ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<gcd_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1c64a,%esi
  mov    $0x70c80,%edi
  call   <gcd@plt>
  mov    %eax,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0xb,%esi
  mov    $0x2e,%edi
  call   <gcd@plt>
  mov    %eax,%esi
  lea    D34(%rip),%rdi
  call   <printf@plt>
  mov    $0x80,%esi
  mov    $0x2e,%edi
  call   <gcd@plt>
  mov    %eax,%esi
  lea    D49(%rip),%rdi
  call   <printf@plt>
  mov    $0x18,%esi
  mov    $0x36,%edi
  call   <gcd@plt>
  mov    %eax,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x1ff378,%esi
  mov    $0x5e000,%edi
  call   <gcd@plt>
  mov    %eax,%esi
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
datatype: string
name: unpack_data
arguments: label: D34
datatype: string
name: unpack_data
arguments: label: D49
datatype: string
name: unpack_data
arguments: label: D5
datatype: string
name: unpack_data
arguments: label: D33
datatype: string

tool: D0: gcd (461952, 116298)=%d


tool: D34: gcd (46, 11)=%d


tool: D49: gcd (4645, 5634)=%d


tool: D5: gcd (54, 24)=%d


tool: D33: gcd (385024, 2093944)=%d


assistant: ```c
int gcd_main(int argc, char **argv) {
  printf("gcd (461952, 116298)=%d\n", gcd(461952, 116298));
  printf("gcd (46, 11)=%d\n", gcd(46, 11));
  printf("gcd (4645, 5634)=%d\n", gcd(46, 128));
  printf("gcd (54, 24)=%d\n", gcd(54, 24));
  printf("gcd (385024, 2093944)=%d\n", gcd(385024, 2093944));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<gcd_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1c64a,%esi
  mov    $0x70c80,%edi
  call   <gcd@plt>
  lea    D31(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xb,%esi
  mov    $0x2e,%edi
  call   <gcd@plt>
  lea    D42(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x80,%esi
  mov    $0x2e,%edi
  call   <gcd@plt>
  lea    D43(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x18,%esi
  mov    $0x36,%edi
  call   <gcd@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1ff378,%esi
  mov    $0x5e000,%edi
  call   <gcd@plt>
  lea    D47(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D31
type_infered: string
name: unpack_data
arguments: label: D42
type_infered: string
name: unpack_data
arguments: label: D43
type_infered: string
name: unpack_data
arguments: label: D1
type_infered: string
name: unpack_data
arguments: label: D47
type_infered: string

tool: D31: gcd (461952, 116298)=%d


tool: D42: gcd (46, 11)=%d


tool: D43: gcd (4645, 5634)=%d


tool: D1: gcd (54, 24)=%d


tool: D47: gcd (385024, 2093944)=%d


assistant: ```c
int gcd_main(int argc, char **argv) {
  printf("gcd (461952, 116298)=%d\n", gcd(461952, 116298));
  printf("gcd (46, 11)=%d\n", gcd(46, 11));
  printf("gcd (4645, 5634)=%d\n", gcd(46, 128));
  printf("gcd (54, 24)=%d\n", gcd(54, 24));
  printf("gcd (385024, 2093944)=%d\n", gcd(385024, 2093944));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<gcd_main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1c64a,%esi
  mov    $0x70c80,%edi
  call   <gcd@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0xb,%esi
  mov    $0x2e,%edi
  call   <gcd@plt>
  lea    D14(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x80,%esi
  mov    $0x2e,%edi
  call   <gcd@plt>
  lea    D15(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x18,%esi
  mov    $0x36,%edi
  call   <gcd@plt>
  lea    D46(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x1ff378,%esi
  mov    $0x5e000,%edi
  call   <gcd@plt>
  lea    D10(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
type_infered: string
name: unpack_data
arguments: label: D14
type_infered: string
name: unpack_data
arguments: label: D15
type_infered: string
name: unpack_data
arguments: label: D46
type_infered: string
name: unpack_data
arguments: label: D10
type_infered: string

