ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<to_bin>:
  endbr64
  xor    %eax,%eax
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L32:
  test   %rax,%rax
  je     L18
  lea    0x1(%rax),%edx
  imul   $0xcccccccd,%edx,%edx
  cmp    $0x33333333,%edx
  ja     L18
  cmpb   $0xa,(%rdi,%rax,1)
  je     L81
L24:
  mov    $0xffffffff,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  movzbl (%rdi,%rax,1),%edx
  cmp    $0x2e,%dl
  je     L52
  cmp    $0x23,%dl
  jne    L24
  lea    0x1(%r8,%r8,1),%r8d
L81:
  add    $0x1,%rax
  cmp    $0x14,%rax
  jne    L32
L8:
  movzbl 0x14(%rdi),%eax
  cmp    $0xa,%al
  je     L20
  test   %al,%al
  jne    L24
L20:
  mov    %r8d,%edi
  jmp    1050 <convert@plt>
  nopl   0x0(%rax)
L52:
  add    $0x1,%rax
  add    %r8d,%r8d
  cmp    $0x14,%rax
  jne    L32
  jmp    L8
```

assistant: ```c
int to_bin(char buff[21]) {
  int val;
  int i;
  i = -1;
  val = 0;
  while (i++ < 20)
    if (i == 20 && (buff[i] == '\n' || buff[i] == '\0'))
      return (convert(val));
    else if (i == 20)
      return (-1);
    else if (i != 0 && (i + 1) % 5 == 0 && buff[i] == '\n')
      continue;
    else if (i != 0 && (i + 1) % 5 == 0)
      return (-1);
    else if (buff[i] == '.')
      val *= 2;
    else if (buff[i] == '#')
      val = val * 2 + 1;
    else
      return (-1);
  return (-1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<to_bin>:
  endbr64
  xor    %eax,%eax
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L77:
  test   %rax,%rax
  je     L57
  lea    0x1(%rax),%edx
  imul   $0xcccccccd,%edx,%edx
  cmp    $0x33333333,%edx
  ja     L57
  cmpb   $0xa,(%rdi,%rax,1)
  je     L61
L44:
  mov    $0xffffffff,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L57:
  movzbl (%rdi,%rax,1),%edx
  cmp    $0x2e,%dl
  je     L18
  cmp    $0x23,%dl
  jne    L44
  lea    0x1(%r8,%r8,1),%r8d
L61:
  add    $0x1,%rax
  cmp    $0x14,%rax
  jne    L77
L78:
  movzbl 0x14(%rdi),%eax
  cmp    $0xa,%al
  je     L10
  test   %al,%al
  jne    L44
L10:
  mov    %r8d,%edi
  jmp    1050 <convert@plt>
  nopl   0x0(%rax)
L18:
  add    $0x1,%rax
  add    %r8d,%r8d
  cmp    $0x14,%rax
  jne    L77
  jmp    L78
```

assistant: ```c
int to_bin(char buff[21]) {
  int val;
  int i;
  i = -1;
  val = 0;
  while (i++ < 20)
    if (i == 20 && (buff[i] == '\n' || buff[i] == '\0'))
      return (convert(val));
    else if (i == 20)
      return (-1);
    else if (i != 0 && (i + 1) % 5 == 0 && buff[i] == '\n')
      continue;
    else if (i != 0 && (i + 1) % 5 == 0)
      return (-1);
    else if (buff[i] == '.')
      val *= 2;
    else if (buff[i] == '#')
      val = val * 2 + 1;
    else
      return (-1);
  return (-1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dissect_mobility_options>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L4
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <tprintf@plt>
L4:
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
type_infered: string

tool: D4: MH Option(s) recognized 

assistant: ```c
void dissect_mobility_options(struct pkt_buff *pkt __attribute__((__unused__)), ssize_t *message_data_len) {
  if (*message_data_len)
    tprintf("MH Option(s) recognized ");
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<dissect_mobility_options>:
  endbr64
  cmpq   $0x0,(%rsi)
  jne    L1
  ret
L1:
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <tprintf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D7
type_infered: string

