user: 这段汇编代码用C语言表示是什么样的?

```asm
<spiralPrint>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,0x14(%rsp)
  mov    %rdx,0x18(%rsp)
  test   %esi,%esi
  jle    L22
  test   %edi,%edi
  jle    L22
  movslq %esi,%rax
  mov    %rdx,%rcx
  movl   $0x0,0xc(%rsp)
  mov    %esi,%ebp
  lea    0x14(,%rax,4),%rdx
  sub    $0x1,%rax
  xor    %r13d,%r13d
  xor    %r15d,%r15d
  add    %rdx,%rcx
  sub    $0x14,%rdx
  mov    %rax,0x38(%rsp)
  lea    D5(%rip),%rbx
  mov    %rcx,0x20(%rsp)
  mov    %rdx,0x30(%rsp)
  nopw   0x0(%rax,%rax,1)
L35:
  mov    0x18(%rsp),%rdx
  lea    0x0(%r13,%r13,2),%rax
  lea    (%rdx,%rax,8),%r14
  movslq %r15d,%rax
  mov    %rax,0x28(%rsp)
  mov    %rax,%r12
  nopl   0x0(%rax)
L129:
  mov    (%r14,%r12,4),%esi
  mov    %rbx,%rdi
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12d,%ebp
  jg     L129
  addl   $0x1,0xc(%rsp)
  mov    0x14(%rsp),%ecx
  lea    -0x1(%rbp),%edx
  mov    0xc(%rsp),%eax
  mov    %edx,0x10(%rsp)
  cmp    %ecx,%eax
  jge    L20
  lea    -0x2(%rcx),%eax
  mov    0x18(%rsp),%rcx
  mov    0x20(%rsp),%r14
  sub    %r13d,%eax
  add    %r13,%rax
  lea    (%rax,%rax,2),%rdx
  mov    0x38(%rsp),%rax
  sub    %r13,%rax
  lea    (%rax,%rdx,2),%rax
  lea    0x30(%rcx,%rax,4),%r12
  nopl   0x0(%rax)
L73:
  mov    (%r14),%esi
  mov    %rbx,%rdi
  add    $0x18,%r14
  call   <printf@plt>
  cmp    %r12,%r14
  jne    L73
  lea    -0x2(%rbp),%eax
  cmp    %eax,%r15d
  jg     L79
  movslq 0x14(%rsp),%rax
  mov    0x18(%rsp),%rcx
  sub    %r15d,%ebp
  lea    (%rax,%rax,2),%rax
  lea    -0x18(%rcx,%rax,8),%rax
  mov    0x30(%rsp),%rcx
  lea    -0x4(%rax,%rcx,1),%rdx
  lea    (%rax,%rcx,1),%r12
  lea    -0x2(%rbp),%eax
  shl    $0x2,%rax
  sub    %rax,%rdx
  mov    %rdx,%rbp
  nopl   (%rax)
L43:
  mov    -0x8(%r12),%esi
  mov    %rbx,%rdi
  sub    $0x4,%r12
  call   <printf@plt>
  cmp    %rbp,%r12
  jne    L43
L79:
  subl   $0x1,0x14(%rsp)
L20:
  cmp    %r15d,0x10(%rsp)
  jg     L89
L22:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L89:
  mov    0xc(%rsp),%edx
  cmp    %edx,0x14(%rsp)
  jle    L37
  movslq 0x14(%rsp),%rax
  mov    0x28(%rsp),%rdx
  mov    %rax,%rcx
  lea    (%rax,%rax,2),%rax
  lea    (%rdx,%rax,2),%rax
  mov    0x18(%rsp),%rdx
  shl    $0x2,%rax
  lea    -0x30(%rdx,%rax,1),%rbp
  lea    -0x18(%rdx,%rax,1),%r12
  mov    %ecx,%eax
  sub    %r13d,%eax
  sub    $0x2,%eax
  lea    (%rax,%rax,2),%rax
  shl    $0x3,%rax
  sub    %rax,%rbp
  nopl   (%rax)
L80:
  mov    (%r12),%esi
  mov    %rbx,%rdi
  sub    $0x18,%r12
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L80
L37:
  addq   $0x14,0x20(%rsp)
  mov    0xc(%rsp),%edx
  add    $0x1,%r15d
  add    $0x1,%r13
  subq   $0x4,0x30(%rsp)
  cmp    %edx,0x14(%rsp)
  jle    L22
  cmp    %r15d,0x10(%rsp)
  jle    L22
  mov    0x10(%rsp),%ebp
  jmp    L35
```

tool_calls: name: unpack_data
arguments: label: D5
type_guessed: string

tool: D5: %d 

assistant: ```c
void spiralPrint(int m, int n, int a[3][6]) {
  int i, k = 0, l = 0;
  while (k < m && l < n) {
    for (i = l; i < n; ++i) {
      printf("%d ", a[k][i]);
    }
    k++;
    for (i = k; i < m; ++i) {
      printf("%d ", a[i][n - 1]);
    }
    n--;
    if (k < m) {
      for (i = n - 1; i >= l; --i) {
        printf("%d ", a[m - 1][i]);
      }
      m--;
    }
    if (l < n) {
      for (i = m - 1; i >= k; --i) {
        printf("%d ", a[i][l]);
      }
      l++;
    }
  }
}
```