ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<mprintf>:
  endbr64
  push   %r14
  mov    %rdi,%rsi
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  lea    0x1(%rax),%r14d
  call   <va_end@plt>
  mov    %r14d,%edi
  call   <xmalloc@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %rax,%r13
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *mprintf(const char *fmt, ...) {
  char *s;
  int sz;
  va_list va;
  va_start(va, fmt);
  sz = vsnprintf(NULL, 0, fmt, va) + 1;
  va_end(va);
  s = xmalloc(sz);
  va_start(va, fmt);
  vsnprintf(s, sz, fmt, va);
  va_end(va);
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mprintf>:
  endbr64
  push   %r14
  mov    %rdi,%rsi
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  lea    0x1(%rax),%r14d
  call   <va_end@plt>
  mov    %r14d,%edi
  call   <xmalloc@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    %rax,%r13
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *mprintf(const char *fmt, ...) {
  char *s;
  int sz;
  va_list va;
  va_start(va, fmt);
  sz = vsnprintf(NULL, 0, fmt, va) + 1;
  va_end(va);
  s = xmalloc(sz);
  va_start(va, fmt);
  vsnprintf(s, sz, fmt, va);
  va_end(va);
  return s;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x22b,%edi
  call   <gen@plt>
  mov    %rax,-0x18(%rbp)
  mov    $0x22b,%edi
  call   <gen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <print@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <addTwoNumbers@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <print@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  struct ListNode *p, *q;
  p = gen(555);
  q = gen(555);
  struct ListNode *res;
  print(p);
  res = addTwoNumbers(p, q);
  print(res);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x22b,%edi
  call   <gen@plt>
  mov    %rax,%rbx
  mov    $0x22b,%edi
  call   <gen@plt>
  mov    %rax,%rbp
  mov    %rbx,%rdi
  call   <print@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <addTwoNumbers@plt>
  mov    %rax,%rdi
  call   <print@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  struct ListNode *p, *q;
  p = gen(555);
  q = gen(555);
  struct ListNode *res;
  print(p);
  res = addTwoNumbers(p, q);
  print(res);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x22b,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <gen@plt>
  mov    $0x22b,%edi
  mov    %rax,%rbp
  call   <gen@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <print@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <addTwoNumbers@plt>
  mov    %rax,%rdi
  call   <print@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  struct ListNode *p, *q;
  p = gen(555);
  q = gen(555);
  struct ListNode *res;
  print(p);
  res = addTwoNumbers(p, q);
  print(res);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x22b,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <gen@plt>
  mov    $0x22b,%edi
  mov    %rax,%rbp
  call   <gen@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <print@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <addTwoNumbers@plt>
  mov    %rax,%rdi
  call   <print@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  struct ListNode *p, *q;
  p = gen(555);
  q = gen(555);
  struct ListNode *res;
  print(p);
  res = addTwoNumbers(p, q);
