ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ReportUsage>:
  endbr64
  sub    $0x8,%rsp
  lea    D53(%rip),%rdi
  call   <printf@plt>
  lea    D109(%rip),%rdi
  call   <printf@plt>
  lea    D127(%rip),%rdi
  call   <printf@plt>
  lea    D125(%rip),%rdi
  call   <printf@plt>
  lea    D81(%rip),%rdi
  call   <printf@plt>
  lea    D52(%rip),%rdi
  call   <printf@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  lea    D91(%rip),%rdi
  call   <printf@plt>
  lea    D88(%rip),%rdi
  call   <printf@plt>
  lea    D41(%rip),%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    D71(%rip),%rdi
  call   <PrintStdOpts@plt>
  lea    D44(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <printf@plt>
```

tool_calls: name: parse_data
arguments: tag: D53
datatype: string
name: parse_data
arguments: tag: D109
datatype: string
name: parse_data
arguments: tag: D127
datatype: string
name: parse_data
arguments: tag: D125
datatype: string
name: parse_data
arguments: tag: D81
datatype: string
name: parse_data
arguments: tag: D52
datatype: string
name: parse_data
arguments: tag: D35
datatype: string
name: parse_data
arguments: tag: D91
datatype: string
name: parse_data
arguments: tag: D88
datatype: string
name: parse_data
arguments: tag: D41
datatype: string
name: parse_data
arguments: tag: D1
datatype: string
name: parse_data
arguments: tag: D71
datatype: string
name: parse_data
arguments: tag: D44
datatype: string

tool: D53: 
USAGE: LGCopy [options] map <[mult] file ....>



tool: D109:  Option                                       Default



tool: D127:  -a n    allow up to 'n' new classes          100


tool: D125:  -b n    set ngram buffer size to 'n'         2000000


tool: D81:  -d s    database directory 's'               current


tool: D52:  -i n    set output gram file start index     0


tool: D35:  -m fn   save new word map to 'fn'            off


tool: D91:  -n n    create n-grams of size 'n'           max


tool: D88:  -o      output class mappings only           off


tool: D41:  -r s    set root data base file name 's'     data


tool: D1:  -w fn   load class map from 'fn'             none


tool: 

tool: D44: 



assistant: ```c
void ReportUsage(void) {
  printf("\nUSAGE: LGCopy [options] map <[mult] file ....>\n\n");
  printf(" Option                                       Default\n\n");
  printf(" -a n    allow up to 'n' new classes          100\n");
  printf(" -b n    set ngram buffer size to 'n'         2000000\n");
  printf(" -d s    database directory 's'               current\n");
  printf(" -i n    set output gram file start index     0\n");
  printf(" -m fn   save new word map to 'fn'            off\n");
  printf(" -n n    create n-grams of size 'n'           max\n");
  printf(" -o      output class mappings only           off\n");
  printf(" -r s    set root data base file name 's'     data\n");
  printf(" -w fn   load class map from 'fn'             none\n");
  PrintStdOpts("");
  printf("\n\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<noencripchar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0xffffffff,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L64
L67:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x15(%rbp)
  cmpb   $0x40,-0x15(%rbp)
  jle    L38
  cmpb   $0x5a,-0x15(%rbp)
  jle    L21
L38:
  cmpb   $0x60,-0x15(%rbp)
  jle    L3
  cmpb   $0x7a,-0x15(%rbp)
  jle    L21
L3:
  cmpb   $0xf1,-0x15(%rbp)
  je     L21
  cmpb   $0xd1,-0x15(%rbp)
  je     L21
  addl   $0x1,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x15(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L21:
  addl   $0x1,-0x10(%rbp)
L64:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L67
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *noencripchar(char *cadena) {
  int tamano = 0;
  int contador;
  int i;
  char tmp;
  char *noEncriptados;
  noEncriptados = (char *)malloc(sizeof(char) * 256);
  tamano = strlen(cadena);
  contador = -1;
  for (i = 0; i < tamano; i++) {
    tmp = cadena[i];