tool: D8: gcd (461952, 116298)=%d


tool: D14: gcd (46, 11)=%d


tool: D15: gcd (4645, 5634)=%d


tool: D46: gcd (54, 24)=%d


tool: D10: gcd (385024, 2093944)=%d


assistant: ```c
int gcd_main(int argc, char **argv) {
  printf("gcd (461952, 116298)=%d\n", gcd(461952, 116298));
  printf("gcd (46, 11)=%d\n", gcd(46, 11));
  printf("gcd (4645, 5634)=%d\n", gcd(46, 128));
  printf("gcd (54, 24)=%d\n", gcd(54, 24));
  printf("gcd (385024, 2093944)=%d\n", gcd(385024, 2093944));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x12(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D12
datatype: string
name: retrieve_data
arguments: label: D6
datatype: string

tool: D12: Endereco de memoria var1: %x


tool: D6: Endereco de memoria var2: %x


assistant: ```c
int main() {
  int var1;
  char var2[10];
  printf("Endereco de memoria var1: %x\n", &var1);
  printf("Endereco de memoria var2: %x\n", &var2);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    0xe(%rsp),%rsi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D1
type: string
name: read_data
arguments: data_tag: D13
type: string

tool: D1: Endereco de memoria var1: %x


tool: D13: Endereco de memoria var2: %x


assistant: ```c
int main() {
  int var1;
  char var2[10];
  printf("Endereco de memoria var1: %x\n", &var1);
  printf("Endereco de memoria var2: %x\n", &var2);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D13(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <printf@plt>
  xor    %eax,%eax
  lea    0xe(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
type_infered: string
name: retrieve_data
arguments: data_tag: D12
type_infered: string

tool: D13: Endereco de memoria var1: %x


tool: D12: Endereco de memoria var2: %x


assistant: ```c
int main() {
  int var1;
  char var2[10];
  printf("Endereco de memoria var1: %x\n", &var1);
  printf("Endereco de memoria var2: %x\n", &var2);
  return 0;
}
```
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
  sub    $0x28,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <printf@plt>
  xor    %eax,%eax
  lea    0xe(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D16
data_type: string
name: extract_symbolic_data
arguments: label: D7
data_type: string

tool: D16: Endereco de memoria var1: %x


tool: D7: Endereco de memoria var2: %x


assistant: ```c
int main() {
  int var1;
  char var2[10];
  printf("Endereco de memoria var1: %x\n", &var1);
  printf("Endereco de memoria var2: %x\n", &var2);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<unicoder_decodeBom>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x18(%rbp)
  jne    L63
  mov    $0xffffffff,%eax
  jmp    L3
L63:
  movl   $0x0,-0x10(%rbp)
  jmp    L22
L193:
  mov    -0x10(%rbp),%eax
  cltq
  movb   $0x0,-0xc(%rbp,%rax,1)
  addl   $0x1,-0x10(%rbp)
L22:
  cmpl   $0x3,-0x10(%rbp)
  jle    L193
  mov    -0x18(%rbp),%rcx
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rcx
  lea    -0xc(%rbp),%rax
  add    $0x1,%rax
  mov    $0x3,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x18(%rbp),%rax
  lea    0x2(%rax),%rcx
  lea    -0xc(%rbp),%rax
  add    $0x2,%rax
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x18(%rbp),%rax
  lea    0x3(%rax),%rcx
  lea    -0xc(%rbp),%rax
  add    $0x3,%rax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movzbl -0xc(%rbp),%eax
  test   %al,%al
  jne    L152
  movzbl -0xb(%rbp),%eax
  test   %al,%al
  jne    L152
  movzbl -0xa(%rbp),%eax
  cmp    $0xfe,%al
  jne    L152
  movzbl -0x9(%rbp),%eax
  cmp    $0xff,%al
  jne    L152
  mov    $0x5,%eax
  jmp    L3
L152:
  movzbl -0xc(%rbp),%eax
  cmp    $0xff,%al
  jne    L143
  movzbl -0xb(%rbp),%eax
  cmp    $0xfe,%al
  jne    L143
  movzbl -0xa(%rbp),%eax
  test   %al,%al
  jne    L143
  movzbl -0x9(%rbp),%eax
  test   %al,%al
  jne    L143
  mov    $0x6,%eax
  jmp    L3
L143:
  movzbl -0xc(%rbp),%eax
  cmp    $0xef,%al
  jne    L32
  movzbl -0xb(%rbp),%eax
  cmp    $0xbb,%al
  jne    L32
  movzbl -0xa(%rbp),%eax
  cmp    $0xbf,%al
  jne    L32
  mov    $0x2,%eax
  jmp    L3
L32:
  movzbl -0xc(%rbp),%eax
  cmp    $0xfe,%al
  jne    L106
  movzbl -0xb(%rbp),%eax
  cmp    $0xff,%al
  jne    L106
  mov    $0x3,%eax
  jmp    L3
L106:
  movzbl -0xc(%rbp),%eax
  cmp    $0xff,%al
  jne    L11
  movzbl -0xb(%rbp),%eax
  cmp    $0xfe,%al
  jne    L11
  mov    $0x4,%eax
  jmp    L3
L11:
  mov    $0x1,%eax
L3:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L109
  call   <__stack_chk_fail@plt>
L109:
  leave
  ret
```

