You are not forced to use a function in the Wolfram Language like you are in other languages. You only "need a function" if you will call this as a function repeatedly. While loops and older "software programming styles" are something you'll learn you rarely need or should use.

You really need to read the Documentation (Help) on functions. It can't be explained quickly.

Your basic strategy for "accepting multiple inputs" is multiple function definitions of the same function (which WL allows) (and is very typically done). Now for function return you can return a List[] and you'll very typical to see things in lists going in and out of WL functions.

fun[x_]:={x,x}
fun[x_,y_]:={x+y,x*y}
fun[xi_List]:={Plus@@xi, Times@@xi}
...

You now have three function of the same name without any templating, three ways to invoke, and the last one may have wide output, you'd use Dimensions[fun[list]]] to know what size the output is.

Davood:

I do not have time to rewrite and test your example code, but here is a simpler example:

a = 1;
b = 2;
c = 30;

While[
 a < c,
 a = a * b;
 ]
{a, c}

Can be turned into the following function:

f[a_Integer, b_Integer, c_Integer] := Module[
   {
    retVal
    },
   retVal = a;
   While[
    retVal < c,
    retVal = retVal * b;
    ];
   {retVal, c}
   ];

And then called as follows:

f[1, 2, 30]

Which returns:

{32, 30}

Note that I had to declare a privately scoped variable (retVal) for Module because parameters to functions are immutable. You do not always have to do this, but it is helpful to know how.

In my Program I have these initial values : Pr, Mr, Er,r then I want to calculate the lines below :

While[Pr > 0, 

 f1 = -(1.47*Mr*
      Er*(1 + Pr/Er)*(1 + 4*\[Pi]*r^3*Pr*.08969/Mr))/(r^2*(1 - 
       2*Mr*1.47/r));
 g1 = 4*\[Pi]*r^2*Er*.08969;
 k1 = h*f1;
 m1 = h*g1;

 f2 = -(1.47*Mr*
      Er*(1 + (Pr + k1/2)/Er)*(1 + 
        4*\[Pi]*(r + h/2)^3*(Pr + k1/2)*.08969/Mr))/((r + h/2)^2*(1 - 
       2*Mr*1.47/(r + h/2)));
 g2 = 4*\[Pi]*(r + h/2)^2*(Er + m1/2)*.08969;
 k2 = h*f2;
 m2 = h*g2;

 f3 = -(1.47*Mr*
      Er*(1 + (Pr + k2/2)/Er)*(1 + 
        4*\[Pi]*(r + h/2)^3*(Pr + k2/2)*.08969/Mr))/((r + h/2)^2*(1 - 
       2*Mr*1.47/(r + h/2)));
 g3 = 4*\[Pi]*(r + h/2)^2*(Er + m2/2)*.08969;
 k3 = h*f3;
 m3 = h*g3;

 f4 = -(1.47*Mr*
      Er*(1 + (Pr + k3)/Er)*(1 + 
        4*\[Pi]*(r + h)^3*(Pr + k3)*.08969/Mr))/((r + h)^2*(1 - 
       2*Mr*1.47/(r + h)));
 g4 = 4*\[Pi]*(r + h)^2*(Er + m3)*.08969;
 k4 = h*f4;
 m4 = h*g4;
 Pr = Pr + 1/6*(k1 + 2*k2 + 2*k3 + k4);
 Mr = Mr + 1/6*(m1 + 2*m2 + 2*m3 + m4);
 Er = fromPressureToEnergy[Pr];
 r = r + h;
 ]

Then the result are the last Mr and r when the loop is finished.

Could you please help me that how I can write a function for this code and calculations?

Thank you very much for your attention and consideration. Very nice and important proverb especially for someone who is a beginner in something.

Hi Davood,

Take a look at this tech note.