      1
      2
      3
      4
      5
      6
      7
      8
      9
     10
     11
     12
     13
     14
     15
     16
     17
     18
     19
     20
     21
     22
     23
     24
     25
     26
     27
     28
     29
     30
     31
     32
     33
     34
     35
     36
     37
     38
     39
     40
     41
     42
     43
     44
     45
     46
     47
     48
     49
     50
     51
     52
     53
     54
     55
     56
     57
     58
     59
     60
     61
     62
     63
     64
     65
     66
     67
     68
     69
     70
     71
     72
     73
     74
     75
     76
     77
     78
     79
     80
     81
     82
     83
     84
     85
     86
     87
     88
     89
     90
     91
     92
     93
     94
     95
     96
     97
     98
     99
    100
    101
    102
    103
    104
    105
    106
    107
    108
    109
    110
    111
    112
    113
    114
    115
    116
    117
    118
    119
    120
    121
    122
    123
    124
    125
    126
    127
    128
    129
    130
    131
    132
    133
    134
    135
    136
    137
    138
    139
    140
    141
    142
    143
    144
    145
    146
    147
    148
    149
    150
    151
    152
    153
    154
    155
    156
    157
    158
    159
    160
    161
    162
    163
    164
    165
    166
    167
    168
    169
    170
    171
    172
    173
    174
    175
    176
    177
    178
    179
    180
    181
    182
    183
    184
    185
    186
    187
    188
    189
    190
    191
    192
    193
    194
    195
    196
    197
    198
    199
    200
    201
    202
    203
    204
    205
    206
    207
    208
    209
    210
    211
    212
    213
    214
    215
    216
    217
    218
    219
    220
    221
    222
    223
    224
    225
    226
    227
    228
    229
    230
    231
    232
    233
    234
    235
    236
    237
    238
    239
    240
    241
    242
    243
    244
    245
    246
    247
    248
    249
    250
    251
    252
    253
    254
    255
    256
    257
    258
    259
    260
    261
    262
    263
    264
    265
    266
    267
    268
    269
    270
    271
    272
    273
    274
    275
    276
    277
    278
    279
    280
    281
    282
    283
    284
    285
    286
    287
    288
    289
    290
    291
    292
    293
    294
    295
    296
    297
    298
    299
    300
    301
    302
    303
    304
    305
    306
    307
    308
    309
    310
    311
    312
    313
    314
    315
    316
    317
    318
    319
    320
    321
    322
    323
    324
    325
    326
    327
    328
    329
    330
    331
    332
    333
    334
    335
    336
    337
    338
    339
    340
    341
    342
    343
    344
    345
    346
    347
    348
    349
    350
    351
    352
    353
    354
    355
    356
    357
    358
    359
    360
    361
    362
    363
    364
    365
    366
    367
    368
    369
    370
    371
    372
    373
    374
    375
    376
    377
    378
    379
    380
    381
    382
    383
    384
    385
    386
    387
    388
    389
    390
    391
    392
    393
    394
    395
    396
    397
    398
    399
    400
    401
    402
    403
    404
    405
    406
    407
    408
    409
    410
    411
    412
    413
    414
    415
    416
    417
    418
    419
    420
    421
    422
    423
    424
    425
    426
    427
    428
    429
    430
    431
    432
    433
    434
    435
    436
    437
    438
    439
    440
    441
    442
    443
    444
    445
    446
    447
    448
    449
    450
    451
    452
    453
    454
    455
    456
    457
    458
    459
    460
    461
    462
    463
    464
    465
    466
    467
    468
    469
    470
    471
    472
    473
    474
    475
    476
    477
    478
    479
    480
    481
    482
    483
    484
    485
    486
    487
    488
    489
    490
    491
    492
    493
    494
    495
    496
    497
    498
    499
    500
    501
    502
    503
    504
    505
    506
    507
    508
    509
    510
    511
    512
    513
    514
    515
    516
    517
    518
    519
    520
    521
    522
    523
    524
    525
    526
    527
    528
    529
    530
    531
    532
    533
    534
    535
    536
    537
    538
    539
    540
    541
    542
    543
    544
    545
    546
    547
    548
    549
    550
    551
    552
    553
    554
    555
    556
    557
    558
    559
    560
    561
    562
    563
    564
    565
    566
    567
    568
    569
    570
    571
    572
    573
    574
    575
    576
    577
    578
    579
    580
    581
    582
    583
    584
    585
    586
    587
    588
    589
    590
    591
    592
    593
    594
    595
    596
    597
    598
    599
    600
    601
    602
    603
    604
    605
    606
    607
    608
    609
    610
    611
    612
    613
    614
    615
    616
    617
    618
    619
    620
    621
    622
    623
    624
    625
    626
    627
    628
    629
    630
    631
    632
    633
    634
    635
    636
    637
    638
    639
    640
    641
    642
    643
    644
    645
    646
    647
    648
    649
    650
    651
    652
    653
    654
    655
    656
    657
    658
    659
    660
    661
    662
    663
    664
    665
    666
    667
    668
    669
    670
    671
    672
    673
    674
    675
    676
    677
    678
    679
    680
    681
    682
    683

