ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<RSHash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%edi
  mov    $0x0,%edx
  mov    $0x0,%eax
  mov    $0xf8c9,%ecx
  jmp    L18
L15:
  imul   %ecx,%eax
  movsbl (%rbx,%rdx,1),%esi
  add    %esi,%eax
  imul   $0x5c6b7,%ecx,%ecx
  add    $0x1,%rdx
L18:
  cmp    %edx,%edi
  jg     L15
  pop    %rbx
  ret
```

assistant: ```c
int32_t RSHash(const char *str) {
  int32_t b = 378551;
  int32_t a = 63689;
  int32_t hash = 0;
  for (int32_t i = 0; i < (int)strlen(str); i++) {
    hash = hash * a + *(str + i);
    a = a * b;
  }
  return hash;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<RSHash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  xor    %edx,%edx
  mov    $0xf8c9,%ecx
  mov    %eax,%r8d
  xor    %eax,%eax
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L11:
  imul   %ecx,%eax
  movsbl (%rbx,%rdx,1),%esi
  add    $0x1,%rdx
  imul   $0x5c6b7,%ecx,%ecx
  add    %esi,%eax
L2:
  cmp    %edx,%r8d
  jg     L11
  pop    %rbx
  ret
```

assistant: ```c
int32_t RSHash(const char *str) {
  int32_t b = 378551;
  int32_t a = 63689;
  int32_t hash = 0;
  for (int32_t i = 0; i < (int)strlen(str); i++) {
    hash = hash * a + *(str + i);
    a = a * b;
  }
  return hash;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<RSHash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L6
  sub    $0x1,%eax
  mov    %rbx,%rdi
  mov    $0xf8c9,%edx
  lea    0x1(%rbx,%rax,1),%rsi
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L12:
  imul   %edx,%eax
  movsbl (%rdi),%ecx
  add    $0x1,%rdi
  imul   $0x5c6b7,%edx,%edx
  add    %ecx,%eax
  cmp    %rsi,%rdi
  jne    L12
  pop    %rbx
  ret
  nopl   0x0(%rax)
L6:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int32_t RSHash(const char *str) {
  int32_t b = 378551;
  int32_t a = 63689;
  int32_t hash = 0;
  for (int32_t i = 0; i < (int)strlen(str); i++) {
    hash = hash * a + *(str + i);
    a = a * b;
  }
  return hash;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sDumpLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  nop
L3:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,%ebx
  cmp    $0xa,%ebx
  jne    L3
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int sDumpLine(FILE *fp) {
  register int c;
  while ((c = fgetc(fp)) != '\n')
    ;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<sDumpLine>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
L0:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%eax
  jne    L0
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int sDumpLine(FILE *fp) {
  register int c;
  while ((c = fgetc(fp)) != '\n')
    ;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<sDumpLine>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%eax
  jne    L1
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int sDumpLine(FILE *fp) {
  register int c;
  while ((c = fgetc(fp)) != '\n')
    ;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sDumpLine>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %rbx,%rdi
  call   <fgetc@plt>
  cmp    $0xa,%eax
  jne    L5
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int sDumpLine(FILE *fp) {
  register int c;
  while ((c = fgetc(fp)) != '\n')
    ;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<doubles_mat44_inv_transform_xyz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %rdi,-0x98(%rbp)
  mov    %rsi,-0xa0(%rbp)
  mov    %rdx,-0xa8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x90(%rbp),%rdx
  mov    -0x98(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <doubles_mat44_inv@plt>
  mov    -0xa8(%rbp),%rdx
  mov    -0xa0(%rbp),%rcx
  lea    -0x90(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <doubles_mat44_transform_xyz@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void doubles_mat44_inv_transform_xyz(const double M[16], const double in[3], double out[3]) {
  double T[16];
  doubles_mat44_inv(M, T);
  doubles_mat44_transform_xyz(T, in, out);
}
```
user: Translate the below assembly code into C programming language:

