ARM NEON:如何实现 256 字节的查找表 [英] ARM NEON: How to implement a 256bytes Look Up table

问题描述

我正在使用内联汇编将我编写的一些代码移植到 NEON.

I am porting some code I wrote to NEON using inline assembly.

我需要做的一件事是将范围 [0..128] 的字节值转换为表中采用完整范围 [0..255] 的其他字节值

One of the things I need is to convert byte values ranging [0..128] to other byte values in a table which take the full range [0..255]

表格很短，但背后的数学计算并不容易，所以我认为不值得每次动态"计算它.所以我想尝试查找表.

The table is short but the math behind this is not easy so I think it is not worth calculating it each time "on the fly". So I want to try Look Up tables.

我在 32 字节的情况下使用了 VTBL，并且按预期工作

I have used VTBL for a 32byte case, and works as expected

对于完整范围，一个想法是首先比较源所在的范围并进行不同的查找(即，有 4 个 32 位查找表).

For the full range, one idea would be to first compare the range where the source is and do different lookups (i.e, having 4 32-bit lookup tables).

我的问题是:有没有更有效的方法来做到这一点?

编辑

经过一些试验，我已经完成了四次查找，(仍未安排)我对结果感到满意.我在这里留下了内联汇编中的一段代码行，以防万一有人发现它有用或认为它可以改进.

After some trials, I have done it with four look-ups and (still not scheduled) I am happy with the results. I leave here a piece of the code lines in inline assembly, just in case someone may find it useful or thinks it can be improved.

// Have the original data in d0
// d1 holds #32 value 
// d6,d7,d8,d9 has the images for the values [0..31] 

    //First we look for the 0..31 images. The values out of range will be 0
    "vtbl.u8 d2,{d6,d7,d8,d9},d0    \n\t"

    // Now we sub #32 to d1 and find the images for [32...63], which have been previously loaded in d10,d11,d12,d13
    "vsub.u8 d0,d0,d1\n\t"              
    "vtbl.u8 d3,{d10,d11,d12,d13},d1    \n\t"

    // Do the same and calculating images for [64..95]
    "vsub.u8 d0,d0,d1\n\t"
    "vtbl.u8 d4,{d14,d15,d16,d17},d0    \n\t"

    // Last step: images for [96..127]
    "vsub.u8 d0,d0,d1\n\t"
    "vtbl.u8 d5,{d18,d19,d20,d21},d0    \n\t"

    // Now we add all. No need to saturate, since only one will be different than zero each time
    "vadd.u8 d2,d2,d3\n\t"
    "vadd.u8 d4,d4,d5\n\t"
    "vadd.u8 d2,d2,d4\n\t"   // Leave the result in d2

推荐答案

正确的顺序是通过

vtbl d0, { d2,d3,d4,d5 }, d1   // first value
vsub d1, d1, d31               // decrement index
vtbx d0, { d6,d7,d8,d9 }, d1   // all the subsequent values
vsub d1, d1, d31               // decrement index
vtbx d0, { q5,q6 }, d1         // q5 = d10,d11
vsub d1, d1, d31
vtbx d0, { q7,q8 }, d1

vtbl 和 vtbx 之间的区别在于 vtbl 将元素 d0 归零，当 d1 >= 32 时，vtbx 将 d0 中的原始值保持不变.因此不需要像我的评论中那样的诡计，也不需要合并部分值.

The difference between vtbl and vtbx is that vtbl zeroes the element d0, when d1 >= 32, where as vtbx leaves the original value in d0 intact. Thus there's no need for the trickery as in my comment and no need to merge the partial values.

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