    //! Rust adaptation of the Grisu3 algorithm described in "Printing Floating-Point Numbers Quickly
    //! and Accurately with Integers"[^1]. It uses about 1KB of precomputed table, and in turn, it's
    //! very quick for most inputs.
    //!
    //! [^1]: Florian Loitsch. 2010. Printing floating-point numbers quickly and
    //!   accurately with integers. SIGPLAN Not. 45, 6 (June 2010), 233-243.

    use num::diy_float::Fp;
    use num::flt2dec::{Decoded, MAX_SIG_DIGITS, round_up};

    // see the comments in `format_shortest_opt` for the rationale.
    #[doc(hidden)] pub const ALPHA: i16 = -60;
    #[doc(hidden)] pub const GAMMA: i16 = -32;

    /*
    # the following Python code generates this table:
    for i in xrange(-308, 333, 8):
        if i >= 0: f = 10**i; e = 0
        else: f = 2**(80-4*i) // 10**-i; e = 4 * i - 80
        l = f.bit_length()
        f = ((f << 64 >> (l-1)) + 1) >> 1; e += l - 64
        print '    (%#018x, %5d, %4d),' % (f, e, i)
    */

    #[doc(hidden)]
    pub static CACHED_POW10: [(u64, i16, i16); 81] = [ // (f, e, k)
        (0xe61acf033d1a45df, -1087, -308),
        (0xab70fe17c79ac6ca, -1060, -300),
        (0xff77b1fcbebcdc4f, -1034, -292),
        (0xbe5691ef416bd60c, -1007, -284),
        (0x8dd01fad907ffc3c,  -980, -276),
        (0xd3515c2831559a83,  -954, -268),
        (0x9d71ac8fada6c9b5,  -927, -260),
        (0xea9c227723ee8bcb,  -901, -252),
        (0xaecc49914078536d,  -874, -244),
        (0x823c12795db6ce57,  -847, -236),
        (0xc21094364dfb5637,  -821, -228),
        (0x9096ea6f3848984f,  -794, -220),
        (0xd77485cb25823ac7,  -768, -212),
        (0xa086cfcd97bf97f4,  -741, -204),
        (0xef340a98172aace5,  -715, -196),
        (0xb23867fb2a35b28e,  -688, -188),
        (0x84c8d4dfd2c63f3b,  -661, -180),
        (0xc5dd44271ad3cdba,  -635, -172),
        (0x936b9fcebb25c996,  -608, -164),
        (0xdbac6c247d62a584,  -582, -156),
        (0xa3ab66580d5fdaf6,  -555, -148),
        (0xf3e2f893dec3f126,  -529, -140),
        (0xb5b5ada8aaff80b8,  -502, -132),
        (0x87625f056c7c4a8b,  -475, -124),
        (0xc9bcff6034c13053,  -449, -116),
        (0x964e858c91ba2655,  -422, -108),
        (0xdff9772470297ebd,  -396, -100),
        (0xa6dfbd9fb8e5b88f,  -369,  -92),
        (0xf8a95fcf88747d94,  -343,  -84),
        (0xb94470938fa89bcf,  -316,  -76),
        (0x8a08f0f8bf0f156b,  -289,  -68),
        (0xcdb02555653131b6,  -263,  -60),
        (0x993fe2c6d07b7fac,  -236,  -52),
        (0xe45c10c42a2b3b06,  -210,  -44),
        (0xaa242499697392d3,  -183,  -36),
        (0xfd87b5f28300ca0e,  -157,  -28),
        (0xbce5086492111aeb,  -130,  -20),
        (0x8cbccc096f5088cc,  -103,  -12),
        (0xd1b71758e219652c,   -77,   -4),
        (0x9c40000000000000,   -50,    4),
        (0xe8d4a51000000000,   -24,   12),
        (0xad78ebc5ac620000,     3,   20),
        (0x813f3978f8940984,    30,   28),
        (0xc097ce7bc90715b3,    56,   36),
        (0x8f7e32ce7bea5c70,    83,   44),
        (0xd5d238a4abe98068,   109,   52),
        (0x9f4f2726179a2245,   136,   60),
        (0xed63a231d4c4fb27,   162,   68),
        (0xb0de65388cc8ada8,   189,   76),
        (0x83c7088e1aab65db,   216,   84),
        (0xc45d1df942711d9a,   242,   92),
        (0x924d692ca61be758,   269,  100),
        (0xda01ee641a708dea,   295,  108),
        (0xa26da3999aef774a,   322,  116),
        (0xf209787bb47d6b85,   348,  124),
        (0xb454e4a179dd1877,   375,  132),
        (0x865b86925b9bc5c2,   402,  140),
        (0xc83553c5c8965d3d,   428,  148),
        (0x952ab45cfa97a0b3,   455,  156),
        (0xde469fbd99a05fe3,   481,  164),
        (0xa59bc234db398c25,   508,  172),
        (0xf6c69a72a3989f5c,   534,  180),
        (0xb7dcbf5354e9bece,   561,  188),
        (0x88fcf317f22241e2,   588,  196),
        (0xcc20ce9bd35c78a5,   614,  204),
        (0x98165af37b2153df,   641,  212),
        (0xe2a0b5dc971f303a,   667,  220),
        (0xa8d9d1535ce3b396,   694,  228),
        (0xfb9b7cd9a4a7443c,   720,  236),
        (0xbb764c4ca7a44410,   747,  244),
        (0x8bab8eefb6409c1a,   774,  252),
        (0xd01fef10a657842c,   800,  260),
        (0x9b10a4e5e9913129,   827,  268),
        (0xe7109bfba19c0c9d,   853,  276),
        (0xac2820d9623bf429,   880,  284),
        (0x80444b5e7aa7cf85,   907,  292),
        (0xbf21e44003acdd2d,   933,  300),
        (0x8e679c2f5e44ff8f,   960,  308),
        (0xd433179d9c8cb841,   986,  316),
        (0x9e19db92b4e31ba9,  1013,  324),
        (0xeb96bf6ebadf77d9,  1039,  332),
    ];

