# 2018q1 第 3 週測驗題

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### 測驗 `1`

延續 [從 $\sqrt{2}$ 的運算談浮點數](https://hackmd.io/s/ryOLwrVtz)，假設 double 為 32-bit 寬度，考慮以下符合 IEEE 754 規範的平方根程式，請補上對應數值，使其得以運作：

#include <stdint.h>

/* A union allowing us to convert between a double and two 32-bit integers.
 * Little-endian representation
 */
typedef union {
    double value;
    struct {
        uint32_t lsw;
        uint32_t msw;
    } parts;
} ieee_double_shape_type;

/* Set a double from two 32 bit ints.  */
#define INSERT_WORDS(d, ix0, ix1)       \
    do {                                \
        ieee_double_shape_type iw_u = { \
            .parts.msw = ix0,           \
            .parts.lsw = ix1,           \
        };                              \
        (d) = iw_u.value;               \
    } while (0)

/* Get two 32 bit ints from a double.  */
#define EXTRACT_WORDS(ix0, ix1, d)   \
    do {                             \
        ieee_double_shape_type ew_u; \
        ew_u.value = (d);            \
        (ix0) = ew_u.parts.msw;      \
        (ix1) = ew_u.parts.lsw;      \
    } while (0)

static const double one = 1.0, tiny = 1.0e-300;

double ieee754_sqrt(double x)
{
    double z;
    int32_t sign = 0x80000000;
    uint32_t r, t1, s1, ix1, q1;
    int32_t ix0, s0, q, m, t, i;

    EXTRACT_WORDS(ix0, ix1, x);

    /* take care of INF and NaN */
    if ((ix0 & KK1) == KK2) {
        /* sqrt(NaN) = NaN, sqrt(+INF) = +INF, sqrt(-INF) = sNaN */
        return x * x + x;
    }
    /* take care of zero */
    if (ix0 <= 0) {
        if (((ix0 & (~sign)) | ix1) == 0)
            return x; /* sqrt(+-0) = +-0 */
        if (ix0 < 0)
            return (x - x) / (x - x); /* sqrt(-ve) = sNaN */
    }
    /* normalize x */
    m = (ix0 >> 20);
    if (m == 0) { /* subnormal x */
        while (ix0 == 0) {
            m -= 21;
            ix0 |= (ix1 >> 11);
            ix1 <<= 21;
        }
        for (i = 0; (ix0 & 0x00100000) == 0; i++)
            ix0 <<= 1;
        m -= i - 1;
        ix0 |= (ix1 >> (32 - i));
        ix1 <<= i;
    }
    m -= KK3; /* unbias exponent */
    ix0 = (ix0 & 0x000fffff) | 0x00100000;
    if (m & 1) { /* odd m, double x to make it even */
        ix0 += ix0 + ((ix1 & sign) >> 31);
        ix1 += ix1;
    }
    m >>= 1; /* m = [m/2] */

    /* generate sqrt(x) bit by bit */
    ix0 += ix0 + ((ix1 & sign) >> 31);
    ix1 += ix1;
    q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */
    r = 0x00200000;       /* r = moving bit from right to left */

    while (r != 0) {
        t = s0 + r;
        if (t <= ix0) {
            s0 = t + r;
            ix0 -= t;
            q += r;
        }
        ix0 += ix0 + ((ix1 & sign) >> 31);
        ix1 += ix1;
        r >>= 1;
    }

    r = sign;
    while (r != 0) {
        t1 = s1 + r;
        t = s0;
        if ((t < ix0) || ((t == ix0) && (t1 <= ix1))) {
            s1 = t1 + r;
            if (((t1 & sign) == sign) && (s1 & sign) == 0)
                s0 += 1;
            ix0 -= t;
            if (ix1 < t1)
                ix0 -= 1;
            ix1 -= t1;
            q1 += r;
        }
        ix0 += ix0 + ((ix1 & sign) >> 31);
        ix1 += ix1;
        r >>= 1;
    }

    /* use floating add to find out rounding direction */
    if ((ix0 | ix1) != 0) {
        z = one - tiny; /* trigger inexact flag */
        if (z >= one) {
            z = one + tiny;
            if (q1 == (uint32_t) 0xffffffff) {
                q1 = 0;
                q += 1;
            } else if (z > one) {
                if (q1 == (uint32_t) KK4)
                    q += 1;
                q1 += 2;
            } else
                q1 += (q1 & 1);
        }
    }
    ix0 = (q >> 1) + 0x3fe00000;
    ix1 = q1 >> 1;
    if ((q & 1) == 1)
        ix1 |= sign;
    ix0 += (m << KK5);
    INSERT_WORDS(z, ix0, ix1);
    return z;
}

