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

```asm
<print_int>:
  endbr64
  mov    %edi,%eax
  neg    %eax
  cmovs  %edi,%eax
  test   %edi,%edi
  je     L3
  xor    %esi,%esi
  mov    $0xcccccccd,%edx
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    %eax,%eax
  add    $0x1,%esi
  imul   %rdx,%rax
  shr    $0x23,%rax
  test   %eax,%eax
  jne    L7
  jmp    1050 <print@plt>
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    $0x1,%esi
  jmp    1050 <print@plt>
```

assistant: ```c
void print_int(int _num) {
  int copy = ((_num < 0) ? -_num : _num);
  int size = 0;
  while (copy != 0) {
    size += 1;
    copy /= 10;
  }
  size = (size != 0) ? size : 1;
  int sign = 0;
  print(_num, sign + size);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<print_int>:
  endbr64
  mov    %edi,%eax
  neg    %eax
  cmovs  %edi,%eax
  test   %edi,%edi
  je     L6
  xor    %esi,%esi
  mov    $0xcccccccd,%edx
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    %eax,%eax
  add    $0x1,%esi
  imul   %rdx,%rax
  shr    $0x23,%rax
  test   %eax,%eax
  jne    L15
  jmp    1050 <print@plt>
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    $0x1,%esi
  jmp    1050 <print@plt>
```

assistant: ```c
void print_int(int _num) {
  int copy = ((_num < 0) ? -_num : _num);
  int size = 0;
  while (copy != 0) {
    size += 1;
    copy /= 10;
  }
  size = (size != 0) ? size : 1;
  int sign = 0;
  print(_num, sign + size);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<irc_client>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    %r8,-0x20(%rbp)
  mov    %r9,-0x28(%rbp)
  mov    $0x0,%edi
  call   <_irc_queue_init@plt>
  test   %rax,%rax
  je     L0
  mov    $0xffffffff,%eax
  jmp    L5
L0:
  mov    -0x20(%rbp),%rdi
  mov    -0x18(%rbp),%rsi
  mov    -0x8(%rbp),%ecx
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  sub    $0x8,%rsp
  push   -0x28(%rbp)
  mov    %rdi,%r9
  mov    %rsi,%r8
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <client@plt>
  add    $0x10,%rsp
L5:
  leave
  ret
```

assistant: ```c
int irc_client(int sid, char *addr, int port, char *nick, char *username, char *realname) {
  if (_irc_queue_init(0) != 0)
    return -1;
  return client(sid, 1, addr, port, nick, username, realname);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<irc_client>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  mov    %rcx,%r13
  mov    %r8,%r14
  mov    %r9,%r15
  mov    $0x0,%edi
  call   <_irc_queue_init@plt>
  test   %rax,%rax
  jne    L16
  sub    $0x8,%rsp
  push   %r15
  mov    %r14,%r9
  mov    %r13,%r8
  mov    %r12d,%ecx
  mov    %rbp,%rdx
  mov    $0x1,%esi
  mov    %ebx,%edi
  call   <client@plt>
  add    $0x10,%rsp
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L16:
  mov    $0xffffffff,%eax
  jmp    L5
```

assistant: ```c
int irc_client(int sid, char *addr, int port, char *nick, char *username, char *realname) {
  if (_irc_queue_init(0) != 0)
    return -1;
  return client(sid, 1, addr, port, nick, username, realname);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<irc_client>:
  endbr64
  push   %r15
  mov    %r9,%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x8,%rsp
  call   <_irc_queue_init@plt>
  test   %rax,%rax
  jne    L1
  sub    $0x8,%rsp
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    %r14,%r9
  push   %r15
  mov    %rbx,%r8
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <client@plt>
  pop    %rdx
  pop    %rcx
L7:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L1:
  mov    $0xffffffff,%eax
  jmp    L7
```

assistant: ```c
int irc_client(int sid, char *addr, int port, char *nick, char *username, char *realname) {
  if (_irc_queue_init(0) != 0)
    return -1;
  return client(sid, 1, addr, port, nick, username, realname);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<irc_client>:
  endbr64
  push   %r15
  mov    %r9,%r15
  push   %r14
  mov    %r8,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x8,%rsp
  call   <_irc_queue_init@plt>
  test   %rax,%rax
  jne    L2
  sub    $0x8,%rsp
  mov    %r13d,%ecx
  mov    %r12,%rdx
  mov    %r14,%r9
  push   %r15
  mov    %rbx,%r8
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <client@plt>
  pop    %rdx
  pop    %rcx
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L2:
  mov    $0xffffffff,%eax
  jmp    L18
```

assistant: ```c
int irc_client(int sid, char *addr, int port, char *nick, char *username, char *realname) {