    #[doc(hidden)] pub const CACHED_POW10_FIRST_E: i16 = -1087;
    #[doc(hidden)] pub const CACHED_POW10_LAST_E: i16 = 1039;

    #[doc(hidden)]
    pub fn cached_power(alpha: i16, gamma: i16) -> (i16, Fp) {
        let offset = CACHED_POW10_FIRST_E as i32;
        let range = (CACHED_POW10.len() as i32) - 1;
        let domain = (CACHED_POW10_LAST_E - CACHED_POW10_FIRST_E) as i32;
        let idx = ((gamma as i32) - offset) * range / domain;
        let (f, e, k) = CACHED_POW10[idx as usize];
        debug_assert!(alpha <= e && e <= gamma);
        (k, Fp { f, e })
    }

    /// Given `x > 0`, returns `(k, 10^k)` such that `10^k <= x < 10^(k+1)`.
    #[doc(hidden)]
    pub fn max_pow10_no_more_than(x: u32) -> (u8, u32) {
        debug_assert!(x > 0);

        const X9: u32 = 10_0000_0000;
        const X8: u32 =  1_0000_0000;
        const X7: u32 =    1000_0000;
        const X6: u32 =     100_0000;
        const X5: u32 =      10_0000;
        const X4: u32 =       1_0000;
        const X3: u32 =         1000;
        const X2: u32 =          100;
        const X1: u32 =           10;

        if x < X4 {
            if x < X2 { if x < X1 {(0,  1)} else {(1, X1)} }
            else      { if x < X3 {(2, X2)} else {(3, X3)} }
        } else {
            if x < X6      { if x < X5 {(4, X4)} else {(5, X5)} }
            else if x < X8 { if x < X7 {(6, X6)} else {(7, X7)} }
            else           { if x < X9 {(8, X8)} else {(9, X9)} }
        }
    }

    /// The shortest mode implementation for Grisu.
    ///
    /// It returns `None` when it would return an inexact representation otherwise.
    pub fn format_shortest_opt(d: &Decoded,
                               buf: &mut [u8]) -> Option<(/*#digits*/ usize, /*exp*/ i16)> {
        assert!(d.mant > 0);
        assert!(d.minus > 0);
        assert!(d.plus > 0);
        assert!(d.mant.checked_add(d.plus).is_some());
        assert!(d.mant.checked_sub(d.minus).is_some());
        assert!(buf.len() >= MAX_SIG_DIGITS);
        assert!(d.mant + d.plus < (1 << 61)); // we need at least three bits of additional precision

        // start with the normalized values with the shared exponent
        let plus = Fp { f: d.mant + d.plus, e: d.exp }.normalize();
        let minus = Fp { f: d.mant - d.minus, e: d.exp }.normalize_to(plus.e);
        let v = Fp { f: d.mant, e: d.exp }.normalize_to(plus.e);

        // find any `cached = 10^minusk` such that `ALPHA <= minusk + plus.e + 64 <= GAMMA`.
        // since `plus` is normalized, this means `2^(62 + ALPHA) <= plus * cached < 2^(64 + GAMMA)`;
        // given our choices of `ALPHA` and `GAMMA`, this puts `plus * cached` into `[4, 2^32)`.
        //
        // it is obviously desirable to maximize `GAMMA - ALPHA`,
        // so that we don't need many cached powers of 10, but there are some considerations:
        //
        // 1. we want to keep `floor(plus * cached)` within `u32` since it needs a costly division.
        //    (this is not really avoidable, remainder is required for accuracy estimation.)
        // 2. the remainder of `floor(plus * cached)` repeatedly gets multiplied by 10,
        //    and it should not overflow.
        //
        // the first gives `64 + GAMMA <= 32`, while the second gives `10 * 2^-ALPHA <= 2^64`;
        // -60 and -32 is the maximal range with this constraint, and V8 also uses them.
        let (minusk, cached) = cached_power(ALPHA - plus.e - 64, GAMMA - plus.e - 64);

