ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<xtoi>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  cmp    $0x30,%dil
  je     L65
L49:
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L4
L62:
  mov    $0x0,%ebp
L23:
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <toupper@plt>
  lea    -0x30(%rax),%edx
  lea    -0x3a(%rax),%ecx
  cmp    $0x6,%cl
  jbe    L43
  cmp    $0x16,%dl
  ja     L43
  sub    $0x37,%eax
  cmp    $0x9,%dl
  cmovle %edx,%eax
  shl    $0x4,%ebp
  movsbl %al,%eax
  add    %eax,%ebp
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L23
L43:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L65:
  cmpb   $0x78,0x1(%rbx)
  jne    L62
  add    $0x2,%rbx
  jmp    L49
L4:
  mov    $0x0,%ebp
  jmp    L43
```

assistant: ```c
int xtoi(char *hexstring) {
  int i = 0;
  if ((*hexstring == '0') && (*(hexstring + 1) == 'x'))
    hexstring += 2;
  while (*hexstring) {
    char c = toupper(*hexstring++);
    if ((c < '0') || (c > 'F') || ((c > '9') && (c < 'A')))
      break;
    c -= '0';
    if (c > 9)
      c -= 7;
    i = (i << 4) + c;
  }
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<xtoi>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  cmp    $0x30,%dil
  je     L72
L23:
  test   %dil,%dil
  je     L8
L47:
  xor    %r12d,%r12d
  nop
L37:
  call   <toupper@plt>
  add    $0x1,%rbx
  lea    -0x3a(%rax),%ecx
  lea    -0x30(%rax),%edx
  cmp    $0x6,%cl
  jbe    L33
  cmp    $0x16,%dl
  ja     L33
  sub    $0x37,%eax
  cmp    $0x9,%dl
  movsbl (%rbx),%edi
  cmova  %eax,%edx
  shl    $0x4,%r12d
  movzbl %dl,%edx
  add    %edx,%r12d
  test   %dil,%dil
  jne    L37
L33:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nop
L72:
  cmpb   $0x78,0x1(%rbx)
  jne    L47
  movsbl 0x2(%rbx),%edi
  add    $0x2,%rbx
  jmp    L23
L8:
  xor    %r12d,%r12d
  jmp    L33
```

assistant: ```c
int xtoi(char *hexstring) {
  int i = 0;
  if ((*hexstring == '0') && (*(hexstring + 1) == 'x'))
    hexstring += 2;
  while (*hexstring) {
    char c = toupper(*hexstring++);
    if ((c < '0') || (c > 'F') || ((c > '9') && (c < 'A')))
      break;
    c -= '0';
    if (c > 9)
      c -= 7;
    i = (i << 4) + c;
  }
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<xtoi>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  cmp    $0x30,%dil
  je     L11
L34:
  test   %dil,%dil
  je     L20
L83:
  xor    %r12d,%r12d
  nop
L37:
  call   <toupper@plt>
  add    $0x1,%rbx
  lea    -0x3a(%rax),%ecx
  lea    -0x30(%rax),%edx
  cmp    $0x6,%cl
  jbe    L10
  cmp    $0x16,%dl
  ja     L10
  shl    $0x4,%r12d
  cmp    $0x9,%dl
  jbe    L99
  sub    $0x37,%eax
  movsbl (%rbx),%edi
  movzbl %al,%eax
  add    %eax,%r12d
  test   %dil,%dil
  jne    L37
L10:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  xchg   %ax,%ax
L99:
  movsbl (%rbx),%edi
  movzbl %dl,%edx
  add    %edx,%r12d
  test   %dil,%dil
  jne    L37
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L11:
  cmpb   $0x78,0x1(%rbx)
  jne    L83
  movsbl 0x2(%rbx),%edi
  add    $0x2,%rbx
  jmp    L34
L20:
  xor    %r12d,%r12d
  jmp    L10
```

assistant: ```c
int xtoi(char *hexstring) {
  int i = 0;
  if ((*hexstring == '0') && (*(hexstring + 1) == 'x'))
    hexstring += 2;
  while (*hexstring) {
    char c = toupper(*hexstring++);
    if ((c < '0') || (c > 'F') || ((c > '9') && (c < 'A')))
      break;
    c -= '0';
    if (c > 9)
      c -= 7;
    i = (i << 4) + c;
  }
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<rmtrail>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L41
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L41
  jmp    L24
L57:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %eax,%eax
  je     L44
L24:
  subl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L57
  jmp    L35
L44:
  nop
L35:
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L41:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *rmtrail(char *str) {
  int i;
  if (str && 0 != (i = strlen(str))) {
    while (--i >= 0) {
      if (!isspace(str[i]))
        break;
    }
    str[++i] = '\0';