  print(res);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<decode_utf8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0xffffffff,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  and    $0x80,%eax
  test   %eax,%eax
  jne    L63
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L31
L63:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  and    $0x40,%eax
  test   %eax,%eax
  je     L31
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  and    $0x20,%eax
  test   %eax,%eax
  jne    L18
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  shl    $0x6,%eax
  xor    $0x30,%ah
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  xor    $0x80,%al
  xor    %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L31
L18:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  and    $0x10,%eax
  test   %eax,%eax
  jne    L69
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  shl    $0xc,%eax
  xor    $0xe0000,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  shl    $0x6,%eax
  xor    $0x20,%ah
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  xor    $0x80,%al
  xor    %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L31
L69:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  and    $0x8,%eax
  test   %eax,%eax
  jne    L31
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  shl    $0x12,%eax
  xor    $0x3c00000,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  shl    $0xc,%eax
  xor    $0x80000,%eax
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  shl    $0x6,%eax
  xor    $0x20,%ah
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  xor    $0x80,%al
  xor    %edx,%eax
  mov    %eax,-0xc(%rbp)
L31:
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int decode_utf8(const char *ibuf) {
  const u_char *buf = (u_char *)ibuf;
  int u = -1;
  if ((buf[0] & 0x80) == 0)
    u = buf[0];
  else if ((buf[0] & 0x40) == 0)
    ;
  else {
    if ((buf[0] & 0x20) == 0)
      u = (buf[0] ^ 0xC0) << 6 ^ (buf[1] ^ 0x80);
    else if ((buf[0] & 0x10) == 0)
      u = (buf[0] ^ 0xE0) << 12 ^ (buf[1] ^ 0x80) << 6 ^ (buf[2] ^ 0x80);
    else if (((buf[0] & 0x08) == 0))
      u = (buf[0] ^ 0xF0) << 18 ^ (buf[1] ^ 0x80) << 12 ^ (buf[2] ^ 0x80) << 6 ^ (buf[3] ^ 0x80);
  }
  return u;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<decode_utf8>:
  endbr64
  mov    (%rdi),%eax
  test   $0x80,%al
  je     L19
  test   $0x40,%al
  je     L28
  test   $0x20,%al
  jne    L11
  shl    $0x6,%eax
  xor    0x4(%rdi),%eax
  xor    $0x3080,%eax
  ret
L11:
  test   $0x10,%al
  jne    L18
  mov    0x4(%rdi),%edx
  shl    $0x6,%edx
  shl    $0xc,%eax
  xor    %eax,%edx
  xor    0x8(%rdi),%edx
  mov    %edx,%eax
  xor    $0xe2080,%eax
  ret
L18:
  test   $0x8,%al
  jne    L33
  mov    0x8(%rdi),%edx
  shl    $0x6,%edx
  shl    $0x12,%eax
  xor    %eax,%edx
  xor    0xc(%rdi),%edx
  mov    0x4(%rdi),%eax
  shl    $0xc,%eax
  xor    %eax,%edx
  mov    %edx,%eax
  xor    $0x3c82080,%eax
  ret
L28:
  mov    $0xffffffff,%eax
  ret
L33:
  mov    $0xffffffff,%eax
L19:
  ret
```

assistant: ```c
int decode_utf8(const char *ibuf) {
  const u_char *buf = (u_char *)ibuf;
  int u = -1;
  if ((buf[0] & 0x80) == 0)
    u = buf[0];
  else if ((buf[0] & 0x40) == 0)
    ;
  else {
    if ((buf[0] & 0x20) == 0)
      u = (buf[0] ^ 0xC0) << 6 ^ (buf[1] ^ 0x80);
    else if ((buf[0] & 0x10) == 0)
