ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  je     L30
  cmp    $0x2,%eax
  jne    L31
L18:
  lea    D11(%rip),%rdi
  call   <printf@plt>
L31:
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L30:
  lea    D16(%rip),%rdi
  call   <printf@plt>
  jmp    L18
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D31
type: string
name: extract_symbolic_data
arguments: data_label: D11
type: string
name: extract_symbolic_data
arguments: data_label: D29
type: string
name: extract_symbolic_data
arguments: data_label: D16
type: string

tool: D31: %d

tool: D11: Vy vveli "2"


tool: D29: Vy vveli ne "1" i ne "2"

tool: D16: Vy vveli "1"


assistant: ```c
int main(void) {
  int a;
  scanf("%d", &a);
  switch (a) {
  case 1:
    printf("Vy vveli \"1\"\n");
  case 2:
    printf("Vy vveli \"2\"\n");
  default:
    printf("Vy vveli ne \"1\" i ne \"2\"");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D32(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  je     L31
  cmp    $0x2,%eax
  jne    L17
L29:
  lea    D36(%rip),%rdi
  call   <printf@plt>
L17:
  lea    D26(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L31:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L29
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D32
datatype: string
name: retrieve_data
arguments: data_label: D36
datatype: string
name: retrieve_data
arguments: data_label: D26
datatype: string
name: retrieve_data
arguments: data_label: D6
datatype: string

tool: D32: %d

tool: D36: Vy vveli "2"


tool: D26: Vy vveli ne "1" i ne "2"

tool: D6: Vy vveli "1"


assistant: ```c
int main(void) {
  int a;
  scanf("%d", &a);
  switch (a) {
  case 1:
    printf("Vy vveli \"1\"\n");
  case 2:
    printf("Vy vveli \"2\"\n");
  default:
    printf("Vy vveli ne \"1\" i ne \"2\"");
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x1,%eax
  je     L24
  cmp    $0x2,%eax
  jne    L16
L20:
  lea    D33(%rip),%rdi
  call   <printf@plt>
L16:
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L24:
  lea    D26(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D14
type: string
name: retrieve_data
arguments: data_tag: D33
type: string
name: retrieve_data
arguments: data_tag: D22
type: string
name: retrieve_data
arguments: data_tag: D26
type: string

tool: D14: %d

tool: D33: Vy vveli "2"


tool: D22: Vy vveli ne "1" i ne "2"

tool: D26: Vy vveli "1"


assistant: ```c
int main(void) {
  int a;
  scanf("%d", &a);
  switch (a) {
  case 1:
    printf("Vy vveli \"1\"\n");
  case 2:
    printf("Vy vveli \"2\"\n");
  default:
    printf("Vy vveli ne \"1\" i ne \"2\"");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<doXor2Arrays>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L9
L3:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%ecx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x28(%rbp),%rax
  add    %rsi,%rax
  xor    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L3
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void doXor2Arrays(char *source, int srcOff, char *result, int resOff, int length) {
  int srcIndex = srcOff;
  int resIndex = resOff;
  int i = 0;
  for (i = 0; i < length; i++) {
    result[resIndex] = (char)(source[srcIndex] ^ result[resIndex]);
    srcIndex++;
    resIndex++;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<doXor2Arrays>:
  endbr64
  test   %r8d,%r8d
  jle    L9
  movslq %ecx,%rcx
  lea    -0x1(%r8),%r8d
  add    %rcx,%r8
  mov    %rcx,%rax
  movslq %esi,%rsi
  sub    %rcx,%rsi
  add    %rdi,%rsi
L17:
  movzbl (%rsi,%rax,1),%ecx
  xor    %cl,(%rdx,%rax,1)
  mov    %rax,%rcx
  add    $0x1,%rax
  cmp    %r8,%rcx
  jne    L17
L9:
  ret
```

assistant: ```c
void doXor2Arrays(char *source, int srcOff, char *result, int resOff, int length) {
  int srcIndex = srcOff;
  int resIndex = resOff;
  int i = 0;
  for (i = 0; i < length; i++) {
    result[resIndex] = (char)(source[srcIndex] ^ result[resIndex]);
    srcIndex++;
    resIndex++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<doXor2Arrays>:
  endbr64
  mov    %rdx,%rax
  test   %r8d,%r8d
  jle    L5
  movslq %ecx,%rcx
  movslq %esi,%rsi
  sub    $0x1,%r8d
  sub    %rcx,%rsi
  add    %rcx,%r8
  add    %rdi,%rsi
  nop
L3:
  movzbl (%rsi,%rcx,1),%edx
  xor    %dl,(%rax,%rcx,1)
  mov    %rcx,%rdx
  add    $0x1,%rcx
  cmp    %rdx,%r8
  jne    L3
L5:
  ret
```

