ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xa0(%rbp)
  movl   $0x0,-0x9c(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <inicializaTabuleiro@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  call   <iniciaLixos@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
L5:
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <mostraTabuleiro@plt>
  lea    -0x98(%rbp),%rax
  mov    %rax,%rdi
  call   <Pescar@plt>
  addl   $0x1,-0xa0(%rbp)
  lea    -0x90(%rbp),%rdx
  lea    -0x98(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <acertou@plt>
  test   %rax,%rax
  je     L17
  mov    -0xa0(%rbp),%edx
  lea    -0x90(%rbp),%rcx
  lea    -0x98(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <dica@plt>
  addl   $0x1,-0x9c(%rbp)
  jmp    L6
L17:
  mov    -0xa0(%rbp),%edx
  lea    -0x90(%rbp),%rcx
  lea    -0x98(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <dica@plt>
L6:
  lea    -0x70(%rbp),%rdx
  lea    -0x90(%rbp),%rcx
  lea    -0x98(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <alteraTabuleiro@plt>
  cmpl   $0x3,-0x9c(%rbp)
  jne    L5
  mov    -0xa0(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <mostraTabuleiro@plt>
  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: D5
type_guessed: string
name: read_data
arguments: data_label: D18
type_guessed: string

tool: D5: 


Jogo terminado. Voce encontrou 3 lixos no oceano em %d tentativas

tool: D18: 


Voce tornou o oceano um pouco mais limpo, as tartarugas e baleias agradecem!!!



assistant: ```c
int main() {
  int tabuleiro[5][5];
  int lixos[3][2];
  int pesca[2];
  int tentativas = 0, acertos = 0;
  inicializaTabuleiro(tabuleiro);
  iniciaLixos(lixos);
  printf("\n");
  do {
    mostraTabuleiro(tabuleiro);
    Pescar(pesca);
    tentativas++;
    if (acertou(pesca, lixos)) {
      dica(pesca, lixos, tentativas);
      acertos++;
    } else
      dica(pesca, lixos, tentativas);
    alteraTabuleiro(pesca, lixos, tabuleiro);
  } while (acertos != 3);
  printf("\n\n\nJogo terminado. Voce encontrou 3 lixos no oceano em %d tentativas", tentativas);
  printf("\n\n\nVoce tornou o oceano um pouco mais limpo, as tartarugas e baleias agradecem!!!\n\n\n");
  mostraTabuleiro(tabuleiro);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rdi
  call   <inicializaTabuleiro@plt>
  lea    0x10(%rsp),%rdi
  call   <iniciaLixos@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%r14d
  mov    $0x0,%ebp
  lea    0x30(%rsp),%r13
  lea    0x8(%rsp),%rbx
  lea    0x10(%rsp),%r12
  jmp    L0
L30:
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <dica@plt>
  add    $0x1,%r14d
L20:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <alteraTabuleiro@plt>
  cmp    $0x3,%r14d
  je     L10
L0:
  mov    %r13,%rdi
  call   <mostraTabuleiro@plt>
  mov    %rbx,%rdi
  call   <Pescar@plt>
  add    $0x1,%ebp
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <acertou@plt>
  test   %rax,%rax
  jne    L30
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <dica@plt>
  jmp    L20
L10:
  mov    %ebp,%esi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <puts@plt>
  lea    0x30(%rsp),%rdi
  call   <mostraTabuleiro@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0xa0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D19
type_infered: string
name: disassemble_data
arguments: data_tag: D12
type_infered: string

tool: D19: 


Jogo terminado. Voce encontrou 3 lixos no oceano em %d tentativas

tool: D12: 


Voce tornou o oceano um pouco mais limpo, as tartarugas e baleias agradecem!!!