      u = (buf[0] ^ 0xE0) << 12 ^ (buf[1] ^ 0x80) << 6 ^ (buf[2] ^ 0x80);
    else if (((buf[0] & 0x08) == 0))
      u = (buf[0] ^ 0xF0) << 18 ^ (buf[1] ^ 0x80) << 12 ^ (buf[2] ^ 0x80) << 6 ^ (buf[3] ^ 0x80);
  }
  return u;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<decode_utf8>:
  endbr64
  mov    (%rdi),%eax
  test   $0x80,%al
  je     L49
  test   $0x40,%al
  je     L38
  test   $0x20,%al
  jne    L32
  shl    $0x6,%eax
  xor    0x4(%rdi),%eax
  xor    $0x3080,%eax
  ret
  xchg   %ax,%ax
L38:
  mov    $0xffffffff,%eax
L49:
  ret
  cs nopw 0x0(%rax,%rax,1)
L32:
  test   $0x10,%al
  je     L29
  test   $0x8,%al
  jne    L38
  mov    0x8(%rdi),%edx
  shl    $0x12,%eax
  shl    $0x6,%edx
  xor    %eax,%edx
  mov    0x4(%rdi),%eax
  xor    0xc(%rdi),%edx
  shl    $0xc,%eax
  xor    %eax,%edx
  mov    %edx,%eax
  xor    $0x3c82080,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L29:
  mov    0x4(%rdi),%edx
  shl    $0xc,%eax
  shl    $0x6,%edx
  xor    %eax,%edx
  xor    0x8(%rdi),%edx
  mov    %edx,%eax
  xor    $0xe2080,%eax
  ret
```

assistant: ```c
int decode_utf8(const char *ibuf) {
  const u_char *buf = (u_char *)ibuf;
  int u = -1;
  if ((buf[0] & 0x80) == 0)
    u = buf[0];
  else if ((buf[0] & 0x40) == 0)
    ;
  else {
    if ((buf[0] & 0x20) == 0)
      u = (buf[0] ^ 0xC0) << 6 ^ (buf[1] ^ 0x80);
    else if ((buf[0] & 0x10) == 0)
      u = (buf[0] ^ 0xE0) << 12 ^ (buf[1] ^ 0x80) << 6 ^ (buf[2] ^ 0x80);
    else if (((buf[0] & 0x08) == 0))
      u = (buf[0] ^ 0xF0) << 18 ^ (buf[1] ^ 0x80) << 12 ^ (buf[2] ^ 0x80) << 6 ^ (buf[3] ^ 0x80);
  }
  return u;
}
```
user: What is the c source code of the assembly code below:

```asm
<decode_utf8>:
  endbr64
  mov    (%rdi),%eax
  test   $0x80,%al
  je     L34
  test   $0x40,%al
  je     L29
  test   $0x20,%al
  jne    L28
  shl    $0x6,%eax
  xor    0x4(%rdi),%eax
  xor    $0x3080,%eax
  ret
  xchg   %ax,%ax
L29:
  mov    $0xffffffff,%eax
L34:
  ret
  cs nopw 0x0(%rax,%rax,1)
L28:
  test   $0x10,%al
  je     L49
  test   $0x8,%al
  jne    L29
  mov    0x8(%rdi),%edx
  shl    $0x12,%eax
  shl    $0x6,%edx
  xor    %eax,%edx
  mov    0x4(%rdi),%eax
  xor    0xc(%rdi),%edx
  shl    $0xc,%eax
  xor    %eax,%edx
  mov    %edx,%eax
  xor    $0x3c82080,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L49:
  mov    0x4(%rdi),%edx
  shl    $0xc,%eax
  shl    $0x6,%edx
  xor    %eax,%edx
  xor    0x8(%rdi),%edx
  mov    %edx,%eax
  xor    $0xe2080,%eax
  ret
```

assistant: ```c
int decode_utf8(const char *ibuf) {
  const u_char *buf = (u_char *)ibuf;
  int u = -1;
  if ((buf[0] & 0x80) == 0)
    u = buf[0];
  else if ((buf[0] & 0x40) == 0)
    ;
  else {
    if ((buf[0] & 0x20) == 0)
      u = (buf[0] ^ 0xC0) << 6 ^ (buf[1] ^ 0x80);
    else if ((buf[0] & 0x10) == 0)
      u = (buf[0] ^ 0xE0) << 12 ^ (buf[1] ^ 0x80) << 6 ^ (buf[2] ^ 0x80);
    else if (((buf[0] & 0x08) == 0))
      u = (buf[0] ^ 0xF0) << 18 ^ (buf[1] ^ 0x80) << 12 ^ (buf[2] ^ 0x80) << 6 ^ (buf[3] ^ 0x80);
  }
  return u;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<loop_g_2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x1014(%rbp)
  movl   $0x0,-0x101c(%rbp)
  jmp    L13
L87:
  mov    -0x101c(%rbp),%eax
  cltq
  mov    -0x101c(%rbp),%edx
  mov    %edx,-0x1010(%rbp,%rax,4)
  addl   $0x1,-0x101c(%rbp)
