vpmax

Vector Pairwise Maximum

VPMAX<c>.<dt> <Dd>, <Dn>, <Dm>

Maximum of adjacent pairs.

Pseudocode Operation

case dt of
  when I8:  for i = 0 to 3: Dd[i] ← max(Dn[2*i], Dn[2*i+1], Dm[2*i], Dm[2*i+1])
  when I16: for i = 0 to 1: Dd[i] ← max(Dn[2*i], Dn[2*i+1], Dm[2*i], Dm[2*i+1])
  when I32: Dd[0] ← max(Dn[0], Dn[1]); Dd[1] ← max(Dm[0], Dm[1])
  when F32: Dd[0] ← max(Dn[0], Dn[1]); Dd[1] ← max(Dm[0], Dm[1])

Example

VPMAX.dt d0, d1, d2

Encoding

Binary Layout

1111001

31:25

size

21:20

19:16

15:12

1010

11:8

3:0

Format NEON 3-Reg

Opcode 0xF2000A00

Extension NEON (SIMD)

Operands

Dd
Destination 64-bit SIMD/FP register
Dn
First source 64-bit SIMD/FP register
Dm
Second source 64-bit SIMD/FP register

Reference

View in Arm A64 ISA Reference ↗ Arm AArch32 ISA

Instruction Forms

Encoding	Instruction	ISA	Bit pattern
0xF3000F00	VPMAX{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>	A32	1111001 \| 1 \| 0 \| D \| 0 \| sz \| Vn \| Vd \| 1111 \| N \| 0 \| M \| 0 \| Vm
0xFF000F00	VPMAX{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>	T32	111 \| 1 \| 11110 \| D \| 0 \| sz \| Vn \| Vd \| 1111 \| N \| 0 \| M \| 0 \| Vm
0xF2000A00	VPMAX{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>	A32	1111001 \| U \| 0 \| D \| size \| Vn \| Vd \| 1010 \| N \| 0 \| M \| 0 \| Vm
0xEF000A00	VPMAX{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>	T32	111 \| U \| 11110 \| D \| size \| Vn \| Vd \| 1010 \| N \| 0 \| M \| 0 \| Vm

Description

Vector Pairwise Maximum compares adjacent pairs of elements in two doubleword vectors, and copies the larger of each pair into the corresponding element in the destination doubleword vector. Depending on settings in the CPACR, NSACR, and HCPTR registers, and the Security state and PE mode in which the instruction is executed, an attempt to execute the instruction might be undefined, or trapped to Hyp mode. For more information see Enabling Advanced SIMD and floating-point support.

Operation

if ConditionPassed() then
    EncodingSpecificOperations();  CheckAdvSIMDEnabled();
    bits(64) dest;
    h = elements DIV 2;

    for e = 0 to h-1
        op1 = Int(Elem[D[n],2*e,esize], unsigned);
        op2 = Int(Elem[D[n],2*e+1,esize], unsigned);
        result = if maximum then Max(op1,op2) else Min(op1,op2);
        Elem[dest,e,esize] = result<esize-1:0>;
        op1 = Int(Elem[D[m],2*e,esize], unsigned);
        op2 = Int(Elem[D[m],2*e+1,esize], unsigned);
        result = if maximum then Max(op1,op2) else Min(op1,op2);
        Elem[dest,e+h,esize] = result<esize-1:0>;

    D[d] = dest;