ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<split>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <str_trim_ht@plt>
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L4
L12:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <str_trim_crlf@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L4:
  cmpq   $0x0,-0x8(%rbp)
  jne    L12
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int split(char *str, char *seg, char array[][100]) {
  int i = 0;
  str_trim_ht(str);
  char *substr = strtok(str, seg);
  while (substr != NULL) {
    str_trim_crlf(substr);
    strcpy(array[i], substr);
    substr = strtok(NULL, seg);
    i++;
  }
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<split>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  call   <str_trim_ht@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L1
  mov    %rax,%rbx
  mov    $0x0,%r13d
L18:
  mov    %rbx,%rdi
  call   <str_trim_crlf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,%rbx
  add    $0x1,%r13d
  add    $0x64,%rbp
  test   %rax,%rax
  jne    L18
L5:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  mov    $0x0,%r13d
  jmp    L5
```

assistant: ```c
int split(char *str, char *seg, char array[][100]) {
  int i = 0;
  str_trim_ht(str);
  char *substr = strtok(str, seg);
  while (substr != NULL) {
    str_trim_crlf(substr);
    strcpy(array[i], substr);
    substr = strtok(NULL, seg);
    i++;
  }
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<split>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <str_trim_ht@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L7
  mov    %rax,%rbp
  xor    %r13d,%r13d
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %rbp,%rdi
  add    $0x1,%r13d
  call   <str_trim_crlf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  add    $0x64,%rbx
  call   <strcpy@plt>
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L17
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  add    $0x8,%rsp
  xor    %r13d,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int split(char *str, char *seg, char array[][100]) {
  int i = 0;
  str_trim_ht(str);
  char *substr = strtok(str, seg);
  while (substr != NULL) {
    str_trim_crlf(substr);
    strcpy(array[i], substr);
    substr = strtok(NULL, seg);
    i++;
  }
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<split>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <str_trim_ht@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strtok@plt>
  test   %rax,%rax
  je     L16
  mov    %rax,%rbp
  xor    %r13d,%r13d
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %rbp,%rdi
  add    $0x1,%r13d
  call   <str_trim_crlf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  add    $0x64,%rbx
  call   <strcpy@plt>
  mov    %r12,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L17
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  add    $0x8,%rsp
  xor    %r13d,%r13d
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int split(char *str, char *seg, char array[][100]) {
  int i = 0;
  str_trim_ht(str);
  char *substr = strtok(str, seg);
  while (substr != NULL) {
    str_trim_crlf(substr);
    strcpy(array[i], substr);
    substr = strtok(NULL, seg);
    i++;
  }
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<gtty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    $0x7408,%esi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <ioctl@plt>
  leave
  ret
```

assistant: ```c
int gtty(fildes, argp)
int fildes;
char *argp;
{ return ioctl(fildes, (('t' << 8) | 8), argp); }
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<gtty>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  mov    $0x7408,%esi
  mov    $0x0,%eax
  call   <ioctl@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int gtty(fildes, argp)
int fildes;
char *argp;
{ return ioctl(fildes, (('t' << 8) | 8), argp); }
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<gtty>:
  endbr64
  mov    %rsi,%rdx
  xor    %eax,%eax
  mov    $0x7408,%esi
  jmp    1050 <ioctl@plt>
```

assistant: ```c
int gtty(fildes, argp)
int fildes;
char *argp;
{ return ioctl(fildes, (('t' << 8) | 8), argp); }
```
user: What would the C source look like for the assembly code provided here?

```asm
<gtty>:
  endbr64
  mov    %rsi,%rdx
  xor    %eax,%eax
  mov    $0x7408,%esi
  jmp    1050 <ioctl@plt>
```