L13:
  cmpl   $0x3ff,-0x101c(%rbp)
  jle    L87
  movl   $0x0,-0x101c(%rbp)
  jmp    L18
L75:
  mov    -0x101c(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x1014(%rbp)
  movl   $0x0,-0x1018(%rbp)
  jmp    L52
L28:
  mov    -0x101c(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0x1018(%rbp),%eax
  add    %edx,%eax
  cltq
  mov    -0x1010(%rbp,%rax,4),%ecx
  mov    -0x101c(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0x1018(%rbp),%eax
  add    %edx,%eax
  mov    -0x1014(%rbp),%edx
  add    %ecx,%edx
  cltq
  mov    %edx,-0x1010(%rbp,%rax,4)
  addl   $0x1,-0x1018(%rbp)
L52:
  cmpl   $0x1f,-0x1018(%rbp)
  jle    L28
  addl   $0x1,-0x101c(%rbp)
L18:
  cmpl   $0x1f,-0x101c(%rbp)
  jle    L75
  movl   $0x0,-0x101c(%rbp)
  jmp    L85
L59:
  mov    -0x101c(%rbp),%eax
  cltq
  mov    -0x1010(%rbp,%rax,4),%edx
  mov    -0x101c(%rbp),%eax
  lea    0x1f(%rax),%ecx
  test   %eax,%eax
  cmovs  %ecx,%eax
  sar    $0x5,%eax
  lea    (%rax,%rax,1),%ecx
  mov    -0x101c(%rbp),%eax
  add    %ecx,%eax
  cmp    %eax,%edx
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  addl   $0x1,-0x101c(%rbp)
L85:
  cmpl   $0x3ff,-0x101c(%rbp)
  jle    L59
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

assistant: ```c
void loop_g_2() {
  int x = 5, i, arr[32 * 32];
  for (i = 0; i < 32 * 32; i++)
    arr[i] = i;
#pragma acc parallel copy(arr) num_gangs(32) num_workers(8) vector_length(32)
  {
#pragma acc loop gang private(x)
    for (i = 0; i < 32; i++) {
      x = i * 2;
#pragma acc loop worker
      for (int j = 0; j < 32; j++)
        arr[i * 32 + j] += x;
    }
  }
  for (i = 0; i < 32 * 32; i++)
    assert(arr[i] == i + (i / 32) * 2);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<loop_g_2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
L8:
  mov    %eax,(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x400,%rax
  jne    L8
  lea    0x80(%rsp),%rdx
  mov    $0x0,%esi
L6:
  mov    %esi,%ecx
  lea    -0x80(%rdx),%rax
L21:
  add    %ecx,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L21
  add    $0x2,%esi
  sub    $0xffffffffffffff80,%rdx
  cmp    $0x40,%esi
  jne    L6
  mov    $0x0,%ebx
  mov    %rsp,%rbp
L29:
  lea    0x1f(%rbx),%eax
  test   %ebx,%ebx
  cmovns %ebx,%eax
  sar    $0x5,%eax
  lea    (%rbx,%rax,2),%eax
  cmp    0x0(%rbp,%rbx,4),%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x1,%rbx
  cmp    $0x400,%rbx
  jne    L29
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x1018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void loop_g_2() {
  int x = 5, i, arr[32 * 32];
  for (i = 0; i < 32 * 32; i++)
    arr[i] = i;
#pragma acc parallel copy(arr) num_gangs(32) num_workers(8) vector_length(32)
  {
#pragma acc loop gang private(x)
    for (i = 0; i < 32; i++) {
      x = i * 2;
#pragma acc loop worker
      for (int j = 0; j < 32; j++)
        arr[i * 32 + j] += x;
    }
  }
  for (i = 0; i < 32 * 32; i++)
    assert(arr[i] == i + (i / 32) * 2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<loop_g_2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nopl   0x0(%rax)
L36:
  mov    %eax,0x0(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    $0x400,%rax
  jne    L36
  lea    0x80(%rsp),%rcx
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L39:
  lea    -0x80(%rcx),%rax
  nopl   0x0(%rax)
