ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<hashWordsInFile>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x3c(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <newHashExt@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <lineFileOpen@plt>
  mov    %rax,-0x28(%rbp)
  jmp    L17
L15:
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <hashAdd@plt>
L21:
  lea    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <nextWord@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L15
L17:
  mov    -0x28(%rbp),%rax
  lea    -0x20(%rbp),%rcx
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <lineFileNext@plt>
  test   %rax,%rax
  jne    L21
  lea    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <lineFileClose@plt>
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
struct hash *hashWordsInFile(char *fileName, int hashSize) {
  struct hash *hash = newHashExt(hashSize, 1);
  struct lineFile *lf = lineFileOpen(fileName, 1);
  char *line, *word;
  while (lineFileNext(lf, &line, NULL)) {
    while ((word = nextWord(&line)) != NULL)
      hashAdd(hash, word, NULL);
  }
  lineFileClose(&lf);
  return hash;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hashWordsInFile>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbx
  mov    %esi,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x1,%esi
  call   <newHashExt@plt>
  mov    %rax,%rbp
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <lineFileOpen@plt>
  mov    %rax,0x8(%rsp)
  lea    0x10(%rsp),%rbx
  jmp    L11
L38:
  mov    $0x0,%edx
  mov    %rbp,%rdi
  call   <hashAdd@plt>
L32:
  mov    %rbx,%rdi
  call   <nextWord@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L38
L11:
  mov    $0x0,%edx
  mov    %rbx,%rsi
  mov    0x8(%rsp),%rdi
  call   <lineFileNext@plt>
  test   %rax,%rax
  jne    L32
  lea    0x8(%rsp),%rdi
  call   <lineFileClose@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    %rbp,%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
struct hash *hashWordsInFile(char *fileName, int hashSize) {
  struct hash *hash = newHashExt(hashSize, 1);
  struct lineFile *lf = lineFileOpen(fileName, 1);
  char *line, *word;
  while (lineFileNext(lf, &line, NULL)) {
    while ((word = nextWord(&line)) != NULL)
      hashAdd(hash, word, NULL);
  }
  lineFileClose(&lf);
  return hash;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<hashWordsInFile>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  mov    $0x1,%esi
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  call   <newHashExt@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <lineFileOpen@plt>
  xor    %edx,%edx
  mov    %rbx,%rsi
  mov    %rax,%rdi
  mov    %rax,0x8(%rsp)
  call   <lineFileNext@plt>
  test   %rax,%rax
  jne    L26
  jmp    L10
  nopl   0x0(%rax)
L29:
  xor    %edx,%edx
  mov    %r12,%rdi
  call   <hashAdd@plt>
L26:
  mov    %rbx,%rdi
  call   <nextWord@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L29
  mov    0x8(%rsp),%rdi
  xor    %edx,%edx
  mov    %rbx,%rsi
  call   <lineFileNext@plt>
  test   %rax,%rax
  jne    L26
L10:
  lea    0x8(%rsp),%rdi
  call   <lineFileClose@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x20,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
struct hash *hashWordsInFile(char *fileName, int hashSize) {
  struct hash *hash = newHashExt(hashSize, 1);
  struct lineFile *lf = lineFileOpen(fileName, 1);
  char *line, *word;
  while (lineFileNext(lf, &line, NULL)) {
    while ((word = nextWord(&line)) != NULL)
      hashAdd(hash, word, NULL);
  }
  lineFileClose(&lf);
  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
<hashWordsInFile>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  mov    $0x1,%esi
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  call   <newHashExt@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <lineFileOpen@plt>
  xor    %edx,%edx
  mov    %rbx,%rsi
  mov    %rax,%rdi
  mov    %rax,0x8(%rsp)
  call   <lineFileNext@plt>
  test   %rax,%rax
  jne    L6
  jmp    L33
  nopl   0x0(%rax)
L17:
  xor    %edx,%edx
  mov    %r12,%rdi
  call   <hashAdd@plt>
L6:
  mov    %rbx,%rdi
  call   <nextWord@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  jne    L17
  mov    0x8(%rsp),%rdi
  xor    %edx,%edx
  mov    %rbx,%rsi
  call   <lineFileNext@plt>
  test   %rax,%rax
  jne    L6
L33:
  lea    0x8(%rsp),%rdi
  call   <lineFileClose@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x20,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L40:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
struct hash *hashWordsInFile(char *fileName, int hashSize) {
  struct hash *hash = newHashExt(hashSize, 1);
  struct lineFile *lf = lineFileOpen(fileName, 1);
  char *line, *word;
  while (lineFileNext(lf, &line, NULL)) {
    while ((word = nextWord(&line)) != NULL)
      hashAdd(hash, word, NULL);
  }
  lineFileClose(&lf);
  return hash;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<keyboard_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D8(%rip),%rax
  mov    $0x8e,%edx
  mov    %eax,%esi
  mov    $0x21,%edi
  call   <idt_add@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void keyboard_init() {
  extern void _keyboard_handler();
  idt_add(0x21, (unsigned int)_keyboard_handler, 0x8e);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<keyboard_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8e,%edx
  mov    D1(%rip),%esi
  mov    $0x21,%edi
  call   <idt_add@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void keyboard_init() {
  extern void _keyboard_handler();
  idt_add(0x21, (unsigned int)_keyboard_handler, 0x8e);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<keyboard_init>:
  endbr64
  mov    D5(%rip),%esi
  mov    $0x8e,%edx
  mov    $0x21,%edi
  jmp    1050 <idt_add@plt>
```

assistant: ```c
void keyboard_init() {
  extern void _keyboard_handler();
  idt_add(0x21, (unsigned int)_keyboard_handler, 0x8e);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<keyboard_init>:
  endbr64
  mov    D0(%rip),%esi
  mov    $0x8e,%edx
  mov    $0x21,%edi
  jmp    1050 <idt_add@plt>
```

