ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_all_digit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L47
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L33
L47:
  mov    $0x0,%eax
  jmp    L18
L8:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L30
  mov    $0x0,%eax
  jmp    L18
L30:
  addq   $0x1,-0x8(%rbp)
L33:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L8
  mov    $0x1,%eax
L18:
  leave
  ret
```

assistant: ```c
int is_all_digit(const char *str) {
  if (!str || !*str) {
    return 0;
  }
  while (*str) {
    if (!isdigit(*str)) {
      return 0;
    }
    ++str;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_all_digit>:
  endbr64
  test   %rdi,%rdi
  je     L7
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  mov    $0x0,%eax
  test   %dil,%dil
  je     L25
L24:
  movsbl %dil,%edi
  call   <isdigit@plt>
  test   %eax,%eax
  je     L25
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L24
  mov    $0x1,%eax
L25:
  pop    %rbx
  ret
L7:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int is_all_digit(const char *str) {
  if (!str || !*str) {
    return 0;
  }
  while (*str) {
    if (!isdigit(*str)) {
      return 0;
    }
    ++str;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_all_digit>:
  endbr64
  test   %rdi,%rdi
  je     L18
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L1
  jmp    L30
  nopw   0x0(%rax,%rax,1)
L37:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L3
L1:
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L37
L30:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  mov    $0x1,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  xor    %eax,%eax
  ret
```

assistant: ```c
int is_all_digit(const char *str) {
  if (!str || !*str) {
    return 0;
  }
  while (*str) {
    if (!isdigit(*str)) {
      return 0;
    }
    ++str;
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_all_digit>:
  endbr64
  test   %rdi,%rdi
  je     L44
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L11
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L8:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L6
L11:
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L8
L23:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  mov    $0x1,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L44:
  xor    %eax,%eax
  ret
```

assistant: ```c
int is_all_digit(const char *str) {
  if (!str || !*str) {
    return 0;
  }
  while (*str) {
    if (!isdigit(*str)) {
      return 0;
    }
    ++str;
  }
  return 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
<nullifyColumn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L4:
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  movslq %eax,%rdx
  imul   %rcx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rdx,%rcx
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  movl   $0x0,(%rcx,%rdx,4)
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%edx
  cmp    -0x14(%rbp),%edx
  jl     L4
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void nullifyColumn(int rows, int cols, int index, int matrix[rows][cols]) {
  for (int i = 0; i < rows; ++i) {
    matrix[i][index] = 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<nullifyColumn>:
  endbr64
  movslq %esi,%rsi
  test   %edi,%edi
  jle    L11
  shl    $0x2,%rsi
  movslq %edx,%rdx
  lea    (%rcx,%rdx,4),%rdx
  mov    $0x0,%eax
L14:
  movl   $0x0,(%rdx)
  add    $0x1,%eax
  add    %rsi,%rdx
  cmp    %eax,%edi
  jne    L14
L11:
  ret
```

assistant: ```c
void nullifyColumn(int rows, int cols, int index, int matrix[rows][cols]) {
  for (int i = 0; i < rows; ++i) {
    matrix[i][index] = 0;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<nullifyColumn>:
  endbr64
  movslq %esi,%rsi
  test   %edi,%edi
  jle    L13
  movslq %edx,%rdx
  shl    $0x2,%rsi
  xor    %eax,%eax
  lea    (%rcx,%rdx,4),%rdx
  nopl   0x0(%rax,%rax,1)
L8:
  add    $0x1,%eax
  movl   $0x0,(%rdx)
  add    %rsi,%rdx
  cmp    %eax,%edi
  jne    L8
L13:
  ret
```

assistant: ```c
void nullifyColumn(int rows, int cols, int index, int matrix[rows][cols]) {
  for (int i = 0; i < rows; ++i) {
    matrix[i][index] = 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<nullifyColumn>:
  endbr64
  movslq %esi,%rsi
  test   %edi,%edi
  jle    L2
  movslq %edx,%rdx
  shl    $0x2,%rsi
  xor    %eax,%eax
  lea    (%rcx,%rdx,4),%rdx
  nopl   0x0(%rax,%rax,1)
L15:
  add    $0x1,%eax
  movl   $0x0,(%rdx)
  add    %rsi,%rdx
  cmp    %eax,%edi
  jne    L15
L2:
  ret
```

