ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Program>:
  endbr64
  sub    $0x18,%rsp
  lea    D3(%rip),%rsi
  mov    $0x1,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <Match@plt>
  lea    D22(%rip),%rsi
  mov    $0x1a,%edi
  call   <Match@plt>
  mov    $0x1b,%edi
  lea    D19(%rip),%rsi
  call   <Match@plt>
  lea    0x4(%rsp),%rdi
  call   <Statement_Block@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D3
type: string
name: retrieve_data
arguments: data_label: D22
type: string
name: retrieve_data
arguments: data_label: D19
type: string

tool: D3: main

tool: D22: (

tool: D19: )

assistant: ```c
void Program(void) {
  int nChain;
  Match(1, "main");
  Match(26, "(");
  Match(27, ")");
  Statement_Block(&nChain);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Program>:
  endbr64
  sub    $0x18,%rsp
  lea    D20(%rip),%rsi
  mov    $0x1,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <Match@plt>
  lea    D8(%rip),%rsi
  mov    $0x1a,%edi
  call   <Match@plt>
  mov    $0x1b,%edi
  lea    D2(%rip),%rsi
  call   <Match@plt>
  lea    0x4(%rsp),%rdi
  call   <Statement_Block@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D20
data_type: string
name: parse_data
arguments: tag: D8
data_type: string
name: parse_data
arguments: tag: D2
data_type: string

tool: D20: main

tool: D8: (

tool: D2: )

assistant: ```c
void Program(void) {
  int nChain;
  Match(1, "main");
  Match(26, "(");
  Match(27, ")");
  Statement_Block(&nChain);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <bar@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0xffffffffffffffff,-0x8(%rbp)
  jne    L3
  mov    $0x0,%eax
  jmp    L13
L3:
  mov    $0x3,%eax
L13:
  leave
  ret
```

assistant: ```c
unsigned long foo() {
  unsigned long retval;
  retval = bar();
  if (retval == -1)
    return 0;
  return 3;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <bar@plt>
  cmp    $0xffffffffffffffff,%rax
  setne  %al
  movzbl %al,%eax
  lea    (%rax,%rax,2),%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
unsigned long foo() {
  unsigned long retval;
  retval = bar();
  if (retval == -1)
    return 0;
  return 3;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <bar@plt>
  cmp    $0xffffffffffffffff,%rax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  lea    (%rax,%rax,2),%rax
  ret
```

assistant: ```c
unsigned long foo() {
  unsigned long retval;
  retval = bar();
  if (retval == -1)
    return 0;
  return 3;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<foo>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <bar@plt>
  cmp    $0xffffffffffffffff,%rax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  lea    (%rax,%rax,2),%rax
  ret
```

assistant: ```c
unsigned long foo() {
  unsigned long retval;
  retval = bar();
  if (retval == -1)
    return 0;
  return 3;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<csum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L6
L2:
  mov    -0x18(%rbp),%rax
  lea    0x2(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzwl (%rax),%eax
  movzwl %ax,%eax
  add    %rax,-0x8(%rbp)
  subl   $0x1,-0x1c(%rbp)
L6:
  cmpl   $0x0,-0x1c(%rbp)
  jg     L2
  mov    -0x8(%rbp),%rax
  shr    $0x10,%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  movzwl %ax,%eax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x10,%rax
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  not    %eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned short csum(unsigned short *buf, int nwords) {
  unsigned long sum;
  for (sum = 0; nwords > 0; nwords--)
    sum += *buf++;
  sum = (sum >> 16) + (sum & 0xffff);
  sum += (sum >> 16);
  return ~sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<csum>:
  endbr64
  test   %esi,%esi
  jle    L17
  mov    %esi,%esi
  lea    (%rdi,%rsi,2),%rcx
  mov    $0x0,%eax
L15:
  add    $0x2,%rdi
  movzwl -0x2(%rdi),%edx
  add    %rdx,%rax
  cmp    %rcx,%rdi
  jne    L15
L7:
  mov    %rax,%rdx
  shr    $0x10,%rdx
  movzwl %ax,%eax
  add    %rax,%rdx
  mov    %rdx,%rax
  shr    $0x10,%rax
  add    %rdx,%rax
  not    %eax
  ret
L17:
  mov    $0x0,%eax
  jmp    L7
```

