ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<get_type>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%ecx
  lea    D21(%rip),%rdx
  lea    D3(%rip),%rsi
  call   <get_token@plt>
  mov    $0x2,%ecx
  lea    D38(%rip),%rdx
  lea    D2(%rip),%rsi
  mov    %rax,%rdi
  call   <get_token@plt>
  movsbl (%rax),%eax
  add    $0x8,%rsp
  sub    $0x30,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: label: D21
type_guessed: string
name: retrieve_data
arguments: label: D3
type_guessed: string
name: retrieve_data
arguments: label: D38
type_guessed: string
name: retrieve_data
arguments: label: D2
type_guessed: string

tool: D21: ;


tool: D3: ;

tool: 

tool: D2: :

assistant: ```c
int get_type(const char *line, int index) {
  const char *head;
  const char *type;
  head = get_token(line, ";", ";\n", index);
  type = get_token(head, ":", "", 2);
  return type[0] - '0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<skip_ws>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  jmp    L8
L30:
  addq   $0x1,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L30
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L30
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L30
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char const *skip_ws(char const *c) {
  while (*c == ' ' || *c == '\t' || *c == '\n')
    ++c;
  return c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<skip_ws>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%edx
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  jbe    L7
  cmp    $0x20,%dl
  jne    L8
L7:
  add    $0x1,%rax
  movzbl (%rax),%edx
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  jbe    L7
  cmp    $0x20,%dl
  je     L7
L8:
  ret
```

assistant: ```c
char const *skip_ws(char const *c) {
  while (*c == ' ' || *c == '\t' || *c == '\n')
    ++c;
  return c;
}
```
user: Translate the below assembly code into C programming language:

```asm
<skip_ws>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  ja     L21
  nopw   0x0(%rax,%rax,1)
L15:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  jbe    L15
L21:
  cmp    $0x20,%dl
  je     L15
  ret
```

assistant: ```c
char const *skip_ws(char const *c) {
  while (*c == ' ' || *c == '\t' || *c == '\n')
    ++c;
  return c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<skip_ws>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  ja     L7
  nopw   0x0(%rax,%rax,1)
L5:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  jbe    L5
L7:
  cmp    $0x20,%dl
  je     L5
  ret
```

assistant: ```c
char const *skip_ws(char const *c) {
  while (*c == ' ' || *c == '\t' || *c == '\n')
    ++c;
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x4(%rbp)
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L38
L27:
  cmpb   $0x6e,-0x5(%rbp)
  jne    L34
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L23
L34:
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L23
L47:
  nop
L23:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L47
L38:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x5(%rbp)
  cmpb   $0x79,-0x5(%rbp)
  jne    L27
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D58
type: string
name: retrieve_data
arguments: label: D52
type: string
name: retrieve_data
arguments: label: D36
type: string
name: retrieve_data
arguments: label: D57
type: string
name: retrieve_data
arguments: label: D20
type: string
name: retrieve_data
arguments: label: D41
type: string
name: retrieve_data
arguments: label: D1
type: string

tool: D58: Pick an integer from 1 to 100. I will try to guess  

tool: D52: it.
 Respond with a y if my guess is right and with

tool: D36: 
an n if it is wrong.

tool: D57: Uh...is your number %d?


tool: D20: Well, then, is it %d?


tool: D41: Sorry, I understand only y or n.

tool: D1: I knew I could do it!

assistant: ```c
int main(int argc, const char *argv[]) {
  int guess = 1;
  char response;
  printf("Pick an integer from 1 to 100. I will try to guess  ");
  printf("it.\n Respond with a y if my guess is right and with");
  printf("\nan n if it is wrong.\n");
  printf("Uh...is your number %d?\n", guess);
  while ((response = getchar()) != 'y') {
    if (response == 'n') {
      printf("Well, then, is it %d?\n", ++guess);
    } else {
      printf("Sorry, I understand only y or n.\n");
    }
    while (getchar() != '\n') {
      continue;
    }
  }
  printf("I knew I could do it!\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D60(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D67(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D35(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%esi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ebx
  lea    D16(%rip),%rbp
  lea    D49(%rip),%r12
  jmp    L5
L15:
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L19:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L19
L5:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0x79,%al
  je     L21
  cmp    $0x6e,%al
  je     L15
  mov    %rbp,%rdi
  call   <puts@plt>
  jmp    L19
L21:
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: label: D60
type: string
name: retrieve_data
arguments: label: D67
type: string
name: retrieve_data
arguments: label: D35
type: string
name: retrieve_data
arguments: label: D20
type: string
name: retrieve_data
arguments: label: D16
type: string
name: retrieve_data
arguments: label: D49
type: string
name: retrieve_data
arguments: label: D34
type: string

tool: D60: Pick an integer from 1 to 100. I will try to guess  

tool: D67: it.