        // scale fps. this gives the maximal error of 1 ulp (proved from Theorem 5.1).
        let plus = plus.mul(&cached);
        let minus = minus.mul(&cached);
        let v = v.mul(&cached);
        debug_assert_eq!(plus.e, minus.e);
        debug_assert_eq!(plus.e, v.e);

        //         +- actual range of minus
        //   | <---|---------------------- unsafe region --------------------------> |
        //   |     |                                                                 |
        //   |  |<--->|  | <--------------- safe region ---------------> |           |
        //   |  |     |  |                                               |           |
        //   |1 ulp|1 ulp|                 |1 ulp|1 ulp|                 |1 ulp|1 ulp|
        //   |<--->|<--->|                 |<--->|<--->|                 |<--->|<--->|
        //   |-----|-----|-------...-------|-----|-----|-------...-------|-----|-----|
        //   |   minus   |                 |     v     |                 |   plus    |
        // minus1     minus0           v - 1 ulp   v + 1 ulp           plus0       plus1
        //
        // above `minus`, `v` and `plus` are *quantized* approximations (error < 1 ulp).
        // as we don't know the error is positive or negative, we use two approximations spaced equally
        // and have the maximal error of 2 ulps.
        //
        // the "unsafe region" is a liberal interval which we initially generate.
        // the "safe region" is a conservative interval which we only accept.
        // we start with the correct repr within the unsafe region, and try to find the closest repr
        // to `v` which is also within the safe region. if we can't, we give up.
        let plus1 = plus.f + 1;
    //  let plus0 = plus.f - 1; // only for explanation
    //  let minus0 = minus.f + 1; // only for explanation
        let minus1 = minus.f - 1;
        let e = -plus.e as usize; // shared exponent

        // divide `plus1` into integral and fractional parts.
        // integral parts are guaranteed to fit in u32, since cached power guarantees `plus < 2^32`
        // and normalized `plus.f` is always less than `2^64 - 2^4` due to the precision requirement.
        let plus1int = (plus1 >> e) as u32;
        let plus1frac = plus1 & ((1 << e) - 1);

        // calculate the largest `10^max_kappa` no more than `plus1` (thus `plus1 < 10^(max_kappa+1)`).
        // this is an upper bound of `kappa` below.
        let (max_kappa, max_ten_kappa) = max_pow10_no_more_than(plus1int);

        let mut i = 0;
        let exp = max_kappa as i16 - minusk + 1;

        // Theorem 6.2: if `k` is the greatest integer s.t. `0 <= y mod 10^k <= y - x`,
        //              then `V = floor(y / 10^k) * 10^k` is in `[x, y]` and one of the shortest
        //              representations (with the minimal number of significant digits) in that range.
        //
        // find the digit length `kappa` between `(minus1, plus1)` as per Theorem 6.2.
        // Theorem 6.2 can be adopted to exclude `x` by requiring `y mod 10^k < y - x` instead.
        // (e.g., `x` = 32000, `y` = 32777; `kappa` = 2 since `y mod 10^3 = 777 < y - x = 777`.)
        // the algorithm relies on the later verification phase to exclude `y`.
        let delta1 = plus1 - minus1;
    //  let delta1int = (delta1 >> e) as usize; // only for explanation
        let delta1frac = delta1 & ((1 << e) - 1);

        // render integral parts, while checking for the accuracy at each step.
        let mut kappa = max_kappa as i16;
        let mut ten_kappa = max_ten_kappa; // 10^kappa
        let mut remainder = plus1int; // digits yet to be rendered
        loop { // we always have at least one digit to render, as `plus1 >= 10^kappa`
            // invariants:
            // - `delta1int <= remainder < 10^(kappa+1)`
            // - `plus1int = d[0..n-1] * 10^(kappa+1) + remainder`
            //   (it follows that `remainder = plus1int % 10^(kappa+1)`)

            // divide `remainder` by `10^kappa`. both are scaled by `2^-e`.
            let q = remainder / ten_kappa;
            let r = remainder % ten_kappa;
            debug_assert!(q < 10);
            buf[i] = b'0' + q as u8;
            i += 1;

            let plus1rem = ((r as u64) << e) + plus1frac; // == (plus1 % 10^kappa) * 2^e
            if plus1rem < delta1 {
                // `plus1 % 10^kappa < delta1 = plus1 - minus1`; we've found the correct `kappa`.
                let ten_kappa = (ten_kappa as u64) << e; // scale 10^kappa back to the shared exponent
                return round_and_weed(&mut buf[..i], exp, plus1rem, delta1, plus1 - v.f, ten_kappa, 1);
            }

            // break the loop when we have rendered all integral digits.
            // the exact number of digits is `max_kappa + 1` as `plus1 < 10^(max_kappa+1)`.
            if i > max_kappa as usize {
                debug_assert_eq!(ten_kappa, 1);
                debug_assert_eq!(kappa, 0);
                break;
            }

            // restore invariants
            kappa -= 1;
            ten_kappa /= 10;
            remainder = r;
        }