==作答區==

KK1 = ?
* `(a)` 0x1ff00000
* `(b)` 0x2ff00000
* `(c)` 0x3ff00000
* `(d)` 0x4ff00000
* `(e)` 0x5ff00000
* `(f)` 0x6ff00000
* `(g)` 0x7ff00000
* `(h)` 0x8ff00000
* `(i)` 0x9ff00000
* `(j)` 0xaff00000

KK2 = ?
* `(a)` 0x1ff00000
* `(b)` 0x2ff00000
* `(c)` 0x3ff00000
* `(d)` 0x4ff00000
* `(e)` 0x5ff00000
* `(f)` 0x6ff00000
* `(g)` 0x7ff00000
* `(h)` 0x8ff00000
* `(i)` 0x9ff00000
* `(j)` 0xaff00000

KK3 = ?
* `(a)` 31
* `(b)` 63
* `(c)` 127
* `(d)` 255
* `(e)` 511
* `(f)` 1023
* `(g)` 2047

KK4 = ?
* `(a)` 0xffffffff
* `(b)` 0xfffffffe
* `(c)` 0xfffffffd
* `(d)` 0xfffffffc
* `(e)` 0xfffffffb
* `(f)` 0xfffffffa
* `(g)` 0xfffffff9
* `(h)` 0xfffffff8

KK5 = ?
* `(a)` 40
* `(b)` 30
* `(c)` 20
* `(d)` 10
* `(e)` 0
* `(f)` -10
* `(g)` -20

Reference:
* [Floating point division and square root algorithms and implementation in the AMD-K7/sup TM/ microprocessor](http://ieeexplore.ieee.org/document/762835/)
* [√Square Roots](http://www.azillionmonkeys.com/qed/sqroot.html)

:::info
延伸題目: 解釋上述程式碼何以運作，並且改為 float (單精度) 版本，注意應該用更短的程式碼來實作
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### 測驗 `2`

考慮到下方函式 `shift_right_arith` 和 `shift_right_logical` 分別表示算術右移和邏輯右移，請嘗試補完程式碼。可由 `sizeof(int) * 8` 得知整數型態 `int` 的位元數 `w`，而位移量 `k` 的有效範圍在 0 到 `w - 1`。

#include <stdio.h>
int shift_right_arith(int x, int k) {
    int xsrl = (unsigned) x >> k;
    int w = sizeof(int) << P1;
    int mask = (int) -1 << (P2);
    if (x < 0)
        return xsrl P3 mask;
    return xsrl;
}

unsigned shift_right_logical(unsigned x, int k) {
    unsigned xsra = (int) x >> k;
    int w = sizeof(int) << P4;
    int mask = (int) -1 << (P5);
    return xsra P6 P7;
}

==作答區==

P1 = ?
* `(a)` 1
* `(b)` 2
* `(c)` 3
* `(d)` -1
* `(e)` -2
* `(f)` -3
* `(g)` 0

P2 = ?
* `(a)` 1
* `(b)` k
* `(c)` k - w
* `(d)` w + k
* `(e)` w
* `(f)` w - k

P3 = ? (為某種 operation)
* `(a)` ^
* `(b)` &
* `(c)` <<
* `(d)` |
* `(e)` >>
* `(f)` +
* `(g)` -

P4 = ?
* `(a)` 1
* `(b)` 2
* `(c)` 3
* `(d)` -1
* `(e)` -2
* `(f)` -3
* `(g)` 0

P5 = ?
* `(a)` 1
* `(b)` k
* `(c)` k - w
* `(d)` w + k
* `(e)` w
* `(f)` w - k

P6 = ? (為某種 operation)
* `(a)` ^
* `(b)` &
* `(c)` <<
* `(d)` |
* `(e)` >>
* `(f)` +
* `(g)` -

P7 = ?
* `(a)` mask
* `(b)` ~mask

Reference:
* [Arithmetic Shift Operations](http://www-mdp.eng.cam.ac.uk/web/library/enginfo/mdp_micro/lecture4/lecture4-3-3.html)
(63)

:::info
延伸題目: 在 x86_64 和 Aarch32/Aarch64 找到對應的指令，並說明其作用和限制
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