assistant: ```c
int main() {
  int tabuleiro[5][5];
  int lixos[3][2];
  int pesca[2];
  int tentativas = 0, acertos = 0;
  inicializaTabuleiro(tabuleiro);
  iniciaLixos(lixos);
  printf("\n");
  do {
    mostraTabuleiro(tabuleiro);
    Pescar(pesca);
    tentativas++;
    if (acertou(pesca, lixos)) {
      dica(pesca, lixos, tentativas);
      acertos++;
    } else
      dica(pesca, lixos, tentativas);
    alteraTabuleiro(pesca, lixos, tabuleiro);
  } while (acertos != 3);
  printf("\n\n\nJogo terminado. Voce encontrou 3 lixos no oceano em %d tentativas", tentativas);
  printf("\n\n\nVoce tornou o oceano um pouco mais limpo, as tartarugas e baleias agradecem!!!\n\n\n");
  mostraTabuleiro(tabuleiro);
  return 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   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%r13
  lea    0x10(%rsp),%r12
  mov    %r13,%rdi
  lea    0x8(%rsp),%rbx
  call   <inicializaTabuleiro@plt>
  mov    %r12,%rdi
  call   <iniciaLixos@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L43
  cs nopw 0x0(%rax,%rax,1)
L32:
  call   <dica@plt>
  add    $0x1,%r14d
L10:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <alteraTabuleiro@plt>
  cmp    $0x3,%r14d
  je     L13
L43:
  mov    %r13,%rdi
  add    $0x1,%ebp
  call   <mostraTabuleiro@plt>
  mov    %rbx,%rdi
  call   <Pescar@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <acertou@plt>
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  test   %rax,%rax
  jne    L32
  call   <dica@plt>
  jmp    L10
  nop
L13:
  mov    %ebp,%esi
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <puts@plt>
  mov    %r13,%rdi
  call   <mostraTabuleiro@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
datatype: string
name: retrieve_data
arguments: data_tag: D3
datatype: string

tool: D9: 


Jogo terminado. Voce encontrou 3 lixos no oceano em %d tentativas

tool: D3: 


Voce tornou o oceano um pouco mais limpo, as tartarugas e baleias agradecem!!!



assistant: ```c
int main() {
  int tabuleiro[5][5];
  int lixos[3][2];
  int pesca[2];
  int tentativas = 0, acertos = 0;
  inicializaTabuleiro(tabuleiro);
  iniciaLixos(lixos);
  printf("\n");
  do {
    mostraTabuleiro(tabuleiro);
    Pescar(pesca);
    tentativas++;
    if (acertou(pesca, lixos)) {
      dica(pesca, lixos, tentativas);
      acertos++;
    } else
      dica(pesca, lixos, tentativas);
    alteraTabuleiro(pesca, lixos, tabuleiro);
  } while (acertos != 3);
  printf("\n\n\nJogo terminado. Voce encontrou 3 lixos no oceano em %d tentativas", tentativas);
  printf("\n\n\nVoce tornou o oceano um pouco mais limpo, as tartarugas e baleias agradecem!!!\n\n\n");
  mostraTabuleiro(tabuleiro);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%r13
  lea    0x10(%rsp),%r12
  mov    %r13,%rdi
  lea    0x8(%rsp),%rbx
  call   <inicializaTabuleiro@plt>
  mov    %r12,%rdi
  call   <iniciaLixos@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L48
  cs nopw 0x0(%rax,%rax,1)
L32:
  call   <dica@plt>
  add    $0x1,%r14d
L5:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <alteraTabuleiro@plt>
  cmp    $0x3,%r14d
  je     L13
L48:
  mov    %r13,%rdi
  add    $0x1,%ebp
  call   <mostraTabuleiro@plt>
  mov    %rbx,%rdi
  call   <Pescar@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <acertou@plt>
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  test   %rax,%rax
  jne    L32
  call   <dica@plt>
  jmp    L5
  nop
L13:
  mov    %ebp,%esi
  lea    D0(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  mov    %r13,%rdi
  call   <mostraTabuleiro@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D0
data_type: string
name: retrieve_data
arguments: tag: D13
data_type: string

tool: D0: 


Jogo terminado. Voce encontrou 3 lixos no oceano em %d tentativas

tool: D13: 


Voce tornou o oceano um pouco mais limpo, as tartarugas e baleias agradecem!!!



