Mantain the original value in a recursive function c++ - c++

Question

I am trying to solve a square root in a recursive way. I do not want the code to solve that! I am almost done, but I do not know how to mantain the original value of the function:
float raizCuadrada(float num, float err) {
float nuevo = num / 2;
float resta=(nuevo*nuevo)-num;
if(resta>err){
return (raizCuadrada(nuevo, err));
}
else if (resta <= err) {
return (nuevo);
}
}
I basicly want to know how to "save" that first "num" call, somewhere, to use it ALWAYS in the "resta", the "- num" should be always the number that 1st put.
NOTES: I cannot input more inputs. I can only input 1 number and the error.

Answer 1

You can define a global variable at the top of your code.
float num_fixed;
float raizCuadrada(float num, float err) {
float nuevo = num / 2;
float resta=(nuevo*nuevo)-num_fixed;
if(resta>err){
return (raizCuadrada(nuevo, err));
}
else {
return (nuevo);
}
}
int main(void){
float num = 2.0;
float err = 0.000001;
float output;
num_fixed = num;
output = raizCuadrada(num, err);
}

Answer 2

I would recommend the two solutions that have been proposed. Either overload a function that takes three arguments or use a global.
If you for some reason don't want to do that, you can use some trickery:
float raizCuadrada(float num, float err) {
static float org = -1;
if(org<0)
org = num;
float nuevo = num / 2;
float resta=(nuevo*nuevo)-org;
float ret;
if(resta>err)
ret = raizCuadrada(nuevo, err);
else
ret = nuevo;
if(org>=0)
org=-1;
return ret;
}
I would not recommend this method unless you really have a special need for it. It is quite easy to do mistakes, and I cannot think of any real benefit. But since it is possible, I wanted to show it.

Answer 3

I basicly want to know how to "save" that first "num" call, somewhere,
to use it ALWAYS in the "resta", the "- num" should be always the
number that 1st put.
NOTES: I cannot input more inputs. I can only input 1 number and the
error. c++
The solution I most often use is to separate the two 'efforts' into two methods:
1) Start with a method (use your original method name) to save the first variable to a safe space. I will leave the 'where' up to you. Here I have simply made up a new variable name (origNum), in a scope available to the code of each method. In C++, I would expect that the class containing these methods would 'save' and use an instance data attribute.
float raizCuadrada(float num, float err) // same name as original
{
origNum = num; // "save" original 'num'
return (raizCuadrada2(num, err)); // continue normally
}
2) Then a mild re-factor of your previous code uses the "saved" value. Since you can not change the parameters, I suggest a slight method name change, here I created raizCuadrada with suffix '2'.
float raizCuadrada2(float num, float err) {
float nuevo = num / 2;
float resta=(nuevo*nuevo)-origNum; // use origNum here
if(resta>err){
return (raizCuadrada(nuevo, err));
}
else if (resta <= err) {
return (nuevo);
}
}
note - not compiled. not tested.