最好的办法是

static BOOL(__cdecl *OriginalVarArgsFunction)(BYTE variable1, char* format, ...)(0x12345678); //TODO: change address lolz

BOOL __cdecl HookedVarArgsFunction(BYTE variable1, char* format, ...)
{
    BOOL res;

    va_list vl;
    va_start(vl, format);

    // Get variable arguments count from disasm. -2 because of existing 'format', 'variable1'
    uint32_t argCount = *((uint8_t*)_ReturnAddress() + 2) / sizeof(void*) - 2;
    printf("arg count = %d\n", argCount);

    // ((int( __cdecl* )(const char*, ...))&oldCode)(fmt, ...);
    __asm
    {
        mov eax, argCount
        test eax, eax
        je noLoop
        mov edx, vl
        loop1 :
        push dword ptr[edx + eax * 4 - 4]
        sub eax, 1
        jnz loop1
        noLoop :
        push format
        push variable1
        //lea eax, [oldCode] // oldCode - original function pointer
        mov eax, OriginalVarArgsFunction
        call eax
        mov res, eax
        mov eax, argCount
        lea eax, [eax * 4 + 8] //+8 because 2 parameters (format and variable1)
        add esp, eax
    }
    return res;
}

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

基本上有三种选择。

一种是不传递它而是使用目标函数的变进实现而不传递椭圆。另一种方法是使用可变宏。第三个选择是我所缺少的所有东西。

我通常选择第一种，因为我觉得这真的很容易处理。选项二有一个缺点，因为调用可变宏有一些限制。

下面是一些示例代码:

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

#define Option_VariadicMacro(f, ...)\
    printf("printing using format: %s", f);\
    printf(f, __VA_ARGS__)

int Option_ResolveVariadicAndPassOn(const char * f, ... )
{
    int r;
    va_list args;

    printf("printing using format: %s", f);
    va_start(args, f);
    r = vprintf(f, args);
    va_end(args);
    return r;
}

void main()
{
    const char * f = "%s %s %s\n";
    const char * a = "One";
    const char * b = "Two";
    const char * c = "Three";
    printf("---- Normal Print ----\n");
    printf(f, a, b, c);
    printf("\n");
    printf("---- Option_VariadicMacro ----\n");
    Option_VariadicMacro(f, a, b, c);
    printf("\n");
    printf("---- Option_ResolveVariadicAndPassOn ----\n");
    Option_ResolveVariadicAndPassOn(f, a, b, c);
    printf("\n");
}

最好的办法是

static BOOL(__cdecl *OriginalVarArgsFunction)(BYTE variable1, char* format, ...)(0x12345678); //TODO: change address lolz

BOOL __cdecl HookedVarArgsFunction(BYTE variable1, char* format, ...)
{
    BOOL res;

    va_list vl;
    va_start(vl, format);

    // Get variable arguments count from disasm. -2 because of existing 'format', 'variable1'
    uint32_t argCount = *((uint8_t*)_ReturnAddress() + 2) / sizeof(void*) - 2;
    printf("arg count = %d\n", argCount);

    // ((int( __cdecl* )(const char*, ...))&oldCode)(fmt, ...);
    __asm
    {
        mov eax, argCount
        test eax, eax
        je noLoop
        mov edx, vl
        loop1 :
        push dword ptr[edx + eax * 4 - 4]
        sub eax, 1
        jnz loop1
        noLoop :
        push format
        push variable1
        //lea eax, [oldCode] // oldCode - original function pointer
        mov eax, OriginalVarArgsFunction
        call eax
        mov res, eax
        mov eax, argCount
        lea eax, [eax * 4 + 8] //+8 because 2 parameters (format and variable1)
        add esp, eax
    }
    return res;
}

没有办法转发这样的函数调用，因为可以检索原始堆栈元素的唯一位置是my_print()。像这样包装调用的通常方法是有两个函数，一个只是将参数转换为各种varargs结构体，另一个实际操作这些结构体。使用这样的双函数模型，您可以(例如)通过使用va_start()初始化my_printf()中的结构来包装printf()，然后将它们传递给vfprintf()。