assistant: ```c
int unicoder_decodeBom(unsigned char *p) {
  int i;
  unsigned int len;
  unsigned char newp[4];
  if (p == NULL)
    return -1;
  for (i = 0; i < 4; i++)
    newp[i] = 0;
  strncpy(newp, p, 4);
  strncpy(newp + 1, p + 1, 3);
  strncpy(newp + 2, p + 2, 2);
  strncpy(newp + 3, p + 3, 1);
  if ((newp[0] == 0x00) && (newp[1] == 0x00) && (newp[2] == 0xfe) && (newp[3] == 0xff))
    return 5;
  if ((newp[0] == 0xff) && (newp[1] == 0xfe) && (newp[2] == 0x00) && (newp[3] == 0x00))
    return 6;
  if ((newp[0] == 0xef) && (newp[1] == 0xbb) && (newp[2] == 0xbf))
    return 2;
  if ((newp[0] == 0xfe) && (newp[1] == 0xff))
    return 3;
  if ((newp[0] == 0xff) && (newp[1] == 0xfe))
    return 4;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<unicoder_decodeBom>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L80
  mov    %rdi,%rbx
  movb   $0x0,0x4(%rsp)
  movb   $0x0,0x5(%rsp)
  movb   $0x0,0x6(%rsp)
  movb   $0x0,0x7(%rsp)
  lea    0x4(%rsp),%rdi
  mov    $0x4,%edx
  mov    %rbx,%rsi
  call   <strncpy@plt>
  lea    0x1(%rbx),%rsi
  lea    0x5(%rsp),%rdi
  mov    $0x3,%edx
  call   <strncpy@plt>
  lea    0x2(%rbx),%rsi
  lea    0x6(%rsp),%rdi
  mov    $0x2,%edx
  call   <strncpy@plt>
  lea    0x3(%rbx),%rsi
  lea    0x7(%rsp),%rdi
  mov    $0x1,%edx
  call   <strncpy@plt>
  movzbl 0x4(%rsp),%eax
  test   %al,%al
  jne    L85
  mov    $0x1,%eax
  cmpb   $0x0,0x5(%rsp)
  jne    L67
  cmpb   $0xfe,0x6(%rsp)
  jne    L67
  cmpb   $0xff,0x7(%rsp)
  sete   %al
  movzbl %al,%eax
  lea    0x1(,%rax,4),%eax
  jmp    L67
L85:
  cmp    $0xff,%al
  je     L136
  cmp    $0xef,%al
  jne    L60
  mov    $0x1,%eax
  cmpb   $0xbb,0x5(%rsp)
  jne    L67
  cmpb   $0xbf,0x6(%rsp)
  sete   %al
  movzbl %al,%eax
  add    $0x1,%eax
  jmp    L67
L136:
  mov    $0x1,%eax
  cmpb   $0xfe,0x5(%rsp)
  jne    L67
  movzbl 0x6(%rsp),%eax
  or     0x7(%rsp),%al
  cmp    $0x1,%al
  sbb    %eax,%eax
  and    $0x2,%eax
  add    $0x4,%eax
  jmp    L67
L60:
  cmp    $0xfe,%al
  jne    L145
  cmpb   $0xff,0x5(%rsp)
  sete   %al
  movzbl %al,%eax
  lea    0x1(%rax,%rax,1),%eax
L67:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L120
  add    $0x10,%rsp
  pop    %rbx
  ret
L80:
  mov    $0xffffffff,%eax
  jmp    L67
L145:
  mov    $0x1,%eax
  jmp    L67
L120:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int unicoder_decodeBom(unsigned char *p) {
  int i;
  unsigned int len;
  unsigned char newp[4];
  if (p == NULL)
    return -1;
  for (i = 0; i < 4; i++)
    newp[i] = 0;
  strncpy(newp, p, 4);
  strncpy(newp + 1, p + 1, 3);
  strncpy(newp + 2, p + 2, 2);
  strncpy(newp + 3, p + 3, 1);
  if ((newp[0] == 0x00) && (newp[1] == 0x00) && (newp[2] == 0xfe) && (newp[3] == 0xff))
    return 5;
  if ((newp[0] == 0xff) && (newp[1] == 0xfe) && (newp[2] == 0x00) && (newp[3] == 0x00))
    return 6;
  if ((newp[0] == 0xef) && (newp[1] == 0xbb) && (newp[2] == 0xbf))
    return 2;
  if ((newp[0] == 0xfe) && (newp[1] == 0xff))
    return 3;
  if ((newp[0] == 0xff) && (newp[1] == 0xfe))
    return 4;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<unicoder_decodeBom>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L85
  mov    %rdi,%rbx
  mov    $0x4,%edx
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  mov    %rbx,%rsi
  call   <strncpy@plt>
  lea    0x1(%rbx),%rsi
  lea    0x5(%rsp),%rdi
  mov    $0x3,%edx
  call   <strncpy@plt>
  lea    0x2(%rbx),%rsi
  lea    0x6(%rsp),%rdi
  mov    $0x2,%edx
  call   <strncpy@plt>
  lea    0x3(%rbx),%rsi
  lea    0x7(%rsp),%rdi
  mov    $0x1,%edx
  call   <strncpy@plt>
  movzbl 0x4(%rsp),%eax
  test   %al,%al
  je     L136
  cmp    $0xff,%al
  jne    L17
  cmpb   $0xfe,0x5(%rsp)
  mov    $0x1,%eax
  jne    L26
  movzbl 0x6(%rsp),%eax
  or     0x7(%rsp),%al
  cmp    $0x1,%al
  sbb    %eax,%eax
  and    $0x2,%eax
  add    $0x4,%eax
  nopl   0x0(%rax)
L26:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L110
  add    $0x10,%rsp
  pop    %rbx
  ret
  xchg   %ax,%ax
L136:
  cmpb   $0x0,0x5(%rsp)
  mov    $0x1,%eax
  jne    L26
  cmpb   $0xfe,0x6(%rsp)
  jne    L26
  xor    %eax,%eax
  cmpb   $0xff,0x7(%rsp)
  sete   %al
  lea    0x1(,%rax,4),%eax
  jmp    L26
  cs nopw 0x0(%rax,%rax,1)
L17:
  cmp    $0xef,%al
  jne    L67
  cmpb   $0xbb,0x5(%rsp)
  mov    $0x1,%eax
  jne    L26
  xor    %eax,%eax
  cmpb   $0xbf,0x6(%rsp)
  sete   %al
  add    $0x1,%eax
  jmp    L26
  nop
L67:
  cmp    $0xfe,%al
  jne    L0
  xor    %eax,%eax
  cmpb   $0xff,0x5(%rsp)
  sete   %al
  lea    0x1(%rax,%rax,1),%eax
  jmp    L26
L85:
  mov    $0xffffffff,%eax
  jmp    L26
L110:
  call   <__stack_chk_fail@plt>
L0:
  mov    $0x1,%eax
  jmp    L26
```

assistant: ```c
int unicoder_decodeBom(unsigned char *p) {
  int i;
  unsigned int len;
  unsigned char newp[4];
  if (p == NULL)
    return -1;
  for (i = 0; i < 4; i++)
    newp[i] = 0;
  strncpy(newp, p, 4);
  strncpy(newp + 1, p + 1, 3);
  strncpy(newp + 2, p + 2, 2);
  strncpy(newp + 3, p + 3, 1);
  if ((newp[0] == 0x00) && (newp[1] == 0x00) && (newp[2] == 0xfe) && (newp[3] == 0xff))
    return 5;
  if ((newp[0] == 0xff) && (newp[1] == 0xfe) && (newp[2] == 0x00) && (newp[3] == 0x00))
    return 6;
  if ((newp[0] == 0xef) && (newp[1] == 0xbb) && (newp[2] == 0xbf))
    return 2;
  if ((newp[0] == 0xfe) && (newp[1] == 0xff))
    return 3;
  if ((newp[0] == 0xff) && (newp[1] == 0xfe))
    return 4;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<unicoder_decodeBom>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L78
  mov    %rdi,%rbx
  mov    $0x4,%edx
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  mov    %rbx,%rsi
  call   <strncpy@plt>
  lea    0x1(%rbx),%rsi
  lea    0x5(%rsp),%rdi
  mov    $0x3,%edx
  call   <strncpy@plt>
  lea    0x2(%rbx),%rsi
  lea    0x6(%rsp),%rdi
  mov    $0x2,%edx
  call   <strncpy@plt>
  lea    0x3(%rbx),%rsi
  lea    0x7(%rsp),%rdi
  mov    $0x1,%edx
  call   <strncpy@plt>
  movzbl 0x4(%rsp),%eax
  test   %al,%al
  je     L11
  cmp    $0xff,%al
  jne    L120
  cmpb   $0xfe,0x5(%rsp)
  mov    $0x1,%eax
  jne    L103
  movzbl 0x6(%rsp),%eax
  or     0x7(%rsp),%al
  cmp    $0x1,%al
  sbb    %eax,%eax
  and    $0x2,%eax
  add    $0x4,%eax
  nopl   0x0(%rax)
L103:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L139
  add    $0x10,%rsp
  pop    %rbx
  ret
  xchg   %ax,%ax
L11:
  cmpb   $0x0,0x5(%rsp)
  mov    $0x1,%eax
  jne    L103
  cmpb   $0xfe,0x6(%rsp)
  jne    L103
  xor    %eax,%eax
  cmpb   $0xff,0x7(%rsp)
  sete   %al
  lea    0x1(,%rax,4),%eax
  jmp    L103
  cs nopw 0x0(%rax,%rax,1)
L120:
  cmp    $0xef,%al
  jne    L0
  cmpb   $0xbb,0x5(%rsp)
  mov    $0x1,%eax
  jne    L103
  xor    %eax,%eax
  cmpb   $0xbf,0x6(%rsp)
  sete   %al
  add    $0x1,%eax
  jmp    L103
  nop
L0:
  cmp    $0xfe,%al
  jne    L63
  xor    %eax,%eax
  cmpb   $0xff,0x5(%rsp)
  sete   %al
  lea    0x1(%rax,%rax,1),%eax
  jmp    L103
L78:
  mov    $0xffffffff,%eax
  jmp    L103
L139:
  call   <__stack_chk_fail@plt>
L63:
  mov    $0x1,%eax
  jmp    L103
```