  }
  return str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rmtrail>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L2
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L2
  cltq
  lea    -0x1(%r12,%rax,1),%rbp
L34:
  mov    %ebx,%r13d
  sub    $0x1,%ebx
  js     L21
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  sub    $0x1,%rbp
  test   %eax,%eax
  jne    L34
L21:
  movslq %r13d,%r13
  movb   $0x0,(%r12,%r13,1)
L2:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *rmtrail(char *str) {
  int i;
  if (str && 0 != (i = strlen(str))) {
    while (--i >= 0) {
      if (!isspace(str[i]))
        break;
    }
    str[++i] = '\0';
  }
  return str;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rmtrail>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L44
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L44
  cltq
  lea    -0x1(%r12,%rax,1),%rbp
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L37:
  movsbl 0x0(%rbp),%edi
  sub    $0x1,%rbp
  call   <isspace@plt>
  test   %eax,%eax
  je     L10
L39:
  movslq %ebx,%r13
  sub    $0x1,%ebx
  jns    L37
L10:
  movb   $0x0,(%r12,%r13,1)
L44:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *rmtrail(char *str) {
  int i;
  if (str && 0 != (i = strlen(str))) {
    while (--i >= 0) {
      if (!isspace(str[i]))
        break;
    }
    str[++i] = '\0';
  }
  return str;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rmtrail>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L4
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  je     L4
  cltq
  lea    -0x1(%r12,%rax,1),%rbp
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L38:
  movsbl 0x0(%rbp),%edi
  sub    $0x1,%rbp
  call   <isspace@plt>
  test   %eax,%eax
  je     L7
L17:
  movslq %ebx,%r13
  sub    $0x1,%ebx
  jns    L38
L7:
  movb   $0x0,(%r12,%r13,1)
L4:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *rmtrail(char *str) {
  int i;
  if (str && 0 != (i = strlen(str))) {
    while (--i >= 0) {
      if (!isspace(str[i]))
        break;
    }
    str[++i] = '\0';
  }
  return str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_sysinfo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
  pop    %rbp
  ret
```

assistant: ```c
void get_sysinfo(void) {
  int32_t param;
  int32_t pushed;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_sysinfo>:
  endbr64
  ret
```

assistant: ```c
void get_sysinfo(void) {
  int32_t param;
  int32_t pushed;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_sysinfo>:
  endbr64
  ret
```

assistant: ```c
void get_sysinfo(void) {
  int32_t param;
  int32_t pushed;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_sysinfo>:
  endbr64
  ret
```

assistant: ```c
void get_sysinfo(void) {
  int32_t param;
  int32_t pushed;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parse_number_unit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  je     L73
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L73
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x19(%rbp)
  movsbl -0x19(%rbp),%eax
  sub    $0x47,%eax
  cmp    $0x26,%eax
  seta   %dl
  test   %dl,%dl
  jne    L23
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rax
  movabs $0x100000001,%rdx
  and    %rax,%rdx
  test   %rdx,%rdx
  setne  %dl
  test   %dl,%dl
  jne    L57
  movabs $0x4000000040,%rdx
  and    %rax,%rdx
  test   %rdx,%rdx
  setne  %dl
  test   %dl,%dl
  jne    L98
  movabs $0x1000000010,%rdx
  and    %rdx,%rax
  test   %rax,%rax
  setne  %al
  test   %al,%al
  je     L23
  shlq   $0xa,-0x10(%rbp)
  jmp    L73
L98:
  shlq   $0x14,-0x10(%rbp)
  jmp    L73
L57:
  shlq   $0x1e,-0x10(%rbp)
  jmp    L73
L23:
  movsbl -0x19(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <print_error@plt>
  nop
L73:
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D0
type: string

tool: D0: num-images has invalid unit (%c), use K, M, G

assistant: ```c
long parse_number_unit(const char *val) {
  char *endptr;
  long n = strtol(val, &endptr, 10);
  if ((endptr != NULL) && (*endptr != '\0')) {
    const char unit = *endptr;
