user: 请从给定的汇编指令推导出C源代码：

```asm
<shellsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,%eax
  shr    $0x1f,%eax
  add    %esi,%eax
  sar    %eax
  mov    %eax,0x1c(%rsp)
  cmp    $0x1,%esi
  jle    L49
  mov    %esi,%r14d
  lea    -0x1(%rsi),%eax
  mov    %eax,0x3c(%rsp)
  mov    %eax,%eax
  lea    0x4(%rdi,%rax,4),%r15
  mov    $0x1,%esi
  jmp    L111
L97:
  lea    D49(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L46:
  add    $0x1,%ebx
  cmp    %ebx,%r14d
  je     L135
L110:
  cmp    %ebx,%r12d
  je     L113
  cmp    %ebp,%ebx
  jne    L97
L113:
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L46
L135:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    0x20(%rsp),%rbp
  sub    0x1c(%rsp),%r12d
  test   %ebp,%ebp
  js     L25
  mov    0x18(%rsp),%esi
L41:
  mov    0x8(%rsp),%rax
  mov    (%rax,%rbp,4),%edx
  mov    0x10(%rsp),%rax
  mov    (%rax,%rbp,4),%eax
  cmp    %eax,%edx
  jle    L134
  mov    0x8(%rsp),%rbx
  mov    %eax,(%rbx,%rbp,4)
  mov    0x10(%rsp),%rax
  mov    %edx,(%rax,%rbp,4)
  lea    0x1(%rsi),%eax
  mov    %eax,0x18(%rsp)
  lea    D48(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L61:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbx,%r15
  jne    L61
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  jmp    L110
L25:
  mov    0x18(%rsp),%esi
L134:
  mov    0x28(%rsp),%rcx
  lea    0x1(%rcx),%rax
  cmp    0x30(%rsp),%rcx
  je     L36
  mov    %rax,0x28(%rsp)
L47:
  mov    0x28(%rsp),%rcx
  test   %ecx,%ecx
  js     L134
  mov    0x1c(%rsp),%eax
  neg    %eax
  cltq
  mov    %rax,0x20(%rsp)
  mov    0x38(%rsp),%r12d
  add    %ecx,%r12d
  mov    %rcx,%rbp
  lea    D0(%rip),%r13
  jmp    L41
L36:
  mov    0x1c(%rsp),%ecx
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  cmp    $0x1,%ecx
  jle    L49
  mov    %eax,0x1c(%rsp)
L111:
  cmp    0x1c(%rsp),%r14d
  jle    L36
  mov    0x1c(%rsp),%eax
  mov    %eax,0x38(%rsp)
  mov    0x3c(%rsp),%ecx
  sub    %eax,%ecx
  mov    %rcx,0x30(%rsp)
  cltq
  mov    0x8(%rsp),%rcx
  lea    (%rcx,%rax,4),%rax
  mov    %rax,0x10(%rsp)
  movq   $0x0,0x28(%rsp)
  jmp    L47
L49:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D49
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D33
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D48
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D15
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D0
type_infered: string

tool: D49:    

tool: D33:  ^ 

tool: D48: %d)	

tool: D15: 
	

tool: D0: %02d 

assistant: ```c
void shellsort(int v[], int n) {
  int gap, temp;
  int e, ee, eee, lncnt = 1;
  for (gap = n / 2; gap > 0; gap /= 2) {
    for (e = gap; e < n; e++) {
      for (ee = e - gap; ee >= 0 && v[ee] > v[ee + gap]; ee -= gap) {
        temp = v[ee];
        v[ee] = v[ee + gap];
        v[ee + gap] = temp;
        printf("%d)\t", lncnt++);
        for (eee = 0; eee < n; ++eee) {
          printf("%02d ", v[eee]);
        }
        printf("\n\t");
        for (eee = 0; eee < n; ++eee) {
          if (eee == ee || eee == ee + gap) {
            printf(" ^ ");
          } else {
            printf("   ");
          }
        }
        printf("\n");
      }
    }
  }
}
```