tool: D7: MH Option(s) recognized 

assistant: ```c
void dissect_mobility_options(struct pkt_buff *pkt __attribute__((__unused__)), ssize_t *message_data_len) {
  if (*message_data_len)
    tprintf("MH Option(s) recognized ");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<dissect_mobility_options>:
  endbr64
  cmpq   $0x0,(%rsi)
  jne    L9
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  lea    D4(%rip),%rdi
  jmp    1050 <tprintf@plt>
```

tool_calls: name: parse_data
arguments: data_label: D4
data_type: string

tool: D4: MH Option(s) recognized 

assistant: ```c
void dissect_mobility_options(struct pkt_buff *pkt __attribute__((__unused__)), ssize_t *message_data_len) {
  if (*message_data_len)
    tprintf("MH Option(s) recognized ");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dissect_mobility_options>:
  endbr64
  cmpq   $0x0,(%rsi)
  jne    L7
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  lea    D0(%rip),%rdi
  jmp    1050 <tprintf@plt>
```

tool_calls: name: parse_data
arguments: data_label: D0
type_infered: string

tool: D0: MH Option(s) recognized 

assistant: ```c
void dissect_mobility_options(struct pkt_buff *pkt __attribute__((__unused__)), ssize_t *message_data_len) {
  if (*message_data_len)
    tprintf("MH Option(s) recognized ");
}
```
user: Translate the below assembly code into C programming language:

```asm
<save_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  mov    $0x0,%eax
  call   <memory_error@plt>
L7:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *save_string(char *string) {
  char *new_string;
  new_string = malloc((unsigned)(strlen(string) + 1));
  if (new_string == NULL)
    memory_error();
  strcpy(new_string, string);
  return (new_string);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<save_string>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L12
L13:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L12:
  mov    $0x0,%eax
  call   <memory_error@plt>
  jmp    L13
```

assistant: ```c
char *save_string(char *string) {
  char *new_string;
  new_string = malloc((unsigned)(strlen(string) + 1));
  if (new_string == NULL)
    memory_error();
  strcpy(new_string, string);
  return (new_string);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<save_string>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L15
L14:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L15:
  xor    %eax,%eax
  call   <memory_error@plt>
  jmp    L14
```

assistant: ```c
char *save_string(char *string) {
  char *new_string;
  new_string = malloc((unsigned)(strlen(string) + 1));
  if (new_string == NULL)
    memory_error();
  strcpy(new_string, string);
  return (new_string);
}
```
user: Translate the below assembly code into C programming language:

```asm
<save_string>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L16
L5:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L16:
  xor    %eax,%eax
  call   <memory_error@plt>
  jmp    L5
```

assistant: ```c
char *save_string(char *string) {
  char *new_string;
  new_string = malloc((unsigned)(strlen(string) + 1));
  if (new_string == NULL)
    memory_error();
  strcpy(new_string, string);
  return (new_string);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<unquote>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L3:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  jne    L23
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L23
  addq   $0x1,-0x18(%rbp)
L23:
  mov    -0x18(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void unquote(char *ptr) {
  char *dest;
  dest = ptr;
  do {
    if (*ptr == '\\' && ptr[1] != '\0') {
      ptr++;
    }
  } while ((*dest++ = *ptr++) != '\0');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unquote>:
  endbr64
  lea    0x1(%rdi),%rax
  jmp    L24
L8:
  lea    0x1(%rdi),%rcx
  movzbl (%rdi),%edx
  mov    %dl,-0x1(%rax)
  test   %dl,%dl
  je     L29
L37:
  add    $0x1,%rax
  mov    %rcx,%rdi
L24:
  cmpb   $0x5c,(%rdi)
  jne    L8
  cmpb   $0x0,0x1(%rdi)
  je     L45
  add    $0x1,%rdi
  jmp    L8
L45:
  lea    0x1(%rdi),%rcx
  movb   $0x5c,-0x1(%rax)
  jmp    L37
L29:
  ret
```

assistant: ```c
void unquote(char *ptr) {
  char *dest;
  dest = ptr;
  do {
    if (*ptr == '\\' && ptr[1] != '\0') {
      ptr++;
    }
  } while ((*dest++ = *ptr++) != '\0');
}
```
user: Translate the below assembly code into C programming language:

```asm
<unquote>:
  endbr64
  mov    %rdi,%rax
  nopw   0x0(%rax,%rax,1)
L36:
  movzbl (%rax),%edx
  add    $0x1,%rdi
  cmp    $0x5c,%dl
  je     L32
  mov    %dl,-0x1(%rdi)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L36
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  movzbl 0x1(%rax),%edx
  lea    0x1(%rax),%rcx
  test   %dl,%dl
  je     L44
  mov    %dl,-0x1(%rdi)
  add    $0x2,%rax
  jmp    L36
  nopl   (%rax)
L44:
  movb   $0x5c,-0x1(%rdi)
  mov    %rcx,%rax
  jmp    L36
```

assistant: ```c
void unquote(char *ptr) {
  char *dest;
  dest = ptr;
  do {
    if (*ptr == '\\' && ptr[1] != '\0') {
      ptr++;
    }
  } while ((*dest++ = *ptr++) != '\0');
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<unquote>:
  endbr64
  mov    %rdi,%rax
  nopw   0x0(%rax,%rax,1)
L11:
  movzbl (%rax),%edx
  add    $0x1,%rdi
  cmp    $0x5c,%dl
  je     L41
  mov    %dl,-0x1(%rdi)
  add    $0x1,%rax
  test   %dl,%dl
  jne    L11
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  movzbl 0x1(%rax),%edx
  lea    0x1(%rax),%rcx
  test   %dl,%dl
  je     L12
  mov    %dl,-0x1(%rdi)
  add    $0x2,%rax
  jmp    L11
  nopl   (%rax)
L12:
  movb   $0x5c,-0x1(%rdi)
  mov    %rcx,%rax
  jmp    L11
```

