ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strfilter2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x68(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x50(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L62:
  cmp    %rdx,%rsp
  je     L21
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L62
L21:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L68
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L68:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x58(%rbp)
  jmp    L67
L26:
  movsbl -0x59(%rbp),%edx
  mov    -0x70(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x40(%rbp)
  cmpq   $0x0,-0x40(%rbp)
  je     L67
  mov    -0x58(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x58(%rbp)
  movzbl -0x59(%rbp),%edx
  mov    %dl,(%rax)
L67:
  mov    -0x68(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x68(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x59(%rbp)
  cmpb   $0x0,-0x59(%rbp)
  jne    L26
  mov    -0x58(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rbx,%rsp
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
char *strfilter2(char *string, char *letters) {
  char buffer[strlen(string) + 1];
  char *dest = buffer;
  char c;
  while (c = *string++) {
    char *ptr = strchr(letters, c);
    if (ptr) {
      *dest++ = c;
    }
  }
  *dest = '\0';
  return strdup(buffer);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strfilter2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r14
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  add    $0x1,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L66:
  cmp    %rdx,%rsp
  je     L50
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L66
L50:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L71
  orq    $0x0,-0x8(%rsp,%rax,1)
L71:
  mov    %rsp,%r15
  lea    0x1(%rbx),%r12
  movzbl (%rbx),%ebx
  test   %bl,%bl
  je     L12
  mov    %r15,%r13
  jmp    L18
L31:
  mov    %bl,0x0(%r13)
  lea    0x1(%r13),%r13
L68:
  add    $0x1,%r12
  movzbl -0x1(%r12),%ebx
  test   %bl,%bl
  je     L48
L18:
  movsbl %bl,%esi
  mov    %r14,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L31
  jmp    L68
L12:
  mov    %r15,%r13
L48:
  movb   $0x0,0x0(%r13)
  mov    %r15,%rdi
  call   <strdup@plt>
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L62
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L62:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *strfilter2(char *string, char *letters) {
  char buffer[strlen(string) + 1];
  char *dest = buffer;
  char c;
  while (c = *string++) {
    char *ptr = strchr(letters, c);
    if (ptr) {
      *dest++ = c;
    }
  }
  *dest = '\0';
  return strdup(buffer);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strfilter2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  add    $0x1,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L37
L64:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L64
L37:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L76
L8:
  lea    0x1(%rbx),%r12
  movzbl (%rbx),%ebx
  mov    %rsp,%r15
  mov    %rsp,%r13
  test   %bl,%bl
  je     L15
  nopl   0x0(%rax)
L54:
  movsbl %bl,%esi
  mov    %r14,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L44
  mov    %bl,0x0(%r13)
  add    $0x1,%r13
L44:
  movzbl (%r12),%ebx
  add    $0x1,%r12
  test   %bl,%bl
  jne    L54
L15:
  mov    %r15,%rdi
  movb   $0x0,0x0(%r13)
  call   <strdup@plt>
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L76:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L8
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *strfilter2(char *string, char *letters) {
  char buffer[strlen(string) + 1];
  char *dest = buffer;
  char c;
  while (c = *string++) {
    char *ptr = strchr(letters, c);
    if (ptr) {
      *dest++ = c;
    }
  }
  *dest = '\0';
  return strdup(buffer);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strfilter2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  add    $0x1,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L13
L106:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L106
L13:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L42
L79:
  lea    0x1(%rbx),%r12
  movzbl (%rbx),%ebx
  mov    %rsp,%r15
  test   %bl,%bl
  je     L25
  mov    %rsp,%r13
L36:
  movsbl %bl,%esi
  mov    %r14,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L81
L53:
  add    $0x1,%r12
  mov    %bl,0x0(%r13)
  lea    0x1(%r13),%rax
  movzbl -0x1(%r12),%ebx
  test   %bl,%bl
  je     L38
  movsbl %bl,%esi
  mov    %r14,%rdi
  mov    %rax,%r13
  call   <strchr@plt>
  test   %rax,%rax
  jne    L53
L81:
  movzbl (%r12),%ebx
  add    $0x1,%r12
  test   %bl,%bl
  jne    L36
  mov    %r13,%rax
L38:
  movb   $0x0,(%rax)
  mov    %r15,%rdi
  call   <strdup@plt>
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L108
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  xchg   %ax,%ax
L42:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L79
L25:
  mov    %rsp,%rax
  jmp    L38
L108:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *strfilter2(char *string, char *letters) {
  char buffer[strlen(string) + 1];
  char *dest = buffer;
  char c;
  while (c = *string++) {
    char *ptr = strchr(letters, c);
    if (ptr) {
      *dest++ = c;
    }
  }
  *dest = '\0';
  return strdup(buffer);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<chkmv1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    %r9d,-0x2c(%rbp)
  mov    %al,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L61
L38:
  mov    -0x20(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L58
  mov    $0x0,%eax
  jmp    L20
L58:
  mov    -0x20(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L117
  mov    -0x4(%rbp),%eax
  jmp    L20
L117:
  cmpl   $0x0,-0x20(%rbp)
  je     L98
  cmpl   $0x7,-0x20(%rbp)
  je     L98
  cmpl   $0x0,-0x24(%rbp)
  je     L98
  cmpl   $0x7,-0x24(%rbp)
  jne    L77
L98:
  addl   $0xa,-0x4(%rbp)
  jmp    L61
L77:
  addl   $0x1,-0x4(%rbp)
L61:
  mov    -0x28(%rbp),%eax
  add    %eax,-0x20(%rbp)
  cmpl   $0x0,-0x20(%rbp)
  js     L54
  cmpl   $0x7,-0x20(%rbp)
  jg     L54
  mov    -0x2c(%rbp),%eax
  add    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  js     L54
  cmpl   $0x7,-0x24(%rbp)
  jle    L38
L54:
  mov    $0x0,%eax
L20:
  pop    %rbp
  ret
```

assistant: ```c
int chkmv1(char b[64], char p, int x, int y, int m, int n) {
  int k;
  k = 0;
  while ((x += m) >= 0 && x < 8 && (y += n) >= 0 && y < 8) {
    if (b[x * 8 + y] == '-')
      return 0;
    if (b[x * 8 + y] == p)
      return k;
    if (x == 0 || x == 7 || y == 0 || y == 7)
      k += 10;
    else
      k++;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<chkmv1>:
  endbr64
  mov    %esi,%r11d
  lea    (%rdx,%r8,1),%eax
  cmp    $0x7,%eax
  ja     L9
  lea    (%rcx,%r9,1),%edx
  cmp    $0x7,%edx
  ja     L39
  lea    0x0(,%r8,8),%r10d
  movslq %r10d,%r10
  movslq %r9d,%rcx
  add    %rcx,%r10
  lea    0x0(,%rax,8),%esi
  movslq %esi,%rsi
  movslq %edx,%rcx
  add    %rcx,%rsi
  lea    (%rdi,%rsi,1),%rcx
  mov    $0x0,%edi
  jmp    L64
L10:
  add    $0xa,%edi
L77:
  add    %r8d,%eax
  cmp    $0x7,%eax
  ja     L3
  add    %r9d,%edx
  add    %r10,%rcx
  cmp    $0x7,%edx
  ja     L12
L64:
  movzbl (%rcx),%esi
  cmp    $0x2d,%sil
  je     L23
  cmp    %r11b,%sil
  je     L89
  test   %eax,%eax
  je     L10
  cmp    $0x7,%eax
  je     L10
  test   %edx,%edx
  je     L10
  cmp    $0x7,%edx
  je     L10
  add    $0x1,%edi
  jmp    L77
L12:
  mov    $0x0,%edi
  jmp    L89
L9:
  mov    $0x0,%edi
  jmp    L89
L39:
  mov    $0x0,%edi
  jmp    L89
L23:
  mov    $0x0,%edi
  jmp    L89
L3:
  mov    $0x0,%edi
L89:
  mov    %edi,%eax
  ret
```