assistant: ```c
void keyboard_init() {
  extern void _keyboard_handler();
  idt_add(0x21, (unsigned int)_keyboard_handler, 0x8e);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<move_X>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L2
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L4
L2:
  cmpl   $0x0,-0x4(%rbp)
  jle    L20
  lea    D22(%rip),%rax
  jmp    L38
L20:
  lea    D23(%rip),%rax
L38:
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L4:
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D26
type_infered: string
name: parse_data
arguments: tag: D22
type_infered: string
name: parse_data
arguments: tag: D23
type_infered: string
name: parse_data
arguments: tag: D27
type_infered: string

tool: D26: Stop X motor

tool: D22: ++

tool: D23: --

tool: D27: Move X to %s


assistant: ```c
void move_X(int dir) {
  if (dir == 0) {
    printf("Stop X motor\n");
  } else {
    printf("Move X to %s\n", (dir > 0) ? "++" : "--");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<move_X>:
  endbr64
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L6
  lea    D1(%rip),%rsi
  lea    D30(%rip),%rax
  cmovle %rax,%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L14:
  add    $0x8,%rsp
  ret
L6:
  lea    D22(%rip),%rdi
  call   <puts@plt>
  jmp    L14
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_infered: string
name: extract_symbolic_data
arguments: tag: D30
type_infered: string
name: extract_symbolic_data
arguments: tag: D3
type_infered: string
name: extract_symbolic_data
arguments: tag: D22
type_infered: string

tool: D1: ++

tool: D30: --

tool: D3: Move X to %s


tool: D22: Stop X motor

assistant: ```c
void move_X(int dir) {
  if (dir == 0) {
    printf("Stop X motor\n");
  } else {
    printf("Move X to %s\n", (dir > 0) ? "++" : "--");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<move_X>:
  endbr64
  test   %edi,%edi
  je     L9
  lea    D13(%rip),%rax
  lea    D14(%rip),%rsi
  cmovle %rax,%rsi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  jmp    1070 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L9:
  lea    D23(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: read_data
arguments: data_label: D13
data_type: string
name: read_data
arguments: data_label: D14
data_type: string
name: read_data
arguments: data_label: D18
data_type: string
name: read_data
arguments: data_label: D23
data_type: string

