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

```asm
<get_next_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L13
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <ft_strclr@plt>
  jmp    L51
L13:
  mov    $0x1,%edi
  call   <ft_strnew@plt>
  mov    -0x20(%rbp),%rdx
  mov    %rax,(%rdx)
L51:
  mov    D75(%rip),%eax
  mov    -0x20(%rbp),%rdx
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <ft_check_buffer@plt>
  mov    %eax,D25(%rip)
  mov    D25(%rip),%eax
  test   %eax,%eax
  js     L17
  mov    $0x1,%eax
  jmp    L93
L41:
  movl   $0x0,D25(%rip)
  mov    D25(%rip),%eax
  mov    -0x20(%rbp),%rdx
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <ft_check_buffer@plt>
  mov    %eax,D25(%rip)
  mov    D25(%rip),%eax
  test   %eax,%eax
  js     L17
  mov    $0x1,%eax
  jmp    L93
L17:
  mov    -0x14(%rbp),%eax
  mov    $0x100,%edx
  lea    D1(%rip),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L41
  mov    D25(%rip),%eax
  test   %eax,%eax
  jne    L72
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <ft_strlen@plt>
  test   %rax,%rax
  je     L72
  movl   $0x1,D25(%rip)
  mov    $0x1,%eax
  jmp    L93
L72:
  mov    -0x4(%rbp),%eax
L93:
  leave
  ret
```

assistant: ```c
int get_next_line(int const fd, char **line) {
  static char buffer[256 + 1];
  static int cursor;
  static int end;
  int status;
  if (*line)
    ft_strclr(*line);
  else
    *line = ft_strnew(1);
  cursor = ft_check_buffer(buffer, cursor, line);
  if (cursor >= 0)
    return (1);
  while ((status = read(fd, buffer, 256)) > 0) {
    cursor = 0;
    cursor = ft_check_buffer(buffer, cursor, line);
    if (cursor >= 0)
      return (1);
  }
  if (!end && ft_strlen(*line)) {
    end = 1;
    return (1);
  }
  return (status);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_next_line>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  mov    %rsi,%r12
  mov    (%rsi),%rdi
  test   %rdi,%rdi
  je     L14
  call   <ft_strclr@plt>
L7:
  mov    %r12,%rdx
  mov    D12(%rip),%esi
  lea    D44(%rip),%rdi
  call   <ft_check_buffer@plt>
  mov    %eax,D18(%rip)
  mov    $0x1,%ebx
  lea    D44(%rip),%rbp
  test   %eax,%eax
  jns    L65
L66:
  mov    $0x100,%edx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L57
  movl   $0x0,D18(%rip)
  mov    %r12,%rdx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <ft_check_buffer@plt>
  mov    %eax,D18(%rip)
  test   %eax,%eax
  js     L66
  mov    $0x1,%ebx
L65:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  mov    $0x1,%edi
  call   <ft_strnew@plt>
  mov    %rax,(%r12)
  jmp    L7
L57:
  cmpl   $0x0,D18(%rip)
  jne    L65
  mov    (%r12),%rdi
  call   <ft_strlen@plt>
  test   %rax,%rax
  je     L65
  movl   $0x1,D18(%rip)
  mov    $0x1,%ebx
  jmp    L65
```

assistant: ```c
int get_next_line(int const fd, char **line) {
  static char buffer[256 + 1];
  static int cursor;
  static int end;
  int status;
  if (*line)
    ft_strclr(*line);
  else
    *line = ft_strnew(1);
  cursor = ft_check_buffer(buffer, cursor, line);
  if (cursor >= 0)
    return (1);
  while ((status = read(fd, buffer, 256)) > 0) {
    cursor = 0;
    cursor = ft_check_buffer(buffer, cursor, line);
    if (cursor >= 0)
      return (1);
  }
  if (!end && ft_strlen(*line)) {
    end = 1;
    return (1);
  }
