How to wrap a function with variable length arguments?

I am looking to do this in C/C++.

I came across Variable Length Arguments but this suggests a solution with Python & C using libffi.

Now, if i want to wrap printf function with myprintf

i do like below:

void myprintf(char* fmt, ...)
{
    va_ list args;
    va_ start(args,fmt);
    printf(fmt,args);
    va_ end(args);
}

int _tmain(int argc, _TCHAR* argv[])
{
    int a = 9;
    int b = 10;
    char v = 'C';
    myprintf("This is a number: %d and \nthis is a character: %c and \n another number: %d\n",a, v, b);
    return 0;
}

But the results are not as expected!

This is a number: 1244780 and
this is a character: h and
another number: 29953463

any pointer where i miss??

Thanks.. Any suggestion is appreciated.

From stackoverflow Prakash

• How do you mean a pure C/C++ solution?

  The rest parameter (...) is supported cross platform in the C runtime.

  http://msdn.microsoft.com/en-us/library/kb57fad8.aspx

  From Mark Ingram

• Are you using C or C++? The next C++ version, C++0x, will support variadic templates which provide a solution to that problem.

  Another workaround can be achieved by clever operator overloading to achieve a syntax like this:

  void f(varargs va) {
      BOOST_FOREACH(varargs::iterator i, va)
          cout << *i << " ";
  }
  
  f(args = 1, 2, 3, "Hello");
  

  In order to get this to work, the class varargs has to be implemented to override operator = that returns a proxy object which, in turn, overrides operator ,. However, making this variant type safe in current C++ isn't possible as far as I know since it would have to work by type erasure.

  From Konrad Rudolph

• I am also unsure what you mean by pure

  In C++ we use

  class Foo
  {   void Write(const char* pMsg, ...);
  };
  
  void Foo::Write( const char* pMsg, ...)
  {
      char buffer[4096];
          std::va_list arg;
          va_start(arg, pMsg);
          std::vsnprintf(buffer, 4096, pMsg, arg);
          va_end(arg);
          ...
  }
  

  From David Sykes

• the problem is that you cannot use 'printf' with va_args. You must use vprintf if you are using variable argument lists. vprint, vsprintf, vfprintf, etc. (there are also 'safe' versions in Microsoft's C runtime that will prevent buffer overruns, etc.)

  You sample works as follows:

  void myprintf(char* fmt, ...)
  {
      va_ list args;
      va_ start(args,fmt);
      vprintf(fmt,args);
      va_ end(args);
  }
  
  int _tmain(int argc, _TCHAR* argv[])
  {
      int a = 9;
      int b = 10;
      char v = 'C'; 
      myprintf("This is a number: %d and \nthis is a character: %c and \n another number: %d\n",a, v, b);
      return 0;
  }
  

  From Mark

• Reading the comments to this question was like reading my son's handwriting. The correct answer is (as Mark Allender pointed out) to use the vprintf function, not the printf function.

  From 1800 INFORMATION