assistant: ```c
void unquote(char *ptr) {
  char *dest;
  dest = ptr;
  do {
    if (*ptr == '\\' && ptr[1] != '\0') {
      ptr++;
    }
  } while ((*dest++ = *ptr++) != '\0');
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L25
  cmpl   $0x0,-0x4(%rbp)
  je     L21
  mov    $0x1,%eax
  jmp    L26
L25:
  mov    $0x2,%eax
  jmp    L26
L21:
  nop
  cmpl   $0x1,-0x4(%rbp)
  jne    L45
  mov    $0x3,%eax
  jmp    L26
L45:
  cmpl   $0x1,-0x4(%rbp)
  jne    L86
  mov    $0x4,%eax
  jmp    L26
L86:
  cmpl   $0x0,-0x4(%rbp)
  je     L55
  cmpl   $0x1,-0x4(%rbp)
  je     L106
  jmp    L79
L55:
  mov    -0x4(%rbp),%eax
  jmp    L26
L106:
  mov    $0x1,%eax
  jmp    L26
L79:
  mov    $0x1,%eax
L26:
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int x;
  x = 0;
  switch (x)
  case 0:;
    switch (x)
    case 0:
      switch (x) {
      case 0:
        goto next;
      default:
        return 1;
      }
  return 2;
next:
  switch (x)
  case 1:
    return 3;
  switch (x) {
    x = 1 + 1;
  foo:
  case 1:
    return 4;
  }
  switch (x) {
  case 0:
    return x;
  case 1:
    return 1;
  default:
    return 1;
  }
}
```
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
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  x = 0;
  switch (x)
  case 0:;
    switch (x)
    case 0:
      switch (x) {
      case 0:
        goto next;
      default:
        return 1;
      }
  return 2;
next:
  switch (x)
  case 1:
    return 3;
  switch (x) {
    x = 1 + 1;
  foo:
  case 1:
    return 4;
  }
  switch (x) {
  case 0:
    return x;
  case 1:
    return 1;
  default:
    return 1;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  x = 0;
  switch (x)
  case 0:;
    switch (x)
    case 0:
      switch (x) {
      case 0:
        goto next;
      default:
        return 1;
      }
  return 2;
next:
  switch (x)
  case 1:
    return 3;
  switch (x) {
    x = 1 + 1;
  foo:
  case 1:
    return 4;
  }
  switch (x) {
  case 0:
    return x;
  case 1:
    return 1;
  default:
    return 1;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int x;
  x = 0;
  switch (x)
  case 0:;
    switch (x)
    case 0:
      switch (x) {
      case 0:
        goto next;
      default:
        return 1;
      }
  return 2;
next:
  switch (x)
  case 1:
    return 3;
  switch (x) {
    x = 1 + 1;
  foo:
  case 1:
    return 4;
  }
  switch (x) {
  case 0:
    return x;
  case 1:
    return 1;
  default:
    return 1;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<th2in>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x110,%rsp
  mov    %rdi,-0xf8(%rbp)
  mov    %rsi,-0x100(%rbp)
  mov    %rdx,-0x108(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xe4(%rbp)
  jmp    L18
L21:
  mov    -0xe4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x100(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xe4(%rbp),%eax
  cltq
  movsd  %xmm0,-0x70(%rbp,%rax,8)
  addl   $0x1,-0xe4(%rbp)
L18:
  cmpl   $0x3,-0xe4(%rbp)
  jle    L21
  mov    -0xf8(%rbp),%rax
  movsd  0x10(%rax),%xmm0
  movsd  %xmm0,-0x60(%rbp)
  lea    -0x90(%rbp),%rdx
  lea    -0x70(%rbp),%rax
  mov    -0x108(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  movl   $0x0,-0xe4(%rbp)
  jmp    L0
L7:
  mov    -0xe4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x100(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0xe4(%rbp),%eax
  cltq
  movsd  %xmm0,-0x50(%rbp,%rax,8)
  addl   $0x1,-0xe4(%rbp)
L0:
  cmpl   $0x3,-0xe4(%rbp)
  jle    L7
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x40(%rbp)
  lea    -0x30(%rbp),%rdx
  lea    -0x50(%rbp),%rax
  mov    -0x108(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  lea    -0xb0(%rbp),%rdx
  mov    -0xf8(%rbp),%rax
  mov    -0x108(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  lea    -0xd0(%rbp),%rdx
  mov    -0x100(%rbp),%rax
  mov    -0x108(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  mov    -0x108(%rbp),%rdx
  mov    -0x100(%rbp),%rcx
  mov    -0xf8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sinth0@plt>
  mov    %eax,%edi
