses is always an int which will represent the total number variable arguments passed.

要实现这样的功能，您需要使用 stdarg.h 头文件，它提供了实现可变参数功能的函数和宏。

To get such a functionality, you need to use the stdarg.h header file which provides the functions and macros to implement the functionality of variable arguments.

按照以下步骤操作 −

Follow the steps given below −

Define a function with its last parameter as ellipses and the one just before the ellipses is always an int which will represent the number of arguments.
Create a va_list type variable in the function definition. This type is defined in stdarg.h header file.
Use int parameter and va_start macro to initialize the va_list variable to an argument list. The macro va_start is defined in stdarg.h header file.
Use va_arg macro and va_list variable to access each item in argument list.
Use a macro va_end to clean up the memory assigned to va_list variable.

Example

现在让我们遵循上述步骤，写一个可以采用可变数量的参数并返回其平均值的简单函数 −

Let us now follow the above steps and write down a simple function which can take the variable number of parameters and return their average −

#include <stdio.h>
#include <stdarg.h>

double average(int num,...) {

   va_list valist;
   double sum = 0.0;
   int i;

   /* initialize valist for num number of arguments */
   va_start(valist, num);

   /* access all the arguments assigned to valist */
   for (i = 0; i < num; i++) {
      sum += va_arg(valist, int);
   }

   /* clean memory reserved for valist */
   va_end(valist);

   return sum/num;
}

int main() {

   printf("Average of 2, 3, 4, 5 = %f\n", average(4, 2,3,4,5));
   printf("Average of 5, 10, 15 = %f\n", average(3, 5,10,15));
}

Output

编译并执行上述代码会产生以下输出：

When the above code is compiled and executed, it produces the following output −

Average of 2, 3, 4, 5 = 3.500000
Average of 5, 10, 15 = 10.000000

应该指出，函数 average() 被调用了两次，而每次第一个参数都表示要传递的可变参数的总数。只使用省略号来传递可变数量的参数。

It should be noted that the function average() has been called twice and each time the first argument represents the total number of variable arguments being passed. Only ellipses are used to pass variable number of arguments.