assistant: ```c
int main() {
  int tabuleiro[5][5];
  int lixos[3][2];
  int pesca[2];
  int tentativas = 0, acertos = 0;
  inicializaTabuleiro(tabuleiro);
  iniciaLixos(lixos);
  printf("\n");
  do {
    mostraTabuleiro(tabuleiro);
    Pescar(pesca);
    tentativas++;
    if (acertou(pesca, lixos)) {
      dica(pesca, lixos, tentativas);
      acertos++;
    } else
      dica(pesca, lixos, tentativas);
    alteraTabuleiro(pesca, lixos, tabuleiro);
  } while (acertos != 3);
  printf("\n\n\nJogo terminado. Voce encontrou 3 lixos no oceano em %d tentativas", tentativas);
  printf("\n\n\nVoce tornou o oceano um pouco mais limpo, as tartarugas e baleias agradecem!!!\n\n\n");
  mostraTabuleiro(tabuleiro);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0xd4(%rbp)
  jmp    L0
L29:
  lea    -0xd0(%rbp),%rcx
  mov    -0xd4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0xd4(%rbp)
L0:
  cmpl   $0x9,-0xd4(%rbp)
  jle    L29
  lea    -0xd0(%rbp),%rax
  mov    %rax,%rdi
  call   <sort@plt>
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0xd4(%rbp)
  jmp    L32
L16:
  lea    -0xd0(%rbp),%rcx
  mov    -0xd4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0xd4(%rbp)
L32:
  cmpl   $0x9,-0xd4(%rbp)
  jle    L16
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
data_type: string
name: unpack_data
arguments: data_label: D0
data_type: string
name: unpack_data
arguments: data_label: D3
data_type: string

tool: D9: input 10 strings with same length:

tool: D0: %s

tool: D3: new strings are:

assistant: ```c
int main() {
  void sort(char *s[]);
  int i;
  char str[10][20];
  printf("input 10 strings with same length:\n");
  for (i = 0; i < 10; i++)
    scanf("%s", str[i]);
  sort(str);
  printf("new strings are:\n");
  for (i = 0; i < 10; i++)
    printf("%s\n", str[i]);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  lea    D3(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  lea    0xc8(%rsp),%r12
  mov    %rbx,%rbp
  lea    D25(%rip),%r13
L11:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x14,%rbp
  cmp    %r12,%rbp
  jne    L11
  mov    %rsp,%rdi
  call   <sort@plt>
  lea    D16(%rip),%rdi
  call   <puts@plt>
L17:
  mov    %rbx,%rdi
  call   <puts@plt>
  add    $0x14,%rbx
  cmp    %r12,%rbx
  jne    L17
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0xd8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: string
name: unpack_data
arguments: data_tag: D25
data_type: string
name: unpack_data
arguments: data_tag: D16
data_type: string

tool: D3: input 10 strings with same length:

tool: D25: %s

tool: D16: new strings are:

assistant: ```c
int main() {
  void sort(char *s[]);
  int i;
  char str[10][20];
  printf("input 10 strings with same length:\n");
  for (i = 0; i < 10; i++)
    scanf("%s", str[i]);
  sort(str);
  printf("new strings are:\n");
  for (i = 0; i < 10; i++)
    printf("%s\n", str[i]);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  lea    D4(%rip),%rdi
  push   %r13
  lea    D8(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0xc8(%rsp),%r12
  call   <puts@plt>
  mov    %r14,%rbx
  mov    %r14,%rbp
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x14,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L20
  mov    %r14,%rdi
  call   <sort@plt>
  lea    D20(%rip),%rdi
  call   <puts@plt>
  nopl   0x0(%rax,%rax,1)
L26:
  mov    %rbx,%rdi
  add    $0x14,%rbx
  call   <puts@plt>
  cmp    %r12,%rbx
  jne    L26
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0xd0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
data_type: string
name: extract_symbolic_data
arguments: tag: D8
data_type: string
name: extract_symbolic_data
arguments: tag: D20
data_type: string

tool: D4: input 10 strings with same length:

tool: D8: %s

tool: D20: new strings are:

