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
<split1>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x7000(%rsp),%r11
L44:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L44
  sub    $0x540,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r13
  mov    %edx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x7538(%rsp)
  xor    %eax,%eax
  lea    0x2710(%rsp),%rdi
  mov    $0x2710,%edx
  mov    %r14,%rsi
  call   <strncpy@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L58
  mov    %eax,%r12d
  mov    %eax,%ecx
  mov    $0x0,%eax
  lea    0x2710(%rsp),%rdx
L67:
  mov    %eax,%ebp
  cmp    %bl,(%rax,%rdx,1)
  je     L66
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L67
L58:
  mov    %r14,%rdi
  call   <strdup@plt>
  mov    %rax,0x0(%r13)
  movq   $0x0,0x8(%r13)
  mov    $0x1,%eax
L10:
  mov    0x7538(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x7540,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L66:
  lea    0x2710(%rsp),%rsi
  lea    0x4e20(%rsp),%rbx
  mov    %eax,%edx
  mov    %rbx,%rdi
  call   <strncpy@plt>
  movslq %ebp,%rax
  movb   $0x0,0x4e20(%rsp,%rax,1)
  mov    %rbx,%rdi
  call   <strdup@plt>
  mov    %rax,0x0(%r13)
  lea    0x1(%rbp),%eax
  sub    %eax,%r12d
  mov    $0x1,%eax
  test   %r12d,%r12d
  jle    L10
  movslq %ebp,%rbp
  lea    0x2711(%rsp,%rbp,1),%rsi
  mov    %rsp,%rbx
  mov    %r12d,%edx
  mov    %rbx,%rdi
  call   <strncpy@plt>
  movslq %r12d,%r12
  movb   $0x0,(%rsp,%r12,1)
  mov    %rbx,%rdi
  call   <strdup@plt>
  mov    %rax,0x8(%r13)
  mov    $0x2,%eax
  jmp    L10
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int split1(char *strin, char *strpt[], char splitc) {
  char rest[10000], str[10000], ww[10000];
  int len, i, l;
  strncpy(str, strin, 10000);
  len = strlen(strin);
  for (i = 0; i < len; i++) {
    if (str[i] == splitc) {
      l = i;
      strncpy(ww, str, l);
      ww[l] = '\0';
      strpt[0] = strdup(ww);
      l = len - (i + 1);
      if (l <= 0)
        return 1;
      strncpy(rest, str + i + 1, l);
      rest[l] = '\0';
      strpt[1] = strdup(rest);
      return 2;
    }
  }
  strpt[0] = strdup(strin);
  strpt[1] = NULL;
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<split1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x7000(%rsp),%r11
L63:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L63
  sub    $0x558,%rsp
  mov    %rsi,0x8(%rsp)
  mov    %rdi,%r12
  mov    %rdi,%rsi
  mov    %edx,%r13d
  lea    0x2720(%rsp),%r15
  mov    $0x2710,%edx
  mov    %r15,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x7548(%rsp)
  xor    %eax,%eax
  call   <strncpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L50
  mov    %eax,%ebp
  xor    %ebx,%ebx
  mov    %r15,%rax
  jmp    L36
  nopl   (%rax)
L48:
  add    $0x1,%rax
  cmp    %ebx,%ebp
  je     L50
L36:
  movslq %ebx,%r14
  add    $0x1,%ebx
  cmp    %r13b,(%rax)
  jne    L48
  lea    0x4e30(%rsp),%r13
  mov    %r14d,%edx
  mov    %r15,%rsi
  sub    %ebx,%ebp
  mov    %r13,%rdi
  call   <strncpy@plt>
  mov    %r13,%rdi
  movb   $0x0,0x4e30(%rsp,%r14,1)
  call   <strdup@plt>
  mov    0x8(%rsp),%rcx
  mov    %rax,(%rcx)
  mov    $0x1,%eax
  test   %ebp,%ebp
  jle    L79
  lea    0x10(%rsp),%r13
  movslq %ebx,%rbx
  mov    %ebp,%edx
  movslq %ebp,%rbp
  lea    (%r15,%rbx,1),%rsi
  mov    %r13,%rdi
  call   <strncpy@plt>
  mov    %r13,%rdi
  movb   $0x0,0x10(%rsp,%rbp,1)
  call   <strdup@plt>
  mov    0x8(%rsp),%rcx
  mov    %rax,0x8(%rcx)
  mov    $0x2,%eax
  jmp    L79
  nopw   0x0(%rax,%rax,1)
L50:
  mov    %r12,%rdi
  call   <strdup@plt>
  mov    0x8(%rsp),%rcx
  movq   $0x0,0x8(%rcx)
  mov    %rax,(%rcx)
  mov    $0x1,%eax
