Struct with the function parameters

Can I suppose that, from the call stack point view, it's the same to call a function like function1

int function1(T1 t1, T2 t2);

than to another like function2?

struct parameters_t
{ 
    Wide<T1>::type t1;
    Wide<T2>::type t2;
}

int function2(parameters_t p);

Where, 开发者_如何学PythonWide template wide T to the processor word length. For example, for 32-bit processors:

template<typename T, bool b = sizeof(T) >=4 >
struct Wide
{
    typedef T type;
};

template<typename T>
struct Wide<T,false>
{
    typedef unsigned int type;
};

I need to do something like this:

typedef int (*function_t)(parameters_t);

function_t function = (function_t) &function1;

parameters_t params;
// initialize params
function(params);

Thanks!

---

Question 1. No the two function calls aren't necessarily the same -- calling conventions that push parameters right to left and left to right are both in wide use.

It sounds like you want to create a function that takes a variable number of a variable type of parameters. To do that, I'd have it take something like an std::vector<boost:any> as its parameter.

---

What you're asking is actually two different questions. I don't think there's any guarantee that the two methods will look the same on the call stack, but for your purposes of needing to pass parameters to a function called by a pointer, it works.

You should probably consider passing it via a reference or pointer though. Passing a large structure around over and over will be inefficient.