user: 这段汇编代码用C语言表示是什么样的？

```asm
<filter_c_2d_median>:
  endbr64
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x78,%rsp
  mov    0xb0(%rsp),%ebx
  mov    %rdi,0x58(%rsp)
  mov    %esi,0x60(%rsp)
  mov    %ebx,%r13d
  imul   %ebx,%r13d
  lea    0x0(,%r13,4),%edi
  call   <malloc@plt>
  mov    %r14d,%esi
  mov    %rax,0x10(%rsp)
  mov    %rax,%rdx
  mov    %ebx,%eax
  shr    $0x1f,%eax
  add    %ebx,%eax
  mov    %eax,%ecx
  mov    %r13d,%eax
  sar    %ecx
  sar    %eax
  sub    %ecx,%esi
  mov    %ecx,0x64(%rsp)
  mov    %esi,0x40(%rsp)
  cmp    %esi,%ecx
  jge    L64
  mov    %r14d,%edi
  cltq
  mov    %rdx,%rsi
  sub    %ecx,%ebp
  lea    (%rdx,%rax,4),%r14
  movslq %edi,%rax
  movslq %ecx,%rdx
  mov    %ebp,0x44(%rsp)
  lea    0x0(,%rax,4),%r10
  mov    %ecx,%eax
  mov    %rdx,0x50(%rsp)
  movslq %ebx,%rbp
  imul   %edi,%eax
  mov    %ecx,0x38(%rsp)
  neg    %rbp
  lea    -0x1(%r13),%ecx
  mov    %r10,0x48(%rsp)
  shl    $0x2,%rbp
  movl   $0x0,0x3c(%rsp)
  lea    0x0(%rbp,%r10,1),%r15
  cltq
  mov    %ecx,0x2c(%rsp)
  add    %rdx,%rax
  neg    %rdx
  lea    (%r12,%rax,4),%rax
  mov    %r14,%r12
  mov    %rbp,%r14
  mov    %rax,0x30(%rsp)
  mov    %rdx,%rax
  lea    -0x1(%rbx),%edx
  lea    0x0(,%rdx,4),%rdi
  shl    $0x2,%rax
  mov    %rdi,0x20(%rsp)
  sub    %rsi,%rax
  lea    0x4(%rsi,%rdi,1),%rdi
  mov    %rdi,0x18(%rsp)
  mov    %rax,0x68(%rsp)
L36:
  mov    0x64(%rsp),%eax
  mov    0x44(%rsp),%esi
  cmp    %esi,%eax
  jge    L12
  mov    0x3c(%rsp),%edi
  mov    0x68(%rsp),%rcx
  mov    0x30(%rsp),%rbp
  add    %edi,%eax
  mov    %eax,%r9d
  movslq %edi,%rax
  add    0x50(%rsp),%rax
  lea    (%rcx,%rax,4),%rcx
  mov    0x58(%rsp),%rax
  lea    (%rax,%rcx,1),%r13
  lea    (%rsi,%rdi,1),%eax
  mov    %eax,0x28(%rsp)
  mov    %r13,%r8
  mov    %r9d,%r13d
  nopw   0x0(%rax,%rax,1)
L69:
  mov    $0xfffffffffffffffc,%r11
  mov    0x18(%rsp),%rdx
  mov    %r8,%rsi
  xor    %edi,%edi
  sub    0x20(%rsp),%r11
  test   %ebx,%ebx
  jle    L29
  nopw   0x0(%rax,%rax,1)
L15:
  lea    (%r11,%rdx,1),%rax
  nopl   0x0(%rax)
L3:
  movss  (%rsi,%rax,1),%xmm0
  add    $0x4,%rax
  movss  %xmm0,-0x4(%rax)
  cmp    %rdx,%rax
  jne    L3
  add    $0x1,%edi
  add    %r15,%rsi
  sub    %r14,%rdx
  cmp    %edi,%ebx
  jne    L15
L29:
  mov    0x2c(%rsp),%edx
  mov    0x10(%rsp),%rdi
  xor    %esi,%esi
  mov    %r8,0x8(%rsp)
  add    $0x4,%rbp
  add    $0x1,%r13d
  call   <inkernel_quicksort@plt>
  movss  (%r12),%xmm0
  mov    0x8(%rsp),%r8
  movss  %xmm0,-0x4(%rbp)
  add    $0x4,%r8
  cmp    0x28(%rsp),%r13d
  jne    L69
L12:
  mov    0x48(%rsp),%rdi
  addl   $0x1,0x38(%rsp)
  add    %rdi,0x30(%rsp)
  mov    0x38(%rsp),%eax
  mov    0x60(%rsp),%edi
  add    %edi,0x3c(%rsp)
  cmp    0x40(%rsp),%eax
  jne    L36
L64:
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void filter_c_2d_median(float *im, int ny, int nx, float *res, int nyr, int nxr, int w) {
  int nwin = w * w;
  float *win = (float *)malloc(nwin * sizeof(float));
  int edgex = w / 2;
  int edgey = w / 2;
  int mpos = nwin / 2;
  int x, y, wx, wy, ind, indy, indw;
  for (y = edgey; y < (ny - edgey); ++y) {
    ind = y * ny;
    for (x = edgex; x < (nx - edgex); ++x) {
      for (wy = 0; wy < w; ++wy) {
        indw = wy * w;
        indy = ny * (y + wy - edgey);
        for (wx = 0; wx < w; ++wx) {
          win[indw + wx] = im[indy + x + wx - edgex];
        }
      }
      inkernel_quicksort(win, 0, nwin - 1);
      res[ind + x] = win[mpos];
    }
  }
}
```