assistant: ```c
void nullifyColumn(int rows, int cols, int index, int matrix[rows][cols]) {
  for (int i = 0; i < rows; ++i) {
    matrix[i][index] = 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   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  js     L16
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L24
L3:
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  je     L7
  subl   $0x1,-0x10(%rbp)
  jmp    L24
L7:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L24:
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L3
L16:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int r, da, db, temp;
  r = __VERIFIER_nondet_int();
  if (r >= 0) {
    da = 2 * r;
    db = 2 * r;
    while (da >= r) {
      if (__VERIFIER_nondet_int() != 0) {
        da = da - 1;
      } else {
        temp = da;
        da = db - 1;
        db = da;
      }
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  js     L42
  mov    %eax,%ebp
  lea    (%rax,%rax,1),%ebx
  cmp    %ebx,%eax
  jg     L42
  mov    %ebx,%r12d
  jmp    L33
L58:
  cmp    %ebx,%ebp
  jg     L42
L33:
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  sub    $0x1,%ebx
  test   %eax,%eax
  jne    L58
  lea    -0x1(%r12),%ebx
  mov    %ebx,%r12d
  jmp    L58
L42:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  int r, da, db, temp;
  r = __VERIFIER_nondet_int();
  if (r >= 0) {
    da = 2 * r;
    db = 2 * r;
    while (da >= r) {
      if (__VERIFIER_nondet_int() != 0) {
        da = da - 1;
      } else {
        temp = da;
        da = db - 1;
        db = da;
      }
    }
  }
  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   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  js     L36
  lea    (%rax,%rax,1),%ebx
  mov    %eax,%ebp
  cmp    %ebx,%eax
  jg     L36
  mov    %ebx,%r12d
  nop
L21:
  xor    %eax,%eax
  sub    $0x1,%ebx
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jne    L14
  lea    -0x1(%r12),%ebx
  mov    %ebx,%r12d
L14:
  cmp    %ebx,%ebp
  jle    L21
L36:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  int r, da, db, temp;
  r = __VERIFIER_nondet_int();
  if (r >= 0) {
    da = 2 * r;
    db = 2 * r;
    while (da >= r) {
      if (__VERIFIER_nondet_int() != 0) {
        da = da - 1;
      } else {
        temp = da;
        da = db - 1;
        db = da;
      }
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  js     L20
  lea    (%rax,%rax,1),%ebx
  mov    %eax,%ebp
  cmp    %ebx,%eax
  jg     L20
  mov    %ebx,%r12d
  nop
L32:
  xor    %eax,%eax
  sub    $0x1,%ebx
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jne    L29
  lea    -0x1(%r12),%ebx
  mov    %ebx,%r12d
L29:
  cmp    %ebx,%ebp
  jle    L32
L20:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  int r, da, db, temp;
  r = __VERIFIER_nondet_int();
  if (r >= 0) {
    da = 2 * r;
    db = 2 * r;
    while (da >= r) {
      if (__VERIFIER_nondet_int() != 0) {
        da = da - 1;
      } else {
        temp = da;
        da = db - 1;
        db = da;
      }
    }
  }
  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   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x6464614d,-0x6a(%rbp)
  movw   $0x6569,-0x66(%rbp)
  movb   $0x0,-0x64(%rbp)
  movl   $0x72646e41,-0x63(%rbp)
  movw   $0x7765,-0x5f(%rbp)
  movb   $0x0,-0x5d(%rbp)
  lea    D20(%rip),%rax
  mov    %al,-0x6b(%rbp)
  movabs $0x616d726564697053,%rax
  mov    $0x6e,%edx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movw   $0x0,-0x10(%rbp)
  movb   $0x4b,-0x5c(%rbp)
  movb   $0x61,-0x5b(%rbp)
  movb   $0x74,-0x5a(%rbp)
  movb   $0x69,-0x59(%rbp)
  movb   $0x65,-0x58(%rbp)
  movb   $0x0,-0x57(%rbp)
  lea    -0x50(%rbp),%rax
  movabs $0x72636e4920656854,%rsi
  movabs $0x4820656c62696465,%rdi
  mov    %rsi,(%rax)
  mov    %rdi,0x8(%rax)
  movl   $0x6b6c75,0x10(%rax)
  movsbl -0x6b(%rbp),%edx
  lea    -0x5c(%rbp),%rax
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rdx
  lea    -0x6a(%rbp),%rax
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x50(%rbp),%rdx
  lea    -0x63(%rbp),%rax
  mov    %rax,%rsi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D20
datatype: string
name: parse_data
arguments: label: D37
datatype: string

tool: D20: Batman

tool: D37: %s's favorite hero is %s.


