Question

下面展开的popcount的实现会产生错误的结果，我设法观察到只有b[0]和b[2]的元素被计算在b[1]和b[3]不是。

#include <stdio.h>

int count_multiple_bits(unsigned long long *b, int size) {
  unsigned long long *d = b;
  int c;
  __asm__("LD2 {v0.D, v1.D}[0], [%1], #16  \n\t"
          "LD2 {v0.D, v1.D}[1], [%1]       \n\t"
          "CNT v0.16b, v0.16b              \n\t"
          "CNT v1.16b, v1.16b              \n\t"
          "UADDLV h2, v0.16b               \n\t"
          "UADDLV h2, v1.16b               \n\t"
          "UMOV %0, v2.d[0]                \n\t"
          : "+r"(c)
          : "r"(d) : "v0", "v1", "v2");
  return c;
}

int main(int argc, const char *argv[]) {
  unsigned long long bits[] = { -1ull, -1ull, -1ull, -1ull };
  printf("Test: %i\n", count_multiple_bits(bits, 4));
  return 0;
}

这个一次计算2个元素的工作正常：

int count_multiple_bits(unsigned long long *b, int size) {
  unsigned long long *d = b;
  int c;
  __asm__("LD1 {v0.D}[0], [%1], #8   \n\t"
          "LD1 {v0.D}[1], [%1]       \n\t"
          "CNT v0.16b, v0.16b        \n\t"
          "UADDLV h1, v0.16b         \n\t"
          "UMOV %0, v1.d[0]          \n\t"
          : "+r"(c)
          : "r"(d) : "v0", "v1");
  return c;
}

在其他条件相同的情况下，我猜测负载是错误的，这是我假设的布局：

  v0.D[0] = b[0]
  v1.D[0] = b[1]
  v0.D[1] = b[2]
  v1.D[1] = b[3]

Answer 1

我的坏。

问题出现在UADDLV中，它清除目标寄存器，这是最终版本，它总结了任意长度的输入：

int count_multiple_bits(unsigned long long *b, unsigned int size) {
  unsigned long long *d = b;
  unsigned int masked = 0, i = 0;
  int c = 0;

  masked = size & ~3;
  for (; i < masked; i += 4)
    __asm__("LD1 {v0.2D, v1.2D}[0], [%1], #32 \n\t"
            "CNT v0.16b, v0.16b               \n\t"
            "CNT v1.16b, v1.16b               \n\t"
            "UADDLV h2, v0.16b                \n\t"
            "UADDLV h3, v1.16b                \n\t"
            "ADD d2, d3, d2                   \n\t"
            "UMOV x0, v2.d[0]                 \n\t"
            "ADD %0, x0, %0                   \n\t"
            : "+r"(c), "+r"(d)
            :: "x0", "v0", "v1", "v2", "v3");

  masked = size & 3;
  for (i = 0; i < masked; ++i)
    __asm__("LD1  {v0.D}[0], [%1], #8 \n\t"
            "CNT  v0.8b, v0.8b        \n\t"
            "UADDLV h1, v0.8b         \n\t"
            "UMOV x0, v1.d[0]         \n\t"
            "ADD %0, x0, %0           \n\t"
            : "+r"(c), "+r"(d)
            : : "x0", "v0", "v1");

  return c;
}

像魅力一样工作：）

Aarch64程序集LD2问题

1 个答案: