ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<py_timestep_leapfrog>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L5
L19:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <py_leapfrog@plt>
  addq   $0x1,-0x8(%rbp)
L5:
  pxor   %xmm1,%xmm1
  cvtsi2sdq -0x8(%rbp),%xmm1
  movsd  -0x20(%rbp),%xmm0
  comisd %xmm1,%xmm0
  ja     L19
  nop
  nop
  leave
  ret
```

assistant: ```c
void py_timestep_leapfrog(struct py_field *f, double Nt) {
  for (long n = 0; n < Nt; ++n)
    py_leapfrog(f);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x1,-0x18(%rbp)
  jg     L60
  mov    $0x0,%eax
  jmp    L79
L60:
  cmpq   $0x2,-0x18(%rbp)
  jne    L52
  mov    $0x1,%eax
  jmp    L79
L52:
  movq   $0x2,-0x8(%rbp)
  jmp    L42
L47:
  mov    -0x18(%rbp),%rax
  cqto
  idivq  -0x8(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L19
  mov    $0x0,%eax
  jmp    L79
L19:
  addq   $0x1,-0x8(%rbp)
L42:
  mov    D5(%rip),%rdx
  mov    -0x18(%rbp),%rax
  movq   %rdx,%xmm0
  mov    %rax,%rdi
  call   <pow@plt>
  cmp    %rax,-0x8(%rbp)
  jle    L47
  mov    $0x1,%eax
L79:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
datatype: f64

tool: D5: 0.5

assistant: ```c
int is_prime(long i) {
  if (i < 2)
    return 0;
  else if (i == 2)
    return 1;
  for (long j = 2; j <= pow(i, 0.5); j++) {
    if (i % j == 0)
      return 0;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_prime>:
  endbr64
  mov    $0x0,%eax
  cmp    $0x1,%rdi
  jle    L36
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x1,%eax
  mov    $0x2,%ebx
  cmp    $0x2,%rdi
  je     L34
L32:
  mov    D7(%rip),%rax
  movq   %rax,%xmm0
  mov    %rbp,%rdi
  call   <pow@plt>
  cmp    %rbx,%rax
  jl     L25
  mov    %rbp,%rax
  cqto
  idiv   %rbx
  test   %rdx,%rdx
  je     L50
  add    $0x1,%rbx
  jmp    L32
L25:
  mov    $0x1,%eax
L34:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L50:
  mov    $0x0,%eax
  jmp    L34
L36:
  ret
```

tool_calls: name: read_data
arguments: tag: D7
type_infered: f64

tool: D7: 0.5

assistant: ```c
int is_prime(long i) {
  if (i < 2)
    return 0;
  else if (i == 2)
    return 1;
  for (long j = 2; j <= pow(i, 0.5); j++) {
    if (i % j == 0)
      return 0;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_prime>:
  endbr64
  cmp    $0x1,%rdi
  jle    L34
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  cmp    $0x2,%rdi
  jne    L18
  jmp    L37
  nopl   0x0(%rax)
L49:
  mov    %r12,%rax
  cqto
  idiv   %rbx
  test   %rdx,%rdx
  je     L17
  add    $0x1,%rbx
L18:
  mov    D2(%rip),%rax
  mov    %r12,%rdi
  movq   %rax,%xmm0
  call   <pow@plt>
  cmp    %rbx,%rax
  jge    L49
L37:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %r12
  ret
  nop
L17:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L34:
  xor    %eax,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type_guessed: f64

tool: D2: 0.5

assistant: ```c
int is_prime(long i) {
  if (i < 2)
    return 0;
  else if (i == 2)
    return 1;
  for (long j = 2; j <= pow(i, 0.5); j++) {
    if (i % j == 0)
      return 0;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_prime>:
  endbr64
  cmp    $0x1,%rdi
  jle    L5
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  mov    $0x2,%ebx
  sub    $0x8,%rsp
  cmp    $0x2,%rdi
  jne    L33
  jmp    L2
  nopl   0x0(%rax)
L9:
  mov    %r12,%rax
  cqto
  idiv   %rbx
  test   %rdx,%rdx
  je     L34
  add    $0x1,%rbx
L33:
  mov    D7(%rip),%rax
  mov    %r12,%rdi
  movq   %rax,%xmm0
  call   <pow@plt>
  cmp    %rbx,%rax
  jge    L9
L2:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %r12
  ret
  nop
L34:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
data_type: f64