assistant: ```c
int main() {
  char Kid1[12];
  char Kid2[] = "Maddie";
  char Kid3[7] = "Andrew";
  char Hero1 = "Batman";
  char Hero2[34] = "Spiderman";
  char Hero3[25];
  Kid1[0] = 'K';
  Kid1[1] = 'a';
  Kid1[2] = 't';
  Kid1[3] = 'i';
  Kid1[4] = 'e';
  Kid1[5] = '\0';
  strcpy(Hero3, "The Incredible Hulk");
  printf("%s\'s favorite hero is %s.\n", Kid1, Hero1);
  printf("%s\'s favorite hero is %s.\n", Kid2, Hero2);
  printf("%s\'s favorite hero is %s.\n", Kid3, Hero3);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movl   $0x6464614d,0x6(%rsp)
  movw   $0x6569,0xa(%rsp)
  movb   $0x0,0xc(%rsp)
  movl   $0x72646e41,0xd(%rsp)
  movw   $0x7765,0x11(%rsp)
  movb   $0x0,0x13(%rsp)
  movabs $0x616d726564697053,%rax
  mov    $0x6e,%edx
  mov    %rax,0x40(%rsp)
  mov    %rdx,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movw   $0x0,0x60(%rsp)
  movb   $0x4b,0x14(%rsp)
  movb   $0x61,0x15(%rsp)
  movb   $0x74,0x16(%rsp)
  movb   $0x69,0x17(%rsp)
  movb   $0x65,0x18(%rsp)
  movb   $0x0,0x19(%rsp)
  lea    0x20(%rsp),%rbp
  movabs $0x72636e4920656854,%rax
  movabs $0x4820656c62696465,%rdx
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  movl   $0x6b6c75,0x30(%rsp)
  lea    D5(%rip),%rdx
  lea    0x14(%rsp),%rsi
  movsbl %dl,%edx
  lea    D1(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x40(%rsp),%rdx
  lea    0x6(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xd(%rsp),%rsi
  mov    %rbp,%rdx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
data_type: string
name: extract_symbolic_data
arguments: data_label: D1
data_type: string

tool: D5: Batman

tool: D1: %s's favorite hero is %s.


assistant: ```c
int main() {
  char Kid1[12];
  char Kid2[] = "Maddie";
  char Kid3[7] = "Andrew";
  char Hero1 = "Batman";
  char Hero2[34] = "Spiderman";
  char Hero3[25];
  Kid1[0] = 'K';
  Kid1[1] = 'a';
  Kid1[2] = 't';
  Kid1[3] = 'i';
  Kid1[4] = 'e';
  Kid1[5] = '\0';
  strcpy(Hero3, "The Incredible Hulk");
  printf("%s\'s favorite hero is %s.\n", Kid1, Hero1);
  printf("%s\'s favorite hero is %s.\n", Kid2, Hero2);
  printf("%s\'s favorite hero is %s.\n", Kid3, Hero3);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x6569,%edx
  xor    %esi,%esi
  mov    $0x65,%edi
  push   %rbp
  lea    D33(%rip),%rbp
  mov    $0x7765,%ecx
  sub    $0x78,%rsp
  movdqa D31(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %dx,0xa(%rsp)
  lea    D38(%rip),%rdx
  lea    0x20(%rsp),%r12
  movaps %xmm0,0x40(%rsp)
  pxor   %xmm0,%xmm0
  movsbl %dl,%edx
  movaps %xmm0,0x50(%rsp)
  movdqa D19(%rip),%xmm0
  mov    %si,0x60(%rsp)
  lea    0x14(%rsp),%rsi
  mov    %di,0x18(%rsp)
  mov    %rbp,%rdi
  mov    %cx,0x11(%rsp)
  movaps %xmm0,0x20(%rsp)
  movl   $0x6464614d,0x6(%rsp)
  movb   $0x0,0xc(%rsp)
  movl   $0x72646e41,0xd(%rsp)
  movb   $0x0,0x13(%rsp)
  movl   $0x6974614b,0x14(%rsp)
  movl   $0x6b6c75,0x30(%rsp)
  call   <printf@plt>
  lea    0x40(%rsp),%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    0x6(%rsp),%rsi
  call   <printf@plt>
  xor    %eax,%eax
  lea    0xd(%rsp),%rsi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D33
type: string
name: unpack_data
arguments: data_label: D31
type: byte[16]
name: unpack_data
arguments: data_label: D38
type: string
name: unpack_data
arguments: data_label: D19
type: byte[16]

tool: D33: %s's favorite hero is %s.


tool: {"D31": [0x53,0x70,0x69,0x64,0x65,0x72,0x6d,0x61,0x6e,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D38: Batman

tool: {"D19": [0x54,0x68,0x65,0x20,0x49,0x6e,0x63,0x72,0x65,0x64,0x69,0x62,0x6c,0x65,0x20,0x48]}

assistant: ```c
int main() {
  char Kid1[12];
  char Kid2[] = "Maddie";
  char Kid3[7] = "Andrew";
  char Hero1 = "Batman";
  char Hero2[34] = "Spiderman";
  char Hero3[25];
  Kid1[0] = 'K';
  Kid1[1] = 'a';
  Kid1[2] = 't';
  Kid1[3] = 'i';
  Kid1[4] = 'e';
  Kid1[5] = '\0';
  strcpy(Hero3, "The Incredible Hulk");
  printf("%s\'s favorite hero is %s.\n", Kid1, Hero1);
  printf("%s\'s favorite hero is %s.\n", Kid2, Hero2);
  printf("%s\'s favorite hero is %s.\n", Kid3, Hero3);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x6569,%edx
  xor    %esi,%esi
  mov    $0x65,%edi
  push   %rbp
  lea    D39(%rip),%rbp
  mov    $0x7765,%ecx
  sub    $0x78,%rsp
  movdqa D31(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %dx,0xa(%rsp)
  lea    D19(%rip),%rdx
  lea    0x20(%rsp),%r12
  movaps %xmm0,0x40(%rsp)
  pxor   %xmm0,%xmm0
  movsbl %dl,%edx
  movaps %xmm0,0x50(%rsp)
  movdqa D5(%rip),%xmm0
  mov    %si,0x60(%rsp)
  lea    0x14(%rsp),%rsi
  mov    %di,0x18(%rsp)
  mov    %rbp,%rdi
  mov    %cx,0x11(%rsp)
  movaps %xmm0,0x20(%rsp)
  movl   $0x6464614d,0x6(%rsp)
  movb   $0x0,0xc(%rsp)
  movl   $0x72646e41,0xd(%rsp)
  movb   $0x0,0x13(%rsp)
  movl   $0x6974614b,0x14(%rsp)
  movl   $0x6b6c75,0x30(%rsp)
  call   <printf@plt>
  lea    0x40(%rsp),%rdx
  mov    %rbp,%rdi
  xor    %eax,%eax
  lea    0x6(%rsp),%rsi
  call   <printf@plt>
  xor    %eax,%eax
  lea    0xd(%rsp),%rsi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D39
type_guessed: string
name: disassemble_data
arguments: data_label: D31
type_guessed: byte[16]
name: disassemble_data
arguments: data_label: D19
type_guessed: string
name: disassemble_data
arguments: data_label: D5
type_guessed: byte[16]

tool: D39: %s's favorite hero is %s.