  return (status);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_next_line>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  test   %rdi,%rdi
  je     L15
  call   <ft_strclr@plt>
L54:
  mov    D42(%rip),%esi
  mov    %rbp,%rdx
  lea    D36(%rip),%rbx
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L36:
  mov    $0x100,%edx
  mov    %rbx,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jle    L0
  movl   $0x0,D42(%rip)
  mov    %rbp,%rdx
  xor    %esi,%esi
L7:
  mov    %rbx,%rdi
  call   <ft_check_buffer@plt>
  mov    %eax,D4(%rip)
  test   %eax,%eax
  js     L36
L62:
  mov    $0x1,%r12d
L48:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L15:
  mov    $0x1,%edi
  call   <ft_strnew@plt>
  mov    %rax,0x0(%rbp)
  jmp    L54
  nopl   0x0(%rax,%rax,1)
L0:
  mov    D4(%rip),%eax
  test   %eax,%eax
  jne    L48
  mov    0x0(%rbp),%rdi
  call   <ft_strlen@plt>
  test   %rax,%rax
  je     L48
  movl   $0x1,D4(%rip)
  jmp    L62
```

assistant: ```c
int get_next_line(int const fd, char **line) {
  static char buffer[256 + 1];
  static int cursor;
  static int end;
  int status;
  if (*line)
    ft_strclr(*line);
  else
    *line = ft_strnew(1);
  cursor = ft_check_buffer(buffer, cursor, line);
  if (cursor >= 0)
    return (1);
  while ((status = read(fd, buffer, 256)) > 0) {
    cursor = 0;
    cursor = ft_check_buffer(buffer, cursor, line);
    if (cursor >= 0)
      return (1);
  }
  if (!end && ft_strlen(*line)) {
    end = 1;
    return (1);
  }
  return (status);
}
```
user: What is the c source code of the assembly code below:

```asm
<get_next_line>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rsi),%rdi
  test   %rdi,%rdi
  je     L40
  call   <ft_strclr@plt>
L78:
  mov    D38(%rip),%esi
  mov    %rbp,%rdx
  lea    D12(%rip),%rbx
  jmp    L50
  nopl   0x0(%rax,%rax,1)
L64:
  mov    $0x100,%edx
  mov    %rbx,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jle    L70
  movl   $0x0,D38(%rip)
  mov    %rbp,%rdx
  xor    %esi,%esi
L50:
  mov    %rbx,%rdi
  call   <ft_check_buffer@plt>
  mov    %eax,D24(%rip)
  test   %eax,%eax
  js     L64
L14:
  mov    $0x1,%r12d
L22:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L40:
  mov    $0x1,%edi
  call   <ft_strnew@plt>
  mov    %rax,0x0(%rbp)
  jmp    L78
  nopl   0x0(%rax,%rax,1)
L70:
  mov    D24(%rip),%eax
  test   %eax,%eax
  jne    L22
  mov    0x0(%rbp),%rdi
  call   <ft_strlen@plt>
  test   %rax,%rax
  je     L22
  movl   $0x1,D24(%rip)
  jmp    L14
```

assistant: ```c
int get_next_line(int const fd, char **line) {
  static char buffer[256 + 1];
  static int cursor;
  static int end;
  int status;
  if (*line)
    ft_strclr(*line);
  else
    *line = ft_strnew(1);
  cursor = ft_check_buffer(buffer, cursor, line);
  if (cursor >= 0)
    return (1);
  while ((status = read(fd, buffer, 256)) > 0) {
    cursor = 0;
    cursor = ft_check_buffer(buffer, cursor, line);
    if (cursor >= 0)
      return (1);
  }
  if (!end && ft_strlen(*line)) {
    end = 1;
    return (1);
  }
  return (status);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<clearcache>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    $0x1,%eax
  cpuid
  mov    %ebx,%esi
  mov    %eax,-0x1c(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x80000,%eax
  test   %eax,%eax
  jne    L11
  mov    $0x0,%eax