  if (_irc_queue_init(0) != 0)
    return -1;
  return client(sid, 1, addr, port, nick, username, realname);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<generate_cube>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    $0x2000,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L6
L2:
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    (%rdx,%rdx,1),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %ax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L6:
  cmpl   $0xfff,-0xc(%rbp)
  jle    L2
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
unsigned short *generate_cube(x, y, z) {
  unsigned short *map;
  int i;
  map = (unsigned short *)malloc(16 * 16 * 16 * 2);
  for (i = 0; i < 16 * 16 * 16; i++) {
    map[i] = (i % 3);
  }
  return map;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<generate_cube>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2000,%edi
  call   <malloc@plt>
  mov    $0x0,%ecx
L7:
  movslq %ecx,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,2),%esi
  mov    %ecx,%edx
  sub    %esi,%edx
  mov    %dx,(%rax,%rcx,2)
  add    $0x1,%rcx
  cmp    $0x1000,%rcx
  jne    L7
  add    $0x8,%rsp
  ret
```

assistant: ```c
unsigned short *generate_cube(x, y, z) {
  unsigned short *map;
  int i;
  map = (unsigned short *)malloc(16 * 16 * 16 * 2);
  for (i = 0; i < 16 * 16 * 16; i++) {
    map[i] = (i % 3);
  }
  return map;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<generate_cube>:
  endbr64
  mov    $0x2000,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  xor    %ecx,%ecx
  mov    $0xaaaaaaab,%edi
  nopl   0x0(%rax)
L6:
  mov    %ecx,%edx
  imul   %rdi,%rdx
  shr    $0x21,%rdx
  lea    (%rdx,%rdx,2),%esi
  mov    %ecx,%edx
  sub    %esi,%edx
  mov    %dx,(%rax,%rcx,2)
  add    $0x1,%rcx
  cmp    $0x1000,%rcx
  jne    L6
  add    $0x8,%rsp
  ret
```

assistant: ```c
unsigned short *generate_cube(x, y, z) {
  unsigned short *map;
  int i;
  map = (unsigned short *)malloc(16 * 16 * 16 * 2);
  for (i = 0; i < 16 * 16 * 16; i++) {
    map[i] = (i % 3);
  }
  return map;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<generate_cube>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2000,%edi
  call   <malloc@plt>
  movdqa D4(%rip),%xmm3
  movdqa D10(%rip),%xmm6
  movdqa D23(%rip),%xmm5
  movdqa D22(%rip),%xmm4
  mov    %rax,%rdx
  lea    0x2000(%rax),%rcx
  nopl   0x0(%rax)
L1:
  movdqa %xmm3,%xmm0
  add    $0x10,%rdx
  paddd  %xmm6,%xmm3
  movdqa %xmm0,%xmm2
  movdqa %xmm0,%xmm1
  paddd  %xmm5,%xmm2
  punpcklwd %xmm2,%xmm0
  punpckhwd %xmm2,%xmm1
  movdqa %xmm0,%xmm2
  punpcklwd %xmm1,%xmm0
  punpckhwd %xmm1,%xmm2
  punpcklwd %xmm2,%xmm0
  movdqa %xmm0,%xmm1
  pmulhuw %xmm4,%xmm1
  psrlw  $0x1,%xmm1
  movdqa %xmm1,%xmm2
  psllw  $0x1,%xmm2
  paddw  %xmm2,%xmm1
  psubw  %xmm1,%xmm0
  movups %xmm0,-0x10(%rdx)
  cmp    %rdx,%rcx
  jne    L1
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type_guessed: byte[16]
name: disassemble_data
arguments: tag: D10
type_guessed: byte[16]
name: disassemble_data
arguments: tag: D23
type_guessed: byte[16]
name: disassemble_data
arguments: tag: D22
type_guessed: byte[16]

tool: {"D4": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D10": [0x08,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D23": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D22": [0xab,0xaa,0xab,0xaa,0xab,0xaa,0xab,0xaa,0xab,0xaa,0xab,0xaa,0xab,0xaa,0xab,0xaa]}

assistant: ```c
unsigned short *generate_cube(x, y, z) {
  unsigned short *map;
  int i;
  map = (unsigned short *)malloc(16 * 16 * 16 * 2);
  for (i = 0; i < 16 * 16 * 16; i++) {
    map[i] = (i % 3);
  }
  return map;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <twiddle1@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <twiddle2@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <twiddle1@plt>
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <twiddle2@plt>
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D12
type_infered: string
name: read_data
arguments: tag: D36
type_infered: string

tool: D12: %d 

tool: D36: %d


assistant: ```c
int main(void) {
  int a, b;
  a = 0;
  b = 2;
  printf("%d ", twiddle1(&a, &b));
  a = 0;
  b = 2;
  printf("%d ", twiddle2(&a, &b));
  a = 0;
  b = 2;
  printf("%d ", twiddle1(&b, &b));
  a = 0;
  b = 2;
  printf("%d\n", twiddle2(&b, &b));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  lea    0x4(%rsp),%rbx
