如何在 C 语言中使用 stdarg 编写该代码

Question

I need to change that code into another, using library stdarg. Code:

int value(int n, int x_1, ...)
{
      int result = 0;
      int* ptr = &x_1;
      for (int i = 0; i < n / 2; i++)
      {
         result += ((*ptr) / (*(ptr + 1)));
         ptr += 2;
      }
      return result;
 }  

Answer 1

The details of how arguments are passed to a function is the "calling convention." Depending on platform, language, and compiler, the rules can be complex. So it is not safe to assume the x_1 is on the stack and *(ptr + 1) is the first argument after x_1 . The purpose of stdarg.h is to provide a portable way to iterate through the variable arguments.

To use stdarg.h, a function needs three things:

1. At least one fixed argument
2. A way to determine the number of variable arguments
3. A way to determine the type of each variable arugment

Functions like printf have a format string that is both a fixed argument and it encodes the number and type of each variable argument.

For value , the the first argument n is a fixed argument and it gives the number of variable arguments. There isn't a clear way to determine the type of each variable argument for value . One option is to make a choice, for example "int", and document the function. Since the operation inside the for-loop is division, maybe float or double makes more sense.

Using stdarg.h is straight-forward in this case. Use va_start to initialize a va_list and then use va_arg to get the value of each variable argument.

/* value inputs n variable arguments, call them x_i, of type int
 * and returns the value
 * 
 * (x_0 / x_1) + (x_2 / x_3) + ...
 *
 * n must be even
 * the division is integer division
 */
int value(int n, ...)
{
  int result = 0;
  va_list ap;

  va_start(ap, n);

  for (int i = 0; i < n/2; ++i) {
    int a = va_arg(ap, int);
    int b = va_arg(ap, int);
    result += a/b;
  }

  va_end(ap);

  return result;
}

Example Calls

This example computes (6/3) + (21/7):

int r = value(4, 6, 3, 21, 7);
printf("%d\n", r);

and results in

5

This second example shows that value can be called by unpacking an array

  int a[] = {49, 7, 64, 8, 121, 11};
  int r = value(6, a[0], a[1], a[2], a[3], a[4], a[5]);

  printf("%d\n", r);

which results in

26