  call   <fallback_clearcache@plt>
L11:
  mov    -0x18(%rbp),%eax
  shr    $0x8,%eax
  movzbl %al,%eax
  shl    $0x3,%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L12
L4:
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  clflush (%rax)
  mov    -0x18(%rbp),%eax
  add    %eax,-0x20(%rbp)
L12:
  mov    -0x20(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jb     L4
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void clearcache(unsigned char *mem, unsigned size) {
  unsigned i, cl, eax, feat;
  asm("cpuid" : "=a"(eax), "=b"(cl), "=d"(feat) : "0"(1) : "cx");
  if (!(feat & (1 << 19)))
    fallback_clearcache();
  cl = ((cl >> 8) & 0xff) * 8;
  for (i = 0; i < size; i += cl)
    asm("clflush %0" ::"m"(mem[i]));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<clearcache>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x1,%eax
  cpuid
  test   $0x80000,%edx
  je     L7
L36:
  movzbl %bh,%ebx
  lea    0x0(,%rbx,8),%ecx
  test   %ebp,%ebp
  je     L8
  mov    $0x0,%eax
L10:
  mov    %eax,%edx
  clflush (%r12,%rdx,1)
  add    %ecx,%eax
  cmp    %eax,%ebp
  ja     L10
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  mov    $0x0,%eax
  call   <fallback_clearcache@plt>
  jmp    L36
```

assistant: ```c
void clearcache(unsigned char *mem, unsigned size) {
  unsigned i, cl, eax, feat;
  asm("cpuid" : "=a"(eax), "=b"(cl), "=d"(feat) : "0"(1) : "cx");
  if (!(feat & (1 << 19)))
    fallback_clearcache();
  cl = ((cl >> 8) & 0xff) * 8;
  for (i = 0; i < size; i += cl)
    asm("clflush %0" ::"m"(mem[i]));
}
```
user: Translate the below assembly code into C programming language:

```asm
<clearcache>:
  endbr64
  push   %r12
  mov    $0x1,%eax
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  cpuid
  and    $0x80000,%edx
  je     L20
L19:
  movzbl %bh,%ebx
  lea    0x0(,%rbx,8),%ecx
  test   %ebp,%ebp
  je     L16
  xor    %eax,%eax
  nopl   0x0(%rax)
L0:
  mov    %eax,%edx
  clflush (%r12,%rdx,1)
  add    %ecx,%eax
  cmp    %eax,%ebp
  ja     L0
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  xor    %eax,%eax
  call   <fallback_clearcache@plt>
  jmp    L19
```

assistant: ```c
void clearcache(unsigned char *mem, unsigned size) {
  unsigned i, cl, eax, feat;
  asm("cpuid" : "=a"(eax), "=b"(cl), "=d"(feat) : "0"(1) : "cx");
  if (!(feat & (1 << 19)))
    fallback_clearcache();
  cl = ((cl >> 8) & 0xff) * 8;
  for (i = 0; i < size; i += cl)
    asm("clflush %0" ::"m"(mem[i]));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<clearcache>:
  endbr64
  push   %r12
  mov    $0x1,%eax
  mov    %rdi,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  cpuid
  and    $0x80000,%edx
  je     L33
L16:
  movzbl %bh,%ebx
  lea    0x0(,%rbx,8),%ecx
  test   %ebp,%ebp
  je     L2
  xor    %eax,%eax
  nopl   0x0(%rax)
L20:
  mov    %eax,%edx
  clflush (%r12,%rdx,1)
  add    %ecx,%eax
  cmp    %eax,%ebp
  ja     L20
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L33:
  xor    %eax,%eax
  call   <fallback_clearcache@plt>
  jmp    L16
```

assistant: ```c
void clearcache(unsigned char *mem, unsigned size) {
  unsigned i, cl, eax, feat;
  asm("cpuid" : "=a"(eax), "=b"(cl), "=d"(feat) : "0"(1) : "cx");
  if (!(feat & (1 << 19)))
    fallback_clearcache();
  cl = ((cl >> 8) & 0xff) * 8;
  for (i = 0; i < size; i += cl)
