ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<PR16931>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0x9(%rbp)
  lea    -0x9(%rbp),%rax
  mov    %rax,%rdi
  call   <PR16931>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void PR16931(int *x) {
  typedef struct {
    _Atomic(_Bool) flag;
  } flag;
  flag flagvar = {0};
  PR16931(&flagvar);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<PR16931>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rdi
  call   <PR16931>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void PR16931(int *x) {
  typedef struct {
    _Atomic(_Bool) flag;
  } flag;
  flag flagvar = {0};
  PR16931(&flagvar);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<PR16931>:
  endbr64
  nopl   0x0(%rax)
L1:
  jmp    L1
```

assistant: ```c
void PR16931(int *x) {
  typedef struct {
    _Atomic(_Bool) flag;
  } flag;
  flag flagvar = {0};
  PR16931(&flagvar);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<PR16931>:
  endbr64
  nopl   0x0(%rax)
L0:
  jmp    L0
```

assistant: ```c
void PR16931(int *x) {
  typedef struct {
    _Atomic(_Bool) flag;
  } flag;
  flag flagvar = {0};
  PR16931(&flagvar);
}
```
user: What is the c source code of the assembly code below:

```asm
<f_nl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,-0x8(%rbp)
  mov    %r8,-0x20(%rbp)
  mov    %r9,-0x28(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L7
  mov    -0x20(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L7:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D7
datatype: string

tool: D7: 


assistant: ```c
int f_nl(int mode, unsigned char **sec, float *data, unsigned int ndata, char *inv_out, void **local) {
  if (mode >= 0)
    strcpy(inv_out, "\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<f_nl>:
  endbr64
  test   %edi,%edi
  jns    L1
  mov    $0x0,%eax
  ret
L1:
  sub    $0x8,%rsp
  lea    D0(%rip),%rsi
  mov    %r8,%rdi
  call   <strcpy@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D0
data_type: string

tool: D0: 


assistant: ```c
int f_nl(int mode, unsigned char **sec, float *data, unsigned int ndata, char *inv_out, void **local) {
  if (mode >= 0)
    strcpy(inv_out, "\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<f_nl>:
  endbr64
  test   %edi,%edi
  jns    L7
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  sub    $0x8,%rsp
  lea    D0(%rip),%rsi
  mov    %r8,%rdi
  call   <strcpy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type_infered: string

tool: D0: 


assistant: ```c
int f_nl(int mode, unsigned char **sec, float *data, unsigned int ndata, char *inv_out, void **local) {
  if (mode >= 0)
    strcpy(inv_out, "\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<f_nl>:
  endbr64
  test   %edi,%edi
  jns    L4
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  sub    $0x8,%rsp
  lea    D2(%rip),%rsi
  mov    %r8,%rdi
  call   <strcpy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type: string

tool: D2: 


assistant: ```c
int f_nl(int mode, unsigned char **sec, float *data, unsigned int ndata, char *inv_out, void **local) {
  if (mode >= 0)
    strcpy(inv_out, "\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<menuRouletteRusse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <titre@plt>
  cmpl   $0x0,-0x4(%rbp)
  jne    L7
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <lireEtAfficherFichier@plt>
  jmp    L21
L7:
  cmpl   $0x1,-0x4(%rbp)
  jne    L21
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <revenirMenuMulti@plt>
L21:
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <appuyezEntree@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D11
type_guessed: string
name: extract_symbolic_data
arguments: label: D39
type_guessed: string
name: extract_symbolic_data
arguments: label: D40
type_guessed: string
name: extract_symbolic_data
arguments: label: D35
type_guessed: string
name: extract_symbolic_data
arguments: label: D2
type_guessed: string

tool: D11: Roulette Russe

tool: D39: Texte de presentation/Roulette russe/rouletteRusseSolo.txt

tool: D40: Ce jeu n'est pas disponible en multijoueur.


