MI.addOperand(MCOperand::CreateImm(Offset));
OpIdx += 1;
} else {
+ // The opcode ARM::LDRT actually corresponds to both Encoding A1 and A2 of
+ // A8.6.86 LDRT. So if Inst{4} != 0 while Inst{25} (getIBit(insn)) == 1,
+ // we should reject this insn as invalid.
+ //
+ // Ditto for LDRBT.
+ if ((Opcode == ARM::LDRT || Opcode == ARM::LDRBT) && (slice(insn,4,4) == 1))
+ return false;
+
// Disassemble the offset reg (Rm), shift type, and immediate shift length.
MI.addOperand(MCOperand::CreateReg(getRegisterEnum(B, ARM::GPRRegClassID,
decodeRm(insn))));
# CHECK: ldr r0, [r2], #15
0x0f 0x00 0x92 0xe4
+# CHECK: ldr r5, [r7, -r10, lsl #2]
+0x0a 0x51 0x17 0xe7
+
# CHECK: ldrh r0, [r2], #0
0xb0 0x00 0xd2 0xe0
--- /dev/null
+# RUN: llvm-mc --disassemble %s -triple=arm-apple-darwin9 |& grep {invalid instruction encoding}
+
+# Opcode=0 Name=PHI Format=(42)
+# 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
+# -------------------------------------------------------------------------------------------------
+# | 1: 1: 1: 0| 0: 1: 1: 0| 0: 0: 1: 1| 0: 1: 1: 1| 0: 1: 0: 1| 0: 0: 0: 1| 0: 0: 0: 1| 0: 0: 0: 0|
+# -------------------------------------------------------------------------------------------------
+#
+# The bytes have Inst{4} = 1, so it's not an LDRT Encoding A2 instruction.
+0x10 0x51 0x37 0xe6
+
+