Fix Long Numeric Integer representation for Python 3

Before Python 3, long integers were input with an L suffix. Since
Python 3, all integers are 64-bit and do not need the L suffix.

bytecode.cpp

@@ -230,7 +230,7 @@ void print_const(std::ostream& pyc_output, PycRef<PycObject> obj, PycModule* mod
         formatted_print(pyc_output, "%d", obj.cast<PycInt>()->value());
         break;
     case PycObject::TYPE_LONG:
-        formatted_print(pyc_output, "%s", obj.cast<PycLong>()->repr().c_str());
+        formatted_print(pyc_output, "%s", obj.cast<PycLong>()->repr(mod).c_str());
         break;
     case PycObject::TYPE_FLOAT:
         formatted_print(pyc_output, "%s", obj.cast<PycFloat>()->value());

pyc_numeric.cpp

@@ -53,13 +53,13 @@ bool PycLong::isEqual(PycRef<PycObject> obj) const
     return true;
 }
 
-std::string PycLong::repr() const
+std::string PycLong::repr(PycModule* mod) const
 {
     // Longs are printed as hex, since it's easier (and faster) to convert
     // arbitrary-length integers to a power of two than an arbitrary base
 
     if (m_size == 0)
-        return "0x0L";
+        return (mod->verCompare(3, 0) >= 0) ? "0x0" : "0x0L";
 
     // Realign to 32 bits, since Python uses only 15
     std::vector<unsigned> bits;
@@ -90,7 +90,8 @@ std::string PycLong::repr() const
     aptr += snprintf(aptr, 9, "%X", *iter++);
     while (iter != bits.rend())
         aptr += snprintf(aptr, 9, "%08X", *iter++);
-    *aptr++ = 'L';
+    if (mod->verCompare(3, 0) < 0)
+        *aptr++ = 'L';
     *aptr = 0;
     return accum;
 }

pyc_numeric.h

@@ -37,7 +37,7 @@ class PycLong : public PycObject {
     int size() const { return m_size; }
     const std::vector<int>& value() const { return m_value; }
 
-    std::string repr() const;
+    std::string repr(PycModule* mod) const;
 
 private:
     int m_size;