user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D57(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rbp
  lea    D66(%rip),%rbx
  lea    D31(%rip),%r13
  lea    D57(%rip),%r12
  jmp    L70
L119:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L119
  mov    %r13,%rdi
  call   <puts@plt>
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L70:
  mov    %rsp,%rsi
  mov    %rbp,%rdx
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmp    $0x2,%eax
  jne    L119
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L119
  movss  (%rsp),%xmm0
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm0,%xmm1
  comisd D100(%rip),%xmm1
  jbe    L55
  movsd  D75(%rip),%xmm2
  comisd %xmm1,%xmm2
  jbe    L55
  cmpl   $0x0,0x4(%rsp)
  jne    L39
  lea    D7(%rip),%rdi
  call   <puts@plt>
  jmp    L32
L39:
  lea    D109(%rip),%rdi
  call   <puts@plt>
  jmp    L32
L55:
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jg     L90
  jne    L75
  lea    D4(%rip),%rdi
  call   <puts@plt>
  jmp    L32
L90:
  call   <my_power@plt>
  mov    %eax,%esi
  lea    D83(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L32:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L101
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L75:
  neg    %edi
  mov    %edi,0x4(%rsp)
  call   <my_power@plt>
  mov    %eax,%ecx
  mov    $0x1,%eax
  cltd
  idiv   %ecx
  mov    %eax,%esi
  lea    D83(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L32
L101:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D57
type_infered: string
name: read_data
arguments: data_tag: D66
type_infered: string
name: read_data
arguments: data_tag: D31
type_infered: string
name: read_data
arguments: data_tag: D100
type_infered: byte[16]
name: read_data
arguments: data_tag: D75
type_infered: byte[8]
name: read_data
arguments: data_tag: D7
type_infered: string
name: read_data
arguments: data_tag: D109
type_infered: string
name: read_data
arguments: data_tag: D4
type_infered: string
name: read_data
arguments: data_tag: D83
type_infered: string

tool: D57: Pls input the base and power number(e.g. 3.14, 2):

tool: D66: %f,%d

tool: D31: input invalid!

tool: {"D100": [0xe4,0x78,0xaa,0x9d,0xfb,0x38,0x0b,0xb8,0xe4,0x78,0xaa,0x9d,0xfb,0x38,0x0b,0x38]}

tool: {"D75": [0xe4,0x78,0xaa,0x9d,0xfb,0x38,0x0b,0x38]}

tool: D7: forbiden!

tool: D109: answer: 0

tool: D4: answer: 1

tool: D83: answer: %.6f


assistant: ```c
int main(void) {
  float num;
  int pow;
  printf("Pls input the base and power number(e.g. 3.14, 2):");
  while (scanf("%f,%d", &num, &pow) != 2 || getchar() != '\n') {
    while (getchar() != '\n')
      ;
    printf("input invalid!\n");
    printf("Pls input the base and power number(e.g. 3.14, 2):");
  }
  if (num > -1e-38 && num < 1e-38) {
    if (pow == 0)
      printf("forbiden!\n");
    else
      printf("answer: 0\n");
  } else {
    if (pow > 0)
      printf("answer: %.6f\n", my_power(num, pow));
    else if (pow == 0)
      printf("answer: 1\n");
    else {
      pow *= -1;
      printf("answer: %.6f\n", 1 / my_power(num, pow));
    }
  }
  return 0;
}
```