assistant: ```c
int chkmv1(char b[64], char p, int x, int y, int m, int n) {
  int k;
  k = 0;
  while ((x += m) >= 0 && x < 8 && (y += n) >= 0 && y < 8) {
    if (b[x * 8 + y] == '-')
      return 0;
    if (b[x * 8 + y] == p)
      return k;
    if (x == 0 || x == 7 || y == 0 || y == 7)
      k += 10;
    else
      k++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<chkmv1>:
  endbr64
  lea    (%rdx,%r8,1),%eax
  mov    %rdi,%r10
  mov    %esi,%r11d
  cmp    $0x7,%eax
  ja     L7
  lea    (%rcx,%r9,1),%edx
  cmp    $0x7,%edx
  ja     L7
  lea    0x0(,%r8,8),%edi
  movslq %r9d,%rcx
  lea    0x0(,%rax,8),%esi
  movslq %edi,%rdi
  movslq %esi,%rsi
  add    %rcx,%rdi
  movslq %edx,%rcx
  add    %rcx,%rsi
  lea    (%r10,%rsi,1),%rcx
  xor    %r10d,%r10d
  jmp    L108
  nopw   0x0(%rax,%rax,1)
L23:
  test   %edx,%edx
  je     L15
  cmp    $0x7,%edx
  je     L15
  add    %r8d,%eax
  add    $0x1,%r10d
  cmp    $0x7,%eax
  ja     L7
L89:
  add    %r9d,%edx
  add    %rdi,%rcx
  cmp    $0x7,%edx
  ja     L7
L108:
  movzbl (%rcx),%esi
  cmp    $0x2d,%sil
  je     L7
  cmp    %r11b,%sil
  je     L30
  test   %eax,%eax
  je     L15
  cmp    $0x7,%eax
  jne    L23
L15:
  add    %r8d,%eax
  add    $0xa,%r10d
  cmp    $0x7,%eax
  jbe    L89
L7:
  xor    %r10d,%r10d
L30:
  mov    %r10d,%eax
  ret
```

assistant: ```c
int chkmv1(char b[64], char p, int x, int y, int m, int n) {
  int k;
  k = 0;
  while ((x += m) >= 0 && x < 8 && (y += n) >= 0 && y < 8) {
    if (b[x * 8 + y] == '-')
      return 0;
    if (b[x * 8 + y] == p)
      return k;
    if (x == 0 || x == 7 || y == 0 || y == 7)
      k += 10;
    else
      k++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<chkmv1>:
  endbr64
  lea    (%rdx,%r8,1),%eax
  cmp    $0x7,%eax
  ja     L191
  lea    (%rcx,%r9,1),%edx
  cmp    $0x7,%edx
  ja     L191
  mov    %esi,%r11d
  cmp    $0x1,%r9d
  jne    L128
  lea    0x0(,%rax,8),%ecx
  lea    0x0(,%r8,8),%r9d
  movslq %edx,%rsi
  xor    %r10d,%r10d
  movslq %ecx,%rcx
  movslq %r9d,%r9
  add    %rsi,%rcx
  add    $0x1,%r9
  add    %rdi,%rcx
  jmp    L159
  nopw   0x0(%rax,%rax,1)
L43:
  test   %edx,%edx
  je     L161
  cmp    $0x7,%edx
  je     L161
  add    %r8d,%eax
  add    $0x1,%r10d
  cmp    $0x7,%eax
  ja     L191
L44:
  add    $0x1,%edx
  add    %r9,%rcx
  cmp    $0x8,%edx
  je     L191
L159:
  movzbl (%rcx),%esi
  cmp    $0x2d,%sil
  je     L191
  cmp    %sil,%r11b
  je     L132
  test   %eax,%eax
  je     L161
  cmp    $0x7,%eax
  jne    L43
L161:
  add    %r8d,%eax
  add    $0xa,%r10d
  cmp    $0x7,%eax
  jbe    L44
L191:
  xor    %r10d,%r10d
L132:
  mov    %r10d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L128:
  push   %rbx
  lea    0x0(,%r8,8),%ebx
  movslq %r9d,%rcx
  movslq %edx,%rsi
  movslq %ebx,%rbx
  xor    %r10d,%r10d
  add    %rcx,%rbx
  lea    0x0(,%rax,8),%ecx
  movslq %ecx,%rcx
  add    %rsi,%rcx
  add    %rdi,%rcx
  jmp    L115
  nopw   0x0(%rax,%rax,1)
L186:
  cmp    $0x7,%eax
  je     L10
  test   %edx,%edx
  je     L10
  cmp    $0x7,%edx
  je     L10
  add    $0x1,%r10d
L227:
  add    %r8d,%eax
  cmp    $0x7,%eax
  ja     L204
  add    %r9d,%edx
  add    %rbx,%rcx
  cmp    $0x7,%edx
  ja     L204
L115:
  movzbl (%rcx),%esi
  cmp    $0x2d,%sil
  je     L204
  cmp    %r11b,%sil
  je     L120
  test   %eax,%eax
  jne    L186
L10:
  add    $0xa,%r10d
  jmp    L227
  nopl   (%rax)
L204:
  xor    %r10d,%r10d
L120:
  mov    %r10d,%eax
  pop    %rbx
  ret
```