assistant: ```c
int gtty(fildes, argp)
int fildes;
char *argp;
{ return ioctl(fildes, (('t' << 8) | 8), argp); }
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %rsi,%r13
  mov    %rdx,%rbp
  lea    D12(%rip),%rdi
  call   <puts@plt>
  test   %r12d,%r12d
  jle    L10
  mov    %r12d,%r12d
  mov    $0x0,%ebx
  lea    D1(%rip),%r14
L1:
  mov    0x0(%r13,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L1
L10:
  mov    0x0(%rbp),%rdx
  test   %rdx,%rdx
  je     L28
  mov    $0x0,%ebx
  lea    D13(%rip),%r12
L33:
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  mov    0x0(%rbp,%rbx,8),%rdx
  test   %rdx,%rdx
  jne    L33
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D12
type: string
name: extract_symbolic_data
arguments: data_tag: D1
type: string
name: extract_symbolic_data
arguments: data_tag: D13
type: string

tool: D12: Content-type: text/plain


tool: D1: argv[%d]: %s


tool: D13: env[%d]: %s


assistant: ```c
void main(int argc, char *argv[], char *envv[]) {
  int i;
  char curdir[BUFSIZ];
  printf("Content-type: text/plain\n\n");
  for (i = 0; i < argc; i++) {
    printf("argv[%d]: %s\n", i, argv[i]);
  }
  for (i = 0; envv[i]; i++) {
    printf("env[%d]: %s\n", i, envv[i]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  movslq %edi,%r12
  lea    D1(%rip),%rdi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  call   <puts@plt>
  test   %r12d,%r12d
  jle    L6
  xor    %ebx,%ebx
  lea    D22(%rip),%r14
  nop
L8:
  mov    0x0(%r13,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L8
L6:
  mov    0x0(%rbp),%rdx
  test   %rdx,%rdx
  je     L32
  xor    %ebx,%ebx
  lea    D28(%rip),%r12
  nopl   0x0(%rax)
L24:
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    0x0(%rbp,%rbx,8),%rdx
  test   %rdx,%rdx
  jne    L24
L32:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: label: D1
type_guessed: string
name: parse_data
arguments: label: D22
type_guessed: string
name: parse_data
arguments: label: D28
type_guessed: string

tool: D1: Content-type: text/plain


tool: D22: argv[%d]: %s


tool: D28: env[%d]: %s


assistant: ```c
void main(int argc, char *argv[], char *envv[]) {
  int i;
  char curdir[BUFSIZ];
  printf("Content-type: text/plain\n\n");
  for (i = 0; i < argc; i++) {
    printf("argv[%d]: %s\n", i, argv[i]);
  }
  for (i = 0; envv[i]; i++) {
    printf("env[%d]: %s\n", i, envv[i]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  movslq %edi,%r12
  lea    D0(%rip),%rdi
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  call   <puts@plt>
  test   %r12d,%r12d
  jle    L23
  xor    %ebx,%ebx
  lea    D26(%rip),%r14
  nop
L36:
  mov    0x0(%r13,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L36
L23:
  mov    0x0(%rbp),%rdx
  test   %rdx,%rdx
  je     L11
  xor    %ebx,%ebx
  lea    D28(%rip),%r12
  nopl   0x0(%rax)
L19:
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    0x0(%rbp,%rbx,8),%rdx
  test   %rdx,%rdx
  jne    L19
L11:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type_infered: string
name: retrieve_data
arguments: data_tag: D26
type_infered: string
name: retrieve_data
arguments: data_tag: D28
type_infered: string