 Respond with a y if my guess is right and with

tool: D35: 
an n if it is wrong.

tool: D20: Uh...is your number %d?


tool: D16: Sorry, I understand only y or n.

tool: D49: Well, then, is it %d?


tool: D34: I knew I could do it!

assistant: ```c
int main(int argc, const char *argv[]) {
  int guess = 1;
  char response;
  printf("Pick an integer from 1 to 100. I will try to guess  ");
  printf("it.\n Respond with a y if my guess is right and with");
  printf("\nan n if it is wrong.\n");
  printf("Uh...is your number %d?\n", guess);
  while ((response = getchar()) != 'y') {
    if (response == 'n') {
      printf("Well, then, is it %d?\n", ++guess);
    } else {
      printf("Sorry, I understand only y or n.\n");
    }
    while (getchar() != '\n') {
      continue;
    }
  }
  printf("I knew I could do it!\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  lea    D42(%rip),%r12
  push   %rbp
  lea    D33(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <printf@plt>
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%esi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L21:
  mov    %rbp,%rdi
  call   <puts@plt>
L40:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L40
L9:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x79,%al
  je     L6
  cmp    $0x6e,%al
  jne    L21
  add    $0x1,%ebx
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  jmp    L40
  nopl   0x0(%rax,%rax,1)
L6:
  lea    D5(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_label: D16
type_infered: string
name: parse_data
arguments: data_label: D42
type_infered: string
name: parse_data
arguments: data_label: D33
type_infered: string
name: parse_data
arguments: data_label: D15
type_infered: string
name: parse_data
arguments: data_label: D13
type_infered: string
name: parse_data
arguments: data_label: D14
type_infered: string
name: parse_data
arguments: data_label: D5
type_infered: string

tool: D16: Pick an integer from 1 to 100. I will try to guess  

tool: D42: Well, then, is it %d?


tool: D33: Sorry, I understand only y or n.

tool: D15: it.
 Respond with a y if my guess is right and with

tool: D13: 
an n if it is wrong.

tool: D14: Uh...is your number %d?


tool: D5: I knew I could do it!

assistant: ```c
int main(int argc, const char *argv[]) {
  int guess = 1;
  char response;
  printf("Pick an integer from 1 to 100. I will try to guess  ");
  printf("it.\n Respond with a y if my guess is right and with");
  printf("\nan n if it is wrong.\n");
  printf("Uh...is your number %d?\n", guess);
  while ((response = getchar()) != 'y') {
    if (response == 'n') {
      printf("Well, then, is it %d?\n", ++guess);
    } else {
      printf("Sorry, I understand only y or n.\n");
    }
    while (getchar() != '\n') {
      continue;
    }
  }
  printf("I knew I could do it!\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  lea    D41(%rip),%r12
  push   %rbp
  lea    D18(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <printf@plt>
  lea    D55(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D45(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%esi
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L47
  nopl   0x0(%rax,%rax,1)
L40:
  mov    %rbp,%rdi
  call   <puts@plt>
L17:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L17
L47:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x79,%al
  je     L25
  cmp    $0x6e,%al
  jne    L40
  add    $0x1,%ebx
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L25:
  lea    D49(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
type_infered: string
name: retrieve_data
arguments: tag: D41
type_infered: string
name: retrieve_data
arguments: tag: D18
type_infered: string
name: retrieve_data
arguments: tag: D55
type_infered: string
name: retrieve_data
arguments: tag: D45
type_infered: string
name: retrieve_data
arguments: tag: D28
type_infered: string
name: retrieve_data
arguments: tag: D49
type_infered: string

tool: D9: Pick an integer from 1 to 100. I will try to guess  

tool: D41: Well, then, is it %d?


