user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x44(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x3c(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x34(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x2c(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x44(%rbp),%edx
  mov    -0x40(%rbp),%eax
  add    %eax,%edx
  mov    -0x3c(%rbp),%eax
  add    %eax,%edx
  mov    -0x38(%rbp),%eax
  add    %eax,%edx
  mov    -0x34(%rbp),%eax
  add    %eax,%edx
  mov    -0x30(%rbp),%eax
  add    %eax,%edx
  mov    -0x2c(%rbp),%eax
  add    %eax,%edx
  mov    -0x28(%rbp),%eax
  add    %eax,%edx
  mov    -0x24(%rbp),%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  D0(%rip),%xmm1
  divsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x1c(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0x1c(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%r10d
  mov    -0x24(%rbp),%r9d
  mov    -0x28(%rbp),%r8d
  mov    -0x2c(%rbp),%edi
  mov    -0x30(%rbp),%ebx
  mov    -0x34(%rbp),%r11d
  mov    -0x38(%rbp),%ecx
  mov    -0x3c(%rbp),%edx
  mov    -0x40(%rbp),%esi
  mov    -0x44(%rbp),%eax
  push   %r10
  push   %r9
  push   %r8
  push   %rdi
  mov    %ebx,%r9d
  mov    %r11d,%r8d
  mov    %eax,%edi
  call   <sort@plt>
  add    $0x20,%rsp
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D19
type_guessed: string
name: disassemble_data
arguments: label: D106
type_guessed: string
name: disassemble_data
arguments: label: D0
type_guessed: f64
name: disassemble_data
arguments: label: D7
type_guessed: string

tool: D19: Please input 10 numbers:

tool: D106: %d

tool: D0: 10

tool: D7: Average: %f


assistant: ```c
int main() {
  int num1, num2, num3, num4, num5, num6;
  int num7, num8, num9, num10;
  float average;
  printf("Please input 10 numbers:\n");
  scanf("%d", &num1);
  scanf("%d", &num2);
  scanf("%d", &num3);
  scanf("%d", &num4);
  scanf("%d", &num5);
  scanf("%d", &num6);
  scanf("%d", &num7);
  scanf("%d", &num8);
  scanf("%d", &num9);
  scanf("%d", &num10);
  average = (num1 + num2 + num3 + num4 + num5 + num6 + num7 + num8 + num9 + num10) / 10.0;
  printf("Average: %f\n", average);
  sort(num1, num2, num3, num4, num5, num6, num7, num8, num9, num10);
}
```