tool: D35: 



tool: D2: Appuyez sur ENTREE pour commencer : 

assistant: ```c
void menuRouletteRusse(int modeDeJeux) {
  titre("Roulette Russe");
  if (modeDeJeux == 0)
    lireEtAfficherFichier("Texte de presentation/Roulette russe/rouletteRusseSolo.txt");
  else if (modeDeJeux == 1) {
    printf("Ce jeu n'est pas disponible en multijoueur.\n");
    revenirMenuMulti();
  }
  printf("\n\n");
  appuyezEntree("Appuyez sur ENTREE pour commencer : ");
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<menuRouletteRusse>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    D46(%rip),%rdi
  call   <titre@plt>
  test   %ebx,%ebx
  je     L22
  cmp    $0x1,%ebx
  je     L30
L5:
  lea    D33(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <appuyezEntree@plt>
  pop    %rbx
  ret
L22:
  lea    D32(%rip),%rdi
  call   <lireEtAfficherFichier@plt>
  jmp    L5
L30:
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <revenirMenuMulti@plt>
  jmp    L5
```

tool_calls: name: disassemble_data
arguments: label: D46
datatype: string
name: disassemble_data
arguments: label: D33
datatype: string
name: disassemble_data
arguments: label: D13
datatype: string
name: disassemble_data
arguments: label: D32
datatype: string
name: disassemble_data
arguments: label: D38
datatype: string

tool: D46: Roulette Russe

tool: D33: 



tool: D13: Appuyez sur ENTREE pour commencer : 

tool: D32: Texte de presentation/Roulette russe/rouletteRusseSolo.txt

tool: D38: Ce jeu n'est pas disponible en multijoueur.


assistant: ```c
void menuRouletteRusse(int modeDeJeux) {
  titre("Roulette Russe");
  if (modeDeJeux == 0)
    lireEtAfficherFichier("Texte de presentation/Roulette russe/rouletteRusseSolo.txt");
  else if (modeDeJeux == 1) {
    printf("Ce jeu n'est pas disponible en multijoueur.\n");
    revenirMenuMulti();
  }
  printf("\n\n");
  appuyezEntree("Appuyez sur ENTREE pour commencer : ");
}
```
user: Translate the below assembly code into C programming language:

```asm
<menuRouletteRusse>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    D30(%rip),%rdi
  call   <titre@plt>
  test   %ebx,%ebx
  je     L3
  cmp    $0x1,%ebx
  je     L20
L6:
  lea    D31(%rip),%rdi
  call   <printf@plt>
  lea    D42(%rip),%rdi
  pop    %rbx
  jmp    10d0 <appuyezEntree@plt>
  nopl   (%rax)
L3:
  lea    D4(%rip),%rdi
  call   <lireEtAfficherFichier@plt>
  jmp    L6
  cs nopw 0x0(%rax,%rax,1)
L20:
  lea    D15(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <revenirMenuMulti@plt>
  jmp    L6
```

tool_calls: name: disassemble_data
arguments: label: D30
data_type: string
name: disassemble_data
arguments: label: D31
data_type: string
name: disassemble_data
arguments: label: D42
data_type: string
name: disassemble_data
arguments: label: D4
data_type: string
name: disassemble_data
arguments: label: D15
data_type: string

tool: D30: Roulette Russe

tool: D31: 



tool: D42: Appuyez sur ENTREE pour commencer : 

tool: D4: Texte de presentation/Roulette russe/rouletteRusseSolo.txt

tool: D15: Ce jeu n'est pas disponible en multijoueur.