tool: {"D31": [0x53,0x70,0x69,0x64,0x65,0x72,0x6d,0x61,0x6e,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D19: Batman

tool: {"D5": [0x54,0x68,0x65,0x20,0x49,0x6e,0x63,0x72,0x65,0x64,0x69,0x62,0x6c,0x65,0x20,0x48]}

assistant: ```c
int main() {
  char Kid1[12];
  char Kid2[] = "Maddie";
  char Kid3[7] = "Andrew";
  char Hero1 = "Batman";
  char Hero2[34] = "Spiderman";
  char Hero3[25];
  Kid1[0] = 'K';
  Kid1[1] = 'a';
  Kid1[2] = 't';
  Kid1[3] = 'i';
  Kid1[4] = 'e';
  Kid1[5] = '\0';
  strcpy(Hero3, "The Incredible Hulk");
  printf("%s\'s favorite hero is %s.\n", Kid1, Hero1);
  printf("%s\'s favorite hero is %s.\n", Kid2, Hero2);
  printf("%s\'s favorite hero is %s.\n", Kid3, Hero3);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<emovi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rbx
  mov    -0x28(%rbp),%rax
  lea    0xa(%rax),%r12
  movzwl (%r12),%eax
  test   %ax,%ax
  jns    L60
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movw   $0xffff,(%rax)
  jmp    L36
L60:
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movw   $0x0,(%rax)
L36:
  mov    %r12,%rax
  lea    -0x2(%rax),%r12
  movzwl (%rax),%eax
  mov    %ax,(%rbx)
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movzwl (%rax),%edx
  and    $0x7fff,%dx
  mov    %dx,(%rax)
  lea    -0x2(%rbx),%rax
  movzwl (%rax),%eax
  movzwl %ax,%eax
  and    $0x7fff,%eax
  cmp    $0x7fff,%eax
  jne    L9
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <eisnan@plt>
  test   %eax,%eax
  je     L10
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movw   $0x0,(%rax)
  movl   $0x3,-0x14(%rbp)
  jmp    L31
L101:
  mov    %r12,%rdx
  lea    -0x2(%rdx),%r12
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movzwl (%rdx),%edx
  mov    %dx,(%rax)
  addl   $0x1,-0x14(%rbp)
L31:
  cmpl   $0x8,-0x14(%rbp)
  jle    L101
  jmp    L61
L10:
  movl   $0x2,-0x14(%rbp)
  jmp    L85
L5:
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movw   $0x0,(%rax)
  addl   $0x1,-0x14(%rbp)
L85:
  cmpl   $0x8,-0x14(%rbp)
  jle    L5
  jmp    L61
L9:
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movw   $0x0,(%rax)
  movl   $0x0,-0x14(%rbp)
  jmp    L27
L48:
  mov    %r12,%rdx
  lea    -0x2(%rdx),%r12
  mov    %rbx,%rax
  lea    0x2(%rax),%rbx
  movzwl (%rdx),%edx
  mov    %dx,(%rax)
  addl   $0x1,-0x14(%rbp)
L27:
  cmpl   $0x4,-0x14(%rbp)
  jle    L48
  movw   $0x0,(%rbx)
L61:
  add    $0x20,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
void emovi(a, b) unsigned short *a, *b;
{
  register unsigned short *p, *q;
  int i;
  q = b;
  p = a + (6 - 1);
  if (*p & 0x8000)
    *q++ = 0xffff;
  else
    *q++ = 0;
  *q = *p--;
  *q++ &= 0x7fff;
  if ((*(q - 1) & 0x7fff) == 0x7fff) {
    if (eisnan(a)) {
      *q++ = 0;
      for (i = 3; i < (6 + 3); i++)
        *q++ = *p--;
      return;
    }
    for (i = 2; i < (6 + 3); i++)
      *q++ = 0;
    return;
  }
  *q++ = 0;
  for (i = 0; i < 6 - 1; i++)
    *q++ = *p--;
  *q = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<emovi>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  lea    0x2(%rsi),%r12