assistant: ```c
void doXor2Arrays(char *source, int srcOff, char *result, int resOff, int length) {
  int srcIndex = srcOff;
  int resIndex = resOff;
  int i = 0;
  for (i = 0; i < length; i++) {
    result[resIndex] = (char)(source[srcIndex] ^ result[resIndex]);
    srcIndex++;
    resIndex++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<CG_DrawStrlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L56
L57:
  cmpq   $0x0,-0x8(%rbp)
  je     L35
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x5e,%al
  jne    L35
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L35
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  mov    %rax,%rdi
  call   <isalnum@plt>
  test   %rax,%rax
  je     L35
  addq   $0x2,-0x8(%rbp)
  jmp    L56
L35:
  addl   $0x1,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L56:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L57
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int CG_DrawStrlen(const char *str) {
  const char *s = str;
  int count = 0;
  while (*s) {
    if (((s) && *(s) == '^' && *((s) + 1) && isalnum(*((s) + 1)))) {
      s += 2;
    } else {
      count++;
      s++;
    }
  }
  return count;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<CG_DrawStrlen>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L48
  mov    %rdi,%rbx
  mov    $0x0,%ebp
  jmp    L68
L71:
  add    $0x1,%ebp
  add    $0x1,%rbx
L57:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L27
L68:
  cmp    $0x5e,%al
  jne    L71
  test   %rbx,%rbx
  je     L71
  movzbl 0x1(%rbx),%edi
  test   %dil,%dil
  je     L71
  movsbq %dil,%rdi
  call   <isalnum@plt>
  test   %rax,%rax
  je     L71
  add    $0x2,%rbx
  jmp    L57
L48:
  mov    $0x0,%ebp
L27:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int CG_DrawStrlen(const char *str) {
  const char *s = str;
  int count = 0;
  while (*s) {
    if (((s) && *(s) == '^' && *((s) + 1) && isalnum(*((s) + 1)))) {
      s += 2;
    } else {
      count++;
      s++;
    }
  }
  return count;
}
```
user: What is the c source code of the assembly code below:

```asm
<CG_DrawStrlen>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L35
  mov    %rdi,%rbx
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L56:
  movsbq 0x1(%rbx),%rdi
  cmp    $0x5e,%al
  je     L61
L21:
  add    $0x1,%r12d
  add    $0x1,%rbx
  mov    %edi,%eax
L9:
  test   %al,%al
  jne    L56
L19:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L61:
  test   %dil,%dil
  jne    L48
  add    $0x1,%r12d
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L48:
  call   <isalnum@plt>
  test   %rax,%rax
  je     L30
  movzbl 0x2(%rbx),%eax
  add    $0x2,%rbx
  jmp    L9
  nopl   0x0(%rax)
L30:
  movzbl 0x1(%rbx),%edi
  jmp    L21
  xchg   %ax,%ax
L35:
  xor    %r12d,%r12d
  jmp    L19
```

assistant: ```c
int CG_DrawStrlen(const char *str) {
  const char *s = str;
  int count = 0;
  while (*s) {
    if (((s) && *(s) == '^' && *((s) + 1) && isalnum(*((s) + 1)))) {
      s += 2;
    } else {
      count++;
      s++;
    }
  }
  return count;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<CG_DrawStrlen>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L11
  mov    %rdi,%rbx
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L50:
  movsbq 0x1(%rbx),%rdi
  cmp    $0x5e,%al
  je     L34
L66:
  add    $0x1,%r12d
  add    $0x1,%rbx
  mov    %edi,%eax
L10:
  test   %al,%al
  jne    L50
L21:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L34:
  test   %dil,%dil
  jne    L29
  add    $0x1,%r12d
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L29:
  call   <isalnum@plt>
  test   %rax,%rax
  je     L68
  movzbl 0x2(%rbx),%eax
  add    $0x2,%rbx
  jmp    L10
  nopl   0x0(%rax)
L68:
  movzbl 0x1(%rbx),%edi
  jmp    L66
  xchg   %ax,%ax
L11:
  xor    %r12d,%r12d
  jmp    L21
```