  call   <asin@plt>
  movq   %xmm0,%rax
  mov    %rax,-0xe0(%rbp)
  movsd  -0xc0(%rbp),%xmm0
  movsd  -0x80(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  movsd  -0x20(%rbp),%xmm1
  movsd  -0x80(%rbp),%xmm3
  movapd %xmm1,%xmm2
  subsd  %xmm3,%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  -0x20(%rbp),%xmm0
  subsd  -0xe0(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0xe0(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0xd8(%rbp)
  movsd  -0xd8(%rbp),%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double th2in(double *X0, double *X, void (*vofx)(double *, double *)) {
  double V[(4)];
  double V0[(4)];
  double Vc0[(4)];
  double Xc0[(4)];
  double Xcmid[(4)];
  double Vcmid[(4)];
  int j;
  double res;
  double th0;
  for (j = 0; j < (4); j++)
    Xc0[j] = X[j];
  Xc0[2] = X0[2];
  vofx(Xc0, Vc0);
  for (j = 0; j < (4); j++)
    Xcmid[j] = X[j];
  Xcmid[2] = 0;
  vofx(Xcmid, Vcmid);
  vofx(X0, V0);
  vofx(X, V);
  th0 = asin(sinth0(X0, X, vofx));
  res = (V[2] - Vc0[2]) / (Vcmid[2] - Vc0[2]) * (Vcmid[2] - th0) + th0;
  return (res);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<th2in>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xd0,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movsd  (%rsi),%xmm0
  movsd  %xmm0,0x60(%rsp)
  movsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x68(%rsp)
  movsd  0x18(%rsi),%xmm0
  movsd  %xmm0,0x78(%rsp)
  movsd  0x10(%rdi),%xmm0
  movsd  %xmm0,0x70(%rsp)
  lea    0x40(%rsp),%rsi
  lea    0x60(%rsp),%rdi
  call   *%rdx
  movsd  (%rbx),%xmm0
  movsd  %xmm0,0x80(%rsp)
  movsd  0x8(%rbx),%xmm0
  movsd  %xmm0,0x88(%rsp)
  movsd  0x18(%rbx),%xmm0
  movsd  %xmm0,0x98(%rsp)
  movq   $0x0,0x90(%rsp)
  lea    0xa0(%rsp),%rsi
  lea    0x80(%rsp),%rdi
  call   *%rbp
  lea    0x20(%rsp),%rsi
  mov    %r12,%rdi
  call   *%rbp
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   *%rbp
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <sinth0@plt>
  mov    %eax,%edi
  call   <asin@plt>
  movapd %xmm0,%xmm1
  movsd  0x50(%rsp),%xmm3
  movsd  0xb0(%rsp),%xmm2
  movsd  0x10(%rsp),%xmm0
  subsd  %xmm3,%xmm0
  movapd %xmm2,%xmm4
  subsd  %xmm3,%xmm4
  divsd  %xmm4,%xmm0
  subsd  %xmm1,%xmm2
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0xd0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double th2in(double *X0, double *X, void (*vofx)(double *, double *)) {
  double V[(4)];
  double V0[(4)];
  double Vc0[(4)];
  double Xc0[(4)];
  double Xcmid[(4)];
  double Vcmid[(4)];
  int j;
  double res;
  double th0;
  for (j = 0; j < (4); j++)
    Xc0[j] = X[j];
  Xc0[2] = X0[2];
  vofx(Xc0, Vc0);
  for (j = 0; j < (4); j++)
    Xcmid[j] = X[j];
  Xcmid[2] = 0;
  vofx(Xcmid, Vcmid);
  vofx(X0, V0);
  vofx(X, V);
  th0 = asin(sinth0(X0, X, vofx));
  res = (V[2] - Vc0[2]) / (Vcmid[2] - Vc0[2]) * (Vcmid[2] - th0) + th0;
  return (res);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<th2in>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0xd0,%rsp
  movdqu 0x10(%rsi),%xmm6
  movdqu (%rsi),%xmm5
  movsd  0x10(%r13),%xmm0
  lea    0x60(%rsp),%rdi
  lea    0x40(%rsp),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movaps %xmm6,0x70(%rsp)
  movaps %xmm5,0x60(%rsp)
  movsd  %xmm0,0x70(%rsp)
  call   *%rdx
  movdqu 0x0(%rbp),%xmm7
  movdqu 0x10(%rbp),%xmm5
  lea    0x80(%rsp),%rdi
  lea    0xa0(%rsp),%rsi
  movaps %xmm5,0x90(%rsp)
  movaps %xmm7,0x80(%rsp)
  movq   $0x0,0x90(%rsp)
  call   *%r12
  lea    0x20(%rsp),%rsi
  mov    %r13,%rdi
  call   *%r12
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   *%r12
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <sinth0@plt>
  mov    %eax,%edi
  call   <asin@plt>
  movsd  0x50(%rsp),%xmm3
  movsd  0xb0(%rsp),%xmm2
  movapd %xmm0,%xmm1
  movsd  0x10(%rsp),%xmm0
  movapd %xmm2,%xmm4
  subsd  %xmm1,%xmm2
  subsd  %xmm3,%xmm0
  subsd  %xmm3,%xmm4
  divsd  %xmm4,%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0xd0,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double th2in(double *X0, double *X, void (*vofx)(double *, double *)) {
