1 files changed, 39 insertions, 0 deletions
diff --git a/2/4_floating_point/special_values.c b/2/4_floating_point/special_values.c
new file mode 100644
index 0000000..cedd2ba
--- /dev/null
+++ b/2/4_floating_point/special_values.c
@@ -0,0 +1,39 @@
+#include <stdio.h>
+
+#define POS_INFINITY 1.8e400
+#define NEG_INFINITY -POS_INFINITY
+#define NEG_ZERO (-1.0/POS_INFINITY)
+
+int main() {
+    /*
+      2.53
+    */
+    printf("%f\n", POS_INFINITY);
+    printf("%f\n", NEG_INFINITY);
+    printf("%f\n", NEG_ZERO);
+
+    /*
+      2.54
+      A. Yes, int->double->int, int->double preserves value, so double->int after that also has the same value, since we know rounding and overflow is not relevant.
+      B. NO, same as A, actually Fmax < Tmax, so casting from int to float can overflow.
+      C. No, double->float->double, float has lower range, so overflow could happen. Float also has lower precision, so when converting to float there may also be round-to-even.
+      D. Yes, float->double->float, exact value can be kept when casting to double.
+      E. Yes, Sign bit can be flipped without influencing the rest of the bit pattern. So the double negation cancels itself and the value is exactly the same after.
+      F. Yes, 1.0/2 == 1/2.0, integers are first cast to float.
+      G. Yes, d*d >= 0.0, this is a result of the monotonicity property of floating point numbers.
+      H. (f+d)-f == d, in the addition rounding errors can occur, for example when adding a small value to a large value. It can also overflow, causing (inf-f) = inf.
+      In this comparison the problem arises when the value that is compared is the small one, since that will get lost in the rounding error of the addition.
+    */
+    int x = (1<<31) - 1;
+    float f = 3.14f;
+    double d = 1e307f;
+    printf("%d\n", x == (int) (double) x);
+    printf("%d\n", x == (int) (float) x);
+    printf("%d, value: %f\n", d == (double) (float) d, (double) (float) d);
+    printf("%d, value: %f, orig: %f\n", f == (float) (double) f, (float) (double) f, f);
+    printf("%d\n", f == -(-f));
+    printf("%d\n", 1.0/2 == 1/2.0);
+    printf("%d\n", d * d >= 0.0);
+    printf("%d\n", (f+d)-f == d);
+    return 0;
+}