Speed up FastAsciiArrayJoin on ia32 by improving hand-written assembly code.

src/ia32/macro-assembler-ia32.cc

Index: src/ia32/macro-assembler-ia32.cc
===================================================================
--- src/ia32/macro-assembler-ia32.cc	(revision 6301)
+++ src/ia32/macro-assembler-ia32.cc	(working copy)
@@ -877,55 +877,51 @@
       Immediate(Factory::cons_ascii_string_map()));
 }
 
-// All registers must be distinct.  Only current_string needs valid contents
-// on entry.  All registers may be invalid on exit.  result_operand is
-// unchanged, padding_chars is updated correctly.
-void MacroAssembler::AppendStringToTopOfNewSpace(
-    Register current_string,  // Tagged pointer to string to copy.
-    Register current_string_length,
-    Register result_pos,
-    Register scratch,
-    Register new_padding_chars,
-    Operand operand_result,
-    Operand operand_padding_chars,
-    Label* bailout) {
-  mov(current_string_length,
-      FieldOperand(current_string, String::kLengthOffset));
-  shr(current_string_length, 1);
-  sub(current_string_length, operand_padding_chars);
-  mov(new_padding_chars, current_string_length);
-  add(Operand(current_string_length), Immediate(kObjectAlignmentMask));
-  and_(Operand(current_string_length), Immediate(~kObjectAlignmentMask));
-  sub(new_padding_chars, Operand(current_string_length));
-  neg(new_padding_chars);
-  // We need an allocation even if current_string_length is 0, to fetch
-  // result_pos.  Consider using a faster fetch of result_pos in that case.
-  AllocateInNewSpace(current_string_length, result_pos, scratch, no_reg,
-                     bailout, NO_ALLOCATION_FLAGS);
-  sub(result_pos, operand_padding_chars);
-  mov(operand_padding_chars, new_padding_chars);
 
-  Register scratch_2 = new_padding_chars;  // Used to compute total length.
-  // Copy string to the end of result.
-  mov(current_string_length,
-      FieldOperand(current_string, String::kLengthOffset));
-  mov(scratch, operand_result);
-  mov(scratch_2, current_string_length);
-  add(scratch_2, FieldOperand(scratch, String::kLengthOffset));
-  mov(FieldOperand(scratch, String::kLengthOffset), scratch_2);
-  shr(current_string_length, 1);
-  lea(current_string,
-      FieldOperand(current_string, SeqAsciiString::kHeaderSize));
-  // Loop condition: while (--current_string_length >= 0).
-  Label copy_loop;
-  Label copy_loop_entry;
-  jmp(&copy_loop_entry);
-  bind(&copy_loop);
-  mov_b(scratch, Operand(current_string, current_string_length, times_1, 0));
-  mov_b(Operand(result_pos, current_string_length, times_1, 0), scratch);
-  bind(&copy_loop_entry);
-  sub(Operand(current_string_length), Immediate(1));
-  j(greater_equal, &copy_loop);
+// Copy memory, byte-by-byte, from source to destination.  Not optimized for
+// long or aligned copies.  The contents of scratch and length are destroyed.
+// Source and destination are incremented by length.
+// Many variants of movsb, loop unrolling, word moves, and indexed operands
+// have been tried here already, and this is fastest.
+// A simpler loop is faster on small copies, but 30% slower on large ones.
+// The cld() instruction must have been emitted, to set the direction flag(),
+// before calling this function.
+void MacroAssembler::CopyBytes(Register source,
+                               Register destination,
+                               Register length,
+                               Register scratch) {
+  Label loop, done, short_string, short_loop;
+  // Experimentation shows that the short string loop is faster if length < 10.
+  cmp(Operand(length), Immediate(10));
+  j(less_equal, &short_string);
+
+  ASSERT(source.is(esi));
+  ASSERT(destination.is(edi));
+  ASSERT(length.is(ecx));
+
+  // Because destination is 4-byte aligned, we keep it aligned for movs.

[Lasse Reichstein, 2011/01/14 10:47:37]

How do we know that destination is 4-byte aligned?
Do you mean "If destination is 4-byte aligned"?

[William Hesse, 2011/01/14 10:54:44]

In our uses, source is 4-byte aligned.
Changed comment.

+  mov(scratch, Operand(source, length, times_1, -4));
+  mov(Operand(destination, length, times_1, -4), scratch);
+  mov(scratch, ecx);
+  shr(ecx, 2);

[Lasse Reichstein, 2011/01/14 10:47:37]

If length was divisible by four, you will copy the last word twice.
Subtract one from ecx before shifting, to copy one less word if (length & 3) ==
0, but not otherwise.

[William Hesse, 2011/01/14 10:54:44]

Long rep.movs averages much less than a cycle per word.  This would save one
word only 1/4 of the time.  Not worth the extra code, I think.

On 2011/01/14 10:47:37, Lasse Reichstein wrote:

+  rep_movs();
+  and_(Operand(scratch), Immediate(0x3));
+  add(destination, Operand(scratch));
+  jmp(&done);
+
+  bind(&short_string);
+  test(length, Operand(length));
+  j(zero, &done);
+
+  bind(&short_loop);
+  mov_b(scratch, Operand(source, 0));
+  mov_b(Operand(destination, 0), scratch);
+  inc(source);
+  inc(destination);
+  dec(length);

[Lasse Reichstein, 2011/01/14 10:47:37]

This won't be faster if you do:

 add(source, length);
 add(destination, length);
 neg(length);
 bind(&short_loop);
 mov_b(scratch, Operand(source, ecx));
 mov_b(Operand(destination, ecx), scratch);
 inc(ecx);
 j(not_zero, &short_loop);

(It's quite possible that it won't. There's probably lots of latency available
during reading and writing for updating all three registers, and it's actually
one instruction longer in total).

[William Hesse, 2011/01/14 10:54:44]

I tried that - it was slower. On 2011/01/14 10:47:37, Lasse Reichstein wrote:

+  j(not_zero, &short_loop);
+
+  bind(&done);
 }