assistant: ```c
int CG_DrawStrlen(const char *str) {
  const char *s = str;
  int count = 0;
  while (*s) {
    if (((s) && *(s) == '^' && *((s) + 1) && isalnum(*((s) + 1)))) {
      s += 2;
    } else {
      count++;
      s++;
    }
  }
  return count;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cbm_raw_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x4(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <cbmarch_raw_write@plt>
  leave
  ret
```

assistant: ```c
int cbm_raw_write(int HandleDevice, const void *Buffer, size_t Count) {
  ;
  return cbmarch_raw_write(HandleDevice, Buffer, Count);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cbm_raw_write>:
  endbr64
  sub    $0x8,%rsp
  call   <cbmarch_raw_write@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int cbm_raw_write(int HandleDevice, const void *Buffer, size_t Count) {
  ;
  return cbmarch_raw_write(HandleDevice, Buffer, Count);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<cbm_raw_write>:
  endbr64
  jmp    1050 <cbmarch_raw_write@plt>
```

assistant: ```c
int cbm_raw_write(int HandleDevice, const void *Buffer, size_t Count) {
  ;
  return cbmarch_raw_write(HandleDevice, Buffer, Count);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<cbm_raw_write>:
  endbr64
  jmp    1050 <cbmarch_raw_write@plt>
```

assistant: ```c
int cbm_raw_write(int HandleDevice, const void *Buffer, size_t Count) {
  ;
  return cbmarch_raw_write(HandleDevice, Buffer, Count);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ring_writeall>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <ring_avail@plt>
  cmp    %eax,-0x14(%rbp)
  jbe    L2
  mov    $0x0,%eax
  jmp    L4
L2:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <ring_write@plt>
L4:
  leave
  ret
```

assistant: ```c
unsigned int ring_writeall(struct ring *ring, void *buffer, unsigned int count) {
  if (count > ring_avail(ring))
    return 0;
  return ring_write(ring, buffer, count);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ring_writeall>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %edx,%ebp
  call   <ring_avail@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    %ebp,%edx
  jae    L12
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <ring_write@plt>
  jmp    L9
```

assistant: ```c
unsigned int ring_writeall(struct ring *ring, void *buffer, unsigned int count) {
  if (count > ring_avail(ring))
    return 0;
  return ring_write(ring, buffer, count);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ring_writeall>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  call   <ring_avail@plt>
  cmp    %r12d,%eax
  jae    L8
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L8:
  mov    %r12d,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <ring_write@plt>
```

assistant: ```c
unsigned int ring_writeall(struct ring *ring, void *buffer, unsigned int count) {
  if (count > ring_avail(ring))
    return 0;
  return ring_write(ring, buffer, count);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ring_writeall>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  call   <ring_avail@plt>
  cmp    %r12d,%eax
  jae    L4
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L4:
  mov    %r12d,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <ring_write@plt>
```

assistant: ```c
unsigned int ring_writeall(struct ring *ring, void *buffer, unsigned int count) {
  if (count > ring_avail(ring))
    return 0;
  return ring_write(ring, buffer, count);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<level_to_pers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x4,%eax
  cmp    $0xe,%eax
  ja     L17
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D15(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D15(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x7,%eax
  jmp    L89
  mov    $0x6,%eax
  jmp    L89
  mov    $0x5,%eax
  jmp    L89
  mov    $0x1,%eax
  jmp    L89
  mov    $0x2,%eax
  jmp    L89
  mov    $0x3,%eax
  jmp    L89
  mov    $0x4,%eax
  jmp    L89
  mov    $0x8,%eax
  jmp    L89
  mov    $0x9,%eax
  jmp    L89
L17:
  mov    $0x0,%eax
L89:
  pop    %rbp
  ret
```