tool: D13: --

tool: D14: ++

tool: D18: Move X to %s


tool: D23: Stop X motor

assistant: ```c
void move_X(int dir) {
  if (dir == 0) {
    printf("Stop X motor\n");
  } else {
    printf("Move X to %s\n", (dir > 0) ? "++" : "--");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<move_X>:
  endbr64
  test   %edi,%edi
  je     L9
  lea    D33(%rip),%rax
  lea    D30(%rip),%rsi
  cmovle %rax,%rsi
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  jmp    1070 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L9:
  lea    D29(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: unpack_data
arguments: tag: D33
type: string
name: unpack_data
arguments: tag: D30
type: string
name: unpack_data
arguments: tag: D26
type: string
name: unpack_data
arguments: tag: D29
type: string

tool: D33: --

tool: D30: ++

tool: D26: Move X to %s


tool: D29: Stop X motor

assistant: ```c
void move_X(int dir) {
  if (dir == 0) {
    printf("Stop X motor\n");
  } else {
    printf("Move X to %s\n", (dir > 0) ? "++" : "--");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<explicit_conversion_interference1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %ax,-0x2(%rbp)
  movzwl -0x2(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char explicit_conversion_interference1(unsigned int c) {
  unsigned short b;
  return (unsigned char)(b = c);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<explicit_conversion_interference1>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
unsigned char explicit_conversion_interference1(unsigned int c) {
  unsigned short b;
  return (unsigned char)(b = c);
}
```
user: Translate the below assembly code into C programming language:

```asm
<explicit_conversion_interference1>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
unsigned char explicit_conversion_interference1(unsigned int c) {
  unsigned short b;
  return (unsigned char)(b = c);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<explicit_conversion_interference1>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
unsigned char explicit_conversion_interference1(unsigned int c) {
  unsigned short b;
  return (unsigned char)(b = c);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getln>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L55
L25:
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L20
  mov    $0x0,%eax
  jmp    L33
L20:
  cmpl   $0xa,-0xc(%rbp)
  jne    L47
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  jmp    L33
L47:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <getc@plt>
  mov    %eax,-0xc(%rbp)
L55:
  mov    -0x1c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  test   %eax,%eax
  jne    L25
  mov    -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
L33:
  leave
  ret
```

assistant: ```c
char *getln(char *s, int n, FILE *fp) {
  int c = getc(fp);
  char *str = s;
  while (n--) {
    if (c == EOF) {
      return NULL;
    } else if (c == '\n') {
      *str++ = '\0';
      return s;
    }
    *str++ = c;
    c = getc(fp);
  }
  s[n - 1] = '\0';
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getln>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %rdx,%r13
  mov    %rdx,%rdi
  call   <getc@plt>
  lea    -0x1(%rbp),%edx
  test   %ebp,%ebp
  je     L46
  mov    %edx,%r15d
  mov    %ebp,%ebp
  add    %r12,%rbp
  mov    %r12,%rbx
  mov    D2(%rip),%r14
L11:
  cmp    %eax,(%r14)
  je     L54
  cmp    $0xa,%eax
  je     L9
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  mov    %r13,%rdi
  call   <getc@plt>
  lea    (%r15,%r12,1),%edx
  sub    %ebx,%edx
  cmp    %rbp,%rbx
  jne    L11
L46:
  movslq %edx,%rdx
  movb   $0x0,-0x1(%r12,%rdx,1)
  mov    %r12,%rax
L25:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  movb   $0x0,(%rbx)
  mov    %r12,%rax
  jmp    L25
L54:
  mov    $0x0,%eax
  jmp    L25
```