assistant: ```c
int main() {
  void sort(char *s[]);
  int i;
  char str[10][20];
  printf("input 10 strings with same length:\n");
  for (i = 0; i < 10; i++)
    scanf("%s", str[i]);
  sort(str);
  printf("new strings are:\n");
  for (i = 0; i < 10; i++)
    printf("%s\n", str[i]);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D16(%rip),%rdi
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xd0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0xc8(%rsp),%r12
  call   <puts@plt>
  mov    %r14,%rbx
  mov    %r14,%rbp
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x14,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L19
  mov    %r14,%rdi
  call   <sort@plt>
  lea    D7(%rip),%rdi
  call   <puts@plt>
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %rbx,%rdi
  add    $0x14,%rbx
  call   <puts@plt>
  cmp    %r12,%rbx
  jne    L11
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0xd0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D16
datatype: string
name: retrieve_data
arguments: data_tag: D1
datatype: string
name: retrieve_data
arguments: data_tag: D7
datatype: string

tool: D16: input 10 strings with same length:

tool: D1: %s

tool: D7: new strings are:

assistant: ```c
int main() {
  void sort(char *s[]);
  int i;
  char str[10][20];
  printf("input 10 strings with same length:\n");
  for (i = 0; i < 10; i++)
    scanf("%s", str[i]);
  sort(str);
  printf("new strings are:\n");
  for (i = 0; i < 10; i++)
    printf("%s\n", str[i]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tty_show_title>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    D2(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
data_type: string

tool: D2: show_title: %s


assistant: ```c
void tty_show_title(void *ui, const char *title) {
  FILE *tty = ui;
  fprintf(tty, "show_title: %s\r\n", title);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<tty_show_title>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  lea    D7(%rip),%rsi
  call   <fprintf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
type_infered: string

tool: D7: show_title: %s


assistant: ```c
void tty_show_title(void *ui, const char *title) {
  FILE *tty = ui;
  fprintf(tty, "show_title: %s\r\n", title);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<tty_show_title>:
  endbr64
  mov    %rsi,%rdx
  lea    D0(%rip),%rsi
  jmp    1050 <fprintf@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
type: string

tool: D0: show_title: %s


assistant: ```c
void tty_show_title(void *ui, const char *title) {
  FILE *tty = ui;
  fprintf(tty, "show_title: %s\r\n", title);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<tty_show_title>:
  endbr64
  mov    %rsi,%rdx
  lea    D5(%rip),%rsi
  jmp    1050 <fprintf@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type: string

tool: D5: show_title: %s


assistant: ```c
void tty_show_title(void *ui, const char *title) {
  FILE *tty = ui;
  fprintf(tty, "show_title: %s\r\n", title);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strdata>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L44:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L21
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jbe    L68
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  ja     L68
L21:
  mov    $0x1,%eax
  jmp    L77
L68:
  mov    $0x0,%eax
L77:
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L45
  mov    -0x4(%rbp),%eax
  cltq
  lea    D14(%rip),%rdx
  movb   $0x2e,(%rax,%rdx,1)
  jmp    L15
L45:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  cltq
  lea    D14(%rip),%rdx
  mov    %cl,(%rax,%rdx,1)
L15:
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L44
  mov    -0x4(%rbp),%eax
  cltq
  lea    D14(%rip),%rdx
  movb   $0x0,(%rax,%rdx,1)
  lea    D14(%rip),%rax
  leave
  ret
```

assistant: ```c
char *strdata(unsigned char *buf, int len) {
  static char _strdata[1 + 100];
  int i;
  for (i = 0; i < len; i++) {
    assert(buf[i] == 0 || (buf[i] >= '0' && buf[i] <= '9'));
    if (buf[i] == 0)
      _strdata[i] = '.';
    else
      _strdata[i] = buf[i];
  }
  _strdata[i] = 0;
  return _strdata;
}
```
user: What is the c source code of the assembly code below?

```asm
<strdata>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L7
  mov    %esi,%r14d
  mov    %rdi,%rbx
  lea    D6(%rip),%rbp
  lea    -0x1(%rsi),%r12d
  lea    0x1(%rbp),%rax
  add    %rax,%r12
  mov    $0x2e,%r13d
L21:
  movzbl (%rbx),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  setbe  %dil
  test   %al,%al
  sete   %al
  or     %eax,%edi
  movzbl %dil,%edi
  call   <assert@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  cmove  %r13d,%eax
  mov    %al,0x0(%rbp)
  add    $0x1,%rbx
  add    $0x1,%rbp
  cmp    %r12,%rbp
  jne    L21
L18:
  lea    D6(%rip),%rax
  movslq %r14d,%r14
  movb   $0x0,(%rax,%r14,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  mov    $0x0,%r14d
  jmp    L18
```