tool: D0: Content-type: text/plain


tool: D26: argv[%d]: %s


tool: D28: env[%d]: %s


assistant: ```c
void main(int argc, char *argv[], char *envv[]) {
  int i;
  char curdir[BUFSIZ];
  printf("Content-type: text/plain\n\n");
  for (i = 0; i < argc; i++) {
    printf("argv[%d]: %s\n", i, argv[i]);
  }
  for (i = 0; envv[i]; i++) {
    printf("env[%d]: %s\n", i, envv[i]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<__stpncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  xor    %rdx,%rax
  and    $0x7,%eax
  test   %rax,%rax
  jne    L117
  jmp    L33
L100:
  subq   $0x1,-0x28(%rbp)
  addq   $0x1,-0x20(%rbp)
  addq   $0x1,-0x18(%rbp)
L33:
  mov    -0x20(%rbp),%rax
  and    $0x7,%eax
  test   %rax,%rax
  je     L114
  cmpq   $0x0,-0x28(%rbp)
  je     L114
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L100
L114:
  cmpq   $0x0,-0x28(%rbp)
  je     L124
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L124
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L136
L119:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,(%rax)
  subq   $0x8,-0x28(%rbp)
  addq   $0x8,-0x8(%rbp)
  addq   $0x8,-0x10(%rbp)
L136:
  cmpq   $0x7,-0x28(%rbp)
  jbe    L60
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rcx
  mov    D3(%rip),%rax
  mov    (%rax),%rsi
  mov    $0xffffffffffffffff,%rax
  mov    $0x0,%edx
  div    %rsi
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  not    %rax
  mov    %rdx,%rcx
  and    %rax,%rcx
  mov    D3(%rip),%rax
  mov    (%rax),%rdi
  mov    $0xffffffffffffffff,%rax
  mov    $0x0,%edx
  div    %rdi
  mov    %rax,%rdx
  mov    D3(%rip),%rax
  mov    (%rax),%rax
  shr    %rax
  add    $0x1,%rax
  imul   %rdx,%rax
  and    %rcx,%rax
  test   %rax,%rax
  je     L119
L60:
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L117
L86:
  subq   $0x1,-0x28(%rbp)
  addq   $0x1,-0x20(%rbp)
  addq   $0x1,-0x18(%rbp)
L117:
  cmpq   $0x0,-0x28(%rbp)
  je     L8
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L86
  jmp    L8
L124:
  nop
  jmp    L101
L8:
  nop
L101:
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *__stpncpy(char *d, const char *s, size_t n) {
  size_t *wd;
  const size_t *ws;
  if (((uintptr_t)s & (sizeof(size_t) - 1)) == ((uintptr_t)d & (sizeof(size_t) - 1))) {
    for (; ((uintptr_t)s & (sizeof(size_t) - 1)) && n && (*d = *s); n--, s++, d++)
      ;
    if (!n || !*s)
      goto tail;
    wd = (void *)d;
    ws = (const void *)s;
    for (; n >= sizeof(size_t) && !((*ws) - ((size_t)-1 / UCHAR_MAX) & ~(*ws) & (((size_t)-1 / UCHAR_MAX) * (UCHAR_MAX / 2 + 1))); n -= sizeof(size_t), ws++, wd++)
      *wd = *ws;
    d = (void *)wd;
    s = (const void *)ws;
  }
  for (; n && (*d = *s); n--, s++, d++)
    ;
tail:
  memset(d, 0, n);
  return d;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__stpncpy>:
  endbr64
  push   %rbx
  mov    %rsi,%rcx
  mov    %rdx,%rsi
  mov    %rcx,%rax
  xor    %rdi,%rax
  test   $0x7,%al
  jne    L77
  test   $0x7,%cl
  je     L60
  test   %rdx,%rdx
  je     L60
L21:
  movzbl (%rcx),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L60
  sub    $0x1,%rsi
  add    $0x1,%rcx
  add    $0x1,%rdi
  test   $0x7,%cl
  je     L60
  test   %rsi,%rsi
  jne    L21
L60:
  mov    %rdi,%rbx
  test   %rsi,%rsi
  je     L88
  cmpb   $0x0,(%rcx)
  je     L88
  cmp    $0x7,%rsi
  jbe    L126
  mov    D4(%rip),%r11
  mov    $0xffffffffffffffff,%r10
L118:
  mov    (%rcx),%r9
  mov    (%r11),%r8
  mov    %r10,%rax
  mov    $0x0,%edx
  div    %r8
  mov    %r9,%rdx
  sub    %rax,%rdx
  mov    %r9,%rbx
  not    %rbx
  and    %rbx,%rdx
  shr    %r8
  add    $0x1,%r8
  imul   %rax,%r8
  test   %r8,%rdx
  jne    L77
  mov    %r9,(%rdi)
  sub    $0x8,%rsi
  add    $0x8,%rcx
  add    $0x8,%rdi
  cmp    $0x7,%rsi
  ja     L118
L77:
  mov    %rdi,%rbx
  test   %rsi,%rsi
  je     L88
L126:
  mov    %rdi,%rbx
L49:
  movzbl (%rcx),%eax
  mov    %al,(%rbx)
  test   %al,%al
  je     L88
  add    $0x1,%rcx
  add    $0x1,%rbx
  sub    $0x1,%rsi
  jne    L49
L88:
  mov    %rsi,%rdx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *__stpncpy(char *d, const char *s, size_t n) {
  size_t *wd;
  const size_t *ws;
