The TLBTR characteristics are:
Provides information about the TLB implementation. The register must define whether the implementation provides separate instruction and data TLBs, or a unified TLB. Normally, the IMPLEMENTATION DEFINED information in this register includes the number of lockable entries in the TLB.
This register is present only when EL1 is capable of using AArch32. Otherwise, direct accesses to TLBTR are UNDEFINED.
TLBTR is a 32-bit register.
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 |
IMPLEMENTATION DEFINED | nU |
IMPLEMENTATION DEFINED.
Not Unified TLB. Indicates whether the implementation has a unified TLB:
nU | Meaning |
---|---|
0b0 |
Unified TLB. |
0b1 |
Separate Instruction and Data TLBs. |
Accesses to this register use the following encodings in the System register encoding space:
coproc | opc1 | CRn | CRm | opc2 |
---|---|---|---|---|
0b1111 | 0b000 | 0b0000 | 0b0000 | 0b011 |
if PSTATE.EL == EL0 then UNDEFINED; elsif PSTATE.EL == EL1 then if EL2Enabled() && !ELUsingAArch32(EL2) && HSTR_EL2.T0 == '1' then AArch64.AArch32SystemAccessTrap(EL2, 0x03); elsif EL2Enabled() && ELUsingAArch32(EL2) && HSTR.T0 == '1' then AArch32.TakeHypTrapException(0x03); elsif EL2Enabled() && !ELUsingAArch32(EL2) && HCR_EL2.TID1 == '1' then AArch64.AArch32SystemAccessTrap(EL2, 0x03); elsif EL2Enabled() && ELUsingAArch32(EL2) && HCR.TID1 == '1' then AArch32.TakeHypTrapException(0x03); else R[t] = TLBTR; elsif PSTATE.EL == EL2 then R[t] = TLBTR; elsif PSTATE.EL == EL3 then R[t] = TLBTR;
04/07/2023 11:24; 1b994cb0b8c6d1ae5a9a15edbc8bd6ce3b5c7d68
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