  mov    %rsp,%r12
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <twiddle1@plt>
  mov    %eax,%esi
  lea    D17(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <twiddle2@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  call   <twiddle1@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  mov    %rbx,%rsi
  mov    %rbx,%rdi
  call   <twiddle2@plt>
  mov    %eax,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D17
type_infered: string
name: read_data
arguments: tag: D10
type_infered: string

tool: D17: %d 

tool: D10: %d


assistant: ```c
int main(void) {
  int a, b;
  a = 0;
  b = 2;
  printf("%d ", twiddle1(&a, &b));
  a = 0;
  b = 2;
  printf("%d ", twiddle2(&a, &b));
  a = 0;
  b = 2;
  printf("%d ", twiddle1(&b, &b));
  a = 0;
  b = 2;
  printf("%d\n", twiddle2(&b, &b));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D18(%rip),%r12
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  mov    %rsp,%r13
  movl   $0x0,(%rsp)
  mov    %rbp,%rsi
  mov    %r13,%rdi
  movl   $0x2,0x4(%rsp)
  call   <twiddle1@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  call   <twiddle2@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbp,%rdi
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  call   <twiddle1@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbp,%rdi
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  call   <twiddle2@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D18
type_guessed: string
name: retrieve_data
arguments: label: D0
type_guessed: string

tool: D18: %d 

tool: D0: %d


assistant: ```c
int main(void) {
  int a, b;
  a = 0;
  b = 2;
  printf("%d ", twiddle1(&a, &b));
  a = 0;
  b = 2;
  printf("%d ", twiddle2(&a, &b));
  a = 0;
  b = 2;
  printf("%d ", twiddle1(&b, &b));
  a = 0;
  b = 2;
  printf("%d\n", twiddle2(&b, &b));
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D17(%rip),%r12
  push   %rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  mov    %rsp,%r13
  movl   $0x0,(%rsp)
  mov    %rbp,%rsi
  mov    %r13,%rdi
  movl   $0x2,0x4(%rsp)
  call   <twiddle1@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  call   <twiddle2@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbp,%rdi
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  call   <twiddle1@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbp,%rdi
  movl   $0x0,(%rsp)
  movl   $0x2,0x4(%rsp)
  call   <twiddle2@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D6
type_guessed: string

tool: D17: %d 

tool: D6: %d


assistant: ```c
int main(void) {
  int a, b;
  a = 0;
  b = 2;
  printf("%d ", twiddle1(&a, &b));
  a = 0;
  b = 2;
  printf("%d ", twiddle2(&a, &b));
  a = 0;
  b = 2;
  printf("%d ", twiddle1(&b, &b));
  a = 0;
  b = 2;
  printf("%d\n", twiddle2(&b, &b));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_nbt_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L18
  mov    $0x0,%eax
  jmp    L4
L18:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
L4:
  leave
  ret
```

assistant: ```c
char *_nbt_strdup(const char *s) {
  char *r = malloc(strlen(s) + 1);
  if (r == NULL)
    return NULL;
  strcpy(r, s);
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<_nbt_strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L6
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L6:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *_nbt_strdup(const char *s) {
  char *r = malloc(strlen(s) + 1);
  if (r == NULL)
    return NULL;
  strcpy(r, s);
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_nbt_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L2:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *_nbt_strdup(const char *s) {
  char *r = malloc(strlen(s) + 1);
  if (r == NULL)
    return NULL;
  strcpy(r, s);
  return r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_nbt_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L3
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L3:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *_nbt_strdup(const char *s) {
  char *r = malloc(strlen(s) + 1);
  if (r == NULL)
    return NULL;
  strcpy(r, s);
  return r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<tolowers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x40,-0x4(%rbp)
  jle    L4
  cmpl   $0x5a,-0x4(%rbp)
  jg     L4
  addl   $0x20,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int tolowers(int c) {