```asm
<doubles_mat44_inv_transform_xyz>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x90,%rsp
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rsi
  call   <doubles_mat44_inv@plt>
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <doubles_mat44_transform_xyz@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x90,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void doubles_mat44_inv_transform_xyz(const double M[16], const double in[3], double out[3]) {
  double T[16];
  doubles_mat44_inv(M, T);
  doubles_mat44_transform_xyz(T, in, out);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<doubles_mat44_inv_transform_xyz>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  mov    %r13,%rsi
  call   <doubles_mat44_inv@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <doubles_mat44_transform_xyz@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x90,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void doubles_mat44_inv_transform_xyz(const double M[16], const double in[3], double out[3]) {
  double T[16];
  doubles_mat44_inv(M, T);
  doubles_mat44_transform_xyz(T, in, out);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<doubles_mat44_inv_transform_xyz>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  mov    %r13,%rsi
  call   <doubles_mat44_inv@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <doubles_mat44_transform_xyz@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x90,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void doubles_mat44_inv_transform_xyz(const double M[16], const double in[3], double out[3]) {
  double T[16];
  doubles_mat44_inv(M, T);
  doubles_mat44_transform_xyz(T, in, out);
}
```
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    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <is_gtin@plt>
  test   %rax,%rax
  je     L5
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L16
L5:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L16:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
type_guessed: string
name: retrieve_data
arguments: data_tag: D3
type_guessed: string

tool: D5: Yes, '%s' is valid


tool: D3: No, '%s' is not valid


assistant: ```c
int main(int argc, char **argv) {
  if (is_gtin(argv[1])) {
    printf("Yes, '%s' is valid\n", argv[1]);
  } else {
    printf("No, '%s' is not valid\n", argv[1]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    0x8(%rsi),%rdi
  call   <is_gtin@plt>
  test   %rax,%rax
  je     L1
  mov    0x8(%rbx),%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
L13:
  mov    $0x0,%eax
  pop    %rbx
  ret
L1:
  mov    0x8(%rbx),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L13
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
data_type: string
name: extract_symbolic_data
arguments: data_tag: D1
data_type: string

tool: D6: Yes, '%s' is valid


tool: D1: No, '%s' is not valid


assistant: ```c
int main(int argc, char **argv) {
  if (is_gtin(argv[1])) {
    printf("Yes, '%s' is valid\n", argv[1]);
  } else {
    printf("No, '%s' is not valid\n", argv[1]);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbx
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  call   <is_gtin@plt>
  mov    0x8(%rbx),%rsi
  test   %rax,%rax
  je     L10
  lea    D19(%rip),%rdi
  call   <printf@plt>
L11:
  xor    %eax,%eax
  pop    %rbx
  ret
L10:
  lea    D18(%rip),%rdi
  call   <printf@plt>
  jmp    L11
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D19
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D18
type_guessed: string

tool: D19: Yes, '%s' is valid


tool: D18: No, '%s' is not valid


assistant: ```c
int main(int argc, char **argv) {
  if (is_gtin(argv[1])) {
    printf("Yes, '%s' is valid\n", argv[1]);
  } else {
    printf("No, '%s' is not valid\n", argv[1]);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    0x8(%rsi),%rdi
  mov    %rsi,%rbx
  call   <is_gtin@plt>
  mov    0x8(%rbx),%rsi
  test   %rax,%rax
  je     L1
  lea    D11(%rip),%rdi
  call   <printf@plt>
L13:
  xor    %eax,%eax
  pop    %rbx
  ret
L1:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L13
```

tool_calls: name: parse_data
arguments: data_tag: D11
type: string
name: parse_data
arguments: data_tag: D4
type: string

