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

```asm
<f3>:
  endbr64
  movdqa D9(%rip),%xmm1
  movdqu (%rdi),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rdi)
  movdqu 0x10(%rdi),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,0x10(%rdi)
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
type: byte[16]

tool: {"D9": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
void f3(int *x) {
  for (int j = 0; j < 8; ++j)
    x[j] += 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_number>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    (%rdi),%rdi
  cmpb   $0x2d,(%rdi)
  sete   %bl
  movzbl %bl,%ebx
  add    %rdi,%rbx
  movzbl (%rbx),%eax
  cmp    $0x30,%al
  je     L21
  sub    $0x30,%eax
  mov    $0x0,%edx
  cmp    $0x9,%al
  ja     L74
L52:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L52
L92:
  cmpb   $0x2e,(%rbx)
  je     L72
L90:
  movzbl (%rbx),%eax
  and    $0xffffffdf,%eax
  cmp    $0x45,%al
  jne    L22
  movzbl 0x1(%rbx),%eax
  sub    $0x2b,%eax
  mov    %rbx,%rdx
  test   $0xfd,%al
  sete   %bl
  movzbl %bl,%ebx
  lea    0x1(%rbx,%rdx,1),%rbx
  movzbl (%rbx),%eax
  sub    $0x30,%eax
  mov    $0x0,%edx
  cmp    $0x9,%al
  ja     L74
L34:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L34
L22:
  test   %r12,%r12
  je     L2
  mov    $0x0,%esi
  call   <strtod@plt>
  movsd  %xmm0,(%r12)
L2:
  mov    %rbx,0x0(%rbp)
  mov    $0x1,%edx
L74:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L21:
  add    $0x1,%rbx
  jmp    L92
L72:
  movzbl 0x1(%rbx),%eax
  sub    $0x30,%eax
  mov    $0x0,%edx
  cmp    $0x9,%al
  ja     L74
  add    $0x1,%rbx
L57:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L57
  jmp    L90
```

assistant: ```c
int parse_number(const char **sp, double *out) {
  const char *s = *sp;
  if (*s == '-')
    s++;
  if (*s == '0') {
    s++;
  } else {
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (*s == '.') {
    s++;
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (*s == 'E' || *s == 'e') {
    s++;
    if (*s == '+' || *s == '-')
      s++;
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (out)
    *out = strtod(*sp, NULL);
  *sp = s;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_number>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    (%rdi),%rdi
  movzbl (%rdi),%eax
  mov    %rdi,%rbx
  cmp    $0x2d,%al
  jne    L52
  movzbl 0x1(%rdi),%eax
  lea    0x1(%rdi),%rbx
L52:
  cmp    $0x30,%al
  je     L2
  sub    $0x30,%eax
  xor    %r8d,%r8d
  cmp    $0x9,%al
  ja     L89
  nopl   (%rax)
L100:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L100
  cmp    $0x2e,%al
  je     L13
L68:
  and    $0xffffffdf,%eax
  cmp    $0x45,%al
  jne    L129
  movzbl 0x1(%rbx),%eax
  lea    -0x2b(%rax),%edx
  and    $0xfd,%edx
  je     L55
  add    $0x1,%rbx
L107:
  sub    $0x30,%eax
  xor    %r8d,%r8d
  cmp    $0x9,%al
  ja     L89
  nopl   0x0(%rax)
L10:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L10
L129:
  test   %r12,%r12
  je     L143
  xor    %esi,%esi
  call   <strtod@plt>
  movsd  %xmm0,(%r12)
L143:
  mov    %rbx,0x0(%rbp)
  mov    $0x1,%r8d
L89:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  cmp    $0x2e,%al
  jne    L68
L13:
  movzbl 0x1(%rbx),%eax
  xor    %r8d,%r8d
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L89
  add    $0x1,%rbx
  xchg   %ax,%ax
L135:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L135
  jmp    L68
  nopl   (%rax)
L55:
  movzbl 0x2(%rbx),%eax
  add    $0x2,%rbx
  jmp    L107
```

assistant: ```c
int parse_number(const char **sp, double *out) {
  const char *s = *sp;
  if (*s == '-')
    s++;
  if (*s == '0') {
    s++;
  } else {
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (*s == '.') {
    s++;
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (*s == 'E' || *s == 'e') {
    s++;
    if (*s == '+' || *s == '-')
      s++;
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (out)
    *out = strtod(*sp, NULL);
  *sp = s;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<parse_number>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    (%rdi),%rdi
  movzbl (%rdi),%eax
  mov    %rdi,%rbx
  cmp    $0x2d,%al
  jne    L49
  movzbl 0x1(%rdi),%eax
  lea    0x1(%rdi),%rbx
L49:
  cmp    $0x30,%al
  je     L97
  sub    $0x30,%eax
  xor    %r8d,%r8d
  cmp    $0x9,%al
  ja     L149
  nopl   (%rax)
L113:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L113
  cmp    $0x2e,%al
  je     L52
L126:
  and    $0xffffffdf,%eax
  cmp    $0x45,%al
  jne    L34
  movzbl 0x1(%rbx),%eax
  lea    -0x2b(%rax),%edx
  and    $0xfd,%edx
  je     L14
  add    $0x1,%rbx