  double V[(4)];
  double V0[(4)];
  double Vc0[(4)];
  double Xc0[(4)];
  double Xcmid[(4)];
  double Vcmid[(4)];
  int j;
  double res;
  double th0;
  for (j = 0; j < (4); j++)
    Xc0[j] = X[j];
  Xc0[2] = X0[2];
  vofx(Xc0, Vc0);
  for (j = 0; j < (4); j++)
    Xcmid[j] = X[j];
  Xcmid[2] = 0;
  vofx(Xcmid, Vcmid);
  vofx(X0, V0);
  vofx(X, V);
  th0 = asin(sinth0(X0, X, vofx));
  res = (V[2] - Vc0[2]) / (Vcmid[2] - Vc0[2]) * (Vcmid[2] - th0) + th0;
  return (res);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<th2in>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0xd0,%rsp
  movdqu 0x10(%rsi),%xmm6
  movdqu (%rsi),%xmm5
  movsd  0x10(%r13),%xmm0
  lea    0x60(%rsp),%rdi
  lea    0x40(%rsp),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movaps %xmm6,0x70(%rsp)
  movaps %xmm5,0x60(%rsp)
  movsd  %xmm0,0x70(%rsp)
  call   *%rdx
  movdqu 0x0(%rbp),%xmm7
  movdqu 0x10(%rbp),%xmm5
  lea    0x80(%rsp),%rdi
  lea    0xa0(%rsp),%rsi
  movaps %xmm5,0x90(%rsp)
  movaps %xmm7,0x80(%rsp)
  movq   $0x0,0x90(%rsp)
  call   *%r12
  lea    0x20(%rsp),%rsi
  mov    %r13,%rdi
  call   *%r12
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   *%r12
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <sinth0@plt>
  mov    %eax,%edi
  call   <asin@plt>
  movsd  0x50(%rsp),%xmm3
  movsd  0xb0(%rsp),%xmm2
  movapd %xmm0,%xmm1
  movsd  0x10(%rsp),%xmm0
  movapd %xmm2,%xmm4
  subsd  %xmm1,%xmm2
  subsd  %xmm3,%xmm0
  subsd  %xmm3,%xmm4
  divsd  %xmm4,%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0xd0,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double th2in(double *X0, double *X, void (*vofx)(double *, double *)) {
  double V[(4)];
  double V0[(4)];
  double Vc0[(4)];
  double Xc0[(4)];
  double Xcmid[(4)];
  double Vcmid[(4)];
  int j;
  double res;
  double th0;
  for (j = 0; j < (4); j++)
    Xc0[j] = X[j];
  Xc0[2] = X0[2];
  vofx(Xc0, Vc0);
  for (j = 0; j < (4); j++)
    Xcmid[j] = X[j];
  Xcmid[2] = 0;
  vofx(Xcmid, Vcmid);
  vofx(X0, V0);
  vofx(X, V);
  th0 = asin(sinth0(X0, X, vofx));
  res = (V[2] - Vc0[2]) / (Vcmid[2] - Vc0[2]) * (Vcmid[2] - th0) + th0;
  return (res);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <get_time@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x1,-0x14(%rbp)
  jmp    L8
L10:
  mov    $0x0,%eax
  call   <sha1@plt>
  addl   $0x1,-0x14(%rbp)
L8:
  cmpl   $0x3e8,-0x14(%rbp)
  jle    L10
  mov    $0x0,%eax
  call   <get_time@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  subsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type: string

tool: D6: %f


assistant: ```c
int main() {
  uint16_t iter;
  double before = get_time();
  for (iter = 1; iter <= 1000; iter++) {
    sha1();
  }
  double diff = get_time() - before;
  printf("%f\n", diff);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r14
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <get_time@plt>
  movq   %xmm0,%r14
  mov    $0x3e8,%ebx
L4:
  mov    $0x0,%eax
  call   <sha1@plt>
  sub    $0x1,%ebx
  jne    L4
  mov    $0x0,%eax
  call   <get_time@plt>
  movq   %r14,%xmm1
  subsd  %xmm1,%xmm0
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: label: D5
datatype: string

tool: D5: %f


assistant: ```c
int main() {
  uint16_t iter;
  double before = get_time();
  for (iter = 1; iter <= 1000; iter++) {
    sha1();
  }
  double diff = get_time() - before;
  printf("%f\n", diff);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    $0x3e8,%ebx
  sub    $0x10,%rsp
  call   <get_time@plt>
  movsd  %xmm0,0x8(%rsp)
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  call   <sha1@plt>
  sub    $0x1,%ebx
  jne    L3
  xor    %eax,%eax
  call   <get_time@plt>
  subsd  0x8(%rsp),%xmm0
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: label: D5
data_type: string

tool: D5: %f


assistant: ```c
int main() {
  uint16_t iter;
  double before = get_time();
  for (iter = 1; iter <= 1000; iter++) {
    sha1();
  }
  double diff = get_time() - before;
  printf("%f\n", diff);
  return 0;
}
```