L79:
  mov    0x7548(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L68
  add    $0x7558,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L68:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int split1(char *strin, char *strpt[], char splitc) {
  char rest[10000], str[10000], ww[10000];
  int len, i, l;
  strncpy(str, strin, 10000);
  len = strlen(strin);
  for (i = 0; i < len; i++) {
    if (str[i] == splitc) {
      l = i;
      strncpy(ww, str, l);
      ww[l] = '\0';
      strpt[0] = strdup(ww);
      l = len - (i + 1);
      if (l <= 0)
        return 1;
      strncpy(rest, str + i + 1, l);
      rest[l] = '\0';
      strpt[1] = strdup(rest);
      return 2;
    }
  }
  strpt[0] = strdup(strin);
  strpt[1] = NULL;
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<split1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x7000(%rsp),%r11
L70:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L70
  sub    $0x558,%rsp
  mov    %rsi,0x8(%rsp)
  mov    %rdi,%r12
  mov    %rdi,%rsi
  mov    %edx,%r13d
  lea    0x2720(%rsp),%r15
  mov    $0x2710,%edx
  mov    %r15,%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x7548(%rsp)
  xor    %eax,%eax
  call   <strncpy@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L44
  mov    %eax,%ebp
  xor    %ebx,%ebx
  mov    %r15,%rax
  jmp    L75
  nopl   (%rax)
L77:
  add    $0x1,%rax
  cmp    %ebx,%ebp
  je     L44
L75:
  movslq %ebx,%r14
  add    $0x1,%ebx
  cmp    %r13b,(%rax)
  jne    L77
  lea    0x4e30(%rsp),%r13
  mov    %r14d,%edx
  mov    %r15,%rsi
  sub    %ebx,%ebp
  mov    %r13,%rdi
  call   <strncpy@plt>
  mov    %r13,%rdi
  movb   $0x0,0x4e30(%rsp,%r14,1)
  call   <strdup@plt>
  mov    0x8(%rsp),%rcx
  mov    %rax,(%rcx)
  mov    $0x1,%eax
  test   %ebp,%ebp
  jle    L4
  lea    0x10(%rsp),%r13
  movslq %ebx,%rbx
  mov    %ebp,%edx
  movslq %ebp,%rbp
  lea    (%r15,%rbx,1),%rsi
  mov    %r13,%rdi
  call   <strncpy@plt>
  mov    %r13,%rdi
  movb   $0x0,0x10(%rsp,%rbp,1)
  call   <strdup@plt>
  mov    0x8(%rsp),%rcx
  mov    %rax,0x8(%rcx)
  mov    $0x2,%eax
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L44:
  mov    %r12,%rdi
  call   <strdup@plt>
  mov    0x8(%rsp),%rcx
  movq   $0x0,0x8(%rcx)
  mov    %rax,(%rcx)
  mov    $0x1,%eax
L4:
  mov    0x7548(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L46
  add    $0x7558,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int split1(char *strin, char *strpt[], char splitc) {
  char rest[10000], str[10000], ww[10000];
  int len, i, l;
  strncpy(str, strin, 10000);
  len = strlen(strin);
  for (i = 0; i < len; i++) {
    if (str[i] == splitc) {
      l = i;
      strncpy(ww, str, l);
      ww[l] = '\0';
      strpt[0] = strdup(ww);
      l = len - (i + 1);
      if (l <= 0)
        return 1;
      strncpy(rest, str + i + 1, l);
      rest[l] = '\0';
      strpt[1] = strdup(rest);
      return 2;
    }
  }
  strpt[0] = strdup(strin);
  strpt[1] = NULL;
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<MapFileUnlink>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x128,%rsp
  mov    %rdi,-0x128(%rbp)
  mov    %rsi,-0x130(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    -0x128(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    -0x130(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %ebx,%eax
  cmp    $0x103,%eax
  jg     L5
  mov    -0x130(%rbp),%rcx
  mov    -0x128(%rbp),%rdx
  lea    -0x120(%rbp),%rax
  lea    D6(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x120(%rbp),%rax
  mov    %rax,%rdi
  call   <unlink@plt>
  jmp    L8
L5:
  nop
L8:
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D6
type_guessed: string