tool: D11: Yes, '%s' is valid


tool: D4: No, '%s' is not valid


assistant: ```c
int main(int argc, char **argv) {
  if (is_gtin(argv[1])) {
    printf("Yes, '%s' is valid\n", argv[1]);
  } else {
    printf("No, '%s' is not valid\n", argv[1]);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strtok>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L109
  mov    -0x18(%rbp),%rax
  mov    %rax,D59(%rip)
L109:
  mov    D59(%rip),%rax
  test   %rax,%rax
  jne    L21
  mov    $0x0,%eax
  jmp    L54
L77:
  mov    D59(%rip),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L87
  mov    D59(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D103(%rip)
L21:
  mov    D103(%rip),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L77
  jmp    L75
L87:
  nop
L75:
  mov    D103(%rip),%rbx
  jmp    L117
L59:
  mov    D103(%rip),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L85
  mov    D103(%rip),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,D103(%rip)
  movb   $0x0,(%rax)
  mov    %rbx,%rax
  jmp    L54
L85:
  mov    D103(%rip),%rax
  add    $0x1,%rax
  mov    %rax,D103(%rip)
L117:
  mov    D103(%rip),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L59
  movq   $0x0,D103(%rip)
  movzbl (%rbx),%eax
  test   %al,%al
  je     L114
  mov    %rbx,%rax
  jmp    L54
L114:
  mov    $0x0,%eax
L54:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *strtok(char *str, const char *control) {
  static char *s;
  register char *s1;
  if (str)
    s = str;
  if (!s)
    return (void *)0;
  while (*s) {
    if (strchr(control, *s))
      s++;
    else
      break;
  }
  s1 = s;
  while (*s) {
    if (strchr(control, *s)) {
      *s++ = '\0';
      return s1;
    }
    s++;
  }
  s = (void *)0;
  if (*s1)
    return s1;
  else
    return (void *)0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strtok>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r14
  test   %rdi,%rdi
  je     L81
  mov    %rdi,%rbx
  mov    %rdi,D60(%rip)
L99:
  movzbl (%rbx),%r13d
  test   %r13b,%r13b
  je     L9
L33:
  movsbl %r13b,%esi
  mov    %r14,%rdi
  call   <strchr@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L58
  add    $0x1,%rbx
  mov    %rbx,D33(%rip)
  movzbl (%rbx),%r13d
  test   %r13b,%r13b
  jne    L33
L9:
  movq   $0x0,D33(%rip)
  mov    $0x0,%ebx
  jmp    L40
L81:
  mov    D33(%rip),%rbx
  test   %rbx,%rbx
  jne    L99
  jmp    L40
L58:
  mov    D33(%rip),%rbp
  movzbl 0x0(%rbp),%esi
  test   %sil,%sil
  je     L41
L78:
  movsbl %sil,%esi
  mov    %r14,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L66
  add    $0x1,%rbp
  mov    %rbp,D33(%rip)
  movzbl 0x0(%rbp),%esi
  test   %sil,%sil
  jne    L78
L41:
  movq   $0x0,D33(%rip)
  test   %r13b,%r13b
  cmove  %r12,%rbx
  jmp    L40
L66:
  lea    0x1(%rbp),%rax
  mov    %rax,D33(%rip)
  movb   $0x0,0x0(%rbp)
L40:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *strtok(char *str, const char *control) {
  static char *s;
  register char *s1;
  if (str)
    s = str;
  if (!s)
    return (void *)0;
  while (*s) {
    if (strchr(control, *s))
      s++;
    else
      break;
  }
  s1 = s;
  while (*s) {
    if (strchr(control, *s)) {
      *s++ = '\0';
      return s1;
    }
    s++;
  }
  s = (void *)0;
  if (*s1)
    return s1;
  else
    return (void *)0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strtok>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  test   %rdi,%rdi
  je     L29
  mov    %rdi,D5(%rip)
  mov    %rdi,%r12
L50:
  movsbl (%r12),%esi
  test   %sil,%sil
  jne    L25
  jmp    L23
  nopl   (%rax)
L72:
  add    $0x1,%r12
  mov    %r12,D66(%rip)
  movsbl (%r12),%esi
  test   %sil,%sil
  je     L23
L25:
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L72
  mov    D66(%rip),%rbx
  nopl   (%rax)
L18:
  movsbl (%rbx),%esi
  test   %sil,%sil
  je     L62
  mov    %rbp,%rdi
  call   <strchr@plt>
  mov    %rbx,%rdx
  add    $0x1,%rbx
  mov    %rbx,D66(%rip)
  test   %rax,%rax
  je     L18
  movb   $0x0,(%rdx)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L62:
  movq   $0x0,D66(%rip)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  mov    D66(%rip),%r12
  test   %r12,%r12
  jne    L50
  xor    %r12d,%r12d
L79:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  movq   $0x0,D66(%rip)
  xor    %r12d,%r12d
  jmp    L79
```

assistant: ```c
char *strtok(char *str, const char *control) {
  static char *s;
  register char *s1;
  if (str)
    s = str;
  if (!s)
    return (void *)0;
  while (*s) {
    if (strchr(control, *s))
      s++;
    else
      break;
  }
  s1 = s;
  while (*s) {
    if (strchr(control, *s)) {
      *s++ = '\0';
      return s1;
    }
    s++;
  }
  s = (void *)0;
  if (*s1)
    return s1;
  else
    return (void *)0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strtok>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  test   %rdi,%rdi
  je     L36
  mov    %rdi,D36(%rip)
  mov    %rdi,%r12
L88:
  movsbl (%r12),%esi
  test   %sil,%sil
  jne    L73
  jmp    L12
  nopl   (%rax)
L62:
  add    $0x1,%r12
  mov    %r12,D3(%rip)
  movsbl (%r12),%esi
  test   %sil,%sil
  je     L12
L73:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L62
  movsbl (%r12),%esi
  test   %sil,%sil
  je     L33
  mov    %r12,%rbp
  jmp    L1
  nopl   (%rax)
L42:
  movsbl 0x0(%rbp),%esi
  test   %sil,%sil
  je     L33
L1:
  mov    %rbx,%rdi
  call   <strchr@plt>
  mov    %rbp,%rdx
  add    $0x1,%rbp
  mov    %rbp,D3(%rip)
  test   %rax,%rax
  je     L42
  movb   $0x0,(%rdx)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  movq   $0x0,D3(%rip)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L36:
  mov    D3(%rip),%r12
  test   %r12,%r12
  jne    L88
  xor    %r12d,%r12d
L18:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  movq   $0x0,D3(%rip)
  xor    %r12d,%r12d
  jmp    L18
```