assistant: ```c
int unicoder_decodeBom(unsigned char *p) {
  int i;
  unsigned int len;
  unsigned char newp[4];
  if (p == NULL)
    return -1;
  for (i = 0; i < 4; i++)
    newp[i] = 0;
  strncpy(newp, p, 4);
  strncpy(newp + 1, p + 1, 3);
  strncpy(newp + 2, p + 2, 2);
  strncpy(newp + 3, p + 3, 1);
  if ((newp[0] == 0x00) && (newp[1] == 0x00) && (newp[2] == 0xfe) && (newp[3] == 0xff))
    return 5;
  if ((newp[0] == 0xff) && (newp[1] == 0xfe) && (newp[2] == 0x00) && (newp[3] == 0x00))
    return 6;
  if ((newp[0] == 0xef) && (newp[1] == 0xbb) && (newp[2] == 0xbf))
    return 2;
  if ((newp[0] == 0xfe) && (newp[1] == 0xff))
    return 3;
  if ((newp[0] == 0xff) && (newp[1] == 0xfe))
    return 4;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<refract>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  movsd  %xmm0,-0x38(%rbp)
  movsd  %xmm1,-0x40(%rbp)
  movsd  %xmm2,-0x48(%rbp)
  mov    %rdi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x28(%rbp),%rdx
  movsd  -0x48(%rbp),%xmm1
  movsd  -0x40(%rbp),%xmm0
  mov    -0x38(%rbp),%rax
  mov    %rdx,%rdi
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <unrefract@plt>
  movsd  -0x28(%rbp),%xmm2
  movsd  -0x48(%rbp),%xmm0
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  movsd  D31(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x48(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
L24:
  movsd  -0x10(%rbp),%xmm0
  addsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  lea    -0x28(%rbp),%rdx
  movsd  -0x10(%rbp),%xmm1
  movsd  -0x40(%rbp),%xmm0
  mov    -0x38(%rbp),%rax
  mov    %rdx,%rdi
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <unrefract@plt>
  movsd  -0x28(%rbp),%xmm1
  movsd  -0x48(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movq   D39(%rip),%xmm1
  andpd  %xmm0,%xmm1
  movsd  D16(%rip),%xmm0
  comisd %xmm1,%xmm0
  jae    L1
  movsd  -0x28(%rbp),%xmm1
  movsd  -0x48(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movq   D8(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  -0x28(%rbp),%xmm2
  movsd  -0x18(%rbp),%xmm1
  subsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  movsd  -0x20(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  jmp    L24
L1:
  nop
  mov    -0x50(%rbp),%rax
  movsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D31
type: f64
name: unpack_data
arguments: label: D39
type: qword
name: unpack_data
arguments: label: D16
type: byte[8]
name: unpack_data
arguments: label: D8
type: byte[8]