assistant: ```c
int level_to_pers(int level) {
  switch (level) {
  case (-4):
    return 7UL;
  case -3:
    return 6UL;
  case -2:
    return 5UL;
  case (-1):
    return 1UL;
  case 0:
    return 2UL;
  case 1:
    return 3UL;
  case 4:
  case 5:
    return 4UL;
  case 6:
    return 8UL;
  case 10:
    return 9UL;
  }
  return 0UL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<level_to_pers>:
  endbr64
  add    $0x4,%edi
  cmp    $0xe,%edi
  ja     L0
  mov    %edi,%edi
  lea    D1(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x6,%eax
  ret
  mov    $0x5,%eax
  ret
  mov    $0x1,%eax
  ret
  mov    $0x2,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x4,%eax
  ret
  mov    $0x8,%eax
  ret
  mov    $0x9,%eax
  ret
L0:
  mov    $0x0,%eax
  ret
  mov    $0x7,%eax
  ret
```

assistant: ```c
int level_to_pers(int level) {
  switch (level) {
  case (-4):
    return 7UL;
  case -3:
    return 6UL;
  case -2:
    return 5UL;
  case (-1):
    return 1UL;
  case 0:
    return 2UL;
  case 1:
    return 3UL;
  case 4:
  case 5:
    return 4UL;
  case 6:
    return 8UL;
  case 10:
    return 9UL;
  }
  return 0UL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<level_to_pers>:
  endbr64
  add    $0x4,%edi
  xor    %eax,%eax
  cmp    $0xe,%edi
  ja     L5
  lea    D8(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L5:
  ret
```

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

tool: D8: 

assistant: ```c
int level_to_pers(int level) {
  switch (level) {
  case (-4):
    return 7UL;
  case -3:
    return 6UL;
  case -2:
    return 5UL;
  case (-1):
    return 1UL;
  case 0:
    return 2UL;
  case 1:
    return 3UL;
  case 4:
  case 5:
    return 4UL;
  case 6:
    return 8UL;
  case 10:
    return 9UL;
  }
  return 0UL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<level_to_pers>:
  endbr64
  add    $0x4,%edi
  xor    %eax,%eax
  cmp    $0xe,%edi
  ja     L0
  lea    D7(%rip),%rax
  mov    (%rax,%rdi,4),%eax
L0:
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
datatype: string

tool: D7: 