  if (((uintptr_t)s & (sizeof(size_t) - 1)) == ((uintptr_t)d & (sizeof(size_t) - 1))) {
    for (; ((uintptr_t)s & (sizeof(size_t) - 1)) && n && (*d = *s); n--, s++, d++)
      ;
    if (!n || !*s)
      goto tail;
    wd = (void *)d;
    ws = (const void *)s;
    for (; n >= sizeof(size_t) && !((*ws) - ((size_t)-1 / UCHAR_MAX) & ~(*ws) & (((size_t)-1 / UCHAR_MAX) * (UCHAR_MAX / 2 + 1))); n -= sizeof(size_t), ws++, wd++)
      *wd = *ws;
    d = (void *)wd;
    s = (const void *)ws;
  }
  for (; n && (*d = *s); n--, s++, d++)
    ;
tail:
  memset(d, 0, n);
  return d;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__stpncpy>:
  endbr64
  mov    %rsi,%rax
  push   %r12
  mov    %rsi,%rcx
  mov    %rdx,%r8
  xor    %rdi,%rax
  test   $0x7,%al
  je     L68
L117:
  test   %r8,%r8
  je     L147
L93:
  mov    %rdi,%r12
  jmp    L73
  nopl   0x0(%rax)
L92:
  add    $0x1,%rcx
  add    $0x1,%r12
  sub    $0x1,%r8
  je     L59
L73:
  movzbl (%rcx),%eax
  mov    %al,(%r12)
  test   %al,%al
  jne    L92
L59:
  mov    %r12,%rdi
  mov    %r8,%rdx
  xor    %esi,%esi
  call   <memset@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L47:
  test   %r8,%r8
  je     L147
  movzbl (%rcx),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L116
  sub    $0x1,%r8
  add    $0x1,%rcx
  add    $0x1,%rdi
L68:
  test   $0x7,%cl
  jne    L47
  test   %r8,%r8
  je     L147
L116:
  cmpb   $0x0,(%rcx)
  mov    %rdi,%r12
  je     L59
  cmp    $0x7,%r8
  jbe    L93
  mov    D7(%rip),%r11
  mov    $0xffffffffffffffff,%r10
  jmp    L86
  xchg   %ax,%ax
L25:
  sub    $0x8,%r8
  mov    %r9,(%rdi)
  add    $0x8,%rcx
  add    $0x8,%rdi
  cmp    $0x7,%r8
  jbe    L117
L86:
  mov    (%r11),%rsi
  mov    %r10,%rax
  xor    %edx,%edx
  mov    (%rcx),%r9
  div    %rsi
  shr    %rsi
  mov    %r9,%rdx
  mov    %r9,%r12
  add    $0x1,%rsi
  not    %r12
  imul   %rax,%rsi
  sub    %rax,%rdx
  and    %r12,%rdx
  test   %rsi,%rdx
  je     L25
  jmp    L93
  nopl   0x0(%rax,%rax,1)
L147:
  mov    %rdi,%r12
  jmp    L59
```

assistant: ```c
char *__stpncpy(char *d, const char *s, size_t n) {
  size_t *wd;
  const size_t *ws;
  if (((uintptr_t)s & (sizeof(size_t) - 1)) == ((uintptr_t)d & (sizeof(size_t) - 1))) {
    for (; ((uintptr_t)s & (sizeof(size_t) - 1)) && n && (*d = *s); n--, s++, d++)
      ;
    if (!n || !*s)
      goto tail;
    wd = (void *)d;
    ws = (const void *)s;
    for (; n >= sizeof(size_t) && !((*ws) - ((size_t)-1 / UCHAR_MAX) & ~(*ws) & (((size_t)-1 / UCHAR_MAX) * (UCHAR_MAX / 2 + 1))); n -= sizeof(size_t), ws++, wd++)
      *wd = *ws;
    d = (void *)wd;
    s = (const void *)ws;
  }
  for (; n && (*d = *s); n--, s++, d++)
    ;
tail:
  memset(d, 0, n);
  return d;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<__stpncpy>:
  endbr64
  mov    %rsi,%rax
  push   %r12
  mov    %rsi,%rcx
  mov    %rdx,%r8
  xor    %rdi,%rax
  test   $0x7,%al
  je     L51
L33:
  test   %r8,%r8
  je     L0
L116:
  mov    %rdi,%r12
  jmp    L26
  nopl   0x0(%rax)
L17:
  add    $0x1,%rcx
  add    $0x1,%r12
  sub    $0x1,%r8
  je     L94
L26:
  movzbl (%rcx),%eax
  mov    %al,(%r12)
  test   %al,%al
  jne    L17
L94:
  mov    %r12,%rdi
  mov    %r8,%rdx
  xor    %esi,%esi
  call   <memset@plt>
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L143:
  test   %r8,%r8
  je     L0
  movzbl (%rcx),%eax
  mov    %al,(%rdi)
  test   %al,%al
  je     L75
  sub    $0x1,%r8
  add    $0x1,%rcx
  add    $0x1,%rdi
L51:
  test   $0x7,%cl
  jne    L143
  test   %r8,%r8
  je     L0
L75:
  cmpb   $0x0,(%rcx)
  mov    %rdi,%r12
  je     L94
  cmp    $0x7,%r8
  jbe    L116
  mov    D6(%rip),%r11
  mov    $0xffffffffffffffff,%r10
  jmp    L55
  xchg   %ax,%ax
L136:
  sub    $0x8,%r8
  mov    %r9,(%rdi)
  add    $0x8,%rcx
  add    $0x8,%rdi
  cmp    $0x7,%r8
  jbe    L33
L55:
  mov    (%r11),%rsi
  mov    %r10,%rax
  xor    %edx,%edx
  mov    (%rcx),%r9
  div    %rsi
  shr    %rsi
  mov    %r9,%rdx
  mov    %r9,%r12
  add    $0x1,%rsi
  not    %r12
  imul   %rax,%rsi
  sub    %rax,%rdx
  and    %r12,%rdx
  test   %rsi,%rdx
  je     L136
  jmp    L116
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %rdi,%r12
  jmp    L94
```

