Introduce line editor library.

[oota-llvm.git] / lib / Support / APFloat.cpp

diff --git a/lib/Support/APFloat.cpp b/lib/Support/APFloat.cpp
index 34bc6b6dd6bde0936947cc99a5454d4aa79f7ec0..deb9b05206b3189f0a71e73c2facf59a33e0d24d 100644 (file)
--- a/lib/Support/APFloat.cpp
+++ b/lib/Support/APFloat.cpp
@@ -3546,11 +3546,14 @@ void APFloat::toString(SmallVectorImpl<char> &Str,
   // Set FormatPrecision if zero.  We want to do this before we
   // truncate trailing zeros, as those are part of the precision.
   if (!FormatPrecision) {
-    // It's an interesting question whether to use the nominal
-    // precision or the active precision here for denormals.
+    // We use enough digits so the number can be round-tripped back to an
+    // APFloat. The formula comes from "How to Print Floating-Point Numbers
+    // Accurately" by Steele and White.
+    // FIXME: Using a formula based purely on the precision is conservative;
+    // we can print fewer digits depending on the actual value being printed.
 
-    // FormatPrecision = ceil(significandBits / lg_2(10))
-    FormatPrecision = (semantics->precision * 59 + 195) / 196;
+    // FormatPrecision = 2 + floor(significandBits / lg_2(10))
+    FormatPrecision = 2 + semantics->precision * 59 / 196;
   }
 
   // Ignore trailing binary zeros.
@@ -3773,7 +3776,7 @@ APFloat::opStatus APFloat::next(bool nextDown) {
     //                     change the payload.
     if (isSignaling()) {
       result = opInvalidOp;
-      // For consistency, propogate the sign of the sNaN to the qNaN.
+      // For consistency, propagate the sign of the sNaN to the qNaN.
       makeNaN(false, isNegative(), 0);
     }
     break;
@@ -3813,7 +3816,7 @@ APFloat::opStatus APFloat::next(bool nextDown) {
       // Decrement the significand.
       //
       // We always do this since:
-      //   1. If we are dealing with a non binade decrement, by definition we
+      //   1. If we are dealing with a non-binade decrement, by definition we
       //   just decrement the significand.
       //   2. If we are dealing with a normal -> normal binade decrement, since
       //   we have an explicit integral bit the fact that all bits but the