// The nearest representable values to +1.0. const ONE_PLUS_EPSILON = 1 + Math.pow(2, -52); // 0.9999999999999999 const ONE_MINUS_EPSILON = 1 - Math.pow(2, -53); // 1.0000000000000002 { var fail = function (msg) { var exc = new Error(msg); try { // Try to improve on exc.fileName and .lineNumber; leave exc.stack // alone. We skip two frames: fail() and its caller, an assertX() // function. var frames = exc.stack.trim().split("\n"); if (frames.length > 2) { var m = /@([^@:]*):([0-9]+)$/.exec(frames[2]); if (m) { exc.fileName = m[1]; exc.lineNumber = +m[2]; } } } catch (ignore) { throw ignore;} throw exc; }; var ENDIAN; // 0 for little-endian, 1 for big-endian. // Return the difference between the IEEE 754 bit-patterns for a and b. // // This is meaningful when a and b are both finite and have the same // sign. Then the following hold: // // * If a === b, then diff(a, b) === 0. // // * If a !== b, then diff(a, b) === 1 + the number of representable values // between a and b. // var f = new Float64Array([0, 0]); var u = new Uint32Array(f.buffer); var diff = function (a, b) { f[0] = a; f[1] = b; //print(u[1].toString(16) + u[0].toString(16) + " " + u[3].toString(16) + u[2].toString(16)); return Math.abs((u[3-ENDIAN] - u[1-ENDIAN]) * 0x100000000 + u[2+ENDIAN] - u[0+ENDIAN]); }; // Set ENDIAN to the platform's endianness. ENDIAN = 0; // try little-endian first if (diff(2, 4) === 0x100000) // exact wrong answer we'll get on a big-endian platform ENDIAN = 1; assertEq(diff(2,4), 0x10000000000000); assertEq(diff(0, Number.MIN_VALUE), 1); assertEq(diff(1, ONE_PLUS_EPSILON), 1); assertEq(diff(1, ONE_MINUS_EPSILON), 1); var assertNear = function assertNear(a, b, tolerance=1) { if (!Number.isFinite(b)) { fail("second argument to assertNear (expected value) must be a finite number"); } else if (Number.isNaN(a)) { fail("got NaN, expected a number near " + b); } else if (!Number.isFinite(a)) { if (b * Math.sign(a) < Number.MAX_VALUE) fail("got " + a + ", expected a number near " + b); } else { // When the two arguments do not have the same sign bit, diff() // returns some huge number. So if b is positive or negative 0, // make target the zero that has the same sign bit as a. var target = b === 0 ? a * 0 : b; var err = diff(a, target); if (err > tolerance) { fail("got " + a + ", expected a number near " + b + " (relative error: " + err + ")"); } } }; }