assistant: ```c
char *__stpncpy(char *d, const char *s, size_t n) {
  size_t *wd;
  const size_t *ws;
  if (((uintptr_t)s & (sizeof(size_t) - 1)) == ((uintptr_t)d & (sizeof(size_t) - 1))) {
    for (; ((uintptr_t)s & (sizeof(size_t) - 1)) && n && (*d = *s); n--, s++, d++)
      ;
    if (!n || !*s)
      goto tail;
    wd = (void *)d;
    ws = (const void *)s;
    for (; n >= sizeof(size_t) && !((*ws) - ((size_t)-1 / UCHAR_MAX) & ~(*ws) & (((size_t)-1 / UCHAR_MAX) * (UCHAR_MAX / 2 + 1))); n -= sizeof(size_t), ws++, wd++)
      *wd = *ws;
    d = (void *)wd;
    s = (const void *)ws;
  }
  for (; n && (*d = *s); n--, s++, d++)
    ;
tail:
  memset(d, 0, n);
  return d;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <tcgetattr@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    D36(%rip),%rax
  mov    (%rax),%eax
  not    %eax
  and    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  not    %eax
  and    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  lea    -0x10(%rbp),%rdx
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <tcsetattr@plt>
  lea    -0x9(%rbp),%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  lea    -0x14(%rbp),%rdx
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <tcsetattr@plt>
  movzbl -0x9(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
char getch(void) {
  char chbuf[1];
  struct termios oldstate, newstate;
  tcgetattr(0, &oldstate);
  newstate = oldstate;
  newstate.c_lflag &= ~ICANON;
  newstate.c_lflag &= ~ECHO;
  tcsetattr(0, TCSANOW, &newstate);
  read(0, &chbuf, 1);
  tcsetattr(0, TCSANOW, &oldstate);
  return chbuf[0];
}
```
user: Translate the below assembly code into C programming language:

```asm
<getch>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbp
  mov    %rbp,%rsi
  mov    $0x0,%edi
  call   <tcgetattr@plt>
  mov    0xc(%rsp),%eax
  mov    %eax,0x10(%rsp)
  mov    D10(%rip),%rax
  mov    (%rax),%eax
  mov    D1(%rip),%rdx
  or     (%rdx),%eax
  not    %eax
  and    %eax,0x10(%rsp)
  lea    0x10(%rsp),%rdx
  mov    D0(%rip),%rbx
  mov    (%rbx),%esi
  mov    $0x0,%edi
  call   <tcsetattr@plt>
  lea    0x17(%rsp),%rsi
  mov    $0x1,%edx
  mov    $0x0,%edi
  call   <read@plt>
  mov    %rbp,%rdx
  mov    (%rbx),%esi
  mov    $0x0,%edi
  call   <tcsetattr@plt>
  movzbl 0x17(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char getch(void) {
  char chbuf[1];
  struct termios oldstate, newstate;
  tcgetattr(0, &oldstate);
  newstate = oldstate;
  newstate.c_lflag &= ~ICANON;
  newstate.c_lflag &= ~ECHO;
  tcsetattr(0, TCSANOW, &newstate);
  read(0, &chbuf, 1);
  tcsetattr(0, TCSANOW, &oldstate);
  return chbuf[0];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getch>:
  endbr64
  push   %r12
  xor    %edi,%edi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r12
  mov    %r12,%rsi
  call   <tcgetattr@plt>
  mov    0xc(%rsp),%eax
  mov    D26(%rip),%rbx
  xor    %edi,%edi
  mov    D14(%rip),%rdx
  mov    %eax,0x10(%rsp)
  mov    D19(%rip),%rax
  mov    (%rbx),%esi
  mov    (%rax),%eax
  or     (%rdx),%eax
  lea    0x10(%rsp),%rdx
  not    %eax
  and    %eax,0x10(%rsp)
  call   <tcsetattr@plt>
  lea    0x17(%rsp),%rsi
  mov    $0x1,%edx
  xor    %edi,%edi
  call   <read@plt>
  mov    (%rbx),%esi
  mov    %r12,%rdx
  xor    %edi,%edi
  call   <tcsetattr@plt>
  movzbl 0x17(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  pop    %rbx
  pop    %r12
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char getch(void) {
  char chbuf[1];
  struct termios oldstate, newstate;
  tcgetattr(0, &oldstate);
  newstate = oldstate;
  newstate.c_lflag &= ~ICANON;
  newstate.c_lflag &= ~ECHO;
  tcsetattr(0, TCSANOW, &newstate);
  read(0, &chbuf, 1);
  tcsetattr(0, TCSANOW, &oldstate);
  return chbuf[0];
}
```