    asm("clflush %0" ::"m"(mem[i]));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cparse_str_cmp>:
  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   <cparse_str_empty@plt>
  test   %rax,%rax
  jne    L20
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <cparse_str_empty@plt>
  test   %rax,%rax
  je     L1
L20:
  mov    $0x0,%eax
  jmp    L9
L1:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
L9:
  leave
  ret
```

assistant: ```c
int cparse_str_cmp(const char *a, const char *b) {
  if (cparse_str_empty(a) || cparse_str_empty(b)) {
    return 0;
  }
  return strcmp(a, b);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cparse_str_cmp>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <cparse_str_empty@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L18
L29:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  mov    %rbp,%rdi
  call   <cparse_str_empty@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L29
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  jmp    L29
```

assistant: ```c
int cparse_str_cmp(const char *a, const char *b) {
  if (cparse_str_empty(a) || cparse_str_empty(b)) {
    return 0;
  }
  return strcmp(a, b);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cparse_str_cmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <cparse_str_empty@plt>
  test   %rax,%rax
  je     L11
L7:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L11:
  mov    %r12,%rdi
  call   <cparse_str_empty@plt>
  test   %rax,%rax
  jne    L7
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <strcmp@plt>
```

assistant: ```c
int cparse_str_cmp(const char *a, const char *b) {
  if (cparse_str_empty(a) || cparse_str_empty(b)) {
    return 0;
  }
  return strcmp(a, b);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<cparse_str_cmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <cparse_str_empty@plt>
  test   %rax,%rax
  je     L4
L17:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L4:
  mov    %r12,%rdi
  call   <cparse_str_empty@plt>
  test   %rax,%rax
  jne    L17
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <strcmp@plt>
```

assistant: ```c
int cparse_str_cmp(const char *a, const char *b) {
  if (cparse_str_empty(a) || cparse_str_empty(b)) {
    return 0;
  }
  return strcmp(a, b);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sub_const_i_2>:
  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    %r9d,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L36
L26:
  movl   $0x0,-0x4(%rbp)
  jmp    L24
L13:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  sub    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L24:
  mov    0x10(%rbp),%eax
  sub    -0x30(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jle    L13
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    0x10(%rbp),%eax
  sub    -0x30(%rbp),%eax
  cltq
  add    $0x1,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    0x10(%rbp),%eax
  sub    -0x30(%rbp),%eax
  add    $0x1,%eax
  imul   -0x1c(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rsi
  mov    -0x18(%rbp),%rax
  add    %rsi,%rax
  mov    (%rax),%rsi
  mov    0x10(%rbp),%eax
  sub    -0x30(%rbp),%eax
  cltq
  add    $0x1,%rax
  shl    $0x2,%rax
  add    %rsi,%rax
  sub    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x8(%rbp)
L36:
  mov    -0x2c(%rbp),%eax
  sub    -0x20(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jle    L26
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int **sub_const_i_2(int **res, int cn, int **vec, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++) {
    for (j = 0; j <= hi2 - lo2; j++)
      res[i][j] = cn - vec[i][j];
    res[i][hi2 - lo2 + 1] -= cn * (hi2 - lo2 + 1);
  }
  return res;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sub_const_i_2>:
  endbr64
  mov    %rdi,%r11
  sub    %ecx,%r8d
  js     L56
  push   %r14
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r10d
  mov    %rdx,%rdi
  mov    0x28(%rsp),%ebx
  sub    %r9d,%ebx
  movslq %ebx,%rax
  lea    0x4(,%rax,4),%r12
  lea    0x1(%rbx),%r9d
  imul   %esi,%r9d
  mov    %r11,%rsi
  mov    %r8d,%r8d
  lea    0x8(%r11,%r8,8),%rbp
  mov    %ebx,%eax
  lea    0x4(,%rax,4),%r8
  jmp    L8
L43:
  mov    (%rsi),%rdx
  mov    (%rdi),%rcx
  mov    %r10d,%r14d
  sub    (%rcx,%rax,1),%r14d
  mov    %r14d,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L43
L23:
  mov    %r12,%rax
  add    (%rsi),%rax
  sub    %r9d,(%rax)
  add    $0x8,%rsi
  add    $0x8,%rdi
  cmp    %rbp,%rsi
  je     L28
L8:
  mov    $0x0,%eax
  test   %ebx,%ebx
  jns    L43
  jmp    L23
L28:
  mov    %r11,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L56:
  mov    %rdi,%rax
  ret
```

assistant: ```c
int **sub_const_i_2(int **res, int cn, int **vec, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++) {
    for (j = 0; j <= hi2 - lo2; j++)
      res[i][j] = cn - vec[i][j];
    res[i][hi2 - lo2 + 1] -= cn * (hi2 - lo2 + 1);
  }
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sub_const_i_2>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    %ecx,%r8d
  js     L1
  mov    0x28(%rsp),%r11d
  movslq %r8d,%r8
  mov    %esi,%edi
  mov    %rdx,%rbp
  lea    0x8(,%r8,8),%r13
  sub    %r9d,%r11d
  xor    %r9d,%r9d
  lea    0x1(%r11),%ebx
  mov    %r11d,%eax
  movslq %ebx,%r10
  imul   %esi,%ebx
  lea    0x4(,%rax,4),%r8
  shl    $0x2,%r10
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    (%r12,%r9,1),%rcx
  test   %r11d,%r11d
  js     L38
  mov    0x0(%rbp,%r9,1),%rsi
  xor    %eax,%eax
L17:
  mov    %edi,%edx
  sub    (%rsi,%rax,1),%edx
  mov    %edx,(%rcx,%rax,1)
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L17
L38:
  add    $0x8,%r9
  sub    %ebx,(%rcx,%r10,1)
  cmp    %r9,%r13
  jne    L10
L1:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int **sub_const_i_2(int **res, int cn, int **vec, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++) {
    for (j = 0; j <= hi2 - lo2; j++)
      res[i][j] = cn - vec[i][j];
    res[i][hi2 - lo2 + 1] -= cn * (hi2 - lo2 + 1);
  }
  return res;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sub_const_i_2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    %ecx,%r8d
  js     L47
  mov    0x38(%rsp),%r11d
  movslq %r8d,%r8
  mov    %esi,%r10d
  mov    %esi,%r15d
  lea    0x8(,%r8,8),%rax
  movd   %r10d,%xmm3
  xor    %esi,%esi
  sub    %r9d,%r11d
  mov    %rax,-0x28(%rsp)
  pshufd $0x0,%xmm3,%xmm1
  lea    0x1(%r11),%r14d
  mov    %r14d,%eax
  imul   %r14d,%r15d
  movslq %r14d,%r9
  mov    %r14d,-0x1c(%rsp)
  shr    $0x2,%eax
  shl    $0x2,%r9
  lea    -0x1(%rax),%edi
  mov    %r14d,%eax
  mov    %rdx,%r14