L6:
  sub    $0x30,%eax
  xor    %r8d,%r8d
  cmp    $0x9,%al
  ja     L149
  nopl   0x0(%rax)
L29:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L29
L34:
  test   %r12,%r12
  je     L130
  xor    %esi,%esi
  call   <strtod@plt>
  movsd  %xmm0,(%r12)
L130:
  mov    %rbx,0x0(%rbp)
  mov    $0x1,%r8d
L149:
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L97:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  cmp    $0x2e,%al
  jne    L126
L52:
  movzbl 0x1(%rbx),%eax
  xor    %r8d,%r8d
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L149
  add    $0x1,%rbx
  xchg   %ax,%ax
L32:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L32
  jmp    L126
  nopl   (%rax)
L14:
  movzbl 0x2(%rbx),%eax
  add    $0x2,%rbx
  jmp    L6
```

assistant: ```c
int parse_number(const char **sp, double *out) {
  const char *s = *sp;
  if (*s == '-')
    s++;
  if (*s == '0') {
    s++;
  } else {
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (*s == '.') {
    s++;
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (*s == 'E' || *s == 'e') {
    s++;
    if (*s == '+' || *s == '-')
      s++;
    if (!((*s) >= '0' && (*s) <= '9'))
      return 0;
    do {
      s++;
    } while (((*s) >= '0' && (*s) <= '9'));
  }
  if (out)
    *out = strtod(*sp, NULL);
  *sp = s;
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<chararrcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  je     L25
  mov    $0xffffffff,%eax
  jmp    L54
L25:
  movl   $0x0,-0x4(%rbp)
  jmp    L50
L48:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L8
  mov    $0xffffffff,%eax
  jmp    L54
L8:
  addl   $0x1,-0x4(%rbp)
L50:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L48
  mov    $0x0,%eax
L54:
  pop    %rbp
  ret
```

assistant: ```c
int chararrcmp(char *s, int n, char *p, int m) {
  int i = 0;
  if (n != m)
    return -1;
  for (i = 0; i < n; i++) {
    if (s[i] != p[i])
      return -1;
  }
  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
<chararrcmp>:
  endbr64
  cmp    %ecx,%esi
  jne    L16
  test   %esi,%esi
  jle    L18
  lea    -0x1(%rsi),%esi
  mov    $0x0,%eax
  jmp    L35
L8:
  mov    %rcx,%rax
L35:
  movzbl (%rdx,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  jne    L40
  lea    0x1(%rax),%rcx
  cmp    %rsi,%rax
  jne    L8
  mov    $0x0,%eax
  ret
L16:
  mov    $0xffffffff,%eax
  ret
L18:
  mov    $0x0,%eax
  ret
L40:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int chararrcmp(char *s, int n, char *p, int m) {
  int i = 0;
  if (n != m)
    return -1;
  for (i = 0; i < n; i++) {
    if (s[i] != p[i])
      return -1;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<chararrcmp>:
  endbr64
  cmp    %ecx,%esi
  jne    L18
  test   %esi,%esi
  jle    L1
  movslq %esi,%rsi
  xor    %eax,%eax
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L41:
  add    $0x1,%rax
  cmp    %rsi,%rax
  je     L1
L33:
  movzbl (%rdx,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  je     L41
L18:
  mov    $0xffffffff,%eax
  ret
L1:
  xor    %eax,%eax
  ret
```

assistant: ```c
int chararrcmp(char *s, int n, char *p, int m) {
  int i = 0;
  if (n != m)
    return -1;
  for (i = 0; i < n; i++) {
    if (s[i] != p[i])
      return -1;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<chararrcmp>:
  endbr64
  cmp    %ecx,%esi
  jne    L48
  test   %esi,%esi
  jle    L34
  movslq %esi,%rsi
  xor    %eax,%eax
  jmp    L29
  nopl   0x0(%rax,%rax,1)
L3:
  add    $0x1,%rax
  cmp    %rsi,%rax
  je     L34
L29:
  movzbl (%rdx,%rax,1),%ecx
  cmp    %cl,(%rdi,%rax,1)
  je     L3
L48:
  mov    $0xffffffff,%eax
  ret
L34:
  xor    %eax,%eax
  ret
```

assistant: ```c
int chararrcmp(char *s, int n, char *p, int m) {
  int i = 0;
  if (n != m)
    return -1;
  for (i = 0; i < n; i++) {
    if (s[i] != p[i])
      return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L29
L5:
  cmpl   $0x0,-0xc(%rbp)
  jne    L2
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x1,-0xc(%rbp)
L2:
  mov    -0x18(%rbp),%eax
  mov    -0x14(%rbp),%edx
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  mov    -0x10(%rbp),%edx
  cmp    %eax,%edx
  cmovle %edx,%eax
  mov    %eax,-0x10(%rbp)
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L29:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jne    L5
  cmpl   $0x0,-0xc(%rbp)
  je     L44
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L44:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D46
data_type: string
name: unpack_data
arguments: data_tag: D35
data_type: string
name: unpack_data
arguments: data_tag: D56
data_type: string
name: unpack_data
