update chapter2 homework

homework/chapter2/95.c

@@ -0,0 +1,45 @@
+#include <assert.h>
+#include <math.h>
+#include <limits.h>
+
+/* Access bit-level representation of floating-point number */
+typedef unsigned float_bits;
+
+/* computs 0.5*f.  If f is NaN, then return f. */
+float_bits float_half(float_bits uf) {
+  float_bits s = uf & 0x80000000U, f = uf & 0x007FFFFF, e = (uf >> 23) & 0xFF;
+
+  if (e == 0xFFU) {
+    return uf;
+  }
+
+  if (e > 1) {
+    e = e - 1;
+    return s | (e << 23) | f;
+  }
+
+  if (e == 1) {
+    f = f | (1 << 23);
+    e = 0;
+  }
+  int t = f;
+  f = f >> 1;
+  if (t & f & 1) {
+    f = f + 1;
+  }
+  return s | (e << 23) | f;
+}
+
+int main() {
+  for (float_bits x = 0; ; x = x + 1) {
+    float *a = (float*) &x;
+    float b = 0.5 * *a;
+    float_bits c = float_half(x);
+    float *d = (float*) &c;
+    assert(b == *d || (isnan(b) && isnan(*d)));
+
+    if (x == UINT_MAX) {
+      break;
+    }
+  }
+}

homework/chapter2/97.c

@@ -0,0 +1,73 @@
+#include <assert.h>
+#include <limits.h>
+
+/* Access bit-level representation of floating-point number */
+typedef unsigned float_bits;
+
+/* Compute (float) i */
+float_bits float_i2f(int i) {
+  if (i == INT_MIN) {
+    return 0xCF000000U;
+  }
+
+  if (i == 0) {
+    return 0;
+  }
+
+  float_bits s = 0, f, e;
+  if (i < 0) {
+    i = -i;
+    s = 0x80000000U;
+  }
+
+  int hb = 0, t = i;
+  if (t >> 16) {
+    t = t >> 16;
+    hb = hb + 16;
+  }
+  if (t >> 8) {
+    t = t >> 8;
+    hb = hb + 8;
+  }
+  if (t >> 4) {
+    t = t >> 4;
+    hb = hb + 4;
+  }
+  if (t >> 2) {
+    t = t >> 2;
+    hb = hb + 2;
+  }
+  if (t >> 1) {
+    t = t >> 1;
+    hb = hb + 1;
+  }
+
+  i = i ^ (1 << hb);
+  e = hb + 127;
+  if (hb > 23) {
+    f = i >> (hb - 23);
+    if ((i & (1 << (hb - 24))) && ((f & 1) || (i & ((1 << (hb - 24)) - 1)))) {
+      f = f + 1;
+      if (f & 0x00800000U) {
+        f = 0;
+        e = e + 1;
+      }
+    }
+  } else {
+    f = i << (23 - hb);
+  }
+
+  return s | (e << 23) | f;
+}
+
+int main() {
+  for (int x = INT_MIN; ; x = x + 1) {
+    float_bits y = float_i2f(x);
+    float *z = (float*) &y;
+    assert(*z == (float) x);
+
+    if (x == INT_MAX) {
+      break;
+    }
+  }
+}

homework/chapter2/README.md

@@ -7,15 +7,30 @@
 
 第二次操作，令 `x = x | (x >> 2)`，依次类推。通过 5 次操作即可实现提示中的转换。
 ## 2.75
-$x,y$ 是无符号整数，$x'=U2T_w(x),y'=U2T_w(y)$。
+$x,y$ 是补码数，$x'=T2U_w(x),y'=T2U_w(y)$。
 
 $$
 \begin{aligned}
-x'\times y'&=(x+x_{w-1}2^w)\times(y+y_{w-1}2^w)\\
-&=x\times y+(x_{w-1}\times y+y_{w-1}\times x)2^w+x_{w-1}y_{w-1}2^{2w}
+x'\cdot y'&=(x+x_{w-1}\cdot 2^w)\times(y+y_{w-1}\cdot 2^w)\\
+&=x\cdot y+(x_{w-1}\cdot y+y_{w-1}\cdot x)2^w+x_{w-1}\cdot y_{w-1}\cdot 2^{2w}
 \end{aligned}$$
 
-因此有 $U2B_{2w}(x'\times y')=T2B_{2w}(x\times y+(x_{w-1}\times y+y_{w-1}\times x)2^w)$。即令 `a = signed_high_prod(x, y), b = unsigned_high_prod(x', y')`，$T2B_w(a+x_{w-1}\times y+y_{w-1}\times x)=U2B_w(b)$。
+进一步，令 `a = signed_high_prod(x, y), b = unsigned_high_prod(x', y')`，有 
+$$b\cdot 2^w+x'*^u_w y'=a\cdot 2^w+T2U_w(x*^t_wy)+(x_{w-1}\cdot y+y_{w-1}\cdot x)2^w+x_{w-1}\cdot y_{w-1}\cdot 2^{2w}$$
+
+化简得
+$$b=a+x_{w-1}\cdot y+y_{w-1}\cdot x+x_{w-1}\cdot y_{w-1}\cdot 2^w$$
+
+即
+$$\begin{aligned}
+b&=(a+x_{w-1}\cdot y+y_{w-1}\cdot x)\bmod 2^w\\
+&=T2U_w(a)+^u_t x_{w-1}\cdot y'+^u_t y_{w-1}\cdot x'
+\end{aligned}$$
 
 ## 2.80
-先得到粗糙的答案 `(x >> 2) + ((x >> 2) << 1)`，再根据 `x & 3` 进行修正。
+先得到粗糙的结果 `(x >> 2) + ((x >> 2) << 1)`，再根据 `x & 3` 以及 $x$ 的符号（决定舍入方向）进行修正。
+
+## 2.97
+注意以下细节：
+- 使用“折半递归法”求 $i$ 的有效位数可以避免对于所有 $2^{32}$ 个 `int` 检验花费极长的时间；
+- 在某些情况下，舍入会导致有效位数增加。