assistant: ```c
unsigned short csum(unsigned short *buf, int nwords) {
  unsigned long sum;
  for (sum = 0; nwords > 0; nwords--)
    sum += *buf++;
  sum = (sum >> 16) + (sum & 0xffff);
  sum += (sum >> 16);
  return ~sum;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<csum>:
  endbr64
  test   %esi,%esi
  jle    L13
  movslq %esi,%rsi
  xor    %eax,%eax
  lea    (%rdi,%rsi,2),%rcx
  nopl   0x0(%rax)
L3:
  movzwl (%rdi),%edx
  add    $0x2,%rdi
  add    %rdx,%rax
  cmp    %rcx,%rdi
  jne    L3
  mov    %rax,%rdx
  movzwl %ax,%eax
  shr    $0x10,%rdx
  add    %rax,%rdx
  mov    %rdx,%rax
  shr    $0x10,%rax
  add    %rdx,%rax
  not    %eax
  ret
  nopl   0x0(%rax)
L13:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
unsigned short csum(unsigned short *buf, int nwords) {
  unsigned long sum;
  for (sum = 0; nwords > 0; nwords--)
    sum += *buf++;
  sum = (sum >> 16) + (sum & 0xffff);
  sum += (sum >> 16);
  return ~sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<csum>:
  endbr64
  mov    %esi,%edx
  test   %esi,%esi
  jle    L16
  cmp    $0x7,%esi
  jle    L96
  mov    %esi,%ecx
  pxor   %xmm2,%xmm2
  pxor   %xmm4,%xmm4
  mov    %rdi,%rax
  shr    $0x3,%ecx
  pxor   %xmm3,%xmm3
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopw   0x0(%rax,%rax,1)
L72:
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  movdqa %xmm0,%xmm1
  punpckhwd %xmm4,%xmm0
  movdqa %xmm0,%xmm5
  punpcklwd %xmm4,%xmm1
  punpckhdq %xmm3,%xmm0
  punpckldq %xmm3,%xmm5
  paddq  %xmm5,%xmm0
  movdqa %xmm1,%xmm5
  punpckldq %xmm3,%xmm1
  punpckhdq %xmm3,%xmm5
  paddq  %xmm5,%xmm1
  paddq  %xmm1,%xmm0
  paddq  %xmm0,%xmm2
  cmp    %rcx,%rax
  jne    L72
  movdqa %xmm2,%xmm0
  mov    %edx,%ecx
  and    $0xfffffff8,%ecx
  psrldq $0x8,%xmm0
  paddq  %xmm0,%xmm2
  mov    %ecx,%r8d
  sub    %ecx,%edx
  movq   %xmm2,%rax
  lea    (%rdi,%r8,2),%rdi
  cmp    %esi,%ecx
  je     L85
L107:
  movzwl (%rdi),%ecx
  add    %rcx,%rax
  cmp    $0x1,%edx
  je     L85
  movzwl 0x2(%rdi),%ecx
  add    %rcx,%rax
  cmp    $0x2,%edx
  je     L85
  movzwl 0x4(%rdi),%ecx
  add    %rcx,%rax
  cmp    $0x3,%edx
  je     L85
  movzwl 0x6(%rdi),%ecx
  add    %rcx,%rax
  cmp    $0x4,%edx
  je     L85
  movzwl 0x8(%rdi),%ecx
  add    %rcx,%rax
  cmp    $0x5,%edx
  je     L85
  movzwl 0xa(%rdi),%ecx
  add    %rcx,%rax
  cmp    $0x6,%edx
  je     L85
  movzwl 0xc(%rdi),%edx
  add    %rdx,%rax
L85:
  mov    %rax,%rdx
  movzwl %ax,%eax
  shr    $0x10,%rdx
  add    %rax,%rdx
  mov    %rdx,%rax
  shr    $0x10,%rax
  add    %rdx,%rax
  not    %eax
  ret
  nopl   (%rax)
L16:
  mov    $0xffffffff,%eax
  ret
L96:
  xor    %eax,%eax
  jmp    L107
```

assistant: ```c
unsigned short csum(unsigned short *buf, int nwords) {
  unsigned long sum;
  for (sum = 0; nwords > 0; nwords--)
    sum += *buf++;
  sum = (sum >> 16) + (sum & 0xffff);
  sum += (sum >> 16);
  return ~sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<gchild_func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <yield@plt>
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <yield@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
datatype: string

tool: D1: k = %d


assistant: ```c
void gchild_func() {
  int k = 0;
  printf("k = %d\n", k);
  yield();
  k++;
  printf("k = %d\n", k);
  yield();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<gchild_func>:
  endbr64
  push   %rbx
  mov    $0x0,%esi
  lea    D8(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <yield@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <yield@plt>
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
type: string

tool: D8: k = %d


assistant: ```c
void gchild_func() {
  int k = 0;
  printf("k = %d\n", k);
  yield();
  k++;
  printf("k = %d\n", k);
  yield();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<gchild_func>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rbp
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <yield@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1060 <yield@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
type_guessed: string

tool: D4: k = %d


assistant: ```c