assistant: ```c
int chkmv1(char b[64], char p, int x, int y, int m, int n) {
  int k;
  k = 0;
  while ((x += m) >= 0 && x < 8 && (y += n) >= 0 && y < 8) {
    if (b[x * 8 + y] == '-')
      return 0;
    if (b[x * 8 + y] == p)
      return k;
    if (x == 0 || x == 7 || y == 0 || y == 7)
      k += 10;
    else
      k++;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<reverse_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x1,-0x1c(%rbp)
  jle    L29
  jmp    L33
L38:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
L33:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L38
  jmp    L14
L29:
  nop
L14:
  pop    %rbp
  ret
```

assistant: ```c
void reverse_array(int a[], int size) {
  int index1 = 0, index2 = size - 1, value1, value2;
  if (size < 2) {
    return;
  }
  while (index1 < index2) {
    value1 = a[index1];
    value2 = a[index2];
    a[index1++] = value2;
    a[index2--] = value1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<reverse_array>:
  endbr64
  lea    -0x1(%rsi),%eax
  cmp    $0x1,%esi
  jle    L0
  cltq
  mov    $0x0,%edx
L5:
  mov    (%rdi,%rdx,4),%ecx
  mov    (%rdi,%rax,4),%esi
  mov    %esi,(%rdi,%rdx,4)
  mov    %ecx,(%rdi,%rax,4)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L5
L0:
  ret
```