    switch (unit) {
    case 'k':
    case 'K':
      n *= 1024L;
      break;
    case 'm':
    case 'M':
      n *= 1024L * 1024L;
      break;
    case 'g':
    case 'G':
      n *= 1024L * 1024L * 1024L;
      break;
    default:
      print_error("num-images has invalid unit (%c), use K, M, G", unit);
      break;
    }
  }
  return n;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse_number_unit>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0xa,%edx
  call   <strtol@plt>
  mov    %rax,%rbx
  mov    (%rsp),%rax
  test   %rax,%rax
  je     L74
  movzbl (%rax),%esi
  test   %sil,%sil
  je     L74
  lea    -0x47(%rsi),%ecx
  cmp    $0x26,%cl
  ja     L71
  mov    $0x1,%eax
  shl    %cl,%rax
  movabs $0x100000001,%rdx
  test   %rdx,%rax
  jne    L56
  movabs $0x4000000040,%rdx
  test   %rdx,%rax
  jne    L89
  movabs $0x1000000010,%rdx
  test   %rdx,%rax
  jne    L25
L71:
  movsbl %sil,%esi
  lea    D2(%rip),%rdi
  call   <print_error@plt>
  jmp    L74
L25:
  shl    $0xa,%rbx
L74:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L86
  mov    %rbx,%rax
  add    $0x10,%rsp
  pop    %rbx
  ret
L89:
  shl    $0x14,%rbx
  jmp    L74
L56:
  shl    $0x1e,%rbx
  jmp    L74
L86:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
data_type: string

tool: D2: num-images has invalid unit (%c), use K, M, G

assistant: ```c
long parse_number_unit(const char *val) {
  char *endptr;
  long n = strtol(val, &endptr, 10);
  if ((endptr != NULL) && (*endptr != '\0')) {
    const char unit = *endptr;
    switch (unit) {
    case 'k':
    case 'K':
      n *= 1024L;
      break;
    case 'm':
    case 'M':
      n *= 1024L * 1024L;
      break;
    case 'g':
    case 'G':
      n *= 1024L * 1024L * 1024L;
      break;
    default:
      print_error("num-images has invalid unit (%c), use K, M, G", unit);
      break;
    }
  }
  return n;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse_number_unit>:
  endbr64
  push   %r12
  mov    $0xa,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtol@plt>
  mov    %rax,%r12
  mov    (%rsp),%rax
  test   %rax,%rax
  je     L87
  movsbl (%rax),%esi
  test   %sil,%sil
  je     L87
  lea    -0x47(%rsi),%ecx
  cmp    $0x26,%cl
  ja     L84
  movabs $0x100000001,%rdx
  mov    $0x1,%eax
  shl    %cl,%rax
  test   %rdx,%rax
  jne    L40
  movabs $0x4000000040,%rdx
  test   %rdx,%rax
  jne    L32
  movabs $0x1000000010,%rdx
  test   %rdx,%rax
  jne    L75
L84:
  lea    D8(%rip),%rdi
  call   <print_error@plt>
  nopl   0x0(%rax)
L87:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L75:
  shl    $0xa,%r12
  jmp    L87
  xchg   %ax,%ax
L32:
  shl    $0x14,%r12
  jmp    L87
  cs nopw 0x0(%rax,%rax,1)
L40:
  shl    $0x1e,%r12
  jmp    L87
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
datatype: string

tool: D8: num-images has invalid unit (%c), use K, M, G

assistant: ```c
long parse_number_unit(const char *val) {
  char *endptr;
  long n = strtol(val, &endptr, 10);
  if ((endptr != NULL) && (*endptr != '\0')) {
    const char unit = *endptr;
    switch (unit) {
    case 'k':
    case 'K':
      n *= 1024L;
      break;
    case 'm':
    case 'M':
      n *= 1024L * 1024L;
      break;
    case 'g':
    case 'G':
      n *= 1024L * 1024L * 1024L;
      break;
    default:
      print_error("num-images has invalid unit (%c), use K, M, G", unit);
      break;
    }
  }
  return n;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_number_unit>:
  endbr64
  push   %r12
  mov    $0xa,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtol@plt>
  mov    %rax,%r12
  mov    (%rsp),%rax
  test   %rax,%rax
  je     L71
  movsbl (%rax),%esi
  test   %sil,%sil
  je     L71
  lea    -0x47(%rsi),%ecx
  cmp    $0x26,%cl
  ja     L95
  movabs $0x100000001,%rdx
  mov    $0x1,%eax
  shl    %cl,%rax
  test   %rdx,%rax
  jne    L5
  movabs $0x4000000040,%rdx
  test   %rdx,%rax
  jne    L9
  movabs $0x1000000010,%rdx
  test   %rdx,%rax
  jne    L67
L95:
  lea    D0(%rip),%rdi
  call   <print_error@plt>
  nopl   0x0(%rax)
