user: 请从给定的汇编指令推导出C源代码：

```asm
<find2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  mov    $0xb,%r12d
  push   %rbx
  lea    D3(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  jmp    L103
  nopl   0x0(%rax)
L96:
  add    $0x1,%r12d
  cmp    $0x989680,%r12d
  je     L48
L103:
  mov    D33(%rip),%rax
  mov    %r12d,%edi
  movq   %rax,%xmm0
  call   <pow@plt>
  pxor   %xmm1,%xmm1
  movsd  D29(%rip),%xmm2
  cvttsd2si %xmm0,%r13d
  cvtsi2sd %r13d,%xmm1
  subsd  %xmm1,%xmm0
  andpd  D11(%rip),%xmm0
  comisd %xmm0,%xmm2
  jbe    L96
  mov    %r12d,%edi
  mov    %rsp,%r14
  call   <log10@plt>
  mov    %rsp,%rcx
  lea    0x1(%rax),%r9d
  mov    %rax,%r8
  movslq %r9d,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L8
L25:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L25
L8:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L30
  orq    $0x0,-0x8(%rsp,%rdx,1)
L30:
  mov    %rsp,%rsi
  test   %r9d,%r9d
  jle    L129
  mov    %r8d,%eax
  mov    %rsi,%rcx
  mov    %r12d,%edx
  lea    0x4(%rsi,%rax,4),%rdi
  nopl   (%rax)
L149:
  movslq %edx,%rax
  mov    %edx,%r10d
  add    $0x4,%rcx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%r10d
  sar    $0x22,%rax
  sub    %r10d,%eax
  lea    (%rax,%rax,4),%r10d
  add    %r10d,%r10d
  sub    %r10d,%edx
  mov    %edx,-0x4(%rcx)
  mov    %eax,%edx
  cmp    %rdi,%rcx
  jne    L149
  mov    %r9d,%ecx
  shr    $0x1f,%ecx
  add    %r9d,%ecx
  sar    %ecx
  cmp    $0x1,%r9d
  jle    L129
  movslq %r8d,%r8
  xor    %eax,%eax
  lea    (%rsi,%r8,4),%rdx
  jmp    L119
  nopl   0x0(%rax)
L22:
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%ecx
  jle    L129
L119:
  mov    (%rdx),%edi
  cmp    %edi,(%rsi,%rax,4)
  je     L22
  mov    %r14,%rsp
L116:
  add    $0x1,%r12d
  cmp    $0x989680,%r12d
  jne    L103
L48:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L131
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L129:
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %r14,%rsp
  jmp    L116
L131:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D3
data_type: string
name: retrieve_data
arguments: data_label: D33
data_type: f64
name: retrieve_data
arguments: data_label: D29
data_type: f64
name: retrieve_data
arguments: data_label: D11
data_type: qword[2]

tool: D3: %7d,  %4d, reverse


tool: D33: 0.5

tool: D29: 1e-05

tool: {"D11": [0x7fffffffffffffff,0x0000000000000000]}

assistant: ```c
void find2() {
  int number;
  for (number = 11; number <= 9999999; number++) {
    double root = pow(number, 0.5);
    if (fabs(root - (double)(int)root) < 0.00001) {
      int tempNumber = number;
      int length = (int)log10(number) + 1;
      int digits[length];
      int i;
      for (i = 0; i < length; i++) {
        digits[i] = tempNumber % 10;
        tempNumber /= 10;
      }
      int match = 1;
      for (i = 0; i < length / 2; i++) {
        if (digits[i] != digits[length - i - 1]) {
          match = 0;
          break;
        }
      }
      if (match == 1) {
        printf("%7d,  %4d, reverse\n", number, (int)root);
      }
    }
  }
}
```