  if ('A' <= c && c <= 'Z') {
    c += 'a' - 'A';
  }
  return c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tolowers>:
  endbr64
  mov    %edi,%eax
  lea    -0x41(%rdi),%ecx
  lea    0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int tolowers(int c) {
  if ('A' <= c && c <= 'Z') {
    c += 'a' - 'A';
  }
  return c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tolowers>:
  endbr64
  lea    -0x41(%rdi),%ecx
  mov    %edi,%eax
  lea    0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int tolowers(int c) {
  if ('A' <= c && c <= 'Z') {
    c += 'a' - 'A';
  }
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<tolowers>:
  endbr64
  lea    -0x41(%rdi),%ecx
  mov    %edi,%eax
  lea    0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int tolowers(int c) {
  if ('A' <= c && c <= 'Z') {
    c += 'a' - 'A';
  }
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  add    $0x51e22cdb,%rax
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  nop
  mov    -0x10(%rbp),%rax
  or     $0x65fb30,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  char copy11;
  unsigned int copy12;
  {
    state[0UL] = input[0UL] + 1373777115UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] == local1) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 4);
        *((char *)(&state[0UL]) + 4) = copy11;
        state[local1] = (state[local1] >> (((state[0UL] >> 2UL) & 15UL) | 1UL)) | (state[local1] << (64 - (((state[0UL] >> 2UL) & 15UL) | 1UL)));
      } else {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
        state[0UL] |= (state[local1] & 15UL) << 3UL;
      }
      if (state[0UL] > local1) {
        state[0UL] *= state[0UL];
        state[0UL] |= ((state[local1] + state[0UL]) & 7UL) << 2UL;
      } else {
        state[0UL] = state[local1] | state[0UL];
        state[local1] = (state[local1] << (((state[local1] >> 4UL) & 15UL) | 1UL)) | (state[local1] >> (64 - (((state[local1] >> 4UL) & 15UL) | 1UL)));
      }
      local1++;
    }
    output[0UL] = state[0UL] | 6683440UL;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  add    $0x51e22cdb,%rax
  or     $0x65fb30,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  char copy11;
  unsigned int copy12;
  {
    state[0UL] = input[0UL] + 1373777115UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] == local1) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 4);
        *((char *)(&state[0UL]) + 4) = copy11;
        state[local1] = (state[local1] >> (((state[0UL] >> 2UL) & 15UL) | 1UL)) | (state[local1] << (64 - (((state[0UL] >> 2UL) & 15UL) | 1UL)));
      } else {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
        state[0UL] |= (state[local1] & 15UL) << 3UL;
      }
      if (state[0UL] > local1) {
        state[0UL] *= state[0UL];
        state[0UL] |= ((state[local1] + state[0UL]) & 7UL) << 2UL;
      } else {
        state[0UL] = state[local1] | state[0UL];
        state[local1] = (state[local1] << (((state[local1] >> 4UL) & 15UL) | 1UL)) | (state[local1] >> (64 - (((state[local1] >> 4UL) & 15UL) | 1UL)));
      }
      local1++;
    }
    output[0UL] = state[0UL] | 6683440UL;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  add    $0x51e22cdb,%rax
  or     $0x65fb30,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned long local1;
  char copy11;
  unsigned int copy12;
  {
    state[0UL] = input[0UL] + 1373777115UL;
    local1 = 0UL;
    while (local1 < 0UL) {
      if (state[0UL] == local1) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 4);
        *((char *)(&state[0UL]) + 4) = copy11;
        state[local1] = (state[local1] >> (((state[0UL] >> 2UL) & 15UL) | 1UL)) | (state[local1] << (64 - (((state[0UL] >> 2UL) & 15UL) | 1UL)));
      } else {
        copy12 = *((unsigned int *)(&state[local1]) + 0);
        *((unsigned int *)(&state[local1]) + 0) = *((unsigned int *)(&state[local1]) + 1);
        *((unsigned int *)(&state[local1]) + 1) = copy12;
        state[0UL] |= (state[local1] & 15UL) << 3UL;
      }
      if (state[0UL] > local1) {
        state[0UL] *= state[0UL];
        state[0UL] |= ((state[local1] + state[0UL]) & 7UL) << 2UL;
      } else {
        state[0UL] = state[local1] | state[0UL];
        state[local1] = (state[local1] << (((state[local1] >> 4UL) & 15UL) | 1UL)) | (state[local1] >> (64 - (((state[local1] >> 4UL) & 15UL) | 1UL)));
      }
      local1++;
    }
    output[0UL] = state[0UL] | 6683440UL;
  }
}
```