L14:
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L14
  add    $0x2,%edx
  lea    0x80(%rax),%rcx
  cmp    $0x40,%edx
  jne    L39
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L16:
  mov    %ebx,%eax
  xor    %edi,%edi
  sar    $0x5,%eax
  lea    (%rbx,%rax,2),%eax
  cmp    0x0(%rbp,%rbx,4),%eax
  sete   %dil
  add    $0x1,%rbx
  call   <assert@plt>
  cmp    $0x400,%rbx
  jne    L16
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x1018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void loop_g_2() {
  int x = 5, i, arr[32 * 32];
  for (i = 0; i < 32 * 32; i++)
    arr[i] = i;
#pragma acc parallel copy(arr) num_gangs(32) num_workers(8) vector_length(32)
  {
#pragma acc loop gang private(x)
    for (i = 0; i < 32; i++) {
      x = i * 2;
#pragma acc loop worker
      for (int j = 0; j < 32; j++)
        arr[i * 32 + j] += x;
    }
  }
  for (i = 0; i < 32 * 32; i++)
    assert(arr[i] == i + (i / 32) * 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<loop_g_2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x18,%rsp
  movdqa D15(%rip),%xmm0
  movdqa D3(%rip),%xmm2
  mov    %fs:0x28,%rax
  mov    %rax,0x1008(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x1000(%rsp),%rcx
  mov    %rbp,%rax
  mov    %rbp,%rdx
  nopw   0x0(%rax,%rax,1)
L9:
  movdqa %xmm0,%xmm1
  add    $0x10,%rdx
  paddd  %xmm2,%xmm0
  movaps %xmm1,-0x10(%rdx)
  cmp    %rcx,%rdx
  jne    L9
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L31:
  movdqa (%rax),%xmm1
  movd   %edx,%xmm3
  sub    $0xffffffffffffff80,%rax
  add    $0x2,%edx
  pshufd $0x0,%xmm3,%xmm0
  paddd  %xmm0,%xmm1
  movaps %xmm1,-0x80(%rax)
  movdqa -0x70(%rax),%xmm1
  paddd  %xmm0,%xmm1
  movaps %xmm1,-0x70(%rax)
  movdqa -0x60(%rax),%xmm1
  paddd  %xmm0,%xmm1
  movaps %xmm1,-0x60(%rax)
  movdqa -0x50(%rax),%xmm1
  paddd  %xmm0,%xmm1
  movaps %xmm1,-0x50(%rax)
  movdqa -0x40(%rax),%xmm1
  paddd  %xmm0,%xmm1
  movaps %xmm1,-0x40(%rax)
  movdqa -0x30(%rax),%xmm1
  paddd  %xmm0,%xmm1
  movaps %xmm1,-0x30(%rax)
  movdqa -0x20(%rax),%xmm1
  paddd  %xmm0,%xmm1
  paddd  -0x10(%rax),%xmm0
  movaps %xmm1,-0x20(%rax)
  movaps %xmm0,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L31
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %ebx,%eax
  xor    %edi,%edi
  sar    $0x5,%eax
  lea    (%rbx,%rax,2),%eax
  cmp    0x0(%rbp,%rbx,4),%eax
  sete   %dil
  add    $0x1,%rbx
  call   <assert@plt>
  cmp    $0x400,%rbx
  jne    L17
  mov    0x1008(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x1018,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D15
data_type: byte[16]
name: disassemble_data
arguments: label: D3
data_type: byte[16]

tool: {"D15": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D3": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
void loop_g_2() {
  int x = 5, i, arr[32 * 32];
  for (i = 0; i < 32 * 32; i++)
    arr[i] = i;
#pragma acc parallel copy(arr) num_gangs(32) num_workers(8) vector_length(32)
  {
#pragma acc loop gang private(x)
    for (i = 0; i < 32; i++) {
      x = i * 2;
#pragma acc loop worker
      for (int j = 0; j < 32; j++)
        arr[i * 32 + j] += x;
    }
  }
  for (i = 0; i < 32 * 32; i++)
    assert(arr[i] == i + (i / 32) * 2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<map_buttons>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  je     L4