assistant: ```c
char *getln(char *s, int n, FILE *fp) {
  int c = getc(fp);
  char *str = s;
  while (n--) {
    if (c == EOF) {
      return NULL;
    } else if (c == '\n') {
      *str++ = '\0';
      return s;
    }
    *str++ = c;
    c = getc(fp);
  }
  s[n - 1] = '\0';
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<getln>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    %rdx,%rdi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %esi,%ebx
  call   <getc@plt>
  test   %ebx,%ebx
  je     L49
  mov    D7(%rip),%r13
  add    %r12,%rbx
  mov    %r12,%r14
  jmp    L33
  nop
L4:
  cmp    $0xa,%eax
  je     L41
  mov    %al,(%r14)
  add    $0x1,%r14
  mov    %rbp,%rdi
  call   <getc@plt>
  cmp    %rbx,%r14
  je     L49
L33:
  cmp    %eax,0x0(%r13)
  jne    L4
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L49:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  movb   $0x0,-0x2(%r12)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  movb   $0x0,(%r14)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *getln(char *s, int n, FILE *fp) {
  int c = getc(fp);
  char *str = s;
  while (n--) {
    if (c == EOF) {
      return NULL;
    } else if (c == '\n') {
      *str++ = '\0';
      return s;
    }
    *str++ = c;
    c = getc(fp);
  }
  s[n - 1] = '\0';
  return s;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getln>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    %rdx,%rdi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %esi,%ebx
  call   <getc@plt>
  test   %ebx,%ebx
  je     L23
  mov    D7(%rip),%r13
  add    %r12,%rbx
  mov    %r12,%r14
  jmp    L6
  nop
L0:
  cmp    $0xa,%eax
  je     L47
  mov    %al,(%r14)
  add    $0x1,%r14
  mov    %rbp,%rdi
  call   <getc@plt>
  cmp    %rbx,%r14
  je     L23
L6:
  cmp    %eax,0x0(%r13)
  jne    L0
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L23:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  movb   $0x0,-0x2(%r12)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L47:
  movb   $0x0,(%r14)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *getln(char *s, int n, FILE *fp) {
  int c = getc(fp);
  char *str = s;
  while (n--) {
    if (c == EOF) {
      return NULL;
    } else if (c == '\n') {
      *str++ = '\0';
      return s;
    }
    *str++ = c;
    c = getc(fp);
  }
  s[n - 1] = '\0';
  return s;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<CollectOptions>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L14
L23:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L27
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L27
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    $0x2,%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  cmp    %eax,-0x20(%rbp)
  jne    L27
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L27:
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L23
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int CollectOptions(char **List, int ListLength, int Stream, int *Options) {
  int OptCnt, i;
  OptCnt = 0;
  for (i = 1; i < ListLength; i++)
    if (*List[i] == '-' && !isdigit(*(List[i] + 1)) && atoi(List[i] + 2) == Stream)
      Options[OptCnt++] = i;
  return (OptCnt);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<CollectOptions>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rcx,0x8(%rsp)
  cmp    $0x1,%esi
  jle    L52
  mov    %rdi,%r12
  mov    %edx,%r14d
  mov    %esi,%ebp
  mov    $0x1,%ebx
  mov    $0x0,%r15d
  jmp    L42
L50:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L44
L42:
  mov    (%r12,%rbx,8),%rax
  cmpb   $0x2d,(%rax)
  jne    L50
  movsbl 0x1(%rax),%edi
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L50
  mov    (%r12,%rbx,8),%rax
  lea    0x2(%rax),%rdi
  call   <atoi@plt>
  cmp    %r14d,%eax
  jne    L50
  movslq %r15d,%rax
  mov    0x8(%rsp),%rdx
  mov    %ebx,(%rdx,%rax,4)
  lea    0x1(%r15),%r15d
  jmp    L50
L52:
  mov    $0x0,%r15d
L44:
  mov    %r15d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int CollectOptions(char **List, int ListLength, int Stream, int *Options) {
  int OptCnt, i;
  OptCnt = 0;
  for (i = 1; i < ListLength; i++)
    if (*List[i] == '-' && !isdigit(*(List[i] + 1)) && atoi(List[i] + 2) == Stream)
      Options[OptCnt++] = i;
  return (OptCnt);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<CollectOptions>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%esi
  jle    L50
  mov    %rdi,%r15
  mov    %edx,%r12d
  mov    %rcx,%r13
  movslq %esi,%rbp
  mov    $0x1,%ebx
  xor    %r14d,%r14d
  jmp    L2
  nopl   (%rax)
L76:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L23
L2:
  mov    (%r15,%rbx,8),%rax
  cmpb   $0x2d,(%rax)
  jne    L76
  movsbl 0x1(%rax),%edi
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L76
  mov    (%r15,%rbx,8),%rax
  lea    0x2(%rax),%rdi
  call   <atoi@plt>
  cmp    %r12d,%eax
  jne    L76
  movslq %r14d,%rax
  add    $0x1,%r14d
  mov    %ebx,0x0(%r13,%rax,4)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L2
L23:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L50:
  xor    %r14d,%r14d
  jmp    L23
```

assistant: ```c
int CollectOptions(char **List, int ListLength, int Stream, int *Options) {
  int OptCnt, i;
  OptCnt = 0;
  for (i = 1; i < ListLength; i++)
    if (*List[i] == '-' && !isdigit(*(List[i] + 1)) && atoi(List[i] + 2) == Stream)
      Options[OptCnt++] = i;
  return (OptCnt);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<CollectOptions>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%esi
  jle    L68
  mov    %rdi,%r15
  mov    %edx,%r12d
  mov    %rcx,%r13
  movslq %esi,%rbp
  mov    $0x1,%ebx
  xor    %r14d,%r14d
  jmp    L39
  nopl   (%rax)
L37:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  je     L70
L39:
  mov    (%r15,%rbx,8),%rax
  cmpb   $0x2d,(%rax)
  jne    L37
  movsbl 0x1(%rax),%edi
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L37
  mov    (%r15,%rbx,8),%rax
  lea    0x2(%rax),%rdi
  call   <atoi@plt>
  cmp    %r12d,%eax
  jne    L37
  movslq %r14d,%rax
  add    $0x1,%r14d
  mov    %ebx,0x0(%r13,%rax,4)
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L39
L70:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L68:
  xor    %r14d,%r14d
  jmp    L70
```