assistant: ```c
void menuRouletteRusse(int modeDeJeux) {
  titre("Roulette Russe");
  if (modeDeJeux == 0)
    lireEtAfficherFichier("Texte de presentation/Roulette russe/rouletteRusseSolo.txt");
  else if (modeDeJeux == 1) {
    printf("Ce jeu n'est pas disponible en multijoueur.\n");
    revenirMenuMulti();
  }
  printf("\n\n");
  appuyezEntree("Appuyez sur ENTREE pour commencer : ");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<menuRouletteRusse>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    D27(%rip),%rdi
  call   <titre@plt>
  test   %ebx,%ebx
  je     L2
  cmp    $0x1,%ebx
  je     L7
L13:
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D2(%rip),%rdi
  pop    %rbx
  jmp    10d0 <appuyezEntree@plt>
  nopl   (%rax)
L2:
  lea    D16(%rip),%rdi
  call   <lireEtAfficherFichier@plt>
  jmp    L13
  cs nopw 0x0(%rax,%rax,1)
L7:
  lea    D21(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <revenirMenuMulti@plt>
  jmp    L13
```

tool_calls: name: read_data
arguments: label: D27
type_infered: string
name: read_data
arguments: label: D8
type_infered: string
name: read_data
arguments: label: D2
type_infered: string
name: read_data
arguments: label: D16
type_infered: string
name: read_data
arguments: label: D21
type_infered: string

tool: D27: Roulette Russe

tool: D8: 



tool: D2: Appuyez sur ENTREE pour commencer : 

tool: D16: Texte de presentation/Roulette russe/rouletteRusseSolo.txt

tool: D21: Ce jeu n'est pas disponible en multijoueur.


assistant: ```c
void menuRouletteRusse(int modeDeJeux) {
  titre("Roulette Russe");
  if (modeDeJeux == 0)
    lireEtAfficherFichier("Texte de presentation/Roulette russe/rouletteRusseSolo.txt");
  else if (modeDeJeux == 1) {
    printf("Ce jeu n'est pas disponible en multijoueur.\n");
    revenirMenuMulti();
  }
  printf("\n\n");
  appuyezEntree("Appuyez sur ENTREE pour commencer : ");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nxt_motor_get_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_S8@plt>
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  leave
  ret
```

assistant: ```c
char nxt_motor_get_count(unsigned char port) {
  char tmp;
  {
    tmp = __VERIFIER_nondet_S8();
    return (tmp);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<nxt_motor_get_count>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_S8@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char nxt_motor_get_count(unsigned char port) {
  char tmp;
  {
    tmp = __VERIFIER_nondet_S8();
    return (tmp);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<nxt_motor_get_count>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <__VERIFIER_nondet_S8@plt>
```

assistant: ```c
char nxt_motor_get_count(unsigned char port) {
  char tmp;
  {
    tmp = __VERIFIER_nondet_S8();
    return (tmp);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<nxt_motor_get_count>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <__VERIFIER_nondet_S8@plt>
```

assistant: ```c
char nxt_motor_get_count(unsigned char port) {
  char tmp;
  {
    tmp = __VERIFIER_nondet_S8();
    return (tmp);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<process_audio>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L4
L3:
  movl   $0x0,-0x4(%rbp)
  jmp    L48
L36:
  mov    -0x8(%rbp),%eax
  imul   $0xfa,%eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  add    %eax,%eax
  cmp    %eax,-0x24(%rbp)
  jle    L10
  mov    -0x8(%rbp),%eax
  imul   $0xfa,%eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  add    %eax,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  imul   $0x1f4,%edx,%ecx
  mov    -0x4(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  jmp    L31
L10:
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  imul   $0x1f4,%eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  imul   $0x1f4,%edx,%ecx
  mov    -0x4(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
L31:
  addl   $0x1,-0x4(%rbp)
L48:
  cmpl   $0x1f3,-0x4(%rbp)
  jle    L36
  addl   $0x1,-0x8(%rbp)
L4:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cmp    %eax,-0x8(%rbp)
  jl     L3
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void process_audio(int16_t *inbuf, int16_t *outbuf, int n) {
  int i, j;
  for (i = 0; i < n / 500; ++i)
    for (j = 0; j < 500; ++j)
      if (i * 500 + 2 * j < n)
        outbuf[i * 500 + j] = inbuf[i * 500 + 2 * j];
      else
        outbuf[i * 500 + j] = outbuf[(i - 1) * 500 + j];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<process_audio>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %edx,%r13
  imul   $0x10624dd3,%r13,%r13
  sar    $0x25,%r13
  mov    %edx,%eax
  sar    $0x1f,%eax
  sub    %eax,%r13d
  cmp    $0x1f3,%edx
  jle    L55
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  mov    %edx,%r9d
  mov    $0x0,%ebx
  mov    $0x1f4,%r8d
  mov    $0x0,%r12d
  jmp    L4
L44:
  lea    (%r11,%rax,1),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r10d
  movslq %eax,%rsi
  mov    %r10d,(%rdi,%rsi,4)
L35:
  add    $0x2,%edx
  add    $0x1,%eax
  add    $0x8,%rcx
  cmp    %r8d,%eax
  je     L66
L41:
  cmp    %edx,%r9d
  jle    L44
  mov    (%rcx),%r10d
  movslq %eax,%rsi
  mov    %r10d,(%rdi,%rsi,4)
  jmp    L35
L66:
  add    $0x1,%r12d
  add    $0x1f4,%r8d
  add    $0x1f4,%ebx
  add    $0x7d0,%rbp
  cmp    %r13d,%r12d
  jge    L55
L4:
  lea    -0x3e8(%r8),%r11d
  mov    %rbp,%rcx
  mov    %ebx,%eax
  mov    %ebx,%edx
  sub    %ebx,%r11d
  jmp    L41
L55:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void process_audio(int16_t *inbuf, int16_t *outbuf, int n) {
  int i, j;
  for (i = 0; i < n / 500; ++i)
    for (j = 0; j < 500; ++j)
      if (i * 500 + 2 * j < n)
        outbuf[i * 500 + j] = inbuf[i * 500 + 2 * j];
      else
        outbuf[i * 500 + j] = outbuf[(i - 1) * 500 + j];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<process_audio>:
  endbr64
  push   %rbp
  movslq %edx,%rbp
  mov    %rdi,%r8
  mov    %rbp,%rdi
  imul   $0x10624dd3,%rbp,%rbp
  push   %rbx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sar    $0x25,%rbp
  sub    %eax,%ebp
  cmp    $0x1f3,%edi
  jle    L12
  mov    %rsi,%r10
  xor    %r11d,%r11d
  xor    %ebx,%ebx
L29:
  lea    -0x1f4(%r11),%r9d
  mov    %r11d,%edx
  xor    %eax,%eax
  jmp    L4
  nop
L32:
  mov    (%r8,%rax,8),%ecx
L53:
  mov    %ecx,(%rsi,%rax,4)
  add    $0x1,%rax
  add    $0x2,%edx
  cmp    $0x1f4,%rax
  je     L26
L4:
  cmp    %edx,%edi
  jg     L32
  lea    (%r9,%rax,1),%ecx
  movslq %ecx,%rcx
  mov    (%r10,%rcx,4),%ecx
  jmp    L53
L26:
  add    $0x1,%ebx
  add    $0x7d0,%rsi
  add    $0x7d0,%r8
  add    $0x1f4,%r11d
  cmp    %ebx,%ebp
  jg     L29
L12:
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void process_audio(int16_t *inbuf, int16_t *outbuf, int n) {
  int i, j;
  for (i = 0; i < n / 500; ++i)
    for (j = 0; j < 500; ++j)
      if (i * 500 + 2 * j < n)
        outbuf[i * 500 + j] = inbuf[i * 500 + 2 * j];
      else
        outbuf[i * 500 + j] = outbuf[(i - 1) * 500 + j];
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<process_audio>:
  endbr64
  push   %rbp
  movslq %edx,%rbp
  mov    %rdi,%r8
  mov    %rbp,%rdi
  imul   $0x10624dd3,%rbp,%rbp
  push   %rbx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sar    $0x25,%rbp
  sub    %eax,%ebp
  cmp    $0x1f3,%edi
  jle    L45
  mov    %rsi,%r10
  xor    %r11d,%r11d
  xor    %ebx,%ebx
L31:
  lea    -0x1f4(%r11),%r9d
  mov    %r11d,%edx
  xor    %eax,%eax
  jmp    L39
  nop
L3:
  mov    (%r8,%rax,8),%ecx
L43:
  mov    %ecx,(%rsi,%rax,4)
  add    $0x1,%rax
  add    $0x2,%edx
  cmp    $0x1f4,%rax
  je     L14
L39:
  cmp    %edx,%edi
  jg     L3
  lea    (%r9,%rax,1),%ecx
  movslq %ecx,%rcx
  mov    (%r10,%rcx,4),%ecx
  jmp    L43
L14:
  add    $0x1,%ebx
  add    $0x7d0,%rsi
  add    $0x7d0,%r8
  add    $0x1f4,%r11d
  cmp    %ebx,%ebp
  jg     L31
L45:
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void process_audio(int16_t *inbuf, int16_t *outbuf, int n) {
  int i, j;
  for (i = 0; i < n / 500; ++i)
    for (j = 0; j < 500; ++j)
      if (i * 500 + 2 * j < n)
        outbuf[i * 500 + j] = inbuf[i * 500 + 2 * j];
      else
        outbuf[i * 500 + j] = outbuf[(i - 1) * 500 + j];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D1(%rip),%rax
  mov    %rax,-0x48(%rbp)
  movl   $0x0,-0x4c(%rbp)
  lea    D104(%rip),%rax
  mov    %rax,-0x40(%rbp)
  lea    D59(%rip),%rax
  mov    %rax,-0x30(%rbp)
  lea    D84(%rip),%rax
  mov    %rax,-0x28(%rbp)
  lea    D11(%rip),%rax
  mov    %rax,-0x20(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x54(%rbp),%edx
  mov    -0x38(%rbp),%rax
  lea    D11(%rip),%r9
  lea    D84(%rip),%r8
  lea    D59(%rip),%rcx
  mov    %rax,%rsi
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  endbr64
  mov    -0x4c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4c(%rbp)
  cltq
  mov    -0x30(%rbp,%rax,8),%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  nop
L8:
  jmp    *%rax
  endbr64
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x40(%rbp),%rax
  jmp    L8
  endbr64
  lea    D63(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x40(%rbp),%rax
  jmp    L8
  endbr64
  lea    D64(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x40(%rbp),%rax
  jmp    L8
  endbr64
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type_infered: string
name: read_data
arguments: data_label: D104
type_infered: string
name: read_data
arguments: data_label: D59
type_infered: string
name: read_data
arguments: data_label: D84
type_infered: string
name: read_data
arguments: data_label: D11
type_infered: string
name: read_data
arguments: data_label: D56
type_infered: string
name: read_data
arguments: data_label: D60
type_infered: string
name: read_data
arguments: data_label: D63
type_infered: string
name: read_data
arguments: data_label: D64
type_infered: string

