Finished exercises 2-7 and 2-8

ch_02/exercise_02-07/main.c

@@ -0,0 +1,69 @@
+/*
+ *
+ * Exercise 2-7. Write a function invert(x,p,n) that returns x with the n bits that
+ * begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the
+ * others unchanged.
+ *
+ * Author: Kun Deng
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+
+
+// Returns x with the n bits that begin at position p inverted 
+// (i.e., 1 changed into 0 and vice versa), leaving the others unchanged.
+int invert(unsigned int x, int p, int n);
+
+
+int main(int argc, char **argv)
+{
+    unsigned int x = 511;
+    int p = 4;
+    int n = 3;
+
+    if (argc == 4)
+    {
+        x = atoi(argv[1]);
+        p = atoi(argv[2]);
+        n = atoi(argv[3]);
+    }
+
+    printf("Calling invert(%d, %d, %d)", x, p, n);
+    printf(" = %d\n", invert(x, p, n));
+
+    return 0;
+}
+
+
+int invert(unsigned int x, int p, int n)
+{
+    int x_inverted = x;
+    printf("%d ^ (~(~0 << %d) << (%d ", x, n, p);
+
+    if (n == 1)
+    {
+        printf(" - 1))\n");
+        printf("%d ^ (~(%d << %d) << (%d - 1))\n", x, ~(0), n, p);
+        printf("%d ^ (~(%d) << (%d - 1))\n", x,  (~(0) << n), p);
+        printf("%d ^ (%d << (%d - 1))\n", x,  ~(~(0) << n), p);
+        printf("%d ^ (%d << %d)\n", x,  ~(~(0) << n), (p - 1));
+        printf("%d ^ %d\n", x,  (~(~(0) << n) << (p - 1)));
+
+        // No need to add 1 when the bit amount is 1 in (p + 1 - n)
+        x_inverted = x ^ (~(~0U << n) << (p - 1));
+    }
+    else
+    {
+        printf(" + 1 - %d))\n", n);
+        printf("%d ^ (~(%d << %d) << (%d + 1 - %d))\n", x, ~(0), n, p, n);
+        printf("%d ^ (~(%d) << (%d + 1 - %d))\n", x, (~(0) << n), p, n);
+        printf("%d ^ (%d << (%d + 1 - %d))\n", x, ~(~(0) << n), p, n);
+        printf("%d ^ (%d << %d)\n", x, ~(~(0) << n), (p + 1 - n));
+        printf("%d ^ %d\n", x, (~(~(0) << n) << (p + 1 - n)));
+
+        x_inverted = x ^ (~(~0U << n) << (p + 1 - n));
+    }
+
+    return x_inverted;
+}

ch_02/exercise_02-08/main.c

@@ -0,0 +1,53 @@
+/*
+ *
+ * Exercise 2-8. Write a function rightrot(x,n) that returns the value of the integer x
+ * rotated to the right by n positions.
+ *
+ * Author: Kun Deng
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <limits.h>
+
+
+// Returns the value of the integer x rotated to the right by n positions
+unsigned int rightrot(unsigned int x, int n);
+
+
+int main(int argc, char **argv)
+{
+    unsigned int x = 511;
+    int n = 3;
+
+    if (argc == 3)
+    {
+        x = atoi(argv[1]);
+        n = atoi(argv[2]);
+    }
+
+    unsigned int rotated = rightrot(x, n);
+
+    printf("rightrot(%d, %d)", x, n);
+    printf(" = %u\n", rotated);
+
+    return 0;
+}
+
+
+unsigned int rightrot(unsigned int x, int n)
+{
+    unsigned int some_bits = CHAR_BIT * sizeof(x) - 1;
+    printf("size of x is %ld some_bits = %d\n", sizeof(x), some_bits);
+
+    unsigned int off = n & some_bits;
+    printf("off is %d\n", off);
+
+    printf("(%d >> %d) | (%d << (%d & %d))\n", x, off, x, off, some_bits);
+    printf("(%d >> %d) | (%d << %d)\n", x, off, x, (off && some_bits));
+    printf("%d | %d\n", (x >> off), (x << (off && some_bits)));
+
+    unsigned int rotated = (x >> off) | (x << (off & some_bits));
+
+    return rotated;
+}

ch_02/getbits/main.c

@@ -0,0 +1,41 @@
+/*
+ * Simple implementation of getbits() function.
+ *
+ */
+
+#include <stdio.h>
+
+
+// Gets n bits from position p
+unsigned getbits(unsigned int x, int p, int n);
+
+int main()
+{
+    // 11111110111 = 1015
+    unsigned int x = 1015;
+    int p = 4;
+    int n = 3;
+
+    // Result is 111 = 7
+    unsigned int some_bits = getbits(x, p, n);
+
+
+    return 0;
+}
+
+
+unsigned getbits(unsigned int x, int p, int n)
+{
+    printf("(%d >> (%d + 1 - %d)) & ~(~0 << %d)\n", x, p, n, n);
+    printf("(%d >> (%d + 1 - %d)) & ~(%d << %d)\n", x, p, n, ~0, n);
+    printf("(%d >> (%d + 1 - %d)) & ~(%d)\n", x, p, n, ((~0) << n));
+    printf("(%d >> (%d + 1 - %d)) & %d\n", x, p, n, ~((~0) << n));
+    printf("(%d >> %d) & %d\n", x, (p + 1 - n), ~((~0) << n));
+    printf("%d & %d\n", (x >> (p + 1 - n)), ~((~0) << n));
+    printf("%d\n", ((x >> (p + 1 - n)) & ~((~0) << n)));
+    // At position 4 and 3 bits from that position of value 1023 is
+    // 11111110111 = 1015
+    //      110    = 6
+
+    return (x >> (p + 1 - n)) & ~(~0 << n);
+}