tool: D6: %s%s

assistant: ```c
void MapFileUnlink(const char *rootpath, const char *filename) {
  char fullname[260];
  if (strlen(rootpath) + strlen(filename) >= 260)
    return;
  sprintf(fullname, "%s%s", rootpath, filename);
  unlink(fullname);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<MapFileUnlink>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%ebx
  mov    %r12,%rdi
  call   <strlen@plt>
  add    %eax,%ebx
  cmp    $0x103,%ebx
  jle    L26
L13:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  mov    %rsp,%rbx
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    D0(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <unlink@plt>
  jmp    L13
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D0
type: string

tool: D0: %s%s

assistant: ```c
void MapFileUnlink(const char *rootpath, const char *filename) {
  char fullname[260];
  if (strlen(rootpath) + strlen(filename) >= 260)
    return;
  sprintf(fullname, "%s%s", rootpath, filename);
  unlink(fullname);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<MapFileUnlink>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  add    %eax,%ebx
  cmp    $0x103,%ebx
  jg     L12
  mov    %rsp,%rbp
  mov    %r13,%rcx
  lea    D5(%rip),%rsi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <unlink@plt>
L12:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D5
type: string

tool: D5: %s%s

assistant: ```c
void MapFileUnlink(const char *rootpath, const char *filename) {
  char fullname[260];
  if (strlen(rootpath) + strlen(filename) >= 260)
    return;
  sprintf(fullname, "%s%s", rootpath, filename);
  unlink(fullname);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<MapFileUnlink>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  add    %eax,%ebx
  cmp    $0x103,%ebx
  jg     L2
  mov    %rsp,%rbp
  mov    %r13,%rcx
  lea    D5(%rip),%rsi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <unlink@plt>
L2:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D5
datatype: string