assistant: ```c
void reverse_array(int a[], int size) {
  int index1 = 0, index2 = size - 1, value1, value2;
  if (size < 2) {
    return;
  }
  while (index1 < index2) {
    value1 = a[index1];
    value2 = a[index2];
    a[index1++] = value2;
    a[index2--] = value1;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse_array>:
  endbr64
  lea    -0x1(%rsi),%eax
  cmp    $0x1,%esi
  jle    L18
  cltq
  xor    %edx,%edx
L9:
  mov    (%rdi,%rdx,4),%ecx
  mov    (%rdi,%rax,4),%esi
  mov    %esi,(%rdi,%rdx,4)
  add    $0x1,%rdx
  mov    %ecx,(%rdi,%rax,4)
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L9
L18:
  ret
```

assistant: ```c
void reverse_array(int a[], int size) {
  int index1 = 0, index2 = size - 1, value1, value2;
  if (size < 2) {
    return;
  }
  while (index1 < index2) {
    value1 = a[index1];
    value2 = a[index2];
    a[index1++] = value2;
    a[index2--] = value1;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse_array>:
  endbr64
  lea    -0x1(%rsi),%r8d
  cmp    $0x1,%esi
  jle    L6
  lea    -0x2(%rsi),%edx
  movslq %esi,%rcx
  mov    %edx,%r10d
  shl    $0x2,%rcx
  shr    %r10d
  mov    %rcx,%rsi
  lea    0x1(%r10),%eax
  mov    %rax,%r9
  shl    $0x2,%rax
  sub    %rax,%rsi
  cmp    %rsi,%rax
  jg     L84
  cmp    $0x1,%edx
  jbe    L84
  cmp    $0x5,%edx
  jbe    L17
  mov    %r9d,%esi
  mov    %rdi,%rax
  lea    -0x10(%rdi,%rcx,1),%rdx
  shr    $0x2,%esi
  sub    $0x1,%esi
  shl    $0x4,%rsi
  lea    0x10(%rdi,%rsi,1),%rsi
  nopw   0x0(%rax,%rax,1)
L16:
  movdqu (%rdx),%xmm2
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  sub    $0x10,%rdx
  pshufd $0x1b,%xmm2,%xmm1
  pshufd $0x1b,%xmm0,%xmm0
  movups %xmm1,-0x10(%rax)
  movups %xmm0,0x10(%rdx)
  cmp    %rsi,%rax
  jne    L16
  mov    %r9d,%esi
  and    $0xfffffffc,%esi
  mov    %esi,%edx
  sub    %esi,%r8d
  cmp    %esi,%r9d
  je     L26
L61:
  sub    %esi,%r9d
  cmp    %esi,%r10d
  je     L97
  shl    $0x2,%rsi
  sub    $0x8,%rcx
  sub    %rsi,%rcx
  lea    (%rdi,%rsi,1),%rax
  add    %rdi,%rcx
  movq   (%rax),%xmm0
  movq   (%rcx),%xmm1
  pshufd $0xe1,%xmm0,%xmm0
  pshufd $0xe1,%xmm1,%xmm1
  movq   %xmm1,(%rax)
  mov    %r9d,%eax
  and    $0xfffffffe,%eax
  movq   %xmm0,(%rcx)
  add    %eax,%edx
  sub    %eax,%r8d
  cmp    %eax,%r9d
  je     L6
L97:
  movslq %edx,%rdx
  movslq %r8d,%rax
  lea    (%rdi,%rdx,4),%rdx
  lea    (%rdi,%rax,4),%rax
  mov    (%rdx),%ecx
  mov    (%rax),%esi
  mov    %esi,(%rdx)
  mov    %ecx,(%rax)
  ret
  cs nopw 0x0(%rax,%rax,1)
L84:
  movslq %r8d,%rax
  xor    %edx,%edx
  nopl   (%rax)
L36:
  mov    (%rdi,%rdx,4),%ecx
  mov    (%rdi,%rax,4),%esi
  mov    %esi,(%rdi,%rdx,4)
  add    $0x1,%rdx
  mov    %ecx,(%rdi,%rax,4)
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L36
L6:
  ret
  nopl   0x0(%rax)
L26:
  ret
  nopl   0x0(%rax)
L17:
  xor    %esi,%esi
  xor    %edx,%edx
  jmp    L61
```