tool: 

tool: 

tool: 

tool: 

tool: 

tool: D56: p = %p (argc=%d), a=%p b=%p c=%p


tool: D60: 1

tool: D63: 2

tool: D64: 3

assistant: ```c
int main(int argc, char **argv) {
  void *where[4];
  void *const fin = &&last;
  void *p;
  int i = 0;
  int *next = &&pick;
  where[0] = &&a;
  where[1] = &&b;
  where[2] = &&c;
  where[3] = &&last;
  printf("p = %p (argc=%d), a=%p b=%p c=%p\n", p, argc, &&a, &&b, &&c);
pick:
  p = where[i++];
  goto *p;
a:
  printf("1\n");
  goto *next;
b:
  printf("2\n");
  goto *next;
c:
  printf("3\n");
  goto *next;
last:
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %edi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rcx
  mov    %rcx,(%rsp)
  lea    D77(%rip),%r8
  mov    %r8,0x8(%rsp)
  lea    D25(%rip),%r9
  mov    %r9,0x10(%rsp)
  lea    D44(%rip),%rax
  mov    %rax,0x18(%rsp)
  mov    $0x0,%esi
  lea    D61(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  endbr64
  mov    %ebx,%eax
  add    $0x1,%ebx
  cltq
  mov    (%rsp,%rax,8),%rax
  lea    D21(%rip),%r12
  lea    D13(%rip),%r13
  lea    D20(%rip),%r14
  lea    -0x2b(%rip),%rbp        # 11c5 <main+0x6c>
  jmp    L19
  endbr64
  mov    %r14,%rdi
  call   <puts@plt>
L21:
  mov    %rbp,%rax
L19:
  jmp    *%rax
  endbr64
  mov    %r13,%rdi
  call   <puts@plt>
  jmp    L21
  endbr64
  mov    %r12,%rdi
  call   <puts@plt>
  jmp    L21
  endbr64
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D14
type_guessed: string
name: retrieve_data
arguments: tag: D77
type_guessed: string
name: retrieve_data
arguments: tag: D25
type_guessed: string
name: retrieve_data
arguments: tag: D44
type_guessed: string
name: retrieve_data
arguments: tag: D61
type_guessed: string
name: retrieve_data
arguments: tag: D21
type_guessed: string
name: retrieve_data
arguments: tag: D13
type_guessed: string
name: retrieve_data
arguments: tag: D20
type_guessed: string