void gchild_func() {
  int k = 0;
  printf("k = %d\n", k);
  yield();
  k++;
  printf("k = %d\n", k);
  yield();
}
```
user: Translate the below assembly code into C programming language:

```asm
<gchild_func>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rbp
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <yield@plt>
  mov    %rbp,%rdi
  mov    $0x1,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  jmp    1060 <yield@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_guessed: string

tool: D0: k = %d


assistant: ```c
void gchild_func() {
  int k = 0;
  printf("k = %d\n", k);
  yield();
  k++;
  printf("k = %d\n", k);
  yield();
}
```
user: What is the c source code of the assembly code below?

```asm
<in_menu_wait>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x1c2,-0x4(%rbp)
  mov    D53(%rip),%eax
  test   %eax,%eax
  je     L88
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
L88:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <in_menu_wait_any@plt>
  mov    %eax,-0xc(%rbp)
  mov    D54(%rip),%eax
  cmp    %eax,-0xc(%rbp)
  jne    L13
  mov    D53(%rip),%eax
  add    $0x1,%eax
  mov    %eax,D25(%rip)
  jmp    L13
L68:
  mov    $0xffffffff,%edi
  call   <in_menu_wait_any@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
L13:
  mov    -0xc(%rbp),%eax
  and    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L68
  cmpl   $0x0,-0x8(%rbp)
  jne    L104
  mov    D54(%rip),%eax
  cmp    %eax,-0xc(%rbp)
  je     L87
L104:
  movl   $0x0,D25(%rip)
L87:
  mov    -0xc(%rbp),%eax
  mov    %eax,D25(%rip)
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L98
  mov    -0xc(%rbp),%eax
  and    $0x4,%eax
  test   %eax,%eax
  je     L98
  andl   $0xfffffffb,-0xc(%rbp)
L98:
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L85
  mov    -0xc(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  je     L85
  andl   $0xfffffff7,-0xc(%rbp)
L85:
  mov    -0xc(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L131
  mov    -0xc(%rbp),%eax
  and    $0x4,%eax
  test   %eax,%eax
  je     L131
  andl   $0xfffffffb,-0xc(%rbp)
L131:
  mov    -0xc(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L50
  mov    -0xc(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  je     L50
  andl   $0xfffffff7,-0xc(%rbp)
L50:
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int in_menu_wait(int interesting, int autorep_delay_ms) {
  static int inp_prev = 0;
  static int repeats = 0;
  int ret, release = 0, wait = 450;
  if (repeats)
    wait = autorep_delay_ms;
  ret = in_menu_wait_any(wait);
  if (ret == inp_prev)
    repeats++;
  while (!(ret & interesting)) {
    ret = in_menu_wait_any(-1);
    release = 1;
  }
  if (release || ret != inp_prev)
    repeats = 0;
  inp_prev = ret;
  if ((ret & (1 << 0)) && (ret & (1 << 2)))
    ret &= ~(1 << 2);
  if ((ret & (1 << 0)) && (ret & (1 << 3)))
    ret &= ~(1 << 3);
  if ((ret & (1 << 1)) && (ret & (1 << 2)))
    ret &= ~(1 << 2);
  if ((ret & (1 << 1)) && (ret & (1 << 3)))
    ret &= ~(1 << 3);
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<in_menu_wait>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmpl   $0x0,D11(%rip)
  mov    $0x1c2,%eax
  cmovne %esi,%eax
  mov    %eax,%edi
  call   <in_menu_wait_any@plt>
  cmp    %eax,D44(%rip)
  je     L42
  test   %ebx,%eax
  jne    L31
L16:
  mov    $0xffffffff,%edi
  call   <in_menu_wait_any@plt>
  test   %eax,%ebx
  je     L16
L31:
  movl   $0x0,D11(%rip)
L32:
  mov    %eax,D34(%rip)
  mov    %eax,%ecx
  and    $0x5,%ecx
  mov    %eax,%edx
  and    $0xfffffffb,%edx
  cmp    $0x5,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  and    $0x9,%ecx
  mov    %eax,%edx
  and    $0xfffffff7,%edx
  cmp    $0x9,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  and    $0x6,%ecx
  mov    %eax,%edx
  and    $0xfffffffb,%edx
  cmp    $0x6,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  and    $0xa,%ecx
  mov    %eax,%edx
  and    $0xfffffff7,%edx
  cmp    $0xa,%ecx
  cmove  %edx,%eax
  pop    %rbx
  ret
L42:
  addl   $0x1,D11(%rip)
  test   %ebx,%eax
  je     L16
  jmp    L32
```