tool: D18: Sorry, I understand only y or n.

tool: D55: it.
 Respond with a y if my guess is right and with

tool: D45: 
an n if it is wrong.

tool: D28: Uh...is your number %d?


tool: D49: I knew I could do it!

assistant: ```c
int main(int argc, const char *argv[]) {
  int guess = 1;
  char response;
  printf("Pick an integer from 1 to 100. I will try to guess  ");
  printf("it.\n Respond with a y if my guess is right and with");
  printf("\nan n if it is wrong.\n");
  printf("Uh...is your number %d?\n", guess);
  while ((response = getchar()) != 'y') {
    if (response == 'n') {
      printf("Well, then, is it %d?\n", ++guess);
    } else {
      printf("Sorry, I understand only y or n.\n");
    }
    while (getchar() != '\n') {
      continue;
    }
  }
  printf("I knew I could do it!\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ov_imul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L23
  mov    D73(%rip),%rax
  mov    (%rax),%eax
  jmp    L29
L23:
  cmpl   $0x0,-0x4(%rbp)
  jle    L22
  cmpl   $0x0,-0x8(%rbp)
  jle    L46
  mov    D76(%rip),%rax
  mov    (%rax),%eax
  cltd
  idivl  -0x8(%rbp)
  cmp    %eax,-0x4(%rbp)
  jle    L45
  mov    -0x10(%rbp),%rax
  movl   $0x1,(%rax)
  mov    D73(%rip),%rax
  mov    (%rax),%eax
  jmp    L29
L46:
  mov    D73(%rip),%rax
  mov    (%rax),%eax
  cltd
  idivl  -0x4(%rbp)
  cmp    %eax,-0x8(%rbp)
  jge    L45
  mov    -0x10(%rbp),%rax
  movl   $0x1,(%rax)
  mov    D73(%rip),%rax
  mov    (%rax),%eax
  jmp    L29
L22:
  cmpl   $0x0,-0x8(%rbp)
  jle    L131
  mov    D73(%rip),%rax
  mov    (%rax),%eax
  cltd
  idivl  -0x8(%rbp)
  cmp    %eax,-0x4(%rbp)
  jge    L45
  mov    -0x10(%rbp),%rax
  movl   $0x1,(%rax)
  mov    D73(%rip),%rax
  mov    (%rax),%eax
  jmp    L29
L131:
  cmpl   $0x0,-0x4(%rbp)
  je     L45
  mov    D76(%rip),%rax
  mov    (%rax),%eax
  cltd
  idivl  -0x4(%rbp)
  cmp    %eax,-0x8(%rbp)
  jge    L45
  mov    -0x10(%rbp),%rax
  movl   $0x1,(%rax)
  mov    D73(%rip),%rax
  mov    (%rax),%eax
  jmp    L29
L45:
  mov    -0x4(%rbp),%eax
  imul   -0x8(%rbp),%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
signed int ov_imul(signed int a, signed int b, int *ov) {
  if (*ov)
    return INT_MIN;
  if (a > 0) {
    if (b > 0) {
      if (a > (INT_MAX / b)) {
        *ov = 1;
        return INT_MIN;
      }
    } else {
      if (b < (INT_MIN / a)) {
        *ov = 1;
        return INT_MIN;
      }
    }
  } else {
    if (b > 0) {
      if (a < (INT_MIN / b)) {
        *ov = 1;
        return INT_MIN;
      }
    } else {
      if ((a != 0) && (b < (INT_MAX / a))) {
        *ov = 1;
        return INT_MIN;
      }
    }
  }
  return a * b;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ov_imul>:
  endbr64
  mov    %edi,%ecx
  mov    %rdx,%rdi
  cmpl   $0x0,(%rdx)
  jne    L41
  test   %ecx,%ecx
  jle    L68
  test   %esi,%esi
  jle    L71
  mov    D37(%rip),%rax
  mov    (%rax),%eax
  cltd
  idiv   %esi
  cmp    %ecx,%eax
  jl     L22
L7:
  mov    %ecx,%eax
  imul   %esi,%eax
  ret
L41:
  mov    D61(%rip),%rax
  mov    (%rax),%eax
  ret
L22:
  movl   $0x1,(%rdi)
  mov    D61(%rip),%rax
  mov    (%rax),%eax
  ret
L71:
  mov    D61(%rip),%rax
  mov    (%rax),%eax
  cltd
  idiv   %ecx
  cmp    %esi,%eax
  jle    L7
  movl   $0x1,(%rdi)
  mov    D61(%rip),%rax
  mov    (%rax),%eax
  ret
L68:
  test   %esi,%esi
  jle    L47
  mov    D61(%rip),%rax
  mov    (%rax),%eax
  cltd
  idiv   %esi
  cmp    %ecx,%eax
  jle    L7
  movl   $0x1,(%rdi)
  mov    D61(%rip),%rax
  mov    (%rax),%eax
  ret
L47:
  test   %ecx,%ecx
  je     L7
  mov    D37(%rip),%rax
  mov    (%rax),%eax
  cltd
  idiv   %ecx
  cmp    %esi,%eax
  jle    L7
  movl   $0x1,(%rdi)
  mov    D61(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
signed int ov_imul(signed int a, signed int b, int *ov) {
  if (*ov)
    return INT_MIN;
  if (a > 0) {
    if (b > 0) {
      if (a > (INT_MAX / b)) {
        *ov = 1;
        return INT_MIN;
      }
    } else {
      if (b < (INT_MIN / a)) {
        *ov = 1;
        return INT_MIN;
      }
    }
  } else {
    if (b > 0) {
      if (a < (INT_MIN / b)) {
        *ov = 1;
        return INT_MIN;
      }
    } else {
      if ((a != 0) && (b < (INT_MAX / a))) {
        *ov = 1;
        return INT_MIN;
      }
    }
  }
  return a * b;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ov_imul>:
  endbr64
  mov    (%rdx),%eax
  mov    %edi,%ecx
  mov    %rdx,%rdi
  test   %eax,%eax
  jne    L58
  test   %ecx,%ecx
  jle    L17
  test   %esi,%esi
  jle    L81
  mov    D41(%rip),%rax
  mov    (%rax),%eax
  cltd
  idiv   %esi
  cmp    %ecx,%eax
  jl     L50
L12:
  mov    %ecx,%eax
  imul   %esi,%eax
  ret
  nopl   (%rax)
L17:
  test   %esi,%esi
  jle    L86
  mov    D37(%rip),%r8
  mov    (%r8),%eax
  cltd
  idiv   %esi
  cmp    %ecx,%eax
  jle    L12
L32:
  movl   $0x1,(%rdi)
  mov    (%r8),%eax
  ret
  nop
L86:
  test   %ecx,%ecx
  je     L12
  mov    D41(%rip),%rax
  mov    (%rax),%eax
  cltd
  idiv   %ecx
  cmp    %esi,%eax
  jle    L12
  nopl   0x0(%rax)
L50:
  movl   $0x1,(%rdi)
L58:
  mov    D37(%rip),%rax
  mov    (%rax),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L81:
  mov    D37(%rip),%r8
  mov    (%r8),%eax
  cltd
  idiv   %ecx
  cmp    %esi,%eax
  jg     L32
  mov    %ecx,%eax
  imul   %esi,%eax
  ret
```

assistant: ```c
signed int ov_imul(signed int a, signed int b, int *ov) {
  if (*ov)
    return INT_MIN;
  if (a > 0) {
    if (b > 0) {
      if (a > (INT_MAX / b)) {
        *ov = 1;
        return INT_MIN;
      }
    } else {
      if (b < (INT_MIN / a)) {
        *ov = 1;
        return INT_MIN;
      }
    }
  } else {
    if (b > 0) {
      if (a < (INT_MIN / b)) {
        *ov = 1;
        return INT_MIN;
      }
    } else {
      if ((a != 0) && (b < (INT_MAX / a))) {
        *ov = 1;
        return INT_MIN;
      }
    }
  }
  return a * b;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ov_imul>:
  endbr64
  mov    (%rdx),%eax
