ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D44(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  cmp    $0x13,%eax
  jle    L16
  imul   0x4(%rsp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  D39(%rip),%xmm0
  divsd  D41(%rip),%xmm0
  cvttsd2si %xmm0,%esi
L10:
  lea    D52(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L16:
  mov    0x4(%rsp),%ecx
  mov    $0x2,%esi
  cltd
  idiv   %esi
  imul   %ecx,%eax
  add    %eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  D7(%rip),%xmm0
  divsd  D41(%rip),%xmm0
  imul   %ecx,%edx
  pxor   %xmm1,%xmm1
  cvtsi2sd %edx,%xmm1
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  jmp    L10
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D44
type_guessed: string
name: extract_symbolic_data
arguments: tag: D39
type_guessed: f64
name: extract_symbolic_data
arguments: tag: D41
type_guessed: byte[8]
name: extract_symbolic_data
arguments: tag: D52
type_guessed: string
name: extract_symbolic_data
arguments: tag: D7
type_guessed: f64

tool: D44: %d %d

tool: D39: 80

tool: {"D41": [0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

tool: D52: %.2f

tool: D7: 90

assistant: ```c
int main() {
  int user, price;
  scanf("%d %d", &user, &price);
  printf("%.2f", (user >= 20) ? price * user * 80.0 / 100 : ((user / 2) * 2) * price * 90.0 / 100 + (user % 2) * price);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D30(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  cmp    $0x13,%eax
  jle    L16
  imul   0x4(%rsp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  D17(%rip),%xmm0
  divsd  D35(%rip),%xmm0
  cvttsd2si %xmm0,%esi
L4:
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L16:
  mov    $0x2,%esi
  cltd
  mov    0x4(%rsp),%ecx
  pxor   %xmm1,%xmm1
  idiv   %esi
  pxor   %xmm0,%xmm0
  imul   %ecx,%edx
  imul   %ecx,%eax
  cvtsi2sd %edx,%xmm1
  add    %eax,%eax
  cvtsi2sd %eax,%xmm0
  mulsd  D31(%rip),%xmm0
  divsd  D35(%rip),%xmm0
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  jmp    L4
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D30
type_guessed: string
name: read_data
arguments: data_label: D17
type_guessed: f64
name: read_data
arguments: data_label: D35
type_guessed: byte[8]
name: read_data
arguments: data_label: D38
type_guessed: string
name: read_data
arguments: data_label: D31
type_guessed: f64

tool: D30: %d %d

tool: D17: 80

tool: {"D35": [0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

tool: D38: %.2f

tool: D31: 90

assistant: ```c
int main() {
  int user, price;
  scanf("%d %d", &user, &price);
  printf("%.2f", (user >= 20) ? price * user * 80.0 / 100 : ((user / 2) * 2) * price * 90.0 / 100 + (user % 2) * price);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D46(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  cmp    $0x13,%eax
  jle    L6
  imul   0x4(%rsp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  D47(%rip),%xmm0
  divsd  D11(%rip),%xmm0
  cvttsd2si %xmm0,%esi
L2:
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L6:
  mov    $0x2,%esi
  cltd
  mov    0x4(%rsp),%ecx
  pxor   %xmm1,%xmm1
  idiv   %esi
  pxor   %xmm0,%xmm0
  imul   %ecx,%edx
  imul   %ecx,%eax
  cvtsi2sd %edx,%xmm1
  add    %eax,%eax
  cvtsi2sd %eax,%xmm0
  mulsd  D9(%rip),%xmm0
  divsd  D11(%rip),%xmm0
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%esi
  jmp    L2
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D46
type: string
name: extract_symbolic_data
arguments: tag: D47
type: f64
name: extract_symbolic_data
arguments: tag: D11
type: byte[8]
name: extract_symbolic_data
arguments: tag: D39
type: string
name: extract_symbolic_data
arguments: tag: D9
type: f64