tool: 

tool: 

tool: 

tool: 

tool: D61: p = %p (argc=%d), a=%p b=%p c=%p


tool: D21: 3

tool: D13: 2

tool: D20: 1

assistant: ```c
int main(int argc, char **argv) {
  void *where[4];
  void *const fin = &&last;
  void *p;
  int i = 0;
  int *next = &&pick;
  where[0] = &&a;
  where[1] = &&b;
  where[2] = &&c;
  where[3] = &&last;
  printf("p = %p (argc=%d), a=%p b=%p c=%p\n", p, argc, &&a, &&b, &&c);
pick:
  p = where[i++];
  goto *p;
a:
  printf("1\n");
  goto *next;
b:
  printf("2\n");
  goto *next;
c:
  printf("3\n");
  goto *next;
last:
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    %edi,%edx
  lea    D19(%rip),%rcx
  xor    %ebx,%ebx
  lea    D21(%rip),%r8
  lea    D54(%rip),%r9
  xor    %esi,%esi
  lea    D71(%rip),%rdi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D62(%rip),%rax
  mov    %rcx,(%rsp)
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %r8,0x8(%rsp)
  mov    %r9,0x10(%rsp)
  call   <printf@plt>
  movslq %ebx,%rax
  add    $0x1,%ebx
  jmp    *(%rsp,%rax,8)
  endbr64
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
  endbr64
  lea    D78(%rip),%rdi
  call   <puts@plt>
  movslq %ebx,%rax
  add    $0x1,%ebx
  jmp    *(%rsp,%rax,8)
  endbr64
  lea    D7(%rip),%rdi
  call   <puts@plt>
  movslq %ebx,%rax
  add    $0x1,%ebx
  jmp    *(%rsp,%rax,8)
  endbr64
  lea    D32(%rip),%rdi
  call   <puts@plt>
  movslq %ebx,%rax
  add    $0x1,%ebx
  jmp    *(%rsp,%rax,8)
  endbr64
  movslq %ebx,%rax
  add    $0x1,%ebx
  jmp    *(%rsp,%rax,8)
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D19
data_type: string
name: parse_data
arguments: data_tag: D21
data_type: string
name: parse_data
arguments: data_tag: D54
data_type: string
name: parse_data
arguments: data_tag: D71
data_type: string
name: parse_data
arguments: data_tag: D62
data_type: string
name: parse_data
arguments: data_tag: D78
data_type: string
name: parse_data
arguments: data_tag: D7
data_type: string
name: parse_data
arguments: data_tag: D32
data_type: string

tool: 

tool: 

tool: 

tool: D71: p = %p (argc=%d), a=%p b=%p c=%p


tool: 

tool: D78: 3

tool: D7: 2

tool: D32: 1

assistant: ```c
int main(int argc, char **argv) {
  void *where[4];
  void *const fin = &&last;
  void *p;
  int i = 0;
  int *next = &&pick;
  where[0] = &&a;
  where[1] = &&b;
  where[2] = &&c;
  where[3] = &&last;
  printf("p = %p (argc=%d), a=%p b=%p c=%p\n", p, argc, &&a, &&b, &&c);
pick:
  p = where[i++];
  goto *p;
a:
  printf("1\n");
  goto *next;
b:
  printf("2\n");
  goto *next;
c:
  printf("3\n");
  goto *next;
last:
  return 0;
}
```