assistant: ```c
char *strtok(char *str, const char *control) {
  static char *s;
  register char *s1;
  if (str)
    s = str;
  if (!s)
    return (void *)0;
  while (*s) {
    if (strchr(control, *s))
      s++;
    else
      break;
  }
  s1 = s;
  while (*s) {
    if (strchr(control, *s)) {
      *s++ = '\0';
      return s1;
    }
    s++;
  }
  s = (void *)0;
  if (*s1)
    return s1;
  else
    return (void *)0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<find_two_chars>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  jmp    L60
L53:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L54
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x20(%rbp)
  jne    L54
  mov    $0x0,%eax
  jmp    L6
L54:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x20(%rbp)
  jne    L60
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L60
  mov    $0x1,%eax
  jmp    L6
L60:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L53
  mov    $0x2,%eax
L6:
  pop    %rbp
  ret
```

assistant: ```c
int find_two_chars(char *str, char a, char b) {
  while (*str != '\0') {
    char *cur = str++;
    if (*cur == a && *str == b)
      return 0;
    if (*cur == b && *str == a)
      return 1;
  }
  return 2;
}
```
user: What is the c source code of the assembly code below?

```asm
<find_two_chars>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L11
  mov    $0x2,%eax
  ret
L4:
  cmp    %dl,(%rdi)
  jne    L37
  mov    $0x0,%eax
  ret
L30:
  movzbl (%rdi),%eax
  test   %al,%al
  je     L54
L11:
  add    $0x1,%rdi
  cmp    %al,%sil
  je     L4
L37:
  cmp    %al,%dl
  jne    L30
  cmp    %sil,(%rdi)
  jne    L30
  mov    $0x1,%eax
  ret
L54:
  mov    $0x2,%eax
  ret
```