tool: D31: 0.8

tool: D39: 9223372036854775807

tool: {"D16": [0x4b,0xd9,0x32,0x8f,0x84,0x44,0xa0,0x3e]}

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
void refract(double pr, double tr, double ta, double *aa) {
  double d, t, t0, a;
  unrefract(pr, tr, ta, &t);
  d = 0.8 * (ta - t);
  t0 = t;
  a = ta;
  while (1) {
    a += d;
    unrefract(pr, tr, a, &t);
    if (fabs(ta - t) <= ((0.1 / 3600.) * 3.141592653589793 / 180.))
      break;
    d *= -(ta - t) / (t0 - t);
    t0 = t;
  }
  *aa = a;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<refract>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  movsd  %xmm0,0x8(%rsp)
  movq   %xmm1,%r13
  movq   %xmm2,%r12
  mov    %rdi,%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  call   <unrefract@plt>
  mov    0x10(%rsp),%rbx
  movq   %r12,%xmm0
  movq   %rbx,%xmm6
  subsd  %xmm6,%xmm0
  mulsd  D10(%rip),%xmm0
  movsd  %xmm0,(%rsp)
  mov    %r12,%rbp
  lea    0x10(%rsp),%r14
  jmp    L28
L17:
  xorpd  D31(%rip),%xmm0
  movq   %rbx,%xmm2
  subsd  %xmm1,%xmm2
  divsd  %xmm2,%xmm0
  mulsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  movq   %xmm1,%rbx
L28:
  movq   %rbp,%xmm3
  addsd  (%rsp),%xmm3
  movq   %xmm3,%rbp
  mov    %r14,%rdi
  movapd %xmm3,%xmm2
  movq   %r13,%xmm1
  movsd  0x8(%rsp),%xmm0
  call   <unrefract@plt>
  movsd  0x10(%rsp),%xmm1
  movq   %r12,%xmm0
  subsd  %xmm1,%xmm0
  movapd %xmm0,%xmm2
  andpd  D1(%rip),%xmm2
  movsd  D33(%rip),%xmm5
  comisd %xmm2,%xmm5
  jb     L17
  mov    %rbp,(%r15)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D10
type_infered: f64
name: parse_data
arguments: data_tag: D31
type_infered: byte[16]
name: parse_data
arguments: data_tag: D1
type_infered: qword[2]
name: parse_data
arguments: data_tag: D33
type_infered: byte[8]

tool: D10: 0.8

tool: {"D31": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D1": [0x7fffffffffffffff,0x0000000000000000]}

tool: {"D33": [0x4b,0xd9,0x32,0x8f,0x84,0x44,0xa0,0x3e]}

assistant: ```c
void refract(double pr, double tr, double ta, double *aa) {
  double d, t, t0, a;
  unrefract(pr, tr, ta, &t);
  d = 0.8 * (ta - t);
  t0 = t;
  a = ta;
  while (1) {
    a += d;
    unrefract(pr, tr, a, &t);
    if (fabs(ta - t) <= ((0.1 / 3600.) * 3.141592653589793 / 180.))
      break;
    d *= -(ta - t) / (t0 - t);
    t0 = t;
  }
  *aa = a;
}
```