ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<detab>:
  endbr64
  cmp    $0x9,%edx
  jne    L12
  mov    D24(%rip),%r8d
  mov    %edi,%ecx
  mov    %r8d,%eax
  cltd
  idiv   %esi
  sub    %edx,%esi
  lea    (%r8,%rsi,1),%eax
  mov    %esi,%edi
  mov    %eax,D18(%rip)
  cmp    %ecx,%eax
  jl     L8
  jmp    1070 <print_spaces@plt>
  nop
L12:
  push   %rbx
  mov    %edx,%edi
  mov    %edx,%ebx
  call   <putchar@plt>
  mov    D18(%rip),%eax
  xor    %edx,%edx
  add    $0x1,%eax
  cmp    $0xa,%ebx
  pop    %rbx
  cmove  %edx,%eax
  mov    %eax,D18(%rip)
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  mov    $0x9,%edi
  jmp    1060 <putchar@plt>
```

assistant: ```c
void detab(int column_start, int tabstop, int character) {
  static int column_count;
  int tab_space;
  if (character == '\t') {
    tab_space = tabstop - column_count % tabstop;
    column_count += tab_space;
    if (column_count < column_start)
      putchar(character);
    else if (column_count - tab_space < column_start)
      print_spaces(tab_space);
    else
      print_spaces(tab_space);
  } else {
    putchar(character);
    ++column_count;
  }
  if (character == '\n')
    column_count = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L14:
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
strlen(char s[]) {
  int i;
  i = 0;
  while (s[i] != '\0')
    ++i;
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L3
  mov    $0x1,%eax
L19:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L19
L14:
  mov    %edx,%eax
  ret
L3:
  mov    $0x0,%edx
  jmp    L14
```

assistant: ```c
strlen(char s[]) {
  int i;
  i = 0;
  while (s[i] != '\0')
    ++i;
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L14
  mov    $0x1,%eax
  xchg   %ax,%ax
L0:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L0
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
strlen(char s[]) {
  int i;
  i = 0;
  while (s[i] != '\0')
    ++i;
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L1
  mov    $0x1,%eax
  xchg   %ax,%ax
L19:
  mov    %eax,%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rdi,%rax,1)
  jne    L19
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
strlen(char s[]) {
  int i;
  i = 0;
  while (s[i] != '\0')
    ++i;
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    $0xf,%esi
  mov    %rax,%rdi
  call   <insert@plt>
  lea    -0x10(%rbp),%rax
  mov    $0x14,%esi
  mov    %rax,%rdi
  call   <insert@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <createloop@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <detectloop@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int main(void) {
  struct node *head = NULL;
  insert(&head, 5);
  insert(&head, 10);
  insert(&head, 15);
  insert(&head, 20);
  createloop(head, head);
  detectloop(head);
  return 0;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rbx
  mov    $0x5,%esi
  mov    %rbx,%rdi
  call   <insert@plt>
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <insert@plt>
  mov    $0xf,%esi
  mov    %rbx,%rdi
  call   <insert@plt>
  mov    $0x14,%esi
  mov    %rbx,%rdi
  call   <insert@plt>
  mov    (%rsp),%rdi
  mov    %rdi,%rsi
  call   <createloop@plt>
  mov    (%rsp),%rdi
  call   <detectloop@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct node *head = NULL;
  insert(&head, 5);
  insert(&head, 10);
  insert(&head, 15);
  insert(&head, 20);
  createloop(head, head);
  detectloop(head);
  return 0;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x5,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movq   $0x0,(%rsp)
  mov    %rbp,%rdi
  call   <insert@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <insert@plt>
  mov    $0xf,%esi
  mov    %rbp,%rdi
  call   <insert@plt>
  mov    $0x14,%esi
  mov    %rbp,%rdi
  call   <insert@plt>
  mov    (%rsp),%rdi