tool: D7: 0.5

assistant: ```c
int is_prime(long i) {
  if (i < 2)
    return 0;
  else if (i == 2)
    return 1;
  for (long j = 2; j <= pow(i, 0.5); j++) {
    if (i % j == 0)
      return 0;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x28(%rbp)
  jmp    L46
L7:
  lea    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x28(%rbp)
L46:
  cmpl   $0x4,-0x28(%rbp)
  jle    L7
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x24(%rbp)
  jmp    L39
L18:
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    $0x3,%esi
  mov    %eax,%edi
  call   <pow@plt>
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x24(%rbp)
L39:
  cmpl   $0x4,-0x24(%rbp)
  jle    L18
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
datatype: string
name: unpack_data
arguments: tag: D35
datatype: string
name: unpack_data
arguments: tag: D34
datatype: string
name: unpack_data
arguments: tag: D49
datatype: string
name: unpack_data
arguments: tag: D3
datatype: string
name: unpack_data
arguments: tag: D23
datatype: string

tool: D7: ===== Exercise 16 =====

tool: D35: Read 5 numbers and get the cube value!

tool: D34: Enter 5 numbers:

tool: D49: %i

tool: D3: --> The cube values:

tool: D23: %d ^ 3 = %lf


assistant: ```c
int main(int argc, const char *argv[]) {
  int numberArray[5];
  printf("===== Exercise 16 =====\n");
  printf("Read 5 numbers and get the cube value!\n");
  printf("Enter 5 numbers:\n");
  for (int i = 0; i < 5; i++) {
    scanf("%i", &numberArray[i]);
  }
  printf("--> The cube values:\n");
  for (int j = 0; j < 5; j++) {
    printf("%d ^ 3 = %lf\n", numberArray[j], pow(numberArray[j], 3));
  }
  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
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <puts@plt>
  lea    D52(%rip),%rdi
  call   <puts@plt>
  lea    D7(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  lea    0x14(%rsp),%r12
  mov    %rbx,%rbp
  lea    D28(%rip),%r13
L15:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x4,%rbp
  cmp    %r12,%rbp
  jne    L15
  lea    D49(%rip),%rdi
  call   <puts@plt>
  lea    D39(%rip),%rbp
L25:
  mov    $0x3,%esi
  mov    (%rbx),%edi
  call   <pow@plt>
  mov    %eax,%edx
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L25
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D15
datatype: string
name: disassemble_data
arguments: data_label: D52
datatype: string
name: disassemble_data
arguments: data_label: D7
datatype: string
name: disassemble_data
arguments: data_label: D28
datatype: string
name: disassemble_data
arguments: data_label: D49
datatype: string
name: disassemble_data
arguments: data_label: D39
datatype: string

tool: D15: ===== Exercise 16 =====

tool: D52: Read 5 numbers and get the cube value!

tool: D7: Enter 5 numbers:

tool: D28: %i

tool: D49: --> The cube values:

tool: D39: %d ^ 3 = %lf


assistant: ```c
int main(int argc, const char *argv[]) {
  int numberArray[5];
  printf("===== Exercise 16 =====\n");
  printf("Read 5 numbers and get the cube value!\n");
  printf("Enter 5 numbers:\n");
  for (int i = 0; i < 5; i++) {
    scanf("%i", &numberArray[i]);
  }
  printf("--> The cube values:\n");
  for (int j = 0; j < 5; j++) {
    printf("%d ^ 3 = %lf\n", numberArray[j], pow(numberArray[j], 3));
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D32(%rip),%rdi
  lea    D23(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x14(%rsp),%r12
  call   <puts@plt>
  lea    D51(%rip),%rdi
  mov    %rbx,%rbp
  call   <puts@plt>
  lea    D54(%rip),%rdi
  call   <puts@plt>
L24:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L24
  lea    D56(%rip),%rdi
  lea    D30(%rip),%rbp
  call   <puts@plt>
L8:
  mov    (%rbx),%edi
  mov    $0x3,%esi
  add    $0x4,%rbx
  call   <pow@plt>
  mov    -0x4(%rbx),%esi
  mov    %rbp,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L8
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D32
datatype: string
name: extract_symbolic_data
arguments: data_label: D23
datatype: string
name: extract_symbolic_data
arguments: data_label: D51
datatype: string
name: extract_symbolic_data
arguments: data_label: D54
datatype: string
name: extract_symbolic_data
arguments: data_label: D56
datatype: string
name: extract_symbolic_data
arguments: data_label: D30
datatype: string

tool: D32: ===== Exercise 16 =====

tool: D23: %i

tool: D51: Read 5 numbers and get the cube value!

tool: D54: Enter 5 numbers:

tool: D56: --> The cube values:

tool: D30: %d ^ 3 = %lf


assistant: ```c
int main(int argc, const char *argv[]) {
  int numberArray[5];
  printf("===== Exercise 16 =====\n");
  printf("Read 5 numbers and get the cube value!\n");
  printf("Enter 5 numbers:\n");
  for (int i = 0; i < 5; i++) {
    scanf("%i", &numberArray[i]);
  }
  printf("--> The cube values:\n");
  for (int j = 0; j < 5; j++) {
    printf("%d ^ 3 = %lf\n", numberArray[j], pow(numberArray[j], 3));
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D6(%rip),%rdi
  lea    D58(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x14(%rsp),%r12
  call   <puts@plt>
  lea    D19(%rip),%rdi
  mov    %rbx,%rbp
  call   <puts@plt>
  lea    D17(%rip),%rdi
  call   <puts@plt>
L14:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L14
  lea    D37(%rip),%rdi
  lea    D25(%rip),%rbp
  call   <puts@plt>
L9:
  mov    (%rbx),%edi
  mov    $0x3,%esi
  add    $0x4,%rbx
  call   <pow@plt>
  mov    -0x4(%rbx),%esi
  mov    %rbp,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L9
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D6
type_infered: string
name: retrieve_data
arguments: data_label: D58
type_infered: string
name: retrieve_data
arguments: data_label: D19
type_infered: string
name: retrieve_data
arguments: data_label: D17
type_infered: string
name: retrieve_data
arguments: data_label: D37
type_infered: string
name: retrieve_data
arguments: data_label: D25
type_infered: string

tool: D6: ===== Exercise 16 =====

tool: D58: %i

tool: D19: Read 5 numbers and get the cube value!

tool: D17: Enter 5 numbers:

tool: D37: --> The cube values:

tool: D25: %d ^ 3 = %lf


assistant: ```c
int main(int argc, const char *argv[]) {
  int numberArray[5];
  printf("===== Exercise 16 =====\n");
  printf("Read 5 numbers and get the cube value!\n");
  printf("Enter 5 numbers:\n");
  for (int i = 0; i < 5; i++) {
    scanf("%i", &numberArray[i]);
  }
  printf("--> The cube values:\n");
  for (int j = 0; j < 5; j++) {
    printf("%d ^ 3 = %lf\n", numberArray[j], pow(numberArray[j], 3));
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Thread_destroy_mutex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <pthread_mutex_destroy@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void Thread_destroy_mutex(pthread_mutex_t *mutex) {
  int rc = 0;
  ;
  rc = pthread_mutex_destroy(mutex);
  free(mutex);
  ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Thread_destroy_mutex>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <pthread_mutex_destroy@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  pop    %rbx
  ret
```

assistant: ```c
void Thread_destroy_mutex(pthread_mutex_t *mutex) {
  int rc = 0;
  ;
  rc = pthread_mutex_destroy(mutex);
  free(mutex);
  ;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Thread_destroy_mutex>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <pthread_mutex_destroy@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

assistant: ```c
void Thread_destroy_mutex(pthread_mutex_t *mutex) {
  int rc = 0;
  ;
  rc = pthread_mutex_destroy(mutex);
  free(mutex);
  ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Thread_destroy_mutex>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <pthread_mutex_destroy@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

assistant: ```c
void Thread_destroy_mutex(pthread_mutex_t *mutex) {
  int rc = 0;
  ;
  rc = pthread_mutex_destroy(mutex);
  free(mutex);
  ;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<http_charset_convert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9d,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int http_charset_convert(char *content_type, char *content_encoding, char *data, int len, char *tocharset, int is_need_compress, char **out) {
  int nout = 0;
  return nout;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int http_charset_convert(char *content_type, char *content_encoding, char *data, int len, char *tocharset, int is_need_compress, char **out) {
  int nout = 0;
  return nout;
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
int http_charset_convert(char *content_type, char *content_encoding, char *data, int len, char *tocharset, int is_need_compress, char **out) {
  int nout = 0;
  return nout;
}
```
user: What would the C source look like for the assembly code provided here:

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

assistant: ```c
int http_charset_convert(char *content_type, char *content_encoding, char *data, int len, char *tocharset, int is_need_compress, char **out) {
  int nout = 0;
  return nout;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<eingabe>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x0,-0xd(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <allocate@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0xd(%rbp)
  jmp    L12
L21:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%eax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <allocate@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0xd(%rbp)
L12:
  cmpb   $0xa,-0xd(%rbp)
  jne    L21
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
void eingabe(unsigned char **ein) {
  unsigned char c = 0;
  unsigned int length = 0;
  allocate(ein, &length);
  c = getchar();
  while (c != '\n') {
    (*ein)[length - 1] = c;
    allocate(ein, &length);
    c = getchar();
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<eingabe>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rsi
  call   <allocate@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  je     L4
  lea    0x4(%rsp),%rbp
L8:
  mov    0x4(%rsp),%edx
  lea    -0x1(%rdx),%edx
  mov    (%rbx),%rcx
  mov    %al,(%rcx,%rdx,1)
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <allocate@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L8
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void eingabe(unsigned char **ein) {
  unsigned char c = 0;
  unsigned int length = 0;
  allocate(ein, &length);
  c = getchar();
  while (c != '\n') {
    (*ein)[length - 1] = c;
    allocate(ein, &length);
    c = getchar();
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<eingabe>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  movl   $0x0,0x4(%rsp)
  mov    %rbp,%rsi
  jmp    L8
  nop
L21:
  mov    0x4(%rsp),%edx
  mov    (%rbx),%rcx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  sub    $0x1,%edx
  mov    %al,(%rcx,%rdx,1)
L8:
  call   <allocate@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L21
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void eingabe(unsigned char **ein) {
  unsigned char c = 0;
  unsigned int length = 0;
  allocate(ein, &length);
  c = getchar();
  while (c != '\n') {
    (*ein)[length - 1] = c;
    allocate(ein, &length);
    c = getchar();
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<eingabe>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  movl   $0x0,0x4(%rsp)
  mov    %rbp,%rsi
  jmp    L28
  nop
L29:
  mov    0x4(%rsp),%edx
  mov    (%rbx),%rcx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  sub    $0x1,%edx
  mov    %al,(%rcx,%rdx,1)
L28:
  call   <allocate@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L29
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void eingabe(unsigned char **ein) {
  unsigned char c = 0;
  unsigned int length = 0;
  allocate(ein, &length);
  c = getchar();
  while (c != '\n') {
    (*ein)[length - 1] = c;
    allocate(ein, &length);
    c = getchar();
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ParseInt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L73
  addq   $0x1,-0x8(%rbp)
  movl   $0xffffffff,-0x18(%rbp)
  jmp    L30
L73:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x30,%al
