ARM Neon指令vst1.8与vst1.32是否存在差异？

对比以下两条ARM Neon指令是否存在差异：

vst1.32 {d12, d13, d14, d15}, [r4]

和

vst1.8 {d12, d13, d14, d15}, [r4]

根据ARM官方文档中的伪代码：

case type of
    when '0111'
        regs = 1;  if align<1> == '1' then UNDEFINED;
    when '1010'
        regs = 2;  if align == '11' then UNDEFINED;
    when '0110'
        regs = 3;  if align<1> == '1' then UNDEFINED;
    when '0010'
        regs = 4;
    otherwise
        SEE "Related encodings";
alignment = if align == '00' then 1 else 4 << UInt(align);
ebytes = 1 << UInt(size);  esize = 8 * ebytes;  elements = 8 DIV ebytes;
d = UInt(D:Vd);  n = UInt(Rn);  m = UInt(Rm);
wback = (m != 15);  register_index = (m != 15 && m != 13);
if n == 15 || d+regs > 32 then UNPREDICTABLE;


if ConditionPassed() then
    EncodingSpecificOperations();  CheckAdvSIMDEnabled();  NullCheckIfThumbEE(n);
    address = R[n];  if (address MOD alignment) != 0 then GenerateAlignmentException();
    if wback then R[n] = R[n] + (if register_index then R[m] else 8*regs);
    for r = 0 to regs-1
        for e = 0 to elements-1
            if ebytes != 8 then 
                MemU[address,ebytes] = Elem[D[d+r],e,esize]; 
            else 
                data =Elem[D[d+r],e,esize]; 
                MemU[address,4] = if BigEndian() then data<63:32> else data<31:0>;
                MemU[address+4,4] = if BigEndian() then data<31:0> else data<63:32>;
            address = address + ebytes;

分析上述伪代码后发现，无论指定的size对应何种指令后缀（.8、.16、.32），D寄存器的完整64位（8字节）数据都会被复制到内存中，看起来vst1.32、vst1.8、vst1.16这几条指令似乎等效。但ARM ISA的设计原则中不应存在冗余的存储语义，因此疑惑自身的推理或假设存在何种漏洞。

内容的提问来源于stack exchange，提问作者0xcaff