  mov    %rdi,%rsi
  call   <createloop@plt>
  mov    (%rsp),%rdi
  call   <detectloop@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct node *head = NULL;
  insert(&head, 5);
  insert(&head, 10);
  insert(&head, 15);
  insert(&head, 20);
  createloop(head, head);
  detectloop(head);
  return 0;
  return 0;
}
```
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    $0x5,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movq   $0x0,(%rsp)
  mov    %rbp,%rdi
  call   <insert@plt>
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <insert@plt>
  mov    $0xf,%esi
  mov    %rbp,%rdi
  call   <insert@plt>
  mov    $0x14,%esi
  mov    %rbp,%rdi
  call   <insert@plt>
  mov    (%rsp),%rdi
  mov    %rdi,%rsi
  call   <createloop@plt>
  mov    (%rsp),%rdi
  call   <detectloop@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  struct node *head = NULL;
  insert(&head, 5);
  insert(&head, 10);
  insert(&head, 15);
  insert(&head, 20);
  createloop(head, head);
  detectloop(head);
  return 0;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<charcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jne    L13
  mov    $0x0,%eax
  jmp    L18
L13:
  mov    $0x1,%eax
L18:
  pop    %rbp
  ret
```

assistant: ```c
char charcmp(char *one, char *two) {
  if (one == two) {
    return 0;
  } else {
    return 1;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<charcmp>:
  endbr64
  cmp    %rdi,%rsi
  setne  %al
  ret
```

assistant: ```c
char charcmp(char *one, char *two) {
  if (one == two) {
    return 0;
  } else {
    return 1;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<charcmp>:
  endbr64
  cmp    %rdi,%rsi
  setne  %al
  ret
```

assistant: ```c
char charcmp(char *one, char *two) {
  if (one == two) {
    return 0;
  } else {
    return 1;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<charcmp>:
  endbr64
  cmp    %rdi,%rsi
  setne  %al
  ret
```

assistant: ```c
char charcmp(char *one, char *two) {
  if (one == two) {
    return 0;
  } else {
    return 1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xilinxcorelib_a_1600189650_0543512595_sub_616253073571835568_3745213680167319624>:
  endbr64
  cmp    $0x3,%dl
  je     L4
  ret
L4:
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    0x7970(%rdi),%rbp
  mov    0xcf94(%rdi),%edx
  mov    0x4828(%rdi),%rsi
  mov    0x79a8(%rdi),%rax
  mov    0x38(%rax),%rdi
  call   <memcpy@plt>
  mov    0xcf94(%rbx),%edx
  mov    $0x64,%ecx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <xsi_driver_first_trans_delta@plt>
  mov    $0x64,%edx
  mov    $0x64,%esi
  mov    %rbp,%rdi
  call   <xsi_driver_intertial_reject@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void xilinxcorelib_a_1600189650_0543512595_sub_616253073571835568_3745213680167319624(char *t0, char *t1, unsigned char t2) {
  char t4[8];
  char *t5;
  unsigned char t6;
  char *t7;
  char *t8;
  char *t9;
  char *t10;
  char *t11;
  char *t12;
  char *t13;
  char *t14;
  unsigned int t15;
  char *t16;
  char *t17;
  unsigned int t18;
  unsigned int t19;
  char *t20;
LAB0:
  t5 = (t4 + 4U);
  *((unsigned char *)t5) = t2;
  t6 = (t2 == (unsigned char)3);
  if (t6 != 0)
    goto LAB2;
LAB4:
LAB3:
LAB1:
  return;
LAB2:
  t7 = (t0 + 18472U);
  t8 = *((char **)t7);
  t7 = (t0 + 31088);
  t9 = (t7 + 56U);
  t10 = *((char **)t9);
  t11 = (t10 + 56U);
  t12 = *((char **)t11);
  t13 = (t0 + 53128U);
  t14 = (t13 + 12U);
  t15 = *((unsigned int *)t14);
  t15 = (t15 * 1U);
  memcpy(t12, t8, t15);
  t16 = (t0 + 53128U);
  t17 = (t16 + 12U);
  t18 = *((unsigned int *)t17);
  t19 = (1U * t18);
  xsi_driver_first_trans_delta(t7, 0U, t19, 100LL);
  t20 = (t0 + 31088);
  xsi_driver_intertial_reject(t20, 100LL, 100LL);
  goto LAB3;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xilinxcorelib_a_1600189650_0543512595_sub_616253073571835568_3745213680167319624>:
  endbr64
  cmp    $0x3,%dl
  je     L1
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  push   %rbp
  lea    0x7970(%rdi),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    0x79a8(%rdi),%rax
  mov    0xcf94(%rdi),%edx
  mov    0x4828(%rdi),%rsi
  mov    0x38(%rax),%rdi
  call   <memcpy@plt>
  mov    0xcf94(%rbx),%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  mov    $0x64,%ecx
  call   <xsi_driver_first_trans_delta@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  mov    $0x64,%edx
  pop    %rbx
  mov    $0x64,%esi
  pop    %rbp
  jmp    1090 <xsi_driver_intertial_reject@plt>
```

assistant: ```c
void xilinxcorelib_a_1600189650_0543512595_sub_616253073571835568_3745213680167319624(char *t0, char *t1, unsigned char t2) {
  char t4[8];
  char *t5;
  unsigned char t6;
  char *t7;
  char *t8;
  char *t9;
  char *t10;
  char *t11;
  char *t12;
  char *t13;
  char *t14;
  unsigned int t15;
  char *t16;
  char *t17;
  unsigned int t18;
  unsigned int t19;
  char *t20;
LAB0:
  t5 = (t4 + 4U);
  *((unsigned char *)t5) = t2;
  t6 = (t2 == (unsigned char)3);
  if (t6 != 0)
    goto LAB2;
LAB4:
LAB3:
LAB1:
  return;
LAB2:
  t7 = (t0 + 18472U);
  t8 = *((char **)t7);
  t7 = (t0 + 31088);
  t9 = (t7 + 56U);
  t10 = *((char **)t9);
  t11 = (t10 + 56U);
  t12 = *((char **)t11);
  t13 = (t0 + 53128U);
  t14 = (t13 + 12U);
  t15 = *((unsigned int *)t14);
  t15 = (t15 * 1U);
  memcpy(t12, t8, t15);
  t16 = (t0 + 53128U);
  t17 = (t16 + 12U);
  t18 = *((unsigned int *)t17);
  t19 = (1U * t18);
  xsi_driver_first_trans_delta(t7, 0U, t19, 100LL);
  t20 = (t0 + 31088);
  xsi_driver_intertial_reject(t20, 100LL, 100LL);
  goto LAB3;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<xilinxcorelib_a_1600189650_0543512595_sub_616253073571835568_3745213680167319624>:
  endbr64
  cmp    $0x3,%dl
  je     L7
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  push   %rbp
  lea    0x7970(%rdi),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    0x79a8(%rdi),%rax
  mov    0xcf94(%rdi),%edx
  mov    0x4828(%rdi),%rsi
  mov    0x38(%rax),%rdi
  call   <memcpy@plt>
  mov    0xcf94(%rbx),%edx
  mov    %rbp,%rdi
  xor    %esi,%esi
  mov    $0x64,%ecx
  call   <xsi_driver_first_trans_delta@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  mov    $0x64,%edx
  pop    %rbx
  mov    $0x64,%esi
  pop    %rbp
  jmp    1090 <xsi_driver_intertial_reject@plt>
```

assistant: ```c
void xilinxcorelib_a_1600189650_0543512595_sub_616253073571835568_3745213680167319624(char *t0, char *t1, unsigned char t2) {
  char t4[8];
  char *t5;
  unsigned char t6;
  char *t7;
  char *t8;
  char *t9;
  char *t10;
  char *t11;
  char *t12;
  char *t13;
  char *t14;
  unsigned int t15;
  char *t16;
  char *t17;
  unsigned int t18;
  unsigned int t19;
  char *t20;
LAB0:
  t5 = (t4 + 4U);
  *((unsigned char *)t5) = t2;
  t6 = (t2 == (unsigned char)3);
  if (t6 != 0)
    goto LAB2;
LAB4:
LAB3:
LAB1:
  return;
LAB2:
  t7 = (t0 + 18472U);
  t8 = *((char **)t7);
  t7 = (t0 + 31088);
  t9 = (t7 + 56U);
  t10 = *((char **)t9);
  t11 = (t10 + 56U);
  t12 = *((char **)t11);
  t13 = (t0 + 53128U);
  t14 = (t13 + 12U);
  t15 = *((unsigned int *)t14);
  t15 = (t15 * 1U);
  memcpy(t12, t8, t15);
  t16 = (t0 + 53128U);
  t17 = (t16 + 12U);
  t18 = *((unsigned int *)t17);
  t19 = (1U * t18);
  xsi_driver_first_trans_delta(t7, 0U, t19, 100LL);
  t20 = (t0 + 31088);
  xsi_driver_intertial_reject(t20, 100LL, 100LL);
  goto LAB3;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x8,-0x20(%rbp)
  movl   $0xb,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x9,-0x14(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L22
L44:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x2c(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x2c(%rbp)