  and    $0xfffffffc,%eax
  add    $0x1,%rdi
  lea    0x1(%rax),%ebx
  movslq %eax,%rbp
  mov    %eax,-0x20(%rsp)
  add    $0x2,%eax
  mov    %eax,-0x14(%rsp)
  cltq
  movslq %ebx,%r13
  shl    $0x4,%rdi
  shl    $0x2,%rax
  mov    %ebx,-0x18(%rsp)
  shl    $0x2,%rbp
  mov    %rax,-0x10(%rsp)
  mov    %r11d,%eax
  shl    $0x2,%r13
  lea    0x4(,%rax,4),%rbx
  nopl   0x0(%rax,%rax,1)
L76:
  mov    (%r12,%rsi,1),%rax
  test   %r11d,%r11d
  js     L55
  mov    (%r14,%rsi,1),%rdx
  mov    %rax,%rcx
  lea    0x4(%rdx),%r8
  sub    %r8,%rcx
  cmp    $0x8,%rcx
  mov    $0x0,%ecx
  jbe    L75
  cmp    $0x2,%r11d
  jbe    L75
L60:
  movdqu (%rdx,%rcx,1),%xmm2
  movdqa %xmm1,%xmm0
  psubd  %xmm2,%xmm0
  movups %xmm0,(%rax,%rcx,1)
  add    $0x10,%rcx
  cmp    %rdi,%rcx
  jne    L60
  mov    -0x1c(%rsp),%r8d
  cmp    %r8d,-0x20(%rsp)
  je     L55
  mov    %r10d,%ecx
  sub    (%rdx,%rbp,1),%ecx
  mov    %ecx,(%rax,%rbp,1)
  cmp    -0x18(%rsp),%r11d
  jl     L55
  mov    %r10d,%ecx
  sub    (%rdx,%r13,1),%ecx
  mov    %ecx,(%rax,%r13,1)
  cmp    -0x14(%rsp),%r11d
  jl     L55
  mov    -0x10(%rsp),%rcx
  mov    %r10d,%r8d
  sub    (%rdx,%rcx,1),%r8d
  mov    %r8d,(%rax,%rcx,1)
L55:
  sub    %r15d,(%rax,%r9,1)
  add    $0x8,%rsi
  cmp    -0x28(%rsp),%rsi
  jne    L76
L47:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L75:
  mov    %r10d,%r8d
  sub    (%rdx,%rcx,1),%r8d
  mov    %r8d,(%rax,%rcx,1)
  add    $0x4,%rcx
  cmp    %rbx,%rcx
  jne    L75
  jmp    L55
```

assistant: ```c
int **sub_const_i_2(int **res, int cn, int **vec, int lo1, int hi1, int lo2, int hi2) {
  int i, j;
  for (i = 0; i <= hi1 - lo1; i++) {
    for (j = 0; j <= hi2 - lo2; j++)
      res[i][j] = cn - vec[i][j];
    res[i][hi2 - lo2 + 1] -= cn * (hi2 - lo2 + 1);
  }
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<getLine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L44:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
L13:
  subl   $0x1,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jle    L4
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L4
  cmpl   $0xa,-0x8(%rbp)
  jne    L44
L4:
  cmpl   $0xa,-0x8(%rbp)
  jne    L1
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
L1:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getLine(char s[], int maxLength) {
  int c, i = 0;
  while (--maxLength > 0 && (c = getchar()) != EOF && c != '\n')
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<getLine>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  cmp    $0x1,%esi
  jle    L3
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%r12d
  mov    $0x0,%ebx
L6:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D5(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L58
  cmp    $0xa,%eax
  je     L58
  add    $0x1,%ebx
  mov    %al,0x0(%rbp)
  add    $0x1,%rbp
  cmp    %r12d,%ebx
  jne    L6
  mov    %r12d,%ebx
  jmp    L35
L3:
  mov    $0x0,%ebx
L58:
  cmp    $0xa,%eax
  je     L16
L35:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  movslq %ebx,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  lea    0x1(%rbx),%ebx
  jmp    L35
```

assistant: ```c
int getLine(char s[], int maxLength) {
  int c, i = 0;
  while (--maxLength > 0 && (c = getchar()) != EOF && c != '\n')
    s[i++] = c;
  if (c == '\n')
    s[i++] = c;
  s[i] = '\0';
  return i;
}
```