        // render fractional parts, while checking for the accuracy at each step.
        // this time we rely on repeated multiplications, as division will lose the precision.
        let mut remainder = plus1frac;
        let mut threshold = delta1frac;
        let mut ulp = 1;
        loop { // the next digit should be significant as we've tested that before breaking out
            // invariants, where `m = max_kappa + 1` (# of digits in the integral part):
            // - `remainder < 2^e`
            // - `plus1frac * 10^(n-m) = d[m..n-1] * 2^e + remainder`

            remainder *= 10; // won't overflow, `2^e * 10 < 2^64`
            threshold *= 10;
            ulp *= 10;

            // divide `remainder` by `10^kappa`.
            // both are scaled by `2^e / 10^kappa`, so the latter is implicit here.
            let q = remainder >> e;
            let r = remainder & ((1 << e) - 1);
            debug_assert!(q < 10);
            buf[i] = b'0' + q as u8;
            i += 1;

            if r < threshold {
                let ten_kappa = 1 << e; // implicit divisor
                return round_and_weed(&mut buf[..i], exp, r, threshold,
                                      (plus1 - v.f) * ulp, ten_kappa, ulp);
            }

            // restore invariants
            kappa -= 1;
            remainder = r;
        }

        // we've generated all significant digits of `plus1`, but not sure if it's the optimal one.
        // for example, if `minus1` is 3.14153... and `plus1` is 3.14158..., there are 5 different
        // shortest representation from 3.14154 to 3.14158 but we only have the greatest one.
        // we have to successively decrease the last digit and check if this is the optimal repr.
        // there are at most 9 candidates (..1 to ..9), so this is fairly quick. ("rounding" phase)
        //
        // the function checks if this "optimal" repr is actually within the ulp ranges,
        // and also, it is possible that the "second-to-optimal" repr can actually be optimal
        // due to the rounding error. in either cases this returns `None`. ("weeding" phase)
        //
        // all arguments here are scaled by the common (but implicit) value `k`, so that:
        // - `remainder = (plus1 % 10^kappa) * k`
        // - `threshold = (plus1 - minus1) * k` (and also, `remainder < threshold`)
        // - `plus1v = (plus1 - v) * k` (and also, `threshold > plus1v` from prior invariants)
        // - `ten_kappa = 10^kappa * k`
        // - `ulp = 2^-e * k`
        fn round_and_weed(buf: &mut [u8], exp: i16, remainder: u64, threshold: u64, plus1v: u64,
                          ten_kappa: u64, ulp: u64) -> Option<(usize, i16)> {
            assert!(!buf.is_empty());

            // produce two approximations to `v` (actually `plus1 - v`) within 1.5 ulps.
            // the resulting representation should be the closest representation to both.
            //
            // here `plus1 - v` is used since calculations are done with respect to `plus1`
            // in order to avoid overflow/underflow (hence the seemingly swapped names).
            let plus1v_down = plus1v + ulp; // plus1 - (v - 1 ulp)
            let plus1v_up = plus1v - ulp; // plus1 - (v + 1 ulp)

            // decrease the last digit and stop at the closest representation to `v + 1 ulp`.
            let mut plus1w = remainder; // plus1w(n) = plus1 - w(n)
            {
                let last = buf.last_mut().unwrap();

                // we work with the approximated digits `w(n)`, which is initially equal to `plus1 -
                // plus1 % 10^kappa`. after running the loop body `n` times, `w(n) = plus1 -
                // plus1 % 10^kappa - n * 10^kappa`. we set `plus1w(n) = plus1 - w(n) =
                // plus1 % 10^kappa + n * 10^kappa` (thus `remainder = plus1w(0)`) to simplify checks.
                // note that `plus1w(n)` is always increasing.
                //
                // we have three conditions to terminate. any of them will make the loop unable to
                // proceed, but we then have at least one valid representation known to be closest to
                // `v + 1 ulp` anyway. we will denote them as TC1 through TC3 for brevity.
                //
                // TC1: `w(n) <= v + 1 ulp`, i.e., this is the last repr that can be the closest one.
                // this is equivalent to `plus1 - w(n) = plus1w(n) >= plus1 - (v + 1 ulp) = plus1v_up`.
                // combined with TC2 (which checks if `w(n+1)` is valid), this prevents the possible
                // overflow on the calculation of `plus1w(n)`.
                //
                // TC2: `w(n+1) < minus1`, i.e., the next repr definitely does not round to `v`.
                // this is equivalent to `plus1 - w(n) + 10^kappa = plus1w(n) + 10^kappa >
                // plus1 - minus1 = threshold`. the left hand side can overflow, but we know
                // `threshold > plus1v`, so if TC1 is false, `threshold - plus1w(n) >
                // threshold - (plus1v - 1 ulp) > 1 ulp` and we can safely test if
                // `threshold - plus1w(n) < 10^kappa` instead.
                //
                // TC3: `abs(w(n) - (v + 1 ulp)) <= abs(w(n+1) - (v + 1 ulp))`, i.e., the next repr is
                // no closer to `v + 1 ulp` than the current repr. given `z(n) = plus1v_up - plus1w(n)`,
                // this becomes `abs(z(n)) <= abs(z(n+1))`. again assuming that TC1 is false, we have
                // `z(n) > 0`. we have two cases to consider:
                //
                // - when `z(n+1) >= 0`: TC3 becomes `z(n) <= z(n+1)`. as `plus1w(n)` is increasing,
                //   `z(n)` should be decreasing and this is clearly false.
                // - when `z(n+1) < 0`:
                //   - TC3a: the precondition is `plus1v_up < plus1w(n) + 10^kappa`. assuming TC2 is
                //     false, `threshold >= plus1w(n) + 10^kappa` so it cannot overflow.
                //   - TC3b: TC3 becomes `z(n) <= -z(n+1)`, i.e., `plus1v_up - plus1w(n) >=
                //     plus1w(n+1) - plus1v_up = plus1w(n) + 10^kappa - plus1v_up`. the negated TC1
                //     gives `plus1v_up > plus1w(n)`, so it cannot overflow or underflow when
                //     combined with TC3a.
                //
                // consequently, we should stop when `TC1 || TC2 || (TC3a && TC3b)`. the following is
                // equal to its inverse, `!TC1 && !TC2 && (!TC3a || !TC3b)`.
                while plus1w < plus1v_up &&
                      threshold - plus1w >= ten_kappa &&
                      (plus1w + ten_kappa < plus1v_up ||
                       plus1v_up - plus1w >= plus1w + ten_kappa - plus1v_up) {
                    *last -= 1;
                    debug_assert!(*last > b'0'); // the shortest repr cannot end with `0`
                    plus1w += ten_kappa;
                }
            }