L71:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  add    $0x10,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L67:
  shl    $0xa,%r12
  jmp    L71
  xchg   %ax,%ax
L9:
  shl    $0x14,%r12
  jmp    L71
  cs nopw 0x0(%rax,%rax,1)
L5:
  shl    $0x1e,%r12
  jmp    L71
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type_infered: string

tool: D0: num-images has invalid unit (%c), use K, M, G

assistant: ```c
long parse_number_unit(const char *val) {
  char *endptr;
  long n = strtol(val, &endptr, 10);
  if ((endptr != NULL) && (*endptr != '\0')) {
    const char unit = *endptr;
    switch (unit) {
    case 'k':
    case 'K':
      n *= 1024L;
      break;
    case 'm':
    case 'M':
      n *= 1024L * 1024L;
      break;
    case 'g':
    case 'G':
      n *= 1024L * 1024L * 1024L;
      break;
    default:
      print_error("num-images has invalid unit (%c), use K, M, G", unit);
      break;
    }
  }
  return n;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<log_wrapper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x138,%rsp
  mov    %rdi,-0x128(%rbp)
  mov    %rsi,-0x130(%rbp)
  mov    %rdx,-0x138(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    -0x130(%rbp),%rdx
  lea    -0x120(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  lea    -0x120(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    $0xfe,%edx
  mov    %edx,%ebx
  sub    %eax,%ebx
  lea    -0x120(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x120(%rbp),%rdx
  lea    (%rdx,%rax,1),%rcx
  mov    -0x138(%rbp),%rax
  mov    %rax,%rdx
  mov    %ebx,%esi
  mov    %rcx,%rdi
  call   <snprintf@plt>
  lea    -0x120(%rbp),%rax
  lea    D15(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcspn@plt>
  movb   $0x0,-0x120(%rbp,%rax,1)
  lea    -0x120(%rbp),%rdx
  mov    -0x128(%rbp),%rax
  lea    D5(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D15
type: string
name: unpack_data
arguments: tag: D5
type: string

tool: D15: 


tool: D5: LOG: %s


assistant: ```c
void log_wrapper(FILE *logf, char *msg, char *filename) {
  char log_buf[255];
  strcpy(log_buf, msg);
  snprintf(log_buf + strlen(log_buf), 255 - strlen(log_buf) - 1, filename);
  log_buf[strcspn(log_buf, "\n")] = '\0';
  fprintf(logf, "LOG: %s\n", log_buf);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<log_wrapper>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %rdi,%rbp
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    $0xfe,%esi
  sub    %r13d,%esi
  movslq %eax,%rdi
  add    %rbx,%rdi
  mov    %r12,%rdx
  call   <snprintf@plt>
  lea    D18(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcspn@plt>
  movb   $0x0,(%rsp,%rax,1)
  mov    %rbx,%rdx
  lea    D7(%rip),%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D18
datatype: string
name: unpack_data
arguments: data_tag: D7
datatype: string

tool: D18: 


tool: D7: LOG: %s


assistant: ```c
void log_wrapper(FILE *logf, char *msg, char *filename) {
  char log_buf[255];
  strcpy(log_buf, msg);
  snprintf(log_buf + strlen(log_buf), 255 - strlen(log_buf) - 1, filename);
  log_buf[strcspn(log_buf, "\n")] = '\0';
  fprintf(logf, "LOG: %s\n", log_buf);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<log_wrapper>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    $0xfe,%esi
  mov    %r13,%rdx
  movslq %eax,%rdi
  sub    %ebx,%esi
  add    %r12,%rdi
  call   <snprintf@plt>
  lea    D12(%rip),%rsi
  mov    %r12,%rdi
  call   <strcspn@plt>
  mov    %r12,%rdx
  lea    D3(%rip),%rsi
  mov    %rbp,%rdi
  movb   $0x0,(%rsp,%rax,1)
  call   <fprintf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D12
type_guessed: string
name: unpack_data
arguments: data_tag: D3
type_guessed: string

tool: D12: 


tool: D3: LOG: %s


assistant: ```c
void log_wrapper(FILE *logf, char *msg, char *filename) {
  char log_buf[255];
  strcpy(log_buf, msg);
  snprintf(log_buf + strlen(log_buf), 255 - strlen(log_buf) - 1, filename);
  log_buf[strcspn(log_buf, "\n")] = '\0';
  fprintf(logf, "LOG: %s\n", log_buf);
}
```
user: What is the c source code of the assembly code below?