  mov    %edi,%ecx
  mov    %rdx,%rdi
  test   %eax,%eax
  jne    L5
  test   %ecx,%ecx
  jle    L84
  test   %esi,%esi
  jle    L8
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  cltd
  idiv   %esi
  cmp    %ecx,%eax
  jl     L70
L61:
  mov    %ecx,%eax
  imul   %esi,%eax
  ret
  nopl   (%rax)
L84:
  test   %esi,%esi
  jle    L11
  mov    D20(%rip),%r8
  mov    (%r8),%eax
  cltd
  idiv   %esi
  cmp    %ecx,%eax
  jle    L61
L12:
  movl   $0x1,(%rdi)
  mov    (%r8),%eax
  ret
  nop
L11:
  test   %ecx,%ecx
  je     L61
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  cltd
  idiv   %ecx
  cmp    %esi,%eax
  jle    L61
  nopl   0x0(%rax)
L70:
  movl   $0x1,(%rdi)
L5:
  mov    D20(%rip),%rax
  mov    (%rax),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  mov    D20(%rip),%r8
  mov    (%r8),%eax
  cltd
  idiv   %ecx
  cmp    %esi,%eax
  jg     L12
  mov    %ecx,%eax
  imul   %esi,%eax
  ret
```

assistant: ```c
signed int ov_imul(signed int a, signed int b, int *ov) {
  if (*ov)
    return INT_MIN;
  if (a > 0) {
    if (b > 0) {
      if (a > (INT_MAX / b)) {
        *ov = 1;
        return INT_MIN;
      }
    } else {
      if (b < (INT_MIN / a)) {
        *ov = 1;
        return INT_MIN;
      }
    }
  } else {
    if (b > 0) {
      if (a < (INT_MIN / b)) {
        *ov = 1;
        return INT_MIN;
      }
    } else {
      if ((a != 0) && (b < (INT_MAX / a))) {
        *ov = 1;
        return INT_MIN;
      }
    }
  }
  return a * b;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cec_msg_system_audio_mode_request>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  cmpl   $0xffff,-0x10(%rbp)
  jne    L22
  mov    $0x2,%edx
  jmp    L30
L22:
  mov    $0x4,%edx
L30:
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rax
  lea    0x4(%rax),%rdx
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rax
  add    $0x8,%rax
  mov    -0x10(%rbp),%edx
  sar    $0x8,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rax
  lea    0xc(%rax),%rdx
  mov    -0x10(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,(%rdx)
  cmpl   $0x0,-0xc(%rbp)
  je     L27
  mov    D13(%rip),%rax
  mov    (%rax),%eax
  jmp    L20
L27:
  mov    $0x0,%eax
L20:
  mov    -0x8(%rbp),%rdx
  mov    %eax,0x10(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cec_msg_system_audio_mode_request(struct cec_msg *msg, int reply, __u16 phys_addr) {
  msg->len = phys_addr == 0xffff ? 2 : 4;
  msg->msg[1] = CEC_MSG_SYSTEM_AUDIO_MODE_REQUEST;
  msg->msg[2] = phys_addr >> 8;
  msg->msg[3] = phys_addr & 0xff;
  msg->reply = reply ? CEC_MSG_SET_SYSTEM_AUDIO_MODE : 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cec_msg_system_audio_mode_request>:
  endbr64
  mov    %rdi,%rax
  cmp    $0xffff,%edx
  setne  %cl
  movzbl %cl,%ecx
  lea    0x2(%rcx,%rcx,1),%ecx
  mov    %ecx,(%rdi)
  mov    0x8(%rdi),%rcx
  mov    D16(%rip),%rdi
  mov    (%rdi),%edi
  mov    %edi,0x4(%rcx)
  mov    0x8(%rax),%rcx
  mov    %edx,%edi
  sar    $0x8,%edi