tool: D5: %s%s

assistant: ```c
void MapFileUnlink(const char *rootpath, const char *filename) {
  char fullname[260];
  if (strlen(rootpath) + strlen(filename) >= 260)
    return;
  sprintf(fullname, "%s%s", rootpath, filename);
  unlink(fullname);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <initscr@plt>
  mov    $0x0,%eax
  call   <cbreak@plt>
  mov    $0x0,%eax
  call   <noecho@plt>
  mov    D31(%rip),%rsi
  mov    D20(%rip),%rax
  mov    (%rax),%edi
  call   <signal@plt>
  mov    D51(%rip),%rax
  movswl (%rax),%r12d
  mov    D4(%rip),%rax
  mov    (%rax),%ebx
  sub    $0x4,%ebx
  movswl %bx,%ebx
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %r12d,%esi
  mov    %ebx,%edi
  call   <newwin@plt>
  mov    %rax,%rbp
  mov    $0x0,%ecx
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    $0x4,%edi
  call   <newwin@plt>
  mov    %rax,%rbx
  mov    $0x2a,%edx
  mov    $0x2a,%esi
  mov    %rbp,%rdi
  call   <box@plt>
  mov    $0x2b,%edx
  mov    $0x2b,%esi
  mov    %rbx,%rdi
  call   <box@plt>
  mov    %rbp,%rdi
  call   <touchwin@plt>
  mov    %rbp,%rdi
  call   <wrefresh@plt>
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
  mov    D52(%rip),%rax
  movswl (%rax),%esi
  mov    %rbx,%rdi
  call   <scrollok@plt>
  mov    D64(%rip),%rax
  movzwl (%rax),%ebp
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <wmove@plt>
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
  test   %bp,%bp
  je     L39
  jmp    L35
L26:
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <waddch@plt>
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
L39:
  mov    $0x0,%eax
  call   <getch@plt>
  cmp    $0x3,%eax
  jne    L26
  mov    D52(%rip),%rax
  movzwl (%rax),%ebp
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
  test   %bp,%bp
  je     L39
L35:
  mov    $0x0,%eax
  call   <endwin@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
main() {
  WINDOW *uwin, *lwin;
  short unl, unc, ubr, ubc;
  short lnl, lnc, lbr, lbc;
  char inbuf[100];
  short finished;
  int c;
  void done();
  initscr();
  cbreak();
  noecho();
  signal(SIGINT, done);
  ubr = 0;
  ubc = 0;
  unc = COLS;
  unl = (LINES - 4);
  lbr = LINES - 4;
  lbc = 0;
  lnc = COLS;
  lnl = 4;
  uwin = newwin(unl, unc, ubr, ubc);
  lwin = newwin(lnl, lnc, lbr, lbc);
  box(uwin, '*', '*');
  box(lwin, '+', '+');
  touchwin(uwin);
  wrefresh(uwin);
  touchwin(lwin);
  wrefresh(lwin);
  scrollok(lwin, TRUE);
  finished = FALSE;
  wmove(lwin, 1, 1);
  touchwin(lwin);
  wrefresh(lwin);
  while (!finished) {
    c = getch();
    switch (c) {
    case ('c' & 037):
      finished = TRUE;
      break;
    default:
      waddch(lwin, c);
      break;
    }
    touchwin(lwin);
    wrefresh(lwin);
  }
  endwin();
  exit(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <initscr@plt>
  xor    %eax,%eax
  call   <cbreak@plt>
  xor    %eax,%eax
  call   <noecho@plt>
  mov    D53(%rip),%rax
  mov    D55(%rip),%rsi
  mov    (%rax),%edi
  call   <signal@plt>
  mov    D33(%rip),%rax
  xor    %ecx,%ecx
  xor    %edx,%edx
  movswl (%rax),%r13d
  mov    D5(%rip),%rax
  mov    (%rax),%r12d
  mov    %r13d,%esi
  sub    $0x4,%r12d
  movswl %r12w,%r12d
  mov    %r12d,%edi
  call   <newwin@plt>
  xor    %ecx,%ecx
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    $0x4,%edi
  mov    %rax,%rbp
  call   <newwin@plt>
  mov    %rbp,%rdi
  mov    $0x2a,%edx
  mov    $0x2a,%esi
  mov    %rax,%rbx
  call   <box@plt>
  mov    $0x2b,%edx
  mov    $0x2b,%esi
  mov    %rbx,%rdi
  call   <box@plt>
  mov    %rbp,%rdi
  call   <touchwin@plt>
  mov    %rbp,%rdi
  call   <wrefresh@plt>
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
  mov    D18(%rip),%r12
  mov    %rbx,%rdi
  movswl (%r12),%esi
  call   <scrollok@plt>
  mov    D37(%rip),%rax
  mov    $0x1,%edx
  mov    %rbx,%rdi
  mov    $0x1,%esi
  movzwl (%rax),%ebp
  call   <wmove@plt>
L23:
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
  test   %bp,%bp
  je     L0
  xor    %eax,%eax
  call   <endwin@plt>
  xor    %edi,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  mov    %rbx,%rdi
  call   <waddch@plt>
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
L0:
  xor    %eax,%eax
  call   <getch@plt>
  mov    %eax,%esi
  cmp    $0x3,%eax
  jne    L24
  movzwl (%r12),%ebp
  jmp    L23
```

assistant: ```c
main() {
  WINDOW *uwin, *lwin;
  short unl, unc, ubr, ubc;
  short lnl, lnc, lbr, lbc;
  char inbuf[100];
  short finished;
  int c;
  void done();
  initscr();
  cbreak();
  noecho();
  signal(SIGINT, done);
  ubr = 0;
  ubc = 0;
  unc = COLS;
  unl = (LINES - 4);
  lbr = LINES - 4;
  lbc = 0;
  lnc = COLS;
  lnl = 4;
  uwin = newwin(unl, unc, ubr, ubc);
  lwin = newwin(lnl, lnc, lbr, lbc);
  box(uwin, '*', '*');
  box(lwin, '+', '+');
  touchwin(uwin);
  wrefresh(uwin);
  touchwin(lwin);
  wrefresh(lwin);
  scrollok(lwin, TRUE);
  finished = FALSE;
  wmove(lwin, 1, 1);
  touchwin(lwin);
  wrefresh(lwin);
  while (!finished) {
    c = getch();
    switch (c) {
    case ('c' & 037):
      finished = TRUE;
      break;
    default:
      waddch(lwin, c);
      break;
    }
    touchwin(lwin);
    wrefresh(lwin);
  }
  endwin();
  exit(0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <initscr@plt>
  xor    %eax,%eax
  call   <cbreak@plt>
  xor    %eax,%eax
  call   <noecho@plt>
  mov    D12(%rip),%rax
  mov    D54(%rip),%rsi
  mov    (%rax),%edi
  call   <signal@plt>
  mov    D35(%rip),%rax
  xor    %ecx,%ecx
  xor    %edx,%edx
  movswl (%rax),%r13d
  mov    D22(%rip),%rax
  mov    (%rax),%r12d
  mov    %r13d,%esi
  sub    $0x4,%r12d
  movswl %r12w,%r12d
  mov    %r12d,%edi
  call   <newwin@plt>
  xor    %ecx,%ecx
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    $0x4,%edi
  mov    %rax,%rbp
  call   <newwin@plt>
  mov    %rbp,%rdi
  mov    $0x2a,%edx
  mov    $0x2a,%esi
  mov    %rax,%rbx
  call   <box@plt>
  mov    $0x2b,%edx
  mov    $0x2b,%esi
  mov    %rbx,%rdi
  call   <box@plt>
  mov    %rbp,%rdi
  call   <touchwin@plt>
  mov    %rbp,%rdi
  call   <wrefresh@plt>
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
  mov    D15(%rip),%r12
  mov    %rbx,%rdi
  movswl (%r12),%esi
  call   <scrollok@plt>
  mov    D1(%rip),%rax
  mov    $0x1,%edx
  mov    %rbx,%rdi
  mov    $0x1,%esi
  movzwl (%rax),%ebp
  call   <wmove@plt>
L1:
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
  test   %bp,%bp
  je     L27
  xor    %eax,%eax
  call   <endwin@plt>
  xor    %edi,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  mov    %rbx,%rdi
  call   <waddch@plt>
  mov    %rbx,%rdi
  call   <touchwin@plt>
  mov    %rbx,%rdi
  call   <wrefresh@plt>
L27:
  xor    %eax,%eax
  call   <getch@plt>
  mov    %eax,%esi
  cmp    $0x3,%eax
  jne    L25
  movzwl (%r12),%ebp
  jmp    L1
```