tool: D46: %d %d

tool: D47: 80

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0x59,0x40]}

tool: D39: %.2f

tool: D9: 90

assistant: ```c
int main() {
  int user, price;
  scanf("%d %d", &user, &price);
  printf("%.2f", (user >= 20) ? price * user * 80.0 / 100 : ((user / 2) * 2) * price * 90.0 / 100 + (user % 2) * price);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %edi,-0xc4(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L5
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L5:
  mov    -0xd0(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xd0(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <vprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0xc4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int _error(int code, char const *format, ...) {
  va_list ap;
  va_start(ap, format);
  vprintf(format, ap);
  va_end(ap);
  putchar('\n');
  return code;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<_error>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%rbx
  mov    $0x0,%r12d
  mov    %r12d,%edi
  call   <va_start@plt>
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <vprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int _error(int code, char const *format, ...) {
  va_list ap;
  va_start(ap, format);
  vprintf(format, ap);
  va_end(ap);
  putchar('\n');
  return code;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<_error>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edi,%r12d
  mov    %r13d,%edi
  push   %rbp
  mov    %rsi,%rbp
  call   <va_start@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <vprintf@plt>
  mov    %r13d,%edi
  call   <va_end@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int _error(int code, char const *format, ...) {
  va_list ap;
  va_start(ap, format);
  vprintf(format, ap);
  va_end(ap);
  putchar('\n');
  return code;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_error>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %edi,%r12d
  mov    %r13d,%edi
  push   %rbp
  mov    %rsi,%rbp
  call   <va_start@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <vprintf@plt>
  mov    %r13d,%edi
  call   <va_end@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int _error(int code, char const *format, ...) {
  va_list ap;
  va_start(ap, format);
  vprintf(format, ap);
  va_end(ap);
  putchar('\n');
  return code;
}
```
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
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
#pragma starpu initialize
#pragma starpu shutdown foo bar
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
#pragma starpu initialize
#pragma starpu shutdown foo bar
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
#pragma starpu initialize
#pragma starpu shutdown foo bar
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
#pragma starpu initialize
#pragma starpu shutdown foo bar
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jle    L16
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L25
L16:
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L25:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D4
datatype: string
name: disassemble_data
arguments: data_label: D2
datatype: string
name: disassemble_data
arguments: data_label: D11
datatype: string

tool: D4: %d %d

tool: D2: violation


tool: D11: ok


assistant: ```c
int main() {
  int currentSpeed, maxSpeed;
  scanf("%d %d", &currentSpeed, &maxSpeed);
  if (currentSpeed > maxSpeed) {
    printf("violation\n");
  } else {
    printf("ok\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %eax,(%rsp)
  jle    L19
  lea    D18(%rip),%rdi
  call   <printf@plt>
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L19:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  jmp    L8
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D21
type_infered: string
name: disassemble_data
arguments: tag: D18
type_infered: string
name: disassemble_data
arguments: tag: D9
type_infered: string

tool: D21: %d %d

tool: D18: violation


tool: D9: ok


assistant: ```c
int main() {
  int currentSpeed, maxSpeed;
  scanf("%d %d", &currentSpeed, &maxSpeed);
  if (currentSpeed > maxSpeed) {
    printf("violation\n");
  } else {
    printf("ok\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D29(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %eax,(%rsp)
  jle    L24
  lea    D1(%rip),%rdi
  call   <printf@plt>
L20:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L24:
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L20
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D29
data_type: string
name: retrieve_data
arguments: tag: D1
data_type: string
name: retrieve_data
arguments: tag: D14
data_type: string

tool: D29: %d %d

tool: D1: violation


tool: D14: ok


assistant: ```c
int main() {
  int currentSpeed, maxSpeed;
  scanf("%d %d", &currentSpeed, &maxSpeed);
  if (currentSpeed > maxSpeed) {
    printf("violation\n");
  } else {
    printf("ok\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D0(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %eax,(%rsp)
  jle    L0
  lea    D20(%rip),%rdi
  call   <printf@plt>
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L0:
  lea    D17(%rip),%rdi
  call   <printf@plt>
  jmp    L22
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D0
datatype: string
name: unpack_data
arguments: label: D20
datatype: string
name: unpack_data
arguments: label: D17
datatype: string

tool: D0: %d %d

tool: D20: violation


tool: D17: ok


assistant: ```c
int main() {
  int currentSpeed, maxSpeed;
  scanf("%d %d", &currentSpeed, &maxSpeed);
  if (currentSpeed > maxSpeed) {
    printf("violation\n");
  } else {
    printf("ok\n");
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<Memory_calloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L16
  mov    -0x8(%rbp),%rax
  jmp    L2
L16:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
L2:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
type_infered: string

tool: D9: [memory.c] Could not allocate %d objects of size %d


assistant: ```c
void *Memory_calloc(size_t num, size_t size) {
  void *ptr;
  ptr = calloc(num, size);
  if (ptr != ((void *)0)) {
    return ptr;
  }
  printf("[memory.c] Could not allocate %d objects of size %d\n", num, size);
  return ((void *)0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<Memory_calloc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  call   <calloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L9
L4:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  jmp    L4
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_guessed: string

tool: D0: [memory.c] Could not allocate %d objects of size %d


assistant: ```c
void *Memory_calloc(size_t num, size_t size) {
  void *ptr;
  ptr = calloc(num, size);
  if (ptr != ((void *)0)) {
    return ptr;
  }
  printf("[memory.c] Could not allocate %d objects of size %d\n", num, size);
  return ((void *)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
<Memory_calloc>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L14
L7:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  jmp    L7
```

tool_calls: name: disassemble_data
arguments: tag: D4
type: string

tool: D4: [memory.c] Could not allocate %d objects of size %d


assistant: ```c
void *Memory_calloc(size_t num, size_t size) {
  void *ptr;
  ptr = calloc(num, size);
  if (ptr != ((void *)0)) {
    return ptr;
  }
  printf("[memory.c] Could not allocate %d objects of size %d\n", num, size);
  return ((void *)0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Memory_calloc>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L17
L6:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L6
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_guessed: string

tool: D6: [memory.c] Could not allocate %d objects of size %d


assistant: ```c
void *Memory_calloc(size_t num, size_t size) {
  void *ptr;
  ptr = calloc(num, size);
  if (ptr != ((void *)0)) {
    return ptr;
  }
  printf("[memory.c] Could not allocate %d objects of size %d\n", num, size);
  return ((void *)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
<present_p>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L23
L22:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  add    %edx,%edx
  mov    %edx,(%rax)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  sar    $0x7,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x18(%rbp),%rax
  add    %rsi,%rax
  or     %ecx,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  add    %edx,%edx
  mov    %edx,(%rax)
  mov    -0x4(%rbp),%eax
  add    $0x2,%eax
  and    $0x7,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0x7,%eax
  cmp    $0x7,%eax
  jne    L27
  addl   $0x1,-0x8(%rbp)
L27:
  cmpl   $0x1f,-0xc(%rbp)
  jne    L9
  addl   $0x1,-0x4(%rbp)
L9:
  addl   $0x1,-0xc(%rbp)
L23:
  cmpl   $0x3f,-0xc(%rbp)
  jle    L22
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void present_p(uint8_t *o, uint8_t *i) {
  uint8_t m, n = 0, idx = 0;
  for (m = 0; m < 64; ++m) {
    o[idx] <<= 1;
    o[idx] |= i[n] >> 7;
    i[n] <<= 1;
    idx = (idx + 2) & 7;
    if ((m & 7) == 7) {
      ++n;
    }
    if (m == 31) {
      idx += 1;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<present_p>:
  endbr64
  push   %rbx
  mov    %rdi,%r11
  mov    %rsi,%rbx
  mov    $0x1,%r10d
  mov    $0x0,%r8d
  mov    $0x0,%edx
  mov    $0x0,%r9d
  jmp    L29
L14:
  add    $0x1,%edx
L21:
  add    $0x1,%r8d
  add    $0x1,%r10d
L29:
  movslq %edx,%rax
  lea    (%r11,%rax,4),%rdi
  mov    (%rdi),%eax
  lea    (%rax,%rax,1),%ecx
  mov    %ecx,(%rdi)
  movslq %r9d,%rax
  lea    (%rbx,%rax,4),%rsi
  mov    (%rsi),%eax
  sar    $0x7,%eax
  or     %ecx,%eax
  mov    %eax,(%rdi)
  shll   (%rsi)
  add    $0x2,%edx
  and    $0x7,%edx
  mov    %r8d,%eax
  and    $0x7,%eax
  cmp    $0x7,%eax
  sete   %al
  movzbl %al,%eax
  add    %eax,%r9d
  cmp    $0x1f,%r8d
  je     L14
  cmp    $0x3f,%r10d
  jle    L21
  pop    %rbx
  ret
```

assistant: ```c
void present_p(uint8_t *o, uint8_t *i) {
  uint8_t m, n = 0, idx = 0;
  for (m = 0; m < 64; ++m) {
    o[idx] <<= 1;
    o[idx] |= i[n] >> 7;
    i[n] <<= 1;
    idx = (idx + 2) & 7;
    if ((m & 7) == 7) {
      ++n;
    }
    if (m == 31) {
      idx += 1;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<present_p>:
  endbr64
  mov    %rdi,%r11
  xor    %edx,%edx
  xor    %r10d,%r10d
  xor    %edi,%edi
  xchg   %ax,%ax
L29:
  movslq %edx,%rax
  add    $0x2,%edx
  lea    (%r11,%rax,4),%r9
  and    $0x7,%edx
  mov    (%r9),%eax
  lea    (%rax,%rax,1),%ecx
  movslq %r10d,%rax
  lea    (%rsi,%rax,4),%r8
  mov    %ecx,(%r9)
  mov    (%r8),%eax
  sar    $0x7,%eax
  or     %ecx,%eax
  mov    %eax,(%r9)
  mov    %edi,%eax
  and    $0x7,%eax
  shll   (%r8)
  cmp    $0x7,%eax
  sete   %al
  add    $0x1,%edi
  movzbl %al,%eax
  add    %eax,%r10d
  cmp    $0x20,%edi
  je     L6
  cmp    $0x40,%edi
  jne    L29
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  add    $0x1,%edx
  jmp    L29
```