  mov    %edi,0x8(%rcx)
  mov    0x8(%rax),%rcx
  and    $0xff,%edx
  mov    %edx,0xc(%rcx)
  test   %esi,%esi
  je     L1
  mov    D11(%rip),%rdx
  mov    (%rdx),%esi
L1:
  mov    %esi,0x10(%rax)
  ret
```

assistant: ```c
void cec_msg_system_audio_mode_request(struct cec_msg *msg, int reply, __u16 phys_addr) {
  msg->len = phys_addr == 0xffff ? 2 : 4;
  msg->msg[1] = CEC_MSG_SYSTEM_AUDIO_MODE_REQUEST;
  msg->msg[2] = phys_addr >> 8;
  msg->msg[3] = phys_addr & 0xff;
  msg->reply = reply ? CEC_MSG_SET_SYSTEM_AUDIO_MODE : 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<cec_msg_system_audio_mode_request>:
  endbr64
  xor    %eax,%eax
  cmp    $0xffff,%edx
  mov    D6(%rip),%rcx
  setne  %al
  lea    0x2(%rax,%rax,1),%eax
  mov    (%rcx),%ecx
  mov    %eax,(%rdi)
  mov    0x8(%rdi),%rax
  mov    %ecx,0x4(%rax)
  mov    %edx,%ecx
  and    $0xff,%edx
  sar    $0x8,%ecx
  mov    %edx,0xc(%rax)
  mov    %ecx,0x8(%rax)
  test   %esi,%esi
  je     L7
  mov    D0(%rip),%rax
  mov    (%rax),%esi
L7:
  mov    %esi,0x10(%rdi)
  ret
```

assistant: ```c
void cec_msg_system_audio_mode_request(struct cec_msg *msg, int reply, __u16 phys_addr) {
  msg->len = phys_addr == 0xffff ? 2 : 4;
  msg->msg[1] = CEC_MSG_SYSTEM_AUDIO_MODE_REQUEST;
  msg->msg[2] = phys_addr >> 8;
  msg->msg[3] = phys_addr & 0xff;
  msg->reply = reply ? CEC_MSG_SET_SYSTEM_AUDIO_MODE : 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<cec_msg_system_audio_mode_request>:
  endbr64
  xor    %eax,%eax
  mov    D6(%rip),%r8
  cmp    $0xffff,%edx
  mov    %edx,%ecx
  setne  %al
  sar    $0x8,%ecx
  and    $0xff,%edx
  lea    0x2(%rax,%rax,1),%eax
  movd   (%r8),%xmm0
  movd   %ecx,%xmm1
  mov    %eax,(%rdi)
  mov    0x8(%rdi),%rax
  punpckldq %xmm1,%xmm0
  mov    %edx,0xc(%rax)
  movq   %xmm0,0x4(%rax)
  test   %esi,%esi
  je     L0
  mov    D0(%rip),%rax
  mov    (%rax),%esi
L0:
  mov    %esi,0x10(%rdi)
  ret
```

assistant: ```c
void cec_msg_system_audio_mode_request(struct cec_msg *msg, int reply, __u16 phys_addr) {
  msg->len = phys_addr == 0xffff ? 2 : 4;
  msg->msg[1] = CEC_MSG_SYSTEM_AUDIO_MODE_REQUEST;
  msg->msg[2] = phys_addr >> 8;
  msg->msg[3] = phys_addr & 0xff;
  msg->reply = reply ? CEC_MSG_SET_SYSTEM_AUDIO_MODE : 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<TimerLoadSet64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <TimerBaseValid@plt>
  mov    %eax,%edi
  call   <DEBUG_ASSERT@plt>
  mov    -0x8(%rbp),%rax
  add    $0x2c,%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  shr    $0x20,%rax
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x28,%rax
  mov    -0x10(%rbp),%rdx
  mov    %edx,%edx
  mov    %rdx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void TimerLoadSet64(unsigned long ulBase, unsigned long long ullValue) {
  DEBUG_ASSERT(TimerBaseValid(ulBase));
  (*((volatile unsigned long *)(ulBase + 0x0000002C))) = ullValue >> 32;
  (*((volatile unsigned long *)(ulBase + 0x00000028))) = ullValue & 0xffffffff;
}
```