How can I wrap a function with variable length arguments?

Question

I am looking to do this in C/C++. I came across Variable Length Arguments, but this suggests a solution with Python and C using libffi.

Now, if I want to wrap the printf function with myprintf.

I do it 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

What did I miss?

Answer 1

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;
}

Answer 2

In C++11, this is one possible solution using variadic templates:

template<typename... Args>
void myprintf(const char* fmt, Args... args)
{
    std::printf(fmt, args...);
}

As rubenvb points out, there are trade-offs to consider. For example, you will be generating code for each instance which will lead to code bloat.

Answer 3

I am also unsure what you mean by pure.

In C++ we use:

#include <cstdarg>
#include <cstdio>

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);
    ...
}

Answer 4

Actually, there's a way to call a function that doesn’t have a va_list version from a wrapper. The idea is to use assembler, do not touch arguments on the stack, and temporarily replace the function return address.

An example for Visual C x86. call addr_printf calls printf():

__declspec( thread ) static void* _tls_ret;

static void __stdcall saveret(void *retaddr) {
    _tls_ret = retaddr;
}

static void* __stdcall _getret() {
    return _tls_ret;
}

__declspec(naked)
static void __stdcall restret_and_return_int(int retval) {
    __asm {
        call _getret
        mov [esp], eax   ; /* replace current retaddr with saved */
        mov eax, [esp+4] ; /* retval */
        ret 4
    }
}

static void __stdcall _dbg_printf_beg(const char *fmt, va_list args) {
    printf("calling printf(\"%s\")\n", fmt);
}

static void __stdcall _dbg_printf_end(int ret) {
    printf("printf() returned %d\n", ret);
}

__declspec(naked)
int dbg_printf(const char *fmt, ...)
{
    static const void *addr_printf = printf;
    /* prolog */
    __asm {
        push ebp
        mov  ebp, esp
        sub  esp, __LOCAL_SIZE
        nop
    }
    {
        va_list args;
        va_start(args, fmt);
        _dbg_printf_beg(fmt, args);
        va_end(args);
    }
    /* epilog */
    __asm {
        mov  esp, ebp
        pop  ebp
    }
    __asm  {
        call saveret
        call addr_printf
        push eax
        push eax
        call _dbg_printf_end
        call restret_and_return_int
    }
}

Answer 5

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.

Answer 6

void myprintf(char* fmt, ...)
{
    va_ list args;
    va_ start(args, fmt);
    printf(fmt, args); // This is the fault. "vprintf(fmt, args);"
                       // should have been used.
    va_ end(args);
}

If you're just trying to call printf, there's a printf variant called vprintf that takes the va_list directly: vprintf(fmt, args);

Answer 7

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

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

va_arg, va_copy, va_end, va_start