assistant: ```c
int in_menu_wait(int interesting, int autorep_delay_ms) {
  static int inp_prev = 0;
  static int repeats = 0;
  int ret, release = 0, wait = 450;
  if (repeats)
    wait = autorep_delay_ms;
  ret = in_menu_wait_any(wait);
  if (ret == inp_prev)
    repeats++;
  while (!(ret & interesting)) {
    ret = in_menu_wait_any(-1);
    release = 1;
  }
  if (release || ret != inp_prev)
    repeats = 0;
  inp_prev = ret;
  if ((ret & (1 << 0)) && (ret & (1 << 2)))
    ret &= ~(1 << 2);
  if ((ret & (1 << 0)) && (ret & (1 << 3)))
    ret &= ~(1 << 3);
  if ((ret & (1 << 1)) && (ret & (1 << 2)))
    ret &= ~(1 << 2);
  if ((ret & (1 << 1)) && (ret & (1 << 3)))
    ret &= ~(1 << 3);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<in_menu_wait>:
  endbr64
  mov    D13(%rip),%eax
  push   %rbx
  mov    %edi,%ebx
  test   %eax,%eax
  mov    $0x1c2,%eax
  cmovne %esi,%eax
  mov    %eax,%edi
  call   <in_menu_wait_any@plt>
  cmp    %eax,D25(%rip)
  jne    L1
  jmp    L4
  nopl   0x0(%rax,%rax,1)
L26:
  mov    $0xffffffff,%edi
  call   <in_menu_wait_any@plt>
L1:
  test   %eax,%ebx
  je     L26
  movl   $0x0,D13(%rip)
L9:
  mov    %eax,%ecx
  mov    %eax,%edx
  mov    %eax,D1(%rip)
  pop    %rbx
  and    $0x5,%ecx
  and    $0xfffffffb,%edx
  cmp    $0x5,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  mov    %eax,%edx
  and    $0x9,%ecx
  and    $0xfffffff7,%edx
  cmp    $0x9,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  mov    %eax,%edx
  and    $0x6,%ecx
  and    $0xfffffffb,%edx
  cmp    $0x6,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  mov    %eax,%edx
  and    $0xa,%ecx
  and    $0xfffffff7,%edx
  cmp    $0xa,%ecx
  cmove  %edx,%eax
  ret
L4:
  addl   $0x1,D13(%rip)
  test   %ebx,%eax
  je     L26
  jmp    L9
```

assistant: ```c
int in_menu_wait(int interesting, int autorep_delay_ms) {
  static int inp_prev = 0;
  static int repeats = 0;
  int ret, release = 0, wait = 450;
  if (repeats)
    wait = autorep_delay_ms;
  ret = in_menu_wait_any(wait);
  if (ret == inp_prev)
    repeats++;
  while (!(ret & interesting)) {
    ret = in_menu_wait_any(-1);
    release = 1;
  }
  if (release || ret != inp_prev)
    repeats = 0;
  inp_prev = ret;
  if ((ret & (1 << 0)) && (ret & (1 << 2)))
    ret &= ~(1 << 2);
  if ((ret & (1 << 0)) && (ret & (1 << 3)))
    ret &= ~(1 << 3);
  if ((ret & (1 << 1)) && (ret & (1 << 2)))
    ret &= ~(1 << 2);
  if ((ret & (1 << 1)) && (ret & (1 << 3)))
    ret &= ~(1 << 3);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<in_menu_wait>:
  endbr64
  mov    D35(%rip),%eax
  push   %rbx
  mov    %edi,%ebx
  test   %eax,%eax
  mov    $0x1c2,%eax
  cmovne %esi,%eax
  mov    %eax,%edi
  call   <in_menu_wait_any@plt>
  cmp    %eax,D44(%rip)
  jne    L40
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L39:
  mov    $0xffffffff,%edi
  call   <in_menu_wait_any@plt>
L40:
  test   %eax,%ebx
  je     L39
  movl   $0x0,D35(%rip)
L16:
  mov    %eax,%ecx
  mov    %eax,%edx
  mov    %eax,D13(%rip)
  pop    %rbx
  and    $0x5,%ecx
  and    $0xfffffffb,%edx
  cmp    $0x5,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  mov    %eax,%edx
  and    $0x9,%ecx
  and    $0xfffffff7,%edx
  cmp    $0x9,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  mov    %eax,%edx
  and    $0x6,%ecx
  and    $0xfffffffb,%edx
  cmp    $0x6,%ecx
  cmove  %edx,%eax
  mov    %eax,%ecx
  mov    %eax,%edx
  and    $0xa,%ecx
  and    $0xfffffff7,%edx
  cmp    $0xa,%ecx
  cmove  %edx,%eax
  ret
L18:
  addl   $0x1,D35(%rip)
  test   %ebx,%eax
  je     L39
  jmp    L16
```

