user: 这段汇编代码用C语言表示是什么样的?

```asm
<heapsort_reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L84
L43:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L31
L37:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jge    L70
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  comiss %xmm1,%xmm0
  jbe    L70
  addl   $0x1,-0xc(%rbp)
L70:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  comiss -0x4(%rbp),%xmm0
  jae    L141
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L31:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L37
  jmp    L7
L141:
  nop
L7:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  -0x4(%rbp),%xmm0
  movss  %xmm0,(%rax)
  subl   $0x1,-0x8(%rbp)
L84:
  cmpl   $0x0,-0x8(%rbp)
  jns    L43
  jmp    L106
L96:
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%rax
  movss  (%rax),%xmm0
  movss  %xmm0,(%rdx)
  subl   $0x1,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L80
L36:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jge    L46
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  comiss %xmm1,%xmm0
  jbe    L46
  addl   $0x1,-0xc(%rbp)
L46:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  comiss -0x4(%rbp),%xmm0
  jae    L1
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L80:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L36
  jmp    L45
L1:
  nop
L45:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  -0x4(%rbp),%xmm0
  movss  %xmm0,(%rax)
L106:
  cmpl   $0x0,-0x14(%rbp)
  jg     L96
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void heapsort_reverse(int n, float a[]) {
  int i, j, k;
  float x;
  for (k = n / 2 - 1; k >= 0; k--) {
    i = k;
    x = a[i];
    j = 2 * i + 1;
    while (j <= n - 1) {
      if (j < n - 1 && a[j] > a[j + 1])
        j++;
      if (x <= a[j])
        break;
      a[i] = a[j];
      i = j;
      j = 2 * i + 1;
    }
    a[i] = x;
  }
  while (n > 0) {
    x = a[n - 1];
    a[n - 1] = a[0];
    n--;
    i = 0;
    j = 2 * i + 1;
    while (j <= n - 1) {
      if (j < n - 1 && a[j] > a[j + 1])
        j++;
      if (x <= a[j])
        break;
      a[i] = a[j];
      i = j;
      j = 2 * i + 1;
    }
    a[i] = x;
  }
}
```