implement divide algorithm without using "/" for floating point - algorithm

Question

please if there is any specific algorithm for implementing the divide operator as a function, guide me about their name. I want to implement a function that takes two floating number and return the result of the divide, but in implementation, I won't use "/".
I have done this in a much simpler version when we want just the q in integer,
function divide(num0, num1) {
if ("bigint" != typeof num0 || "bigint" != typeof num1) {
throw new TypeError("The arguments should be bigint.");
}
if (num1 > num0) {
return 0;
}
for (var i = 0n; num0 >= num1; i++) {
num0 -= num1;
}
return i;
}
"I use bigint numeric type just two restrict to integer"
but I think this approach couldn't extend two return floating results. my guess is I approach binary level operation or so; thanks if learning me about any flowchart, pseudo-code, or code-snippet "in any language" to deal with this problem

Answer 1

I wrote this one for this question in js:
function justIntegerDivide(num0, num1) {
for (var q = 0; num0 >= num1; q++) {
num0 -= num1;
}
return [q, num0];
}
const divide = (n0, n1, afterPoint = 10) => {
if ((0 == n1 || 0n == n1) && 0 < n0) return Infinity;
if ((0 == n1 || 0n == n1) && 0 > n0) return -Infinity;
if ((0 == n1 || 0n == n1) && 0 == n0) return NaN;
if ("number" == typeof n0 && "number" == typeof n1) {
let sign = Math.sign(n0) * Math.sign(n1);
let num0 = Math.abs(n0),
num1 = Math.abs(n1);
let counter = 0;
let [q, r] = justIntegerDivide(num0, num1);
result = `${q}.`;
for (counter = 1; counter < afterPoint; counter++) {
var newReminder;
let qAfter;
previousReminder = 1 == counter ? r : newReminder;
[qAfter, newReminder] = justIntegerDivide(previousReminder * 10, num1);
result += qAfter;
if (0 == newReminder) {
return +result * sign;
}
}
return +result * sign;
} else if ("bigint" == typeof n0 && "bigint" == typeof n1) {
let sign = (n0 > 0 && n1 > 0) || (n0 < 0 && n1 < 0) ? 1n : -1n;
let num0 = n0 > 0 ? n0 : -n0;
let num1 = n1 > 0 ? n1 : -n1;
if (num0 < num1) {
return 0n;
}
for (var i = 0n; num0 >= num1; i++) {
num0 -= num1;
}
return i * sign;
} else {
throw new TypeError("Both arguments should be number or bigint");
}
};