update chapter2 homework

homework/chapter2/65.c

@@ -1,16 +1,23 @@
-#include <stdio.h>
+#include <assert.h>
+#include <limits.h>
 
 /* Return 1 when x contains an odd number of 1s; 0 otherwise.
    Assume w=32 */
 int odd_ones(unsigned x) {
-  x = x ^ (x >> 16);
-  x = x ^ (x >> 8);
-  x = x ^ (x >> 4);
-  x = x ^ (x >> 2);
-  x = x ^ (x >> 1);
+  x ^= x >> 16;
+  x ^= x >> 8;
+  x ^= x >> 4;
+  x ^= x >> 2;
+  x ^= x >> 1;
   return x & 1;
 }
 
 int main() {
-  printf("%d %d\n", odd_ones(0xF1F1F0F1), odd_ones(0xFF00AAAA));
+  for (unsigned x = 0; ; x++) {
+    assert(odd_ones(x) == __builtin_parity(x));
+
+    if (x == UINT_MAX) {
+      break;
+    }
+  }
 }

homework/chapter2/66.c

@@ -1,4 +1,5 @@
-#include <stdio.h>
+#include <assert.h>
+#include <limits.h>
 
 /*
  * Generate mask indicating leftmost 1 in x.  Assume w=32.
@@ -7,20 +8,23 @@
  */
 int leftmost_one(unsigned x) {
   unsigned t = x;
-  t = t >> 1;
+  t >>= 1;
 
-  t = t | (t >> 1);
-  t = t | (t >> 2);
-  t = t | (t >> 4);
-  t = t | (t >> 8);
-  t = t | (t >> 16);
+  t |= t >> 1;
+  t |= t >> 2;
+  t |= t >> 4;
+  t |= t >> 8;
+  t |= t >> 16;
 
   return (t + 1) & x;
 }
 
 int main() {
-  printf("%.8X %.8X\n", 0x06050403, leftmost_one(0x06050403));
-  printf("%.8X %.8X\n", 0xF0F0F0F0, leftmost_one(0xF0F0F0F0));
-  printf("%.8X %.8X\n", 0x00000000, leftmost_one(0x00000000));
-  printf("%.8X %.8X\n", 0x00000001, leftmost_one(0x00000001));
+  for (unsigned x = 0; ; x++) {
+    assert((x == 0 && leftmost_one(x) == 0) || (x != 0 && leftmost_one(x) == (1 << (31 - __builtin_clz(x)))));
+
+    if (x == UINT_MAX) {
+      break;
+    }
+  }
 }

homework/chapter2/75.c

@@ -1,5 +1,6 @@
 #include <limits.h>
 #include <assert.h>
+#include <stdlib.h>
 
 int signed_high_prod(int x, int y) {
   long long p = (long long) x * y;
@@ -11,13 +12,20 @@ unsigned unsigned_high_prod(unsigned x, unsigned y) {
   return signed_high_prod(x, y) + (((int) y >> (w - 1)) & x) + (((int) x >> (w - 1)) & y);
 }
 
-int main() {
-  unsigned x = 0x71827364, y = 0xF0E1D2C3;
-  assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
-  x = 0x91827364, y = 0xF0E1D2C3;
-  assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
-  x = 0x71827364, y = 0x60E1D2C3;
-  assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
-  x = 0x91827364, y = 0x60E1D2C3;
-  assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
+unsigned randu() {
+  if (rand() & 1) {
+    return rand();
+  } else {
+    return ~rand();
+  }
+}
+
+int main() {
+  assert(RAND_MAX == INT_MAX);
+
+  const int test_cases = 1e8;
+  for (int i = 0; i < test_cases; i++) {
+    unsigned x = randu(), y = randu();
+    assert((unsigned long long) x * y == ((unsigned long long) unsigned_high_prod(x, y) << 32) + x * y);
+  }
 }