assistant: ```c
int in_menu_wait(int interesting, int autorep_delay_ms) {
  static int inp_prev = 0;
  static int repeats = 0;
  int ret, release = 0, wait = 450;
  if (repeats)
    wait = autorep_delay_ms;
  ret = in_menu_wait_any(wait);
  if (ret == inp_prev)
    repeats++;
  while (!(ret & interesting)) {
    ret = in_menu_wait_any(-1);
    release = 1;
  }
  if (release || ret != inp_prev)
    repeats = 0;
  inp_prev = ret;
  if ((ret & (1 << 0)) && (ret & (1 << 2)))
    ret &= ~(1 << 2);
  if ((ret & (1 << 0)) && (ret & (1 << 3)))
    ret &= ~(1 << 3);
  if ((ret & (1 << 1)) && (ret & (1 << 2)))
    ret &= ~(1 << 2);
  if ((ret & (1 << 1)) && (ret & (1 << 3)))
    ret &= ~(1 << 3);
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <test@plt>
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  js     L14
  addl   $0x1,-0xc(%rbp)
  jmp    L3
L14:
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x10(%rbp)
L3:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a = 0;
  int b = 1;
  test(&a);
  if (a >= 0) {
    b++;
  } else {
    a++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <test@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a = 0;
  int b = 1;
  test(&a);
  if (a >= 0) {
    b++;
  } else {
    a++;
  }
  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
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  call   <test@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a = 0;
  int b = 1;
  test(&a);
  if (a >= 0) {
    b++;
  } else {
    a++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  call   <test@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char *argv[]) {
  int a = 0;
  int b = 1;
  test(&a);
  if (a >= 0) {
    b++;
  } else {
    a++;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<parse_name>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L4:
  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)
L7:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void parse_name(char *buff, char *name, int n) {
  for (int i = 0; i < n; i++) {
    name[i] = buff[i];
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<parse_name>:
  endbr64
  test   %edx,%edx
  jle    L7
  mov    %edx,%edx
  mov    $0x0,%eax
L19:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L19
L7:
  ret
```

assistant: ```c
void parse_name(char *buff, char *name, int n) {
  for (int i = 0; i < n; i++) {
    name[i] = buff[i];
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<parse_name>:
  endbr64
  test   %edx,%edx
  jle    L0
  movslq %edx,%rdx
  xor    %eax,%eax
  nopl   (%rax)
L7:
  movzbl (%rdi,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L7
L0:
  ret
```