assistant: ```c
void present_p(uint8_t *o, uint8_t *i) {
  uint8_t m, n = 0, idx = 0;
  for (m = 0; m < 64; ++m) {
    o[idx] <<= 1;
    o[idx] |= i[n] >> 7;
    i[n] <<= 1;
    idx = (idx + 2) & 7;
    if ((m & 7) == 7) {
      ++n;
    }
    if (m == 31) {
      idx += 1;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<present_p>:
  endbr64
  mov    %rdi,%r11
  xor    %edx,%edx
  xor    %r10d,%r10d
  xor    %edi,%edi
  xchg   %ax,%ax
L2:
  movslq %edx,%rax
  add    $0x2,%edx
  lea    (%r11,%rax,4),%r9
  and    $0x7,%edx
  mov    (%r9),%eax
  lea    (%rax,%rax,1),%ecx
  movslq %r10d,%rax
  lea    (%rsi,%rax,4),%r8
  mov    %ecx,(%r9)
  mov    (%r8),%eax
  sar    $0x7,%eax
  or     %ecx,%eax
  mov    %eax,(%r9)
  mov    %edi,%eax
  and    $0x7,%eax
  shll   (%r8)
  cmp    $0x7,%eax
  sete   %al
  add    $0x1,%edi
  movzbl %al,%eax
  add    %eax,%r10d
  cmp    $0x20,%edi
  je     L8
  cmp    $0x40,%edi
  jne    L2
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  add    $0x1,%edx
  jmp    L2
```

assistant: ```c
void present_p(uint8_t *o, uint8_t *i) {
  uint8_t m, n = 0, idx = 0;
  for (m = 0; m < 64; ++m) {
    o[idx] <<= 1;
    o[idx] |= i[n] >> 7;
    i[n] <<= 1;
    idx = (idx + 2) & 7;
    if ((m & 7) == 7) {
      ++n;
    }
    if (m == 31) {
      idx += 1;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ComputeClass>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbx
  mov    %edx,%r12d
  mov    %ecx,%ebp
  lea    0x0(,%rcx,4),%edi
  call   <malloc@plt>
  mov    %rax,%r9
  movslq 0x0(%r13),%rax
  movl   $0x0,(%r9,%rax,4)
  cmp    $0x1,%ebp
  jle    L30
  mov    %r13,%rcx
  lea    -0x2(%rbp),%eax
  lea    0x4(%r13,%rax,4),%r10
  jmp    L41
L7:
  add    %r12d,%eax
  cltd
  idiv   %ebp
  movslq %edx,%r11
  lea    (%rsi,%r12,1),%eax
  cltd
  idiv   %ebp
  movslq %edx,%rdx
  mov    (%rbx,%rdx,4),%eax
  cmp    %eax,(%rbx,%r11,4)
  je     L27
  mov    (%r9,%r8,4),%eax
  add    $0x1,%eax
  mov    %eax,(%r9,%rdi,4)
L1:
  add    $0x4,%rcx
  cmp    %r10,%rcx
  je     L30
L41:
  mov    0x4(%rcx),%eax
  movslq %eax,%rdi
  mov    (%rcx),%esi
  movslq %esi,%r8
  mov    (%rbx,%r8,4),%edx
  cmp    %edx,(%rbx,%rdi,4)
  je     L7
  mov    (%r9,%r8,4),%eax
  add    $0x1,%eax
  mov    %eax,(%r9,%rdi,4)
  jmp    L1
L27:
  mov    (%r9,%r8,4),%eax
  mov    %eax,(%r9,%rdi,4)
  jmp    L1
L30:
  mov    %r9,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int *ComputeClass(int *order, int *clas, int l, int str_len) {
  int *newClass = (int *)malloc(str_len * sizeof(int));
  newClass[order[0]] = 0;
  for (int i = 1; i <= str_len - 1; i++) {
    if (clas[order[i]] != clas[order[i - 1]])
      newClass[order[i]] = newClass[order[i - 1]] + 1;
    else {
      if (clas[(order[i] + l) % str_len] != clas[(order[i - 1] + l) % str_len])
        newClass[order[i]] = newClass[order[i - 1]] + 1;
      else
        newClass[order[i]] = newClass[order[i - 1]];
    }
  }
  return newClass;
}
```