homework/chapter2/80.c

@@ -1,19 +1,26 @@
 #include <assert.h>
+#include <limits.h>
+#include <stdlib.h>
 
 int threefourths(int x) {
   int p = x >> 2, q = x & 3;
-  return p + (p << 1) + q + ((!q | (x >> 31)) & 1) - 1;
+  return p + (p << 1) + q - ((!!q & (~x >> 31)) & 1);
+}
+
+int randi() {
+  if (rand() & 1) {
+    return rand();
+  } else {
+    return ~rand();
+  }
 }
 
 int main() {
-  int x = 0x88659612;
-  for (int i = 0; i < 100; i++) {
-    x++;
-    assert(threefourths(x) == (long long) x * 3 / 4);
-  }
-  x = 0x79642695;
-  for (int i = 0; i < 100; i++) {
-    x++;
+  assert(RAND_MAX == INT_MAX);
+
+  const int test_cases = 1e8;
+  for (int i = 0; i < test_cases; i++) {
+    int x = randi();
     assert(threefourths(x) == (long long) x * 3 / 4);
   }
 }

homework/chapter2/95.c

@@ -7,31 +7,31 @@ 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;
+  float_bits s = uf & 0x80000000, f = uf & 0x007FFFFF, e = (uf >> 23) & 0xFF;
 
-  if (e == 0xFFU) {
+  if (e == 0xFF) {
     return uf;
   }
 
   if (e > 1) {
-    e = e - 1;
+    e--;
     return s | (e << 23) | f;
   }
 
   if (e == 1) {
-    f = f | (1 << 23);
+    f |= 1 << 23;
     e = 0;
   }
   int t = f;
-  f = f >> 1;
+  f >>= 1;
   if (t & f & 1) {
-    f = f + 1;
+    f++;
   }
   return s | (e << 23) | f;
 }
 
 int main() {
-  for (float_bits x = 0; ; x = x + 1) {
+  for (float_bits x = 0; ; x++) {
     float *a = (float*) &x;
     float b = 0.5 * *a;
     float_bits c = float_half(x);

homework/chapter2/97.c

@@ -7,7 +7,7 @@ typedef unsigned float_bits;
 /* Compute (float) i */
 float_bits float_i2f(int i) {
   if (i == INT_MIN) {
-    return 0xCF000000U;
+    return 0xCF000000;
   }
 
   if (i == 0) {
@@ -17,23 +17,23 @@ float_bits float_i2f(int i) {
   float_bits s = 0, f, e;
   if (i < 0) {
     i = -i;
-    s = 0x80000000U;
+    s = 0x80000000;
   }
 
   int hb = 30;
   while (!(i >> hb)) {
-    hb = hb - 1;
+    hb--;
   }
 
-  i = i ^ (1 << hb);
+  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++;
+      if (f & 0x00800000) {
         f = 0;
-        e = e + 1;
+        e++;
       }
     }
   } else {
@@ -44,7 +44,7 @@ float_bits float_i2f(int i) {
 }
 
 int main() {
-  for (int x = INT_MIN; ; x = x + 1) {
+  for (int x = INT_MIN; ; x++) {
     float_bits y = float_i2f(x);
     float *z = (float*) &y;
     assert(*z == (float) x);

homework/chapter2/README.md

@@ -1,5 +1,5 @@
 ## 2.65
-一次操作将 $w$ 位的整数 $x$ 从中间分成两个 $\frac{w}{2}$ 位的整数 $y$ 和 $z$，接着令 `x = y ^ z`。进行该操作 5 次后得到答案。
+每一次操作，将 $w$ 位的整数 $x$ 从中间分成两个 $\frac{w}{2}$ 位的整数 $y$ 和 $z$，接着令 `x = y ^ z`。进行 5 次操作后即得到答案。
 
 我们将这样的思路称为“折半递归法”。
 
@@ -31,4 +31,4 @@ b&=(a+x_{w-1}\cdot y+y_{w-1}\cdot x)\bmod 2^w\\
 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
-注意在某些情况下，舍入会导致有效位数增加。
+在某些情况下，舍入会导致最高位变化。对于这种情况，我们要将最低有效位提高一位。