vpmin

Vector Pairwise Minimum

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

Minimum of adjacent pairs.

Pseudocode Operation

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

Example

VPMIN.dt d0, d1, d2

Encoding

Binary Layout

1111001

31:25

U

24

0

23

D

22

size

21:20

Vn

19:16

Vd

15:12

1010

11:8

N

7

0

6

M

5

1

4

Vm

3:0

 

Format NEON 3-Reg

Opcode 0xF2000A10

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

Related

More in NEON (SIMD)

Reference

View in Arm A64 ISA Reference ↗ Arm AArch32 ISA

Instruction Forms

Encoding	Instruction		ISA		Bit pattern
0xF3200F00	VPMIN{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>		A32		1111001 | 1 | 0 | D | 1 | sz | Vn | Vd | 1111 | N | 0 | M | 0 | Vm
0xFF200F00	VPMIN{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>		T32		111 | 1 | 11110 | D | 1 | sz | Vn | Vd | 1111 | N | 0 | M | 0 | Vm
0xF2000A10	VPMIN{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>		A32		1111001 | U | 0 | D | size | Vn | Vd | 1010 | N | 0 | M | 1 | Vm
0xEF000A10	VPMIN{<c>}{<q>}.<dt> {<Dd>, }<Dn>, <Dm>		T32		111 | U | 11110 | D | size | Vn | Vd | 1010 | N | 0 | M | 1 | Vm

Description

Vector Pairwise Minimum compares adjacent pairs of elements in two doubleword vectors, and copies the smaller 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;