L22:
  cmpl   $0x3,-0x2c(%rbp)
  jle    L44
  movl   $0x0,-0x2c(%rbp)
  jmp    L15
L17:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x2c(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addq   $0x4,-0x28(%rbp)
  addl   $0x1,-0x2c(%rbp)
L15:
  cmpl   $0x3,-0x2c(%rbp)
  jle    L17
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
type_guessed: string

tool: D14: Nums[%i] = %i


assistant: ```c
int main() {
  int i, nums[4] = {8, 11, 3, 9};
  int *ptr;
  ptr = nums;
  for (i = 0; i < 4; i++)
    printf("Nums[%i] = %i\n", i, nums[i]);
  for (i = 0; i < 4; i++) {
    printf("Nums[%i] = %i\n", i, *ptr);
    ptr++;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x8,(%rsp)
  movl   $0xb,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x9,0xc(%rsp)
  mov    $0x0,%ebx
  lea    D18(%rip),%rbp
L13:
  mov    (%rsp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x4,%rbx
  jne    L13
  mov    $0x0,%ebx
  lea    D18(%rip),%rbp
L21:
  mov    (%rsp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x4,%rbx
  jne    L21
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D18
type: string

tool: D18: Nums[%i] = %i


assistant: ```c
int main() {
  int i, nums[4] = {8, 11, 3, 9};
  int *ptr;
  ptr = nums;
  for (i = 0; i < 4; i++)
    printf("Nums[%i] = %i\n", i, nums[i]);
  for (i = 0; i < 4; i++) {
    printf("Nums[%i] = %i\n", i, *ptr);
    ptr++;
  }
  return 0;
}
```
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    $0x8,%edx
  lea    D1(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0xb00000008,%rax
  mov    %rax,(%rsp)
  movabs $0x900000003,%rax
  mov    %rax,0x8(%rsp)
L0:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  je     L33
  mov    (%rsp,%rbx,4),%edx
  jmp    L0
  nopl   0x0(%rax)
L33:
  xor    %ebx,%ebx
  mov    $0x8,%edx
L19:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  je     L17
  mov    (%rsp,%rbx,4),%edx
  jmp    L19
L17:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type: string

tool: D1: Nums[%i] = %i


assistant: ```c
int main() {
  int i, nums[4] = {8, 11, 3, 9};
  int *ptr;
  ptr = nums;
  for (i = 0; i < 4; i++)
    printf("Nums[%i] = %i\n", i, nums[i]);
  for (i = 0; i < 4; i++) {
    printf("Nums[%i] = %i\n", i, *ptr);
    ptr++;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x20,%rsp
  movdqa D15(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movaps %xmm0,(%rsp)
L6:
  mov    0x0(%rbp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  jne    L6
  xor    %ebx,%ebx
L23:
  mov    0x0(%rbp,%rbx,4),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x4,%rbx
  jne    L23
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D10
type: string
name: read_data
arguments: data_label: D15
type: byte[16]

tool: D10: Nums[%i] = %i


tool: {"D15": [0x08,0x00,0x00,0x00,0x0b,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x09,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int i, nums[4] = {8, 11, 3, 9};
  int *ptr;
  ptr = nums;
  for (i = 0; i < 4; i++)
    printf("Nums[%i] = %i\n", i, nums[i]);
  for (i = 0; i < 4; i++) {
    printf("Nums[%i] = %i\n", i, *ptr);
    ptr++;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<apply_mask>:
  endbr64
  push   %rbx
  cmp    $0x2,%edi
  je     L25
  jg     L34
  test   %edi,%edi
  je     L159
  cmp    $0x1,%edi
  jne    L126
  movb   $0x1,(%rdx)
  test   %esi,%esi
  jle    L126
  mov    %esi,%r11d
  mov    $0x0,%r9d
L107:
  movzbl (%rcx,%r9,1),%edi
  lea    (%r9,%r9,2),%eax
  add    %eax,%edi
  movslq %edi,%rax
  imul   $0x487ede05,%rax,%rax
  sar    $0x25,%rax
  mov    %edi,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%eax
  imul   $0x71,%eax,%eax
  sub    %eax,%edi
  mov    %dil,0x1(%rdx,%r9,1)
  add    $0x1,%r9
  cmp    %r9,%r11
  jne    L107
  jmp    L126
L34:
  cmp    $0x3,%edi