assistant: ```c
void parse_name(char *buff, char *name, int n) {
  for (int i = 0; i < n; i++) {
    name[i] = buff[i];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_name>:
  endbr64
  mov    %rdi,%rcx
  test   %edx,%edx
  jle    L144
  lea    0x1(%rdi),%r9
  mov    %rsi,%rdi
  lea    -0x1(%rdx),%eax
  mov    %edx,%r8d
  sub    %r9,%rdi
  cmp    $0xe,%rdi
  jbe    L105
  cmp    $0x6,%eax
  jbe    L105
  cmp    $0xe,%eax
  jbe    L17
  mov    %edx,%edi
  xor    %eax,%eax
  shr    $0x4,%edi
  shl    $0x4,%rdi
  cs nopw 0x0(%rax,%rax,1)
L52:
  movdqu (%rcx,%rax,1),%xmm0
  movups %xmm0,(%rsi,%rax,1)
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L52
  mov    %edx,%edi
  and    $0xfffffff0,%edi
  mov    %edi,%eax
  cmp    %edi,%edx
  je     L144
  mov    %edx,%r8d
  sub    %edi,%r8d
  lea    -0x1(%r8),%r9d
  cmp    $0x6,%r9d
  jbe    L154
L2:
  mov    (%rcx,%rdi,1),%r9
  mov    %r9,(%rsi,%rdi,1)
  mov    %r8d,%edi
  and    $0xfffffff8,%edi
  add    %edi,%eax
  cmp    %edi,%r8d
  je     L144
L154:
  movslq %eax,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x1(%rax),%edi
  cmp    %edi,%edx
  jle    L144
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x2(%rax),%edi
  cmp    %edi,%edx
  jle    L144
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x3(%rax),%edi
  cmp    %edi,%edx
  jle    L144
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x4(%rax),%edi
  cmp    %edi,%edx
  jle    L144
  movslq %edi,%rdi
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  lea    0x5(%rax),%edi
  cmp    %edi,%edx
  jle    L144
  movslq %edi,%rdi
  add    $0x6,%eax
  movzbl (%rcx,%rdi,1),%r8d
  mov    %r8b,(%rsi,%rdi,1)
  cmp    %eax,%edx
  jle    L144
  cltq
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%rsi,%rax,1)
  ret
  nopw   0x0(%rax,%rax,1)
L105:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L155:
  movzbl (%rcx,%rax,1),%edi
  mov    %dil,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L155
L144:
  ret
L17:
  xor    %edi,%edi
  xor    %eax,%eax
  jmp    L2
```

assistant: ```c
void parse_name(char *buff, char *name, int n) {
  for (int i = 0; i < n; i++) {
    name[i] = buff[i];
  }
}
```