  jne    L30
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x78,%al
  jne    L30
  addq   $0x2,-0x8(%rbp)
  movl   $0x1,-0x14(%rbp)
L30:
  cmpl   $0x0,-0x14(%rbp)
  je     L116
  mov    $0x10,%eax
  jmp    L65
L116:
  mov    $0xa,%eax
L65:
  mov    %eax,-0xc(%rbp)
  jmp    L33
L31:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  mov    %eax,-0x10(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L19
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L19
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x46,%al
  jg     L19
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x37,%eax
  mov    %eax,-0x10(%rbp)
L19:
  mov    -0x1c(%rbp),%eax
  imul   -0xc(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  add    %eax,-0x1c(%rbp)
  addq   $0x1,-0x8(%rbp)
L33:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L81
  mov    -0x8(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  cmp    %rax,%rdx
  jl     L31
L81:
  mov    -0x1c(%rbp),%eax
  imul   -0x18(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int ParseInt(const char *str, int len) {
  const char *cursor = str;
  int val = 0;
  int sign = 1;
  int isHex = 0;
  if (*cursor == '-') {
    cursor++;
    sign = -1;
  } else if (*cursor == '0' && cursor[1] == 'x') {
    cursor += 2;
    isHex = 1;
  }
  const int base = (isHex ? 16 : 10);
  while (*cursor && (cursor - str) < len) {
    int digit = (*cursor - '0');
    if (isHex && *cursor >= 'A' && *cursor <= 'F') {
      digit = (*cursor - 'A') + 10;
    }
    val *= base;
    val += digit;
    cursor++;
  }
  return val * sign;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ParseInt>:
  endbr64
  push   %rbx
  mov    %rdi,%r9
  movzbl (%rdi),%eax
  lea    0x1(%rdi),%rcx
  mov    $0xffffffff,%ebx
  cmp    $0x2d,%al
  je     L48
  mov    %rdi,%rcx
  mov    $0x1,%ebx
  cmp    $0x30,%al
  je     L78
L48:
  mov    $0x0,%r11d
  mov    $0xa,%r10d
L32:
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L110
L12:
  movslq %esi,%rsi
  mov    %rcx,%rax
  sub    %r9,%rax
  cmp    %rsi,%rax
  jge    L35
  mov    $0x0,%eax
  jmp    L70
L78:
  cmpb   $0x78,0x1(%rdi)
  je     L28
  movzbl (%rdi),%edx
  mov    $0x0,%r11d
  mov    $0x1,%ebx
  mov    $0xa,%r10d
  jmp    L12
L28:
  lea    0x2(%rdi),%rcx
  mov    $0x1,%r11d
  mov    $0x10,%r10d
  jmp    L32
L10:
  imul   %r10d,%eax
  add    %r8d,%eax
  add    $0x1,%rcx
  movzbl (%rcx),%edx
  test   %dl,%dl
  je     L30
  mov    %rcx,%rdi
  sub    %r9,%rdi
  cmp    %rsi,%rdi
  jge    L30
L70:
  movsbl %dl,%edi
  lea    -0x30(%rdi),%r8d
  sub    $0x41,%edx
  cmp    $0x5,%dl
  ja     L10
  sub    $0x37,%edi
  test   %r11d,%r11d
  cmovne %edi,%r8d
  jmp    L10
L110:
  mov    $0x0,%eax
L30:
  imul   %ebx,%eax
  pop    %rbx
  ret
L35:
  mov    $0x0,%eax
  jmp    L30
```

assistant: ```c
int ParseInt(const char *str, int len) {
  const char *cursor = str;
  int val = 0;
  int sign = 1;
  int isHex = 0;
  if (*cursor == '-') {
    cursor++;
    sign = -1;
  } else if (*cursor == '0' && cursor[1] == 'x') {
    cursor += 2;
    isHex = 1;
  }
  const int base = (isHex ? 16 : 10);
  while (*cursor && (cursor - str) < len) {
    int digit = (*cursor - '0');
    if (isHex && *cursor >= 'A' && *cursor <= 'F') {
      digit = (*cursor - 'A') + 10;
    }
    val *= base;
    val += digit;
    cursor++;
  }
  return val * sign;
}
```