assistant: ```c
void reverse_array(int a[], int size) {
  int index1 = 0, index2 = size - 1, value1, value2;
  if (size < 2) {
    return;
  }
  while (index1 < index2) {
    value1 = a[index1];
    value2 = a[index2];
    a[index1++] = value2;
    a[index2--] = value1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %edi,-0x84(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
L5:
  lea    -0x70(%rbp),%rax
  mov    $0x50,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  lea    -0x70(%rbp),%rcx
  mov    -0x84(%rbp),%eax
  mov    $0x50,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    $0x50,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  movl   $0x0,-0x74(%rbp)
  nop
L25:
  mov    -0x74(%rbp),%ebx
  lea    0x1(%rbx),%eax
  mov    %eax,-0x74(%rbp)
  mov    $0x0,%eax
  call   <getchar@plt>
  movslq %ebx,%rdx
  mov    %al,-0x70(%rbp,%rdx,1)
  movslq %ebx,%rax
  movzbl -0x70(%rbp,%rax,1),%eax
  cmp    $0xa,%al
  jne    L25
  lea    -0x70(%rbp),%rcx
  mov    -0x84(%rbp),%eax
  mov    $0x50,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  lea    -0x70(%rbp),%rax
  mov    $0x4,%edx
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L5
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  nop
  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: unpack_data
arguments: tag: D18
datatype: string
name: unpack_data
arguments: tag: D12
datatype: string
name: unpack_data
arguments: tag: D29
datatype: string

tool: D18: From client: %s	 To client : 

tool: D12: exit

tool: D29: Server Exit...

assistant: ```c
void func(int sockfd) {
  char buff[80];
  int n;
  for (;;) {
    bzero(buff, 80);
    read(sockfd, buff, sizeof(buff));
    printf("From client: %s\t To client : ", buff);
    bzero(buff, 80);
    n = 0;
    while ((buff[n++] = getchar()) != '\n')
      ;
    write(sockfd, buff, sizeof(buff));
    if (strncmp("exit", buff, 4) == 0) {
      printf("Server Exit...\n");
      break;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<func>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %edi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    D3(%rip),%r14
  lea    D18(%rip),%r13
L3:
  mov    $0x50,%esi
  mov    %rbp,%rdi
  call   <bzero@plt>
  mov    $0x50,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  mov    %rbp,%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x50,%esi
  mov    %rbp,%rdi
  call   <bzero@plt>
  mov    %rbp,%rbx
L28:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    $0xa,%al
  jne    L28
  mov    $0x50,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <write@plt>
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L3
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D3
datatype: string
name: parse_data
arguments: label: D18
datatype: string
name: parse_data
arguments: label: D1
datatype: string

tool: D3: From client: %s	 To client : 

tool: D18: exit

tool: D1: Server Exit...

assistant: ```c
void func(int sockfd) {
  char buff[80];
  int n;
  for (;;) {
    bzero(buff, 80);
    read(sockfd, buff, sizeof(buff));
    printf("From client: %s\t To client : ", buff);
    bzero(buff, 80);
    n = 0;
    while ((buff[n++] = getchar()) != '\n')
      ;
    write(sockfd, buff, sizeof(buff));
    if (strncmp("exit", buff, 4) == 0) {
      printf("Server Exit...\n");
      break;
    }
  }
}
```