    if (!(((tmp >= 'A') && (tmp <= 'Z')) || ((tmp >= 'a') && (tmp <= 'z')) || ((tmp == -15) || (tmp == -47)))) {
      contador++;
      noEncriptados[contador] = tmp;
      noEncriptados[contador + 1] = '\0';
    }
  }
  return noEncriptados;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<noencripchar>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L4
  mov    %rbp,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbp,%rax,1),%rsi
  mov    $0xffffffff,%edi
  jmp    L8
L53:
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  je     L4
L8:
  movzbl (%rdx),%ecx
  mov    %ecx,%eax
  and    $0xffffffdf,%eax
  sub    $0x41,%eax
  cmp    $0x19,%al
  jbe    L53
  mov    %ecx,%eax
  and    $0xffffffdf,%eax
  cmp    $0xd1,%al
  je     L53
  add    $0x1,%edi
  movslq %edi,%rax
  mov    %cl,(%rbx,%rax,1)
  movb   $0x0,0x1(%rbx,%rax,1)
  jmp    L53
L4:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *noencripchar(char *cadena) {
  int tamano = 0;
  int contador;
  int i;
  char tmp;
  char *noEncriptados;
  noEncriptados = (char *)malloc(sizeof(char) * 256);
  tamano = strlen(cadena);
  contador = -1;
  for (i = 0; i < tamano; i++) {
    tmp = cadena[i];
    if (!(((tmp >= 'A') && (tmp <= 'Z')) || ((tmp >= 'a') && (tmp <= 'z')) || ((tmp == -15) || (tmp == -47)))) {
      contador++;
      noEncriptados[contador] = tmp;
      noEncriptados[contador + 1] = '\0';
    }
  }
  return noEncriptados;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<noencripchar>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x100,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  test   %eax,%eax
  jle    L36
  sub    $0x1,%eax
  mov    %rbx,%rdi
  mov    $0xffffffff,%r8d
  lea    0x1(%rbx,%rax,1),%rsi
  nopl   0x0(%rax,%rax,1)
L38:
  movzbl (%rdi),%edx
  mov    %edx,%eax
  and    $0xffffffdf,%eax
  lea    -0x41(%rax),%ecx
  cmp    $0x19,%cl
  jbe    L3
  cmp    $0xd1,%al
  je     L3
  add    $0x1,%r8d
  movslq %r8d,%rax
  mov    %dl,(%r12,%rax,1)
  movb   $0x0,0x1(%r12,%rax,1)
L3:
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L38
L36:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *noencripchar(char *cadena) {
  int tamano = 0;
  int contador;
  int i;
  char tmp;
  char *noEncriptados;
  noEncriptados = (char *)malloc(sizeof(char) * 256);
  tamano = strlen(cadena);
  contador = -1;
  for (i = 0; i < tamano; i++) {
    tmp = cadena[i];
    if (!(((tmp >= 'A') && (tmp <= 'Z')) || ((tmp >= 'a') && (tmp <= 'z')) || ((tmp == -15) || (tmp == -47)))) {
      contador++;
      noEncriptados[contador] = tmp;
      noEncriptados[contador + 1] = '\0';
    }
  }
  return noEncriptados;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<noencripchar>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x100,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rbx,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  test   %eax,%eax
  jle    L0
  sub    $0x1,%eax
  mov    %rbx,%rdi
  mov    $0xffffffff,%r8d
  lea    0x1(%rbx,%rax,1),%rsi
  nopl   0x0(%rax,%rax,1)
L18:
  movzbl (%rdi),%edx
  mov    %edx,%eax
  and    $0xffffffdf,%eax
  lea    -0x41(%rax),%ecx
  cmp    $0x19,%cl
  jbe    L2
  cmp    $0xd1,%al
  je     L2
  add    $0x1,%r8d
  movslq %r8d,%rax
  mov    %dl,(%r12,%rax,1)
  movb   $0x0,0x1(%r12,%rax,1)
L2:
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L18
L0:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *noencripchar(char *cadena) {
  int tamano = 0;
  int contador;
  int i;
  char tmp;
  char *noEncriptados;
  noEncriptados = (char *)malloc(sizeof(char) * 256);
  tamano = strlen(cadena);
  contador = -1;
  for (i = 0; i < tamano; i++) {
    tmp = cadena[i];
    if (!(((tmp >= 'A') && (tmp <= 'Z')) || ((tmp >= 'a') && (tmp <= 'z')) || ((tmp == -15) || (tmp == -47)))) {
      contador++;
      noEncriptados[contador] = tmp;
      noEncriptados[contador + 1] = '\0';
    }
  }
  return noEncriptados;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<assignSemDestroyError>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    $0x83,%eax
