数据类型与 NEON 内在函数的兼容性 [英] Data type compatibility with NEON intrinsics
问题描述
我正在使用 C++ 代码中的 NEON 内在函数进行 ARM 优化.我了解并掌握了大部分打字问题，但我仍然坚持这个问题:
指令vzip_u8 返回一个uint8x8x2_t 值(实际上是两个uint8x8_t 的数组).我想将返回的值分配给一个普通的 uint16x8_t.我认为没有合适的 vreinterpretq 内在来实现这一点，并且拒绝简单的强制转换.
一些定义回答清楚...
霓虹灯 有 32 个寄存器，64 位宽(双视图为 16 个寄存器，128 位宽).
<块引用><块引用>NEON 单元可以查看与以下内容相同的寄存器组:
- 16 个 128 位四字寄存器，Q0-Q15
- 32 个 64 位双字寄存器，D0-D31.
uint16x8_t 是一种需要 128 位存储的类型，因此它需要位于 quadword 寄存器中.
ARM NEON Intrinsics 在 ARM® C 语言扩展:
<块引用><块引用>... 用于加载和存储操作，在查表操作，作为返回一对向量的操作的结果类型.
vzip说明
<块引用><块引用>... 交错两个向量的元素.
vzip Dd, Dm
并且有一个 intrinsic 喜欢
uint8x8x2_t vzip_u8 (uint8x8_t, uint8x8_t)从这些我们可以得出结论，uint8x8x2_t实际上是两个随机编号的双字寄存器的列表，因为vzip指令对输入寄存器的顺序没有任何要求.
现在答案是...
uint8x8x2_t 可以包含不连续的两个双字寄存器，而 uint16x8_t 是由两个连续双字寄存器组成的数据结构，其中第一个具有偶数索引 (D0-D31 -> Q0-Q15).
因此，您无法轻松地将带有两个双字寄存器的向量数组数据类型转换为一个四字寄存器......
编译器可能足够聪明来为您提供帮助，或者您可以强制转换，但我会检查生成的程序集的正确性和性能.
I am working on ARM optimizations using the NEON intrinsics, from C++ code. I understand and master most of the typing issues, but I am stuck on this one:
The instruction vzip_u8 returns a uint8x8x2_t value (in fact an array of two uint8x8_t). I want to assign the returned value to a plain uint16x8_t. I see no appropriate vreinterpretq intrinsic to achieve that, and simple casts are rejected.
Some definitions to answer clearly...
NEON has 32 registers, 64-bits wide (dual view as 16 registers, 128-bits wide).
The NEON unit can view the same register bank as:
- sixteen 128-bit quadword registers, Q0-Q15
- thirty-two 64-bit doubleword registers, D0-D31.
uint16x8_t is a type which requires 128-bit storage thus it needs to be in an quadword register.
ARM NEON Intrinsics has a definition called vector array data type in ARM® C Language Extensions:
... for use in load and store operations, in table-lookup operations, and as the result type of operations that return a pair of vectors.
vzip instruction
... interleaves the elements of two vectors.
vzip Dd, Dm
and has an intrinsic like
uint8x8x2_t vzip_u8 (uint8x8_t, uint8x8_t)
from these we can conclude that uint8x8x2_t is actually a list of two random numbered doubleword registers, because vzip instructions doesn't have any requirement on order of input registers.
Now the answer is...
uint8x8x2_t can contain non-consecutive two dualword registers while uint16x8_t is a data structure consisting of two consecutive dualword registers which first one has an even index (D0-D31 -> Q0-Q15).
Because of this you can't cast vector array data type with two double word registers to a quadword register... easily.
Compiler may be smart enough to assist you, or you can just force conversion however I would check the resulting assembly for correctness as well as performance.
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