SmallVector's growth policies don't like starting from zero capacity.

I think there are good reasons to change this, but in the interests
of short-term stability, make SmallVector<...,0> reserve non-zero
capacity in its constructors.  This means that SmallVector<...,0>
uses more memory than SmallVector<...,1> and should really only be
used (unless/until this workaround is removed) by clients that
care about using SmallVector with an incomplete type.



git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@112147 91177308-0d34-0410-b5e6-96231b3b80d8

include/llvm/ADT/SmallVector.h

@@ -712,25 +712,33 @@ public:
 /// members are required.
 template <typename T>
 class SmallVector<T,0> : public SmallVectorImpl<T> {
+  // SmallVector doesn't like growing from zero capacity.  As a
+  // temporary workaround, avoid changing the growth algorithm by
+  // forcing capacity to be at least 1 in the constructors.
+
 public:
   SmallVector() : SmallVectorImpl<T>(0) {
+    this->reserve(1); // workaround
   }
 
   explicit SmallVector(unsigned Size, const T &Value = T())
     : SmallVectorImpl<T>(0) {
-    this->reserve(Size);
+    this->reserve(Size ? Size : 1); // workaround
     while (Size--)
       this->push_back(Value);
   }
 
   template<typename ItTy>
   SmallVector(ItTy S, ItTy E) : SmallVectorImpl<T>(0) {
+    if (S == E) this->reserve(1); // workaround
     this->append(S, E);
   }
 
   SmallVector(const SmallVector &RHS) : SmallVectorImpl<T>(0) {
     if (!RHS.empty())
       SmallVectorImpl<T>::operator=(RHS);
+    else
+      this->reserve(1); // workaround
   }
 
   const SmallVector &operator=(const SmallVector &RHS) {