  cmpl   $0x4,-0x4(%rbp)
  jne    L29
  mov    $0x0,%eax
  jmp    L10
L4:
  mov    $0x1,%eax
  jmp    L10
L29:
  mov    $0x2,%eax
L10:
  pop    %rbp
  ret
```

assistant: ```c
int map_buttons(int b) {
  switch (b) {
  case 0x04:
    return (0);
  case 0x02:
    return (1);
  default:
    return (2);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<map_buttons>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x2,%edi
  je     L8
  cmp    $0x4,%edi
  setne  %al
  movzbl %al,%eax
  add    %eax,%eax
L8:
  ret
```

assistant: ```c
int map_buttons(int b) {
  switch (b) {
  case 0x04:
    return (0);
  case 0x02:
    return (1);
  default:
    return (2);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<map_buttons>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x2,%edi
  je     L1
  xor    %eax,%eax
  cmp    $0x4,%edi
  setne  %al
  add    %eax,%eax
L1:
  ret
```

assistant: ```c
int map_buttons(int b) {
  switch (b) {
  case 0x04:
    return (0);
  case 0x02:
    return (1);
  default:
    return (2);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<map_buttons>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x2,%edi
  je     L1
  xor    %eax,%eax
  cmp    $0x4,%edi
  setne  %al
  add    %eax,%eax
L1:
  ret
```

assistant: ```c
int map_buttons(int b) {
  switch (b) {
  case 0x04:
    return (0);
  case 0x02:
    return (1);
  default:
    return (2);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<msr_list_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L19
  mov    $0x1,%eax
  jmp    L0
L19:
  mov    -0x8(%rbp),%rax
  mov    0x8(%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
L0:
  pop    %rbp
  ret
```

assistant: ```c
int msr_list_index(void *data, int n, __u32 *index) {
  struct kvm_msr_list *msrs = (struct kvm_msr_list *)data;
  if (n >= msrs->nmsrs) {
    return 1;
  }
  *index = msrs->indices[n];
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<msr_list_index>:
  endbr64
  mov    $0x1,%eax
  cmp    %esi,(%rdi)
  jle    L8
  movslq %esi,%rsi
  mov    0x8(%rdi),%rax
  mov    (%rax,%rsi,4),%eax
  mov    %eax,(%rdx)
  mov    $0x0,%eax
L8:
  ret
```

assistant: ```c
int msr_list_index(void *data, int n, __u32 *index) {
  struct kvm_msr_list *msrs = (struct kvm_msr_list *)data;
  if (n >= msrs->nmsrs) {
    return 1;
  }
  *index = msrs->indices[n];
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<msr_list_index>:
  endbr64
  mov    $0x1,%eax
  cmp    %esi,(%rdi)
  jle    L9
  mov    0x8(%rdi),%rax
  movslq %esi,%rsi
  mov    (%rax,%rsi,4),%eax
  mov    %eax,(%rdx)
  xor    %eax,%eax
L9:
  ret
```

assistant: ```c
int msr_list_index(void *data, int n, __u32 *index) {
  struct kvm_msr_list *msrs = (struct kvm_msr_list *)data;
  if (n >= msrs->nmsrs) {
    return 1;
  }
  *index = msrs->indices[n];
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<msr_list_index>:
  endbr64
  mov    $0x1,%eax
  cmp    %esi,(%rdi)
  jle    L2
  mov    0x8(%rdi),%rax
  movslq %esi,%rsi
  mov    (%rax,%rsi,4),%eax
  mov    %eax,(%rdx)
  xor    %eax,%eax
L2:
  ret
```

assistant: ```c
int msr_list_index(void *data, int n, __u32 *index) {
  struct kvm_msr_list *msrs = (struct kvm_msr_list *)data;
  if (n >= msrs->nmsrs) {
    return 1;
  }
  *index = msrs->indices[n];
  return 0;
}
```