```asm
<log_wrapper>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    $0xfe,%esi
  mov    %r13,%rdx
  movslq %eax,%rdi
  sub    %ebx,%esi
  add    %r12,%rdi
  call   <snprintf@plt>
  lea    D15(%rip),%rsi
  mov    %r12,%rdi
  call   <strcspn@plt>
  mov    %r12,%rdx
  lea    D12(%rip),%rsi
  mov    %rbp,%rdi
  movb   $0x0,(%rsp,%rax,1)
  call   <fprintf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D15
datatype: string
name: parse_data
arguments: data_tag: D12
datatype: string

tool: D15: 


tool: D12: LOG: %s


assistant: ```c
void log_wrapper(FILE *logf, char *msg, char *filename) {
  char log_buf[255];
  strcpy(log_buf, msg);
  snprintf(log_buf + strlen(log_buf), 255 - strlen(log_buf) - 1, filename);
  log_buf[strcspn(log_buf, "\n")] = '\0';
  fprintf(logf, "LOG: %s\n", log_buf);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<stdlib_ndarray_wrap_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L39
  mov    -0x4(%rbp),%eax
  add    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L38
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  imul   -0x4(%rbp),%eax
  sub    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L38
  mov    -0x4(%rbp),%eax
  add    %eax,-0x14(%rbp)
L38:
  mov    -0x14(%rbp),%eax
  jmp    L60
L39:
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L5
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L8
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,-0x14(%rbp)
L8:
  mov    -0x14(%rbp),%eax
  jmp    L60
L5:
  mov    -0x14(%rbp),%eax
L60:
  pop    %rbp
  ret
```

assistant: ```c
int64_t stdlib_ndarray_wrap_index(int64_t idx, int64_t max) {
  int64_t mp1 = max + 1;
  if (idx < 0) {
    idx += mp1;
    if (idx < 0) {
      idx -= mp1 * ((int64_t)(idx / mp1));
      if (idx != 0) {
        idx += mp1;
      }
    }
    return idx;
  }
  if (idx > max) {
    idx -= mp1;
    if (idx > max) {
      idx %= mp1;
    }
    return idx;
  }
  return idx;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<stdlib_ndarray_wrap_index>:
  endbr64
  lea    0x1(%rsi),%ecx
  test   %edi,%edi
  js     L39
  mov    %edi,%edx
  cmp    %edi,%esi
  jl     L40
L27:
  mov    %edx,%eax
  ret
L39:
  mov    %edi,%edx
  add    %ecx,%edx
  jns    L27
  mov    %edx,%eax
  cltd
  idiv   %ecx
  add    %edx,%ecx
  test   %edx,%edx
  cmovne %ecx,%edx
  jmp    L27
L40:
  sub    %ecx,%edx
  cmp    %edx,%esi
  jge    L27
  mov    %edx,%eax
  cltd
  idiv   %ecx
  jmp    L27
```

assistant: ```c
int64_t stdlib_ndarray_wrap_index(int64_t idx, int64_t max) {
  int64_t mp1 = max + 1;
  if (idx < 0) {
    idx += mp1;
    if (idx < 0) {
      idx -= mp1 * ((int64_t)(idx / mp1));
      if (idx != 0) {
        idx += mp1;
      }
    }
    return idx;
  }
  if (idx > max) {
    idx -= mp1;
    if (idx > max) {
      idx %= mp1;
    }
    return idx;
  }
  return idx;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<stdlib_ndarray_wrap_index>:
  endbr64
  mov    %edi,%r8d
  lea    0x1(%rsi),%ecx
  test   %edi,%edi
  js     L26
  cmp    %edi,%esi
  jl     L7
L8:
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L7:
  sub    %ecx,%r8d
  cmp    %r8d,%esi
  jge    L8
  mov    %r8d,%eax
  cltd
  idiv   %ecx
  mov    %edx,%r8d
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L26:
  add    %ecx,%r8d
  jns    L8
  mov    %r8d,%eax
  cltd
  idiv   %ecx
  add    %edx,%ecx
  test   %edx,%edx
  cmovne %ecx,%edx
  mov    %edx,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int64_t stdlib_ndarray_wrap_index(int64_t idx, int64_t max) {
  int64_t mp1 = max + 1;
  if (idx < 0) {
    idx += mp1;
    if (idx < 0) {
      idx -= mp1 * ((int64_t)(idx / mp1));
      if (idx != 0) {
        idx += mp1;
      }
    }
    return idx;
  }
  if (idx > max) {
    idx -= mp1;
    if (idx > max) {
      idx %= mp1;
    }
    return idx;
  }
  return idx;
}
```