assistant: ```c
main() {
  WINDOW *uwin, *lwin;
  short unl, unc, ubr, ubc;
  short lnl, lnc, lbr, lbc;
  char inbuf[100];
  short finished;
  int c;
  void done();
  initscr();
  cbreak();
  noecho();
  signal(SIGINT, done);
  ubr = 0;
  ubc = 0;
  unc = COLS;
  unl = (LINES - 4);
  lbr = LINES - 4;
  lbc = 0;
  lnc = COLS;
  lnl = 4;
  uwin = newwin(unl, unc, ubr, ubc);
  lwin = newwin(lnl, lnc, lbr, lbc);
  box(uwin, '*', '*');
  box(lwin, '+', '+');
  touchwin(uwin);
  wrefresh(uwin);
  touchwin(lwin);
  wrefresh(lwin);
  scrollok(lwin, TRUE);
  finished = FALSE;
  wmove(lwin, 1, 1);
  touchwin(lwin);
  wrefresh(lwin);
  while (!finished) {
    c = getch();
    switch (c) {
    case ('c' & 037):
      finished = TRUE;
      break;
    default:
      waddch(lwin, c);
      break;
    }
    touchwin(lwin);
    wrefresh(lwin);
  }
  endwin();
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ft_add_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x3,-0x8(%rbp)
  movl   $0x2,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <ft_is_prime@plt>
  cmp    $0x1,%eax
  jne    L8
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L8
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x2,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int ft_add_prime(int nb) {
  int sum;
  int i;
  i = 3;
  sum = 2;
  if (ft_is_prime(i) == 1 && i <= nb) {
    sum = sum + i;
    i = i + 2;
  }
  return (i);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_add_prime>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x3,%edi
  call   <ft_is_prime@plt>
  cmp    $0x2,%ebx
  jle    L5
  cmp    $0x1,%eax
  je     L15
L5:
  mov    $0x3,%eax
L3:
  pop    %rbx
  ret
L15:
  mov    $0x5,%eax
  jmp    L3
```

assistant: ```c
int ft_add_prime(int nb) {
  int sum;
  int i;
  i = 3;
  sum = 2;
  if (ft_is_prime(i) == 1 && i <= nb) {
    sum = sum + i;
    i = i + 2;
  }
  return (i);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_add_prime>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x3,%edi
  call   <ft_is_prime@plt>
  cmp    $0x2,%ebx
  jle    L8
  cmp    $0x1,%eax
  je     L0
L8:
  mov    $0x3,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    $0x5,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ft_add_prime(int nb) {
  int sum;
  int i;
  i = 3;
  sum = 2;
  if (ft_is_prime(i) == 1 && i <= nb) {
    sum = sum + i;
    i = i + 2;
  }
  return (i);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_add_prime>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x3,%edi
  call   <ft_is_prime@plt>
  cmp    $0x2,%ebx
  jle    L10
  cmp    $0x1,%eax
  je     L17
L10:
  mov    $0x3,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  mov    $0x5,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ft_add_prime(int nb) {
  int sum;
  int i;
  i = 3;
  sum = 2;
  if (ft_is_prime(i) == 1 && i <= nb) {
    sum = sum + i;
    i = i + 2;
  }
  return (i);
}
```
user: What is the c source code of the assembly code below:

```asm
<command_statistics>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <rule_statistics@plt>
  nop
  leave
  ret
```