assistant: ```c
char *strdata(unsigned char *buf, int len) {
  static char _strdata[1 + 100];
  int i;
  for (i = 0; i < len; i++) {
    assert(buf[i] == 0 || (buf[i] >= '0' && buf[i] <= '9'));
    if (buf[i] == 0)
      _strdata[i] = '.';
    else
      _strdata[i] = buf[i];
  }
  _strdata[i] = 0;
  return _strdata;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strdata>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  movslq %esi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %r12d,%r12d
  jle    L27
  lea    D12(%rip),%r13
  lea    -0x1(%r12),%ebx
  mov    %rdi,%r14
  mov    $0x2e,%ebp
  lea    0x1(%r13),%rax
  mov    %r13,%r15
  add    %rax,%rbx
  nopl   0x0(%rax,%rax,1)
L14:
  movzbl (%r14),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  setbe  %dil
  test   %al,%al
  sete   %al
  or     %eax,%edi
  movzbl %dil,%edi
  call   <assert@plt>
  movzbl (%r14),%eax
  test   %al,%al
  cmove  %ebp,%eax
  add    $0x1,%r15
  add    $0x1,%r14
  mov    %al,-0x1(%r15)
  cmp    %rbx,%r15
  jne    L14
L13:
  movb   $0x0,0x0(%r13,%r12,1)
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L27:
  xor    %r12d,%r12d
  lea    D12(%rip),%r13
  jmp    L13
```

assistant: ```c
char *strdata(unsigned char *buf, int len) {
  static char _strdata[1 + 100];
  int i;
  for (i = 0; i < len; i++) {
    assert(buf[i] == 0 || (buf[i] >= '0' && buf[i] <= '9'));
    if (buf[i] == 0)
      _strdata[i] = '.';
    else
      _strdata[i] = buf[i];
  }
  _strdata[i] = 0;
  return _strdata;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strdata>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L59
  lea    D16(%rip),%r12
  lea    -0x1(%rbp),%ebx
  mov    %rdi,%r13
  lea    0x1(%r12),%rax
  mov    %r12,%r14
  add    %rax,%rbx
  nop
L29:
  movzbl 0x0(%r13),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  setbe  %dil
  test   %al,%al
  sete   %al
  or     %eax,%edi
  movzbl %dil,%edi
  call   <assert@plt>
  movzbl 0x0(%r13),%eax
  test   %al,%al
  je     L2
  mov    %al,(%r14)
  add    $0x1,%r14
  add    $0x1,%r13
  cmp    %r14,%rbx
  jne    L29
L17:
  pop    %rbx
  movb   $0x0,(%r12,%rbp,1)
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L2:
  movb   $0x2e,(%r14)
  add    $0x1,%r14
  add    $0x1,%r13
  cmp    %rbx,%r14
  jne    L29
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L59:
  xor    %ebp,%ebp
  lea    D16(%rip),%r12
  jmp    L17
```

assistant: ```c
char *strdata(unsigned char *buf, int len) {
  static char _strdata[1 + 100];
  int i;
  for (i = 0; i < len; i++) {
    assert(buf[i] == 0 || (buf[i] >= '0' && buf[i] <= '9'));
    if (buf[i] == 0)
      _strdata[i] = '.';
    else
      _strdata[i] = buf[i];
  }
  _strdata[i] = 0;
  return _strdata;
}
```