  je     L47
  cmp    $0x83,%eax
  jg     L71
  cmp    $0x82,%eax
  je     L51
  cmp    $0x82,%eax
  jg     L71
  cmp    $0x80,%eax
  je     L45
  cmp    $0x81,%eax
  je     L75
  jmp    L71
L47:
  mov    D1(%rip),%rax
  movl   $0x82,(%rax)
  jmp    L71
L75:
  mov    D1(%rip),%rax
  movl   $0x82,(%rax)
  jmp    L71
L51:
  nop
  jmp    L71
L45:
  nop
L71:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void assignSemDestroyError() {
  switch (errno) {
  case EACCES:
    errno = EINVAL;
    break;
  case EINVAL:
    break;
  case EPERM:
    errno = EINVAL;
    break;
  case ERANGE:
    break;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<assignSemDestroyError>:
  endbr64
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  cmp    $0x81,%eax
  je     L12
  cmp    $0x83,%eax
  jne    L18
  mov    D15(%rip),%rax
  movl   $0x82,(%rax)
  ret
L12:
  mov    D15(%rip),%rax
  movl   $0x82,(%rax)
L18:
  ret
```

assistant: ```c
void assignSemDestroyError() {
  switch (errno) {
  case EACCES:
    errno = EINVAL;
    break;
  case EINVAL:
    break;
  case EPERM:
    errno = EINVAL;
    break;
  case ERANGE:
    break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<assignSemDestroyError>:
  endbr64
  mov    D8(%rip),%rdx
  mov    (%rdx),%eax
  and    $0xfffffffd,%eax
  cmp    $0x81,%eax
  je     L2
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  movl   $0x82,(%rdx)
  ret
```

assistant: ```c
void assignSemDestroyError() {
  switch (errno) {
  case EACCES:
    errno = EINVAL;
    break;
  case EINVAL:
    break;
  case EPERM:
    errno = EINVAL;
    break;
  case ERANGE:
    break;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<assignSemDestroyError>:
  endbr64
  mov    D6(%rip),%rdx
  mov    (%rdx),%eax
  and    $0xfffffffd,%eax
  cmp    $0x81,%eax
  je     L0
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  movl   $0x82,(%rdx)
  ret
```

assistant: ```c
void assignSemDestroyError() {
  switch (errno) {
  case EACCES:
    errno = EINVAL;
    break;
  case EINVAL:
    break;
  case EPERM:
    errno = EINVAL;
    break;
  case ERANGE:
    break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<nowhite_strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L55
L106:
  addq   $0x1,-0x18(%rbp)
L55:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L106
  jmp    L21
L104:
  addq   $0x1,-0x20(%rbp)
L21:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L104
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L9
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L93
L9:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  jmp    L77
L93:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L81
L73:
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
L81:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L73
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L32
L6:
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
L32:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L6
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
L77:
  leave
  ret
```

assistant: ```c
int nowhite_strcmp(char *s, char *t) {
  char *p;
  while (isWhite(*s))
    s++;
  while (isWhite(*t))
    t++;
  if ((*s == 0) || (*t == 0))
    return strcmp(s, t);
  p = s + strlen(s) - 1;
  while (isWhite(*p))
    *(p--) = 0;
  p = t + strlen(t) - 1;
  while (isWhite(*p))
    *(p--) = 0;
  return strcmp(s, t);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<nowhite_strcmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  jmp    L92
L108:
  add    $0x1,%rbx
L92:
  movsbl (%rbx),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L108
  jmp    L33
L80:
  add    $0x1,%rbp
L33:
  movsbl 0x0(%rbp),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L80
  cmpb   $0x0,(%rbx)
  je     L47
  cmpb   $0x0,0x0(%rbp)
  jne    L78
L47:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
L34:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L78:
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%r12
  jmp    L84
L16:
  sub    $0x1,%r12
  movb   $0x0,0x1(%r12)
L84:
  movsbl (%r12),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L16
  mov    %rbp,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%r12
  jmp    L65
L59:
  sub    $0x1,%r12
  movb   $0x0,0x1(%r12)
L65:
  movsbl (%r12),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L59
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  jmp    L34
```

assistant: ```c
int nowhite_strcmp(char *s, char *t) {
  char *p;
  while (isWhite(*s))
    s++;
  while (isWhite(*t))
    t++;
  if ((*s == 0) || (*t == 0))
    return strcmp(s, t);
  p = s + strlen(s) - 1;
  while (isWhite(*p))
    *(p--) = 0;
  p = t + strlen(t) - 1;
  while (isWhite(*p))
    *(p--) = 0;
  return strcmp(s, t);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<nowhite_strcmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  jmp    L52
L59:
  add    $0x1,%rbp
L52:
  movsbl 0x0(%rbp),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L59
  movsbl (%r12),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  je     L72
  nopl   0x0(%rax)
L86:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L86
L72:
  cmpb   $0x0,0x0(%rbp)
  je     L34
  cmpb   $0x0,(%r12)
  jne    L61
L34:
  pop    %rbx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <strcmp@plt>
  nopl   0x0(%rax,%rax,1)
L61:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  jmp    L56
  nopl   0x0(%rax)
L20:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
L56:
  movsbl (%rbx),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L20
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%r12,%rax,1),%rbx
  jmp    L81
  nopl   (%rax)
L27:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
L81:
  movsbl (%rbx),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L27
  pop    %rbx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <strcmp@plt>
```

assistant: ```c
int nowhite_strcmp(char *s, char *t) {
  char *p;
  while (isWhite(*s))
    s++;
  while (isWhite(*t))
    t++;
  if ((*s == 0) || (*t == 0))
    return strcmp(s, t);
  p = s + strlen(s) - 1;
  while (isWhite(*p))
    *(p--) = 0;
  p = t + strlen(t) - 1;
  while (isWhite(*p))
    *(p--) = 0;
  return strcmp(s, t);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<nowhite_strcmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  jmp    L3
L31:
  add    $0x1,%rbp
L3:
  movsbl 0x0(%rbp),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L31
  movsbl (%r12),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  je     L4
  nopl   0x0(%rax)
L22:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L22
L4:
  cmpb   $0x0,0x0(%rbp)
  je     L44
  cmpb   $0x0,(%r12)
  jne    L15
L44:
  pop    %rbx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <strcmp@plt>
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbp,%rax,1),%rbx
  jmp    L30
  nopl   0x0(%rax)
L97:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
L30:
  movsbl (%rbx),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L97
  mov    %r12,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%r12,%rax,1),%rbx
  jmp    L49
  nopl   (%rax)
L56:
  movb   $0x0,(%rbx)
  sub    $0x1,%rbx
L49:
  movsbl (%rbx),%edi
  call   <isWhite@plt>
  test   %rax,%rax
  jne    L56
  pop    %rbx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <strcmp@plt>
```

assistant: ```c
int nowhite_strcmp(char *s, char *t) {
  char *p;
  while (isWhite(*s))
    s++;
  while (isWhite(*t))
    t++;
  if ((*s == 0) || (*t == 0))
    return strcmp(s, t);
  p = s + strlen(s) - 1;
  while (isWhite(*p))
    *(p--) = 0;
  p = t + strlen(t) - 1;
  while (isWhite(*p))
    *(p--) = 0;
  return strcmp(s, t);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<kernel__Int__lshift>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%ecx
  shl    %cl,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  nop
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long kernel__Int__lshift(long self, long p0) {