            // check if this representation is also the closest representation to `v - 1 ulp`.
            //
            // this is simply same to the terminating conditions for `v + 1 ulp`, with all `plus1v_up`
            // replaced by `plus1v_down` instead. overflow analysis equally holds.
            if plus1w < plus1v_down &&
               threshold - plus1w >= ten_kappa &&
               (plus1w + ten_kappa < plus1v_down ||
                plus1v_down - plus1w >= plus1w + ten_kappa - plus1v_down) {
                return None;
            }

            // now we have the closest representation to `v` between `plus1` and `minus1`.
            // this is too liberal, though, so we reject any `w(n)` not between `plus0` and `minus0`,
            // i.e., `plus1 - plus1w(n) <= minus0` or `plus1 - plus1w(n) >= plus0`. we utilize the facts
            // that `threshold = plus1 - minus1` and `plus1 - plus0 = minus0 - minus1 = 2 ulp`.
            if 2 * ulp <= plus1w && plus1w <= threshold - 4 * ulp {
                Some((buf.len(), exp))
            } else {
                None
            }
        }
    }

    /// The shortest mode implementation for Grisu with Dragon fallback.
    ///
    /// This should be used for most cases.
    pub fn format_shortest(d: &Decoded, buf: &mut [u8]) -> (/*#digits*/ usize, /*exp*/ i16) {
        use num::flt2dec::strategy::dragon::format_shortest as fallback;
        match format_shortest_opt(d, buf) {
            Some(ret) => ret,
            None => fallback(d, buf),
        }
    }

    /// The exact and fixed mode implementation for Grisu.
    ///
    /// It returns `None` when it would return an inexact representation otherwise.
    pub fn format_exact_opt(d: &Decoded, buf: &mut [u8], limit: i16)
                                    -> Option<(/*#digits*/ usize, /*exp*/ i16)> {
        assert!(d.mant > 0);
        assert!(d.mant < (1 << 61)); // we need at least three bits of additional precision
        assert!(!buf.is_empty());

        // normalize and scale `v`.
        let v = Fp { f: d.mant, e: d.exp }.normalize();
        let (minusk, cached) = cached_power(ALPHA - v.e - 64, GAMMA - v.e - 64);
        let v = v.mul(&cached);

        // divide `v` into integral and fractional parts.
        let e = -v.e as usize;
        let vint = (v.f >> e) as u32;
        let vfrac = v.f & ((1 << e) - 1);

        // both old `v` and new `v` (scaled by `10^-k`) has an error of < 1 ulp (Theorem 5.1).
        // as we don't know the error is positive or negative, we use two approximations
        // spaced equally and have the maximal error of 2 ulps (same to the shortest case).
        //
        // the goal is to find the exactly rounded series of digits that are common to
        // both `v - 1 ulp` and `v + 1 ulp`, so that we are maximally confident.
        // if this is not possible, we don't know which one is the correct output for `v`,
        // so we give up and fall back.
        //
        // `err` is defined as `1 ulp * 2^e` here (same to the ulp in `vfrac`),
        // and we will scale it whenever `v` gets scaled.
        let mut err = 1;

        // calculate the largest `10^max_kappa` no more than `v` (thus `v < 10^(max_kappa+1)`).
        // this is an upper bound of `kappa` below.
        let (max_kappa, max_ten_kappa) = max_pow10_no_more_than(vint);

        let mut i = 0;
        let exp = max_kappa as i16 - minusk + 1;

