Yes you can do it, but it is somewhat ugly and you have to know the maximal number of arguments. Furthermore if you are on an architecture where the arguments aren't passed on the stack like the x86 (for instance, PowerPC), you will have to know if "special" types (double, floats, altivec etc.) are used and if so, deal with them accordingly. It can be painful quickly but if you are on x86 or if the original function has a well defined and limited perimeter, it can work. It still will be a hack, use it for debugging purpose. Do not build you software around that. Anyway, here's a working example on x86:

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

int old_variadic_function(int n, ...)
{
  va_list args;
  int i = 0;

  va_start(args, n);

  if(i++<n) printf("arg %d is 0x%x\n", i, va_arg(args, int));
  if(i++<n) printf("arg %d is %g\n",   i, va_arg(args, double));
  if(i++<n) printf("arg %d is %g\n",   i, va_arg(args, double));

  va_end(args);

  return n;
}

int old_variadic_function_wrapper(int n, ...)
{
  va_list args;
  int a1;
  int a2;
  int a3;
  int a4;
  int a5;
  int a6;
  int a7;
  int a8;

  /* Do some work, possibly with another va_list to access arguments */

  /* Work done */

  va_start(args, n);

  a1 = va_arg(args, int);
  a2 = va_arg(args, int);
  a3 = va_arg(args, int);
  a4 = va_arg(args, int);
  a5 = va_arg(args, int);
  a6 = va_arg(args, int);
  a7 = va_arg(args, int);

  va_end(args);

  return old_variadic_function(n, a1, a2, a3, a4, a5, a6, a7, a8);
}

int main(void)
{
  printf("Call 1: 1, 0x123\n");
  old_variadic_function(1, 0x123);
  printf("Call 2: 2, 0x456, 1.234\n");
  old_variadic_function(2, 0x456, 1.234);
  printf("Call 3: 3, 0x456, 4.456, 7.789\n");
  old_variadic_function(3, 0x456, 4.456, 7.789);
  printf("Wrapped call 1: 1, 0x123\n");
  old_variadic_function_wrapper(1, 0x123);
  printf("Wrapped call 2: 2, 0x456, 1.234\n");
  old_variadic_function_wrapper(2, 0x456, 1.234);
  printf("Wrapped call 3: 3, 0x456, 4.456, 7.789\n");
  old_variadic_function_wrapper(3, 0x456, 4.456, 7.789);

  return 0;
}

For some reason, you can't use floats with va_arg, gcc says they are converted to double but the program crashes. That alone demonstrates that this solution is a hack and that there is no general solution. In my example I assumed that the maximum number of arguments was 8, but you can increase that number. The wrapped function also only used integers but it works the same way with other 'normal' parameters since they always cast to integers. The target function will know their types but your intermediary wrapper doesn't need to. The wrapper also doesn't need to know the right number of arguments since the target function will also know it. To do useful work (except just logging the call), you probably will have to know both though.

如果可以使用c++ 11或更高版本的编译器编译代码，则可以使用可变参数函数模板:

#include <stdio.h>

template<typename... Targs>
int my_printf(const char *fmt, Targs... Fargs) {
    fprintf(stderr, "Calling printf with fmt %s", fmt);
    return printf(fmt, Fargs...);;
}

int main() {
    my_printf("test %d\n", 1);
    return 0;
}

Demo

C99支持可变参数宏;取决于你的编译器，你可以声明一个宏来做你想做的事情:

#define my_printf(format, ...) \
    do { \
        fprintf(stderr, "Calling printf with fmt %s\n", format); \
        some_other_variadac_function(format, ##__VA_ARGS__); \
    } while(0)

不过，一般来说，最好的解决方案是使用您试图包装的函数的va_list形式(如果存在的话)。

使用函数:

int my_printf(char *fmt, ...) {
    va_list va;
    int ret;

    va_start(va, fmt);
    ret = vfprintf(stderr, fmt, va);
    va_end(va);
    return ret;
}

Not sure if this helps to answer OP's question since I do not know why the restriction for using a helper function akin to vfprintf in the wrapper function applies. I think the key problem here is that forwarding the variadic argument list without interpreting them is difficult. What is possible, is to perform the formatting (using a helper function akin to vfprintf: vsnprintf) and forward the formatted output to the wrapped function with variadic arguments (i.e. not modifying the definition of the wrapped function). So, here we go:

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

int my_printf(char *fmt, ...)
{
    if (fmt == NULL) {
        /* Invalid format pointer */
        return -1;
    } else {
        va_list args;
        int len;

        /* Initialize a variable argument list */
        va_start(args, fmt);

        /* Get length of format including arguments */
        len = vsnprintf(NULL, 0, fmt, args);

        /* End using variable argument list */
        va_end(args);
        
        if (len < 0) {
            /* vsnprintf failed */
            return -1;
        } else {
            /* Declare a character buffer for the formatted string */
            char formatted[len + 1];

            /* Initialize a variable argument list */
            va_start(args, fmt);
            
            /* Write the formatted output */
            vsnprintf(formatted, sizeof(formatted), fmt, args);
            
            /* End using variable argument list */
            va_end(args);

            /* Call the wrapped function using the formatted output and return */
            fprintf(stderr, "Calling printf with fmt %s", fmt);
            return printf("%s", formatted);
        }
    }
}

int main()
{
    /* Expected output: Test
     * Expected error: Calling printf with fmt Test
     */
    my_printf("Test\n");
    //printf("Test\n");

    /* Expected output: Test
     * Expected error: Calling printf with fmt %s
     */
    my_printf("%s\n", "Test");
    //printf("%s\n", "Test");

    /* Expected output: %s
     * Expected error: Calling printf with fmt %s
     */
    my_printf("%s\n", "%s");
    //printf("%s\n", "%s");

    return 0;
}

我遇到了这个解。

编辑:修正由egmont指出的错误

差不多，使用<stdarg.h>中提供的功能:

#include <stdarg.h>
int my_printf(char *format, ...)
{
   va_list args;
   va_start(args, format);
   int r = vprintf(format, args);
   va_end(args);
   return r;
}

注意，您需要使用vprintf版本，而不是普通的printf。在这种情况下，如果不使用va_list，就无法直接调用变进函数。