arguments: data_tag: D42
data_type: string
name: unpack_data
arguments: data_tag: D10
data_type: string

tool: D46: Escriba 0 cuando desee imprimir el maximo y minimo


tool: D35: Ingrese un numero: 

tool: D56: %d

tool: D42: Numero maximo: %d


tool: D10: Numero minimo: %d


assistant: ```c
int main() {
  int max, min;
  int num;
  int init = 0;
  printf("Escriba 0 cuando desee imprimir el maximo y minimo\n\n");
  printf("Ingrese un numero: ");
  scanf("%d", &num);
  while (num != 0) {
    if (!init) {
      max = min = num;
      init = 1;
    }
    max = (num > max) ? num : max;
    min = (num < min) ? num : min;
    printf("Ingrese un numero: ");
    scanf("%d", &num);
  }
  if (init) {
    printf("Numero maximo: %d\n", max);
    printf("Numero minimo: %d\n", min);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  call   <puts@plt>
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D57(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  test   %ebx,%ebx
  je     L9
  mov    %ebx,%eax
  mov    %ebx,%ebp
  lea    D26(%rip),%r14
  lea    0x4(%rsp),%r13
  lea    D57(%rip),%r12
L19:
  cmp    %eax,%ebx
  cmovl  %eax,%ebx
  cmp    %eax,%ebp
  cmovg  %eax,%ebp
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L19
  mov    %ebx,%esi
  lea    D59(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D45(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D1
type_guessed: string
name: read_data
arguments: data_label: D26
type_guessed: string
name: read_data
arguments: data_label: D57
type_guessed: string
name: read_data
arguments: data_label: D59
type_guessed: string
name: read_data
arguments: data_label: D45
type_guessed: string

tool: D1: Escriba 0 cuando desee imprimir el maximo y minimo


tool: D26: Ingrese un numero: 

tool: D57: %d

tool: D59: Numero maximo: %d


tool: D45: Numero minimo: %d


assistant: ```c
int main() {
  int max, min;
  int num;
  int init = 0;
  printf("Escriba 0 cuando desee imprimir el maximo y minimo\n\n");
  printf("Ingrese un numero: ");
  scanf("%d", &num);
  while (num != 0) {
    if (!init) {
      max = min = num;
      init = 1;
    }
    max = (num > max) ? num : max;
    min = (num < min) ? num : min;
    printf("Ingrese un numero: ");
    scanf("%d", &num);
  }
  if (init) {
    printf("Numero maximo: %d\n", max);
    printf("Numero minimo: %d\n", min);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  lea    D44(%rip),%rdi
  lea    D42(%rip),%r14
  push   %r13
  lea    D25(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  call   <puts@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebp
  test   %ebp,%ebp
  je     L5
  mov    %ebp,%eax
  mov    %ebp,%r12d
  nopl   0x0(%rax)
L1:
  cmp    %eax,%ebp
  mov    %r13,%rdi
  cmovl  %eax,%ebp
  cmp    %eax,%r12d
  cmovg  %eax,%r12d
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L1
  mov    %ebp,%esi
  lea    D40(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L5:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D44
type: string
name: retrieve_data
arguments: tag: D42
type: string
name: retrieve_data
arguments: tag: D25
type: string
name: retrieve_data
arguments: tag: D40
type: string
name: retrieve_data
arguments: tag: D38
type: string

tool: D44: Escriba 0 cuando desee imprimir el maximo y minimo


tool: D42: %d

tool: D25: Ingrese un numero: 

tool: D40: Numero maximo: %d


tool: D38: Numero minimo: %d


assistant: ```c
int main() {
  int max, min;
  int num;
  int init = 0;
  printf("Escriba 0 cuando desee imprimir el maximo y minimo\n\n");
  printf("Ingrese un numero: ");
  scanf("%d", &num);
  while (num != 0) {
    if (!init) {
      max = min = num;
      init = 1;
    }
    max = (num > max) ? num : max;
    min = (num < min) ? num : min;
    printf("Ingrese un numero: ");
    scanf("%d", &num);
  }
  if (init) {
    printf("Numero maximo: %d\n", max);
    printf("Numero minimo: %d\n", min);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D21(%rip),%rdi
  lea    D41(%rip),%r14
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  call   <puts@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebp
  test   %ebp,%ebp
  je     L28
  mov    %ebp,%eax
  mov    %ebp,%r12d
  nopl   0x0(%rax)
L5:
  cmp    %eax,%ebp
  mov    %r13,%rdi
  cmovl  %eax,%ebp
  cmp    %eax,%r12d
  cmovg  %eax,%r12d
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L5
  mov    %ebp,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D21
datatype: string
name: extract_symbolic_data
arguments: label: D41
datatype: string
name: extract_symbolic_data
arguments: label: D7
datatype: string