assistant: ```c
int level_to_pers(int level) {
  switch (level) {
  case (-4):
    return 7UL;
  case -3:
    return 6UL;
  case -2:
    return 5UL;
  case (-1):
    return 1UL;
  case 0:
    return 2UL;
  case 1:
    return 3UL;
  case 4:
  case 5:
    return 4UL;
  case 6:
    return 8UL;
  case 10:
    return 9UL;
  }
  return 0UL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<TEST_HOST_REG_IMM>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  je     L4
  mov    $0x44,%edi
  call   <addbyte@plt>
  mov    $0x89,%edi
  call   <addbyte@plt>
  mov    -0x4(%rbp),%eax
  shl    $0x3,%eax
  and    $0x38,%eax
  or     $0xc2,%al
  mov    %eax,%edi
  call   <addbyte@plt>
  mov    -0x4(%rbp),%eax
  and    $0x10,%eax
  or     $0x2,%eax
  mov    %eax,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  and    $0x10,%eax
  test   %eax,%eax
  je     L5
  mov    $0x66,%edi
  call   <addbyte@plt>
  mov    $0x81,%edi
  call   <addbyte@plt>
  mov    -0x4(%rbp),%eax
  and    $0x7,%eax
  or     $0xe0,%al
  mov    %eax,%edi
  call   <addbyte@plt>
  mov    -0x8(%rbp),%eax
  shl    $0x8,%eax
  or     $0xff,%al
  mov    %eax,%edi
  call   <addword@plt>
  jmp    L8
L5:
  mov    $0x81,%edi
  call   <addbyte@plt>
  mov    -0x4(%rbp),%eax
  and    $0x7,%eax
  or     $0xe0,%al
  mov    %eax,%edi
  call   <addbyte@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <addlong@plt>
L8:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int TEST_HOST_REG_IMM(int host_reg, uint32_t imm) {
  if (host_reg & 8) {
    addbyte(0x44);
    addbyte(0x89);
    addbyte(0xc0 | 2 | ((host_reg & 7) << 3));
    host_reg = 2 | (host_reg & 0x10);
  }
  if (host_reg & 0x10) {
    addbyte(0x66);
    addbyte(0x81);
    addbyte(0xe0 | (host_reg & 7));
    addword((imm << 8) | 0xff);
  } else {
    addbyte(0x81);
    addbyte(0xe0 | (host_reg & 7));
    addlong(imm);
  }
  return host_reg;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<TEST_HOST_REG_IMM>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  test   $0x8,%dil
  jne    L14
L3:
  test   $0x10,%bl
  je     L4
  mov    $0x66,%edi
  call   <addbyte@plt>
  mov    $0x81,%edi
  call   <addbyte@plt>
  mov    %ebx,%edi
  and    $0x7,%edi
  or     $0xe0,%dil
  call   <addbyte@plt>
  mov    %ebp,%edi
  shl    $0x8,%edi
  or     $0xff,%dil
  call   <addword@plt>
L26:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  mov    $0x44,%edi
  call   <addbyte@plt>
  mov    $0x89,%edi
  call   <addbyte@plt>
  lea    0x0(,%rbx,8),%edi
  and    $0x38,%edi
  or     $0xc2,%dil
  call   <addbyte@plt>
  and    $0x10,%ebx
  or     $0x2,%ebx
  jmp    L3
L4:
  mov    $0x81,%edi
  call   <addbyte@plt>
  mov    %ebx,%edi
  and    $0x7,%edi
  or     $0xe0,%dil
  call   <addbyte@plt>
  mov    %ebp,%edi
  call   <addlong@plt>
  jmp    L26
```

assistant: ```c
int TEST_HOST_REG_IMM(int host_reg, uint32_t imm) {
  if (host_reg & 8) {
    addbyte(0x44);
    addbyte(0x89);
    addbyte(0xc0 | 2 | ((host_reg & 7) << 3));
    host_reg = 2 | (host_reg & 0x10);
  }
  if (host_reg & 0x10) {
    addbyte(0x66);
    addbyte(0x81);
    addbyte(0xe0 | (host_reg & 7));
    addword((imm << 8) | 0xff);
  } else {
    addbyte(0x81);
    addbyte(0xe0 | (host_reg & 7));
    addlong(imm);
  }
  return host_reg;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<TEST_HOST_REG_IMM>:
  endbr64
  push   %r14
  mov    %edi,%r14d
  push   %r13
  and    $0x10,%r14d
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %esi,%ebp
  sub    $0x8,%rsp
  and    $0x8,%edi
  jne    L20
  mov    %r12d,%r13d
  and    $0x7,%r13d
  or     $0xe0,%r13b
L22:
  test   %r14d,%r14d
  je     L10
  mov    $0x66,%edi
  call   <addbyte@plt>
  mov    $0x81,%edi
  call   <addbyte@plt>
  mov    %r13d,%edi
  call   <addbyte@plt>
  mov    %ebp,%edi
  shl    $0x8,%edi
  or     $0xff,%dil
  call   <addword@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L10:
  mov    $0x81,%edi
  call   <addbyte@plt>
  mov    %r13d,%edi
  call   <addbyte@plt>
  mov    %ebp,%edi
  call   <addlong@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  mov    $0x44,%edi
  mov    $0xe2,%r13d
  call   <addbyte@plt>
  mov    $0x89,%edi
  call   <addbyte@plt>
  lea    0x0(,%r12,8),%edi
  mov    %r14d,%r12d
  and    $0x38,%edi
  or     $0x2,%r12d
  or     $0xc2,%dil
  call   <addbyte@plt>
  jmp    L22
```

assistant: ```c
int TEST_HOST_REG_IMM(int host_reg, uint32_t imm) {
  if (host_reg & 8) {
    addbyte(0x44);
    addbyte(0x89);
    addbyte(0xc0 | 2 | ((host_reg & 7) << 3));
    host_reg = 2 | (host_reg & 0x10);
  }
  if (host_reg & 0x10) {
    addbyte(0x66);
    addbyte(0x81);
    addbyte(0xe0 | (host_reg & 7));
    addword((imm << 8) | 0xff);
  } else {
    addbyte(0x81);
    addbyte(0xe0 | (host_reg & 7));
    addlong(imm);
  }
  return host_reg;
}
```