assistant: ```c
int CollectOptions(char **List, int ListLength, int Stream, int *Options) {
  int OptCnt, i;
  OptCnt = 0;
  for (i = 1; i < ListLength; i++)
    if (*List[i] == '-' && !isdigit(*(List[i] + 1)) && atoi(List[i] + 2) == Stream)
      Options[OptCnt++] = i;
  return (OptCnt);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<savestr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  test   %rbx,%rbx
  jne    L23
  mov    $0x0,%eax
  jmp    L15
L23:
  mov    %rbx,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%r12
  test   %r12,%r12
  jne    L16
  mov    $0x0,%eax
  jmp    L15
L16:
  nop
L13:
  mov    %rbx,%rdx
  lea    0x1(%rdx),%rbx
  mov    %r12,%rax
  lea    0x1(%rax),%r12
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L13
  mov    -0x18(%rbp),%rax
L15:
  add    $0x10,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
char *savestr(s)
register char *s;
{
  char *p;
  register char *t;
  if (s == 0)
    return (0);
  t = p = (char *)calloc(strlen(s) + 1, 1);
  if (t == 0)
    return (0);
  while ((*t++ = *s++))
    ;
  return (p);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<savestr>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L16
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    $0x1,%esi
  call   <calloc@plt>
  test   %rax,%rax
  je     L28
  mov    $0x0,%edx
L30:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L30
L28:
  pop    %rbx
  ret
L16:
  mov    %rdi,%rax
  jmp    L28
```

assistant: ```c
char *savestr(s)
register char *s;
{
  char *p;
  register char *t;
  if (s == 0)
    return (0);
  t = p = (char *)calloc(strlen(s) + 1, 1);
  if (t == 0)
    return (0);
  while ((*t++ = *s++))
    ;
  return (p);
}
```