Float constants never compare equal in the JS simplifier: x == 2.5 && x == 2.5 compiles to false #9099

Found by a bug hunt over the 16 commits from eb087382 to cd641cbc.

let a = (x: float) => x == 2.5 && x == 2.5
let b = (x: float) => x != 2.5 && x == 2.5

function a(x) {
  return false;
}

function b(x) {
  return x === 2.5;
}

a(2.5) returns false (expected true), and b(2.5) returns true (expected false).

An int constant and a float constant with the same value are also treated as different. In an untagged variant this sends a literal case to the wrong arm:

@unboxed type t = @as(1) L1 | N(float) | S(string)
let f = (v: t) => switch v { | N(1.0) => 4 | _ => 6 }

The compiler warns that 1 is always matched as L1, never as N, yet emits:

function f(v) {
  if (v !== 1.0) {
    return 6;
  } else {
    return 4;
  }
}

so f(L1) returns 4 at run time, where the warning says it should return 6. @as(2.5) L | N(float) with N(2.5) behaves the same way. All of these reproduce at b123b7c3 as well.

Cause

Js_analyzer.eq_expression returns false for any two Number (Float _), and compares Number (Int _) only with another Int. The rules in simplify_and_ take "not eq_expression" to mean "different constants". So x === c1 && x === c2 becomes false, and x !== c1 && x === c2 becomes x === c2. Under JS ===, equal float literals are equal, and so are 1 and 1.0.

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