assistant: ```c
void command_statistics(int id, char *input) {
  (void)id;
  rule_statistics(input);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<command_statistics>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdi
  call   <rule_statistics@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void command_statistics(int id, char *input) {
  (void)id;
  rule_statistics(input);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<command_statistics>:
  endbr64
  mov    %rsi,%rdi
  jmp    1050 <rule_statistics@plt>
```

assistant: ```c
void command_statistics(int id, char *input) {
  (void)id;
  rule_statistics(input);
}
```
user: What is the c source code of the assembly code below?

```asm
<command_statistics>:
  endbr64
  mov    %rsi,%rdi
  jmp    1050 <rule_statistics@plt>
```

assistant: ```c
void command_statistics(int id, char *input) {
  (void)id;
  rule_statistics(input);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<memory_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L10:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void memory_copy(char *src, char *dst, int num_bytes) {
  int i;
  for (i = 0; i < num_bytes; i++)
    dst[i] = src[i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memory_copy>:
  endbr64
  test   %edx,%edx
  jle    L1
  mov    %edx,%edx
  mov    $0x0,%eax
L17:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L17
L1:
  ret
```

assistant: ```c
void memory_copy(char *src, char *dst, int num_bytes) {
  int i;
  for (i = 0; i < num_bytes; i++)
    dst[i] = src[i];
}
```