        // if we are working with the last-digit limitation, we need to shorten the buffer
        // before the actual rendering in order to avoid double rounding.
        // note that we have to enlarge the buffer again when rounding up happens!
        let len = if exp <= limit {
            // oops, we cannot even produce *one* digit.
            // this is possible when, say, we've got something like 9.5 and it's being rounded to 10.
            //
            // in principle we can immediately call `possibly_round` with an empty buffer,
            // but scaling `max_ten_kappa << e` by 10 can result in overflow.
            // thus we are being sloppy here and widen the error range by a factor of 10.
            // this will increase the false negative rate, but only very, *very* slightly;
            // it can only matter noticeably when the mantissa is bigger than 60 bits.
            return possibly_round(buf, 0, exp, limit, v.f / 10, (max_ten_kappa as u64) << e, err << e);
        } else if ((exp as i32 - limit as i32) as usize) < buf.len() {
            (exp - limit) as usize
        } else {
            buf.len()
        };
        debug_assert!(len > 0);

        // render integral parts.
        // the error is entirely fractional, so we don't need to check it in this part.
        let mut kappa = max_kappa as i16;
        let mut ten_kappa = max_ten_kappa; // 10^kappa
        let mut remainder = vint; // digits yet to be rendered
        loop { // we always have at least one digit to render
            // invariants:
            // - `remainder < 10^(kappa+1)`
            // - `vint = d[0..n-1] * 10^(kappa+1) + remainder`
            //   (it follows that `remainder = vint % 10^(kappa+1)`)

            // divide `remainder` by `10^kappa`. both are scaled by `2^-e`.
            let q = remainder / ten_kappa;
            let r = remainder % ten_kappa;
            debug_assert!(q < 10);
            buf[i] = b'0' + q as u8;
            i += 1;

            // is the buffer full? run the rounding pass with the remainder.
            if i == len {
                let vrem = ((r as u64) << e) + vfrac; // == (v % 10^kappa) * 2^e
                return possibly_round(buf, len, exp, limit, vrem, (ten_kappa as u64) << e, err << e);
            }

            // break the loop when we have rendered all integral digits.
            // the exact number of digits is `max_kappa + 1` as `plus1 < 10^(max_kappa+1)`.
            if i > max_kappa as usize {
                debug_assert_eq!(ten_kappa, 1);
                debug_assert_eq!(kappa, 0);
                break;
            }

            // restore invariants
            kappa -= 1;
            ten_kappa /= 10;
            remainder = r;
        }

        // render fractional parts.
        //
        // in principle we can continue to the last available digit and check for the accuracy.
        // unfortunately we are working with the finite-sized integers, so we need some criterion
        // to detect the overflow. V8 uses `remainder > err`, which becomes false when
        // the first `i` significant digits of `v - 1 ulp` and `v` differ. however this rejects
        // too many otherwise valid input.
        //
        // since the later phase has a correct overflow detection, we instead use tighter criterion:
        // we continue til `err` exceeds `10^kappa / 2`, so that the range between `v - 1 ulp` and
        // `v + 1 ulp` definitely contains two or more rounded representations. this is same to
        // the first two comparisons from `possibly_round`, for the reference.
        let mut remainder = vfrac;
        let maxerr = 1 << (e - 1);
        while err < maxerr {
            // invariants, where `m = max_kappa + 1` (# of digits in the integral part):
            // - `remainder < 2^e`
            // - `vfrac * 10^(n-m) = d[m..n-1] * 2^e + remainder`
            // - `err = 10^(n-m)`

            remainder *= 10; // won't overflow, `2^e * 10 < 2^64`
            err *= 10; // won't overflow, `err * 10 < 2^e * 5 < 2^64`

            // divide `remainder` by `10^kappa`.
            // both are scaled by `2^e / 10^kappa`, so the latter is implicit here.
            let q = remainder >> e;
            let r = remainder & ((1 << e) - 1);
            debug_assert!(q < 10);
            buf[i] = b'0' + q as u8;
            i += 1;

            // is the buffer full? run the rounding pass with the remainder.
            if i == len {
                return possibly_round(buf, len, exp, limit, r, 1 << e, err);
            }

            // restore invariants
            remainder = r;
        }

        // further calculation is useless (`possibly_round` definitely fails), so we give up.
        return None;

        // we've generated all requested digits of `v`, which should be also same to corresponding
        // digits of `v - 1 ulp`. now we check if there is a unique representation shared by
        // both `v - 1 ulp` and `v + 1 ulp`; this can be either same to generated digits, or
        // to the rounded-up version of those digits. if the range contains multiple representations
        // of the same length, we cannot be sure and should return `None` instead.
        //
        // all arguments here are scaled by the common (but implicit) value `k`, so that:
        // - `remainder = (v % 10^kappa) * k`
        // - `ten_kappa = 10^kappa * k`
        // - `ulp = 2^-e * k`
        fn possibly_round(buf: &mut [u8], mut len: usize, mut exp: i16, limit: i16,
                          remainder: u64, ten_kappa: u64, ulp: u64) -> Option<(usize, i16)> {
            debug_assert!(remainder < ten_kappa);