assistant: ```c
int find_two_chars(char *str, char a, char b) {
  while (*str != '\0') {
    char *cur = str++;
    if (*cur == a && *str == b)
      return 0;
    if (*cur == b && *str == a)
      return 1;
  }
  return 2;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<find_two_chars>:
  endbr64
  movzbl (%rdi),%eax
  nopw   0x0(%rax,%rax,1)
L45:
  test   %al,%al
  je     L49
  add    $0x1,%rdi
  mov    %eax,%ecx
  movzbl (%rdi),%eax
  cmp    %cl,%sil
  je     L2
L0:
  cmp    %cl,%dl
  jne    L45
  cmp    %al,%sil
  jne    L45
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L2:
  cmp    %al,%dl
  jne    L0
  xor    %eax,%eax
  ret
  nop
L49:
  mov    $0x2,%eax
  ret
```

assistant: ```c
int find_two_chars(char *str, char a, char b) {
  while (*str != '\0') {
    char *cur = str++;
    if (*cur == a && *str == b)
      return 0;
    if (*cur == b && *str == a)
      return 1;
  }
  return 2;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<find_two_chars>:
  endbr64
  movzbl (%rdi),%eax
  nopw   0x0(%rax,%rax,1)
L4:
  test   %al,%al
  je     L36
  add    $0x1,%rdi
  mov    %eax,%ecx
  movzbl (%rdi),%eax
  cmp    %cl,%sil
  je     L43
L34:
  cmp    %cl,%dl
  jne    L4
  cmp    %al,%sil
  jne    L4
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L43:
  cmp    %al,%dl
  jne    L34
  xor    %eax,%eax
  ret
  nop
L36:
  mov    $0x2,%eax
  ret
```

assistant: ```c
int find_two_chars(char *str, char a, char b) {
  while (*str != '\0') {
    char *cur = str++;
    if (*cur == a && *str == b)
      return 0;
    if (*cur == b && *str == a)
      return 1;
  }
  return 2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<binary_search>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L5
L28:
  mov    -0x8(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jne    L21
  mov    -0x4(%rbp),%eax
  jmp    L17
L21:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jge    L56
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L5
L56:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L5:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L28
  mov    $0xffffffff,%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
int binary_search(const int array[], const int size, const int value) {
  int left = 0;
  int right = size - 1;
  while (left <= right) {
    int mid = left + (right - left) / 2;
    if (value == array[mid]) {
      return mid;
    } else if (value < array[mid]) {
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<binary_search>:
  endbr64
  sub    $0x1,%esi
  js     L14
  mov    $0x0,%ecx
  jmp    L38
L50:
  lea    0x1(%rax),%ecx
L20:
  cmp    %esi,%ecx
  jg     L7
L38:
  mov    %esi,%r8d
  sub    %ecx,%r8d
  mov    %r8d,%eax
  shr    $0x1f,%eax
  add    %r8d,%eax
  sar    %eax
  add    %ecx,%eax
  movslq %eax,%r8
  mov    (%rdi,%r8,4),%r8d
  cmp    %edx,%r8d
  je     L10
  jle    L50
  lea    -0x1(%rax),%esi
  jmp    L20
L7:
  mov    $0xffffffff,%eax
  ret
L14:
  mov    $0xffffffff,%eax
L10:
  ret
```

assistant: ```c
int binary_search(const int array[], const int size, const int value) {
  int left = 0;
  int right = size - 1;
  while (left <= right) {
    int mid = left + (right - left) / 2;
    if (value == array[mid]) {
      return mid;
    } else if (value < array[mid]) {
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<binary_search>:
  endbr64
  sub    $0x1,%esi
  js     L15
  xor    %ecx,%ecx
  jmp    L31
  nopl   (%rax)
L8:
  lea    -0x1(%rax),%esi
  cmp    %ecx,%esi
  jl     L15
L31:
  mov    %esi,%eax
  sub    %ecx,%eax
  sar    %eax
  add    %ecx,%eax
  movslq %eax,%r8
  cmp    %edx,(%rdi,%r8,4)
  je     L33
  jg     L8
  lea    0x1(%rax),%ecx
  cmp    %ecx,%esi
  jge    L31
L15:
  mov    $0xffffffff,%eax
L33:
  ret
```

assistant: ```c
int binary_search(const int array[], const int size, const int value) {
  int left = 0;
  int right = size - 1;
  while (left <= right) {
    int mid = left + (right - left) / 2;
    if (value == array[mid]) {
      return mid;
    } else if (value < array[mid]) {
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }
  return -1;
}
```