如何获得小于/大于给定数字的最接近的浮点数

Question

Here is a sample function:

function step(x, min, max) {
  return x >= min && x <= max ? x : 0;
}

console.log(step(-3 - Number.EPSILON, -3, 5)); // Expected 0, actual -3
console.log(step(5 + Number.EPSILON, -3, 5)); // Expected 0, actual 5

I need to check, that it returns zero for values outside [min, max] interval. For sure I can subtract/add a bigger number, for example 1. But I'm pretty sure, that there should exist a function returning previous/next floating point number. Could you please suggest the function or how to implement it?

Answer 1

Not all adjacent representable numbers are the same mathematical distance from one another. Floating point arcana isn't my strong suit, but if you want to find the next representable number, I think you need to keep increasing what you add/subtract from it by Number.EPSILON for as long as you keep getting the same number.

The very naive, simplistic approach would look like this (but keep reading):

 // DON'T USE THIS function next(x) { const ep = x < 0 ? -Number.EPSILON : Number.EPSILON; let adder = ep; let result; do { result = x + adder; adder += ep; } while (result === x); return result; } console.log(`Next for -3: ${next(-3)}`); console.log(`Next for 5: ${next(5)}`);

(That's assuming direction based on the sign of the number given, which is probably not what you really want, but is easily switched up.)

But , that would take hours (at least) to handle next(Number.MAX_SAFE_INTEGER) .

When I posted my caveat on the above originally, I said a better approach would take the magnitude of x into account "...or do bit twiddling (which definitely takes us into floating point arcana land)..." and you pointed to Java's Math.nextAfter operation, so I had to find out what they do. And indeed, it's bit twiddling, and it's wonderfully simple. Here's a re-implementation of the OpenJDK's version from here (the line number in that link will rot):

 // A JavaScript implementation of OpenJDK's `Double.nextAfter` method. function nextAfter(start, direction) { // These arrays share their underlying memory, letting us use them to do what // Java's `Double.doubleToRawLongBits` and `Double.longBitsToDouble` do. const f64 = new Float64Array(1); const b64 = new BigInt64Array(f64.buffer); // Comments from https://github.com/openjdk/jdk/blob/master/src/java.base/share/classes/java/lang/Math.java: /* * The cases: * * nextAfter(+infinity, 0) == MAX_VALUE * nextAfter(+infinity, +infinity) == +infinity * nextAfter(-infinity, 0) == -MAX_VALUE * nextAfter(-infinity, -infinity) == -infinity * * are naturally handled without any additional testing */ /* * IEEE 754 floating-point numbers are lexicographically * ordered if treated as signed-magnitude integers. * Since Java's integers are two's complement, * incrementing the two's complement representation of a * logically negative floating-point value *decrements* * the signed-magnitude representation. Therefore, when * the integer representation of a floating-point value * is negative, the adjustment to the representation is in * the opposite direction from what would initially be expected. */ // Branch to descending case first as it is more costly than ascending // case due to start != 0.0d conditional. if (start > direction) { // descending if (start !== 0) { f64[0] = start; const transducer = b64[0]; b64[0] = transducer + (transducer > 0n ? -1n : 1n); return f64[0]; } else { // start == 0.0d && direction < 0.0d return -Number.MIN_VALUE; } } else if (start < direction) { // ascending // Add +0.0 to get rid of a -0.0 (+0.0 + -0.0 => +0.0) // then bitwise convert start to integer. f64[0] = start + 0; const transducer = b64[0]; b64[0] = transducer + (transducer >= 0n ? 1n : -1n); return f64[0]; } else if (start == direction) { return direction; } else { // isNaN(start) || isNaN(direction) return start + direction; } } function test(start, direction) { const result = nextAfter(start, direction); console.log(`${start} ${direction > 0 ? "up" : "down"} is ${result}`); } test(-3, -Infinity); test(5, Infinity); test(Number.MAX_SAFE_INTEGER, Infinity); test(Number.MAX_SAFE_INTEGER + 2, Infinity);