name: extract_symbolic_data
arguments: label: D16
datatype: string
name: extract_symbolic_data
arguments: label: D38
datatype: string

tool: D21: Escriba 0 cuando desee imprimir el maximo y minimo


tool: D41: %d

tool: D7: Ingrese un numero: 

tool: D16: Numero maximo: %d


tool: D38: Numero minimo: %d


assistant: ```c
int main() {
  int max, min;
  int num;
  int init = 0;
  printf("Escriba 0 cuando desee imprimir el maximo y minimo\n\n");
  printf("Ingrese un numero: ");
  scanf("%d", &num);
  while (num != 0) {
    if (!init) {
      max = min = num;
      init = 1;
    }
    max = (num > max) ? num : max;
    min = (num < min) ? num : min;
    printf("Ingrese un numero: ");
    scanf("%d", &num);
  }
  if (init) {
    printf("Numero maximo: %d\n", max);
    printf("Numero minimo: %d\n", min);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<operator>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <lex_error@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int operator(void) {
  lex_error();
  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
<operator>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <lex_error@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int operator(void) {
  lex_error();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<operator>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <lex_error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int operator(void) {
  lex_error();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<operator>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <lex_error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int operator(void) {
  lex_error();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<c_data_abort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <uart_strln@plt>
  mov    $0x0,%eax
  call   <hang@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
datatype: string

tool: D8: Exception: data abort

assistant: ```c
void c_data_abort() {
  uart_strln("Exception: data abort");
  hang();
}
```
user: Translate the below assembly code into C programming language:

```asm
<c_data_abort>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <uart_strln@plt>
  mov    $0x0,%eax
  call   <hang@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D7: Exception: data abort

assistant: ```c
void c_data_abort() {
  uart_strln("Exception: data abort");
  hang();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<c_data_abort>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <uart_strln@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <hang@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
type: string

tool: D5: Exception: data abort

assistant: ```c
void c_data_abort() {
  uart_strln("Exception: data abort");
  hang();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<c_data_abort>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <uart_strln@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <hang@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
type: string

tool: D2: Exception: data abort

assistant: ```c
void c_data_abort() {
  uart_strln("Exception: data abort");
  hang();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<check_path>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fnmatch@plt>
  test   %rax,%rax
  jne    L3
  mov    $0x1,%eax
  jmp    L2
L3:
  mov    $0x0,%eax
L2:
  leave
  ret
```

assistant: ```c
int check_path(const char *file_name, const char *const *parms, int parm_pos) {
  if (fnmatch(parms[parm_pos + 1], file_name, FNM_NOESCAPE) == 0) {
    return 1;
  } else {
    return 0;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<check_path>:
  endbr64
  sub    $0x8,%rsp
  movslq %edx,%rdx
  mov    0x8(%rsi,%rdx,8),%rax
  mov    D4(%rip),%rdx
  mov    (%rdx),%edx
  mov    %rdi,%rsi
  mov    %rax,%rdi
  call   <fnmatch@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int check_path(const char *file_name, const char *const *parms, int parm_pos) {
  if (fnmatch(parms[parm_pos + 1], file_name, FNM_NOESCAPE) == 0) {
    return 1;
  } else {
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<check_path>:
  endbr64
  sub    $0x8,%rsp
  mov    D0(%rip),%rax
  movslq %edx,%rdx
  mov    %rdi,%r8
  mov    0x8(%rsi,%rdx,8),%rdi
  mov    %r8,%rsi
  mov    (%rax),%edx
  call   <fnmatch@plt>
  test   %rax,%rax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int check_path(const char *file_name, const char *const *parms, int parm_pos) {
  if (fnmatch(parms[parm_pos + 1], file_name, FNM_NOESCAPE) == 0) {
    return 1;
  } else {
    return 0;
  }
}
```