            //           10^kappa
            //    :   :   :<->:   :
            //    :   :   :   :   :
            //    :|1 ulp|1 ulp|  :
            //    :|<--->|<--->|  :
            // ----|-----|-----|----
            //     |     v     |
            // v - 1 ulp   v + 1 ulp
            //
            // (for the reference, the dotted line indicates the exact value for
            // possible representations in given number of digits.)
            //
            // error is too large that there are at least three possible representations
            // between `v - 1 ulp` and `v + 1 ulp`. we cannot determine which one is correct.
            if ulp >= ten_kappa { return None; }

            //    10^kappa
            //   :<------->:
            //   :         :
            //   : |1 ulp|1 ulp|
            //   : |<--->|<--->|
            // ----|-----|-----|----
            //     |     v     |
            // v - 1 ulp   v + 1 ulp
            //
            // in fact, 1/2 ulp is enough to introduce two possible representations.
            // (remember that we need a unique representation for both `v - 1 ulp` and `v + 1 ulp`.)
            // this won't overflow, as `ulp < ten_kappa` from the first check.
            if ten_kappa - ulp <= ulp { return None; }

            //     remainder
            //       :<->|                           :
            //       :   |                           :
            //       :<--------- 10^kappa ---------->:
            //     | :   |                           :
            //     |1 ulp|1 ulp|                     :
            //     |<--->|<--->|                     :
            // ----|-----|-----|------------------------
            //     |     v     |
            // v - 1 ulp   v + 1 ulp
            //
            // if `v + 1 ulp` is closer to the rounded-down representation (which is already in `buf`),
            // then we can safely return. note that `v - 1 ulp` *can* be less than the current
            // representation, but as `1 ulp < 10^kappa / 2`, this condition is enough:
            // the distance between `v - 1 ulp` and the current representation
            // cannot exceed `10^kappa / 2`.
            //
            // the condition equals to `remainder + ulp < 10^kappa / 2`.
            // since this can easily overflow, first check if `remainder < 10^kappa / 2`.
            // we've already verified that `ulp < 10^kappa / 2`, so as long as
            // `10^kappa` did not overflow after all, the second check is fine.
            if ten_kappa - remainder > remainder && ten_kappa - 2 * remainder >= 2 * ulp {
                return Some((len, exp));
            }

            //   :<------- remainder ------>|   :
            //   :                          |   :
            //   :<--------- 10^kappa --------->:
            //   :                    |     |   : |
            //   :                    |1 ulp|1 ulp|
            //   :                    |<--->|<--->|
            // -----------------------|-----|-----|-----
            //                        |     v     |
            //                    v - 1 ulp   v + 1 ulp
            //
            // on the other hands, if `v - 1 ulp` is closer to the rounded-up representation,
            // we should round up and return. for the same reason we don't need to check `v + 1 ulp`.
            //
            // the condition equals to `remainder - ulp >= 10^kappa / 2`.
            // again we first check if `remainder > ulp` (note that this is not `remainder >= ulp`,
            // as `10^kappa` is never zero). also note that `remainder - ulp <= 10^kappa`,
            // so the second check does not overflow.
            if remainder > ulp && ten_kappa - (remainder - ulp) <= remainder - ulp {
                if let Some(c) = round_up(buf, len) {
                    // only add an additional digit when we've been requested the fixed precision.
                    // we also need to check that, if the original buffer was empty,
                    // the additional digit can only be added when `exp == limit` (edge case).
                    exp += 1;
                    if exp > limit && len < buf.len() {
                        buf[len] = c;
                        len += 1;
                    }
                }
                return Some((len, exp));
            }

            // otherwise we are doomed (i.e., some values between `v - 1 ulp` and `v + 1 ulp` are
            // rounding down and others are rounding up) and give up.
            None
        }
    }

    /// The exact and fixed mode implementation for Grisu with Dragon fallback.
    ///
    /// This should be used for most cases.
    pub fn format_exact(d: &Decoded, buf: &mut [u8], limit: i16) -> (/*#digits*/ usize, /*exp*/ i16) {
        use num::flt2dec::strategy::dragon::format_exact as fallback;
        match format_exact_opt(d, buf, limit) {
            Some(ret) => ret,
            None => fallback(d, buf, limit),
        }
    }

# Help

## Keyboard Shortcuts

`?`
  Show this help dialog
`S`
  Focus the search field
`↑`
  Move up in search results
`↓`
  Move down in search results
`↹`
  Switch tab
`⏎`
  Go to active search result
`+`
  Expand all sections
`-`
  Collapse all sections

## Search Tricks

Prefix searches with a type followed by a colon (e.g., `fn:`) to restrict the search to a given type.

Accepted types are: `fn`, `mod`, `struct`, `enum`, `trait`, `type`, `macro`, and `const`.

Search functions by type signature (e.g., `vec -> usize` or `* -> vec`)

Search multiple things at once by splitting your query with comma (e.g., `str,u8` or `String,struct:Vec,test`)