  jne    L126
  movb   $0x3,(%rdx)
  test   %esi,%esi
  jle    L126
  mov    %esi,%r11d
  shl    $0x4,%r11d
  add    %esi,%r11d
  mov    $0x0,%r10d
  mov    $0x0,%r9d
L2:
  movzbl (%rcx,%r10,1),%edi
  add    %r9d,%edi
  movslq %edi,%rax
  imul   $0x487ede05,%rax,%rax
  sar    $0x25,%rax
  mov    %edi,%ebx
  sar    $0x1f,%ebx
  sub    %ebx,%eax
  imul   $0x71,%eax,%eax
  sub    %eax,%edi
  mov    %dil,0x1(%rdx,%r10,1)
  add    $0x11,%r9d
  add    $0x1,%r10
  cmp    %r11d,%r9d
  jne    L2
  jmp    L126
L159:
  movb   $0x0,(%rdx)
  test   %esi,%esi
  jle    L126
  mov    %esi,%r9d
  mov    $0x0,%eax
L6:
  movzbl (%rcx,%rax,1),%edi
  mov    %dil,0x1(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%r9
  jne    L6
L126:
  lea    0x1(%rsi),%edi
  mov    %r8d,%esi
  call   <rsencode@plt>
  pop    %rbx
  ret
L25:
  movb   $0x2,(%rdx)
  test   %esi,%esi
  jle    L126
  lea    0x0(,%rsi,8),%r11d
  sub    %esi,%r11d
  mov    $0x0,%r10d
  mov    $0x0,%r9d
L93:
  movzbl (%rcx,%r10,1),%edi
  add    %r9d,%edi
  movslq %edi,%rax
  imul   $0x487ede05,%rax,%rax
  sar    $0x25,%rax
  mov    %edi,%ebx
  sar    $0x1f,%ebx
  sub    %ebx,%eax
  imul   $0x71,%eax,%eax
  sub    %eax,%edi
  mov    %dil,0x1(%rdx,%r10,1)
  add    $0x7,%r9d
  add    $0x1,%r10
  cmp    %r11d,%r9d
  jne    L93
  jmp    L126
```

assistant: ```c
void apply_mask(int mask, int data_length, unsigned char *masked_codeword_array, unsigned char *codeword_array, int ecc_length) {
  int weight = 0;
  int j;
  switch (mask) {
  case 0:
    masked_codeword_array[0] = 0;
    for (j = 0; j < data_length; j++) {
      masked_codeword_array[j + 1] = codeword_array[j];
    }
    break;
  case 1:
    masked_codeword_array[0] = 1;
    for (j = 0; j < data_length; j++) {
      masked_codeword_array[j + 1] = (weight + codeword_array[j]) % 113;
      weight += 3;
    }
    break;
  case 2:
    masked_codeword_array[0] = 2;
    for (j = 0; j < data_length; j++) {
      masked_codeword_array[j + 1] = (weight + codeword_array[j]) % 113;
      weight += 7;
    }
    break;
  case 3:
    masked_codeword_array[0] = 3;
    for (j = 0; j < data_length; j++) {
      masked_codeword_array[j + 1] = (weight + codeword_array[j]) % 113;
      weight += 17;
    }
    break;
  }
  rsencode(data_length + 1, ecc_length, masked_codeword_array);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <megaInit@plt>
  cmpl   $0x2,-0x24(%rbp)
  je     L23
  mov    $0x1,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L23:
  movl   $0x0,-0x14(%rbp)
  jmp    L59
L45:
  mov    -0x14(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %ax,-0x16(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  movzwl -0x16(%rbp),%edx
  mov    %dx,-0xc(%rbp,%rax,2)
  addl   $0x1,-0x14(%rbp)
L59:
  cmpl   $0x0,-0x14(%rbp)
  jle    L45
  lea    -0xa(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <RandomFunc@plt>
  movzwl -0xa(%rbp),%eax
  cmp    $0xc1b5,%ax
  jne    L33
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L33:
  movl   $0x0,-0x10(%rbp)
  jmp    L26
L29:
  mov    -0x10(%rbp),%eax
  cltq
  movzwl -0xa(%rbp,%rax,2),%eax
  movzwl %ax,%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L26:
  cmpl   $0x0,-0x10(%rbp)
  jle    L29
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type_guessed: string
name: unpack_data
arguments: data_tag: D5
type_guessed: string
name: unpack_data
arguments: data_tag: D11
type_guessed: string

tool: D3: Call this program with %i arguments


tool: D5: You win!

tool: D11: %u


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned short input[1];
  unsigned short output[1];
  int randomFuns_i5;
  unsigned short randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned short)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 49589) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```