  long var;
  long var1;
  var1 = self << p0;
  var = var1;
  goto RET_LABEL;
RET_LABEL:;
  return var;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<kernel__Int__lshift>:
  endbr64
  mov    %rdi,%rax
  mov    %esi,%ecx
  shl    %cl,%rax
  ret
```

assistant: ```c
long kernel__Int__lshift(long self, long p0) {
  long var;
  long var1;
  var1 = self << p0;
  var = var1;
  goto RET_LABEL;
RET_LABEL:;
  return var;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<kernel__Int__lshift>:
  endbr64
  mov    %rdi,%rax
  mov    %esi,%ecx
  shl    %cl,%rax
  ret
```

assistant: ```c
long kernel__Int__lshift(long self, long p0) {
  long var;
  long var1;
  var1 = self << p0;
  var = var1;
  goto RET_LABEL;
RET_LABEL:;
  return var;
}
```
user: Translate the below assembly code into C programming language:

```asm
<kernel__Int__lshift>:
  endbr64
  mov    %rdi,%rax
  mov    %esi,%ecx
  shl    %cl,%rax
  ret
```

assistant: ```c
long kernel__Int__lshift(long self, long p0) {
  long var;
  long var1;
  var1 = self << p0;
  var = var1;
  goto RET_LABEL;
RET_LABEL:;
  return var;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3039,-0x24(%rbp)
  lea    -0x20(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x13,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <itoa@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0xffffcfc7,-0x24(%rbp)
  lea    -0x20(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x13,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <itoa@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D14
type_guessed: string

tool: D14: %s


assistant: ```c
int main() {
  int i = 12345;
  char str[20];
  itoa(i, str, 19);
  printf("%s\n", str);
  i = -12345;
  itoa(i, str, 19);
  printf("%s\n", str);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x13,%edx
  mov    %rbx,%rsi
  mov    $0x3039,%edi
  call   <itoa@plt>
  mov    %rbx,%rsi
  lea    D5(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    $0x13,%edx
  mov    %rbx,%rsi
  mov    $0xffffcfc7,%edi
  call   <itoa@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type_guessed: string

tool: D5: %s


assistant: ```c
int main() {
  int i = 12345;
  char str[20];
  itoa(i, str, 19);
  printf("%s\n", str);
  i = -12345;
  itoa(i, str, 19);
  printf("%s\n", str);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x13,%edx
  mov    $0x3039,%edi
  lea    D6(%rip),%r12
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <itoa@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    $0xffffcfc7,%edi
  mov    $0x13,%edx
  call   <itoa@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
data_type: string

tool: D6: %s


assistant: ```c
int main() {
  int i = 12345;
  char str[20];
  itoa(i, str, 19);
  printf("%s\n", str);
  i = -12345;
  itoa(i, str, 19);
  printf("%s\n", str);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x13,%edx
  mov    $0x3039,%edi
  lea    D7(%rip),%r12
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rsi
  call   <itoa@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    $0xffffcfc7,%edi
  mov    $0x13,%edx
  call   <itoa@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: string

tool: D7: %s


assistant: ```c
int main() {
  int i = 12345;
  char str[20];
  itoa(i, str, 19);
  printf("%s\n", str);
  i = -12345;
  itoa(i, str, 19);
  printf("%s\n", str);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<drop_piece>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  nop
L5:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <down_piece@plt>
  test   %eax,%eax
  je     L5
  nop
  nop
  leave
  ret
```

assistant: ```c
void drop_piece(int player) {
  while (!down_piece(player)) {
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<drop_piece>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
L4:
  mov    %ebx,%edi
  call   <down_piece@plt>
  test   %eax,%eax
  je     L4
  pop    %rbx
  ret
```

assistant: ```c
void drop_piece(int player) {
  while (!down_piece(player)) {
  }
}
```