  mov    $0x0,%eax
  cmpw   $0x0,0xa(%rdi)
  js     L6
L42:
  mov    %ax,(%rsi)
  lea    0x8(%rbp),%rbx
  movzwl 0xa(%rbp),%eax
  and    $0x7fff,%ax
  mov    %ax,(%r12)
  cmp    $0x7fff,%ax
  je     L34
  lea    0x4(%r12),%rax
  movw   $0x0,0x2(%r12)
  sub    $0x2,%rbp
L37:
  sub    $0x2,%rbx
  add    $0x2,%rax
  movzwl 0x2(%rbx),%edx
  mov    %dx,-0x2(%rax)
  cmp    %rbp,%rbx
  jne    L37
  movw   $0x0,0xe(%r12)
L43:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    $0xffffffff,%eax
  jmp    L42
L34:
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <eisnan@plt>
  test   %eax,%eax
  jne    L72
  lea    0x2(%r12),%rax
  add    $0x10,%r12
L45:
  add    $0x2,%rax
  movw   $0x0,-0x2(%rax)
  cmp    %rax,%r12
  jne    L45
  jmp    L43
L72:
  lea    0x4(%r12),%rax
  movw   $0x0,0x2(%r12)
  sub    $0x4,%rbp
L2:
  sub    $0x2,%rbx
  add    $0x2,%rax
  movzwl 0x2(%rbx),%edx
  mov    %dx,-0x2(%rax)
  cmp    %rbx,%rbp
  jne    L2
  jmp    L43
```

assistant: ```c
void emovi(a, b) unsigned short *a, *b;
{
  register unsigned short *p, *q;
  int i;
  q = b;
  p = a + (6 - 1);
  if (*p & 0x8000)
    *q++ = 0xffff;
  else
    *q++ = 0;
  *q = *p--;
  *q++ &= 0x7fff;
  if ((*(q - 1) & 0x7fff) == 0x7fff) {
    if (eisnan(a)) {
      *q++ = 0;
      for (i = 3; i < (6 + 3); i++)
        *q++ = *p--;
      return;
    }
    for (i = 2; i < (6 + 3); i++)
      *q++ = 0;
    return;
  }
  *q++ = 0;
  for (i = 0; i < 6 - 1; i++)
    *q++ = *p--;
  *q = 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<emovi>:
  endbr64
  movzwl 0xa(%rdi),%eax
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sar    $0xf,%ax
  push   %rbx
  lea    0x8(%rdi),%rbx
  mov    %ax,(%rsi)
  movzwl 0xa(%rdi),%eax
  and    $0x7fff,%ax
  mov    %ax,0x2(%rsi)
  cmp    $0x7fff,%ax
  je     L37
  xor    %edx,%edx
  lea    0x6(%rsi),%rax
  sub    $0x2,%rbp
  mov    %dx,0x4(%rsi)
L11:
  movzwl (%rbx),%edx
  sub    $0x2,%rbx
  add    $0x2,%rax
  mov    %dx,-0x2(%rax)
  cmp    %rbp,%rbx
  jne    L11
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  mov    %ax,0x10(%r12)
  pop    %r12
  ret
  xchg   %ax,%ax
L37:
  xor    %eax,%eax
  call   <eisnan@plt>
  test   %eax,%eax
  je     L19
  xor    %esi,%esi
  lea    0x6(%r12),%rax
  sub    $0x4,%rbp
  mov    %si,0x4(%r12)
  nopl   0x0(%rax)
L24:
  movzwl (%rbx),%edx
  sub    $0x2,%rbx
  add    $0x2,%rax
  mov    %dx,-0x2(%rax)
  cmp    %rbp,%rbx
  jne    L24
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L19:
  movq   $0x0,0x4(%r12)
  xor    %ecx,%ecx
  pop    %rbx
  movl   $0x0,0xc(%r12)
  pop    %rbp
  mov    %cx,0x10(%r12)
  pop    %r12
  ret
```

assistant: ```c
void emovi(a, b) unsigned short *a, *b;
{
  register unsigned short *p, *q;
  int i;
  q = b;
  p = a + (6 - 1);
  if (*p & 0x8000)
    *q++ = 0xffff;
  else
    *q++ = 0;
  *q = *p--;
  *q++ &= 0x7fff;
  if ((*(q - 1) & 0x7fff) == 0x7fff) {
    if (eisnan(a)) {
      *q++ = 0;
      for (i = 3; i < (6 + 3); i++)
        *q++ = *p--;
      return;
    }
    for (i = 2; i < (6 + 3); i++)
      *q++ = 0;
    return;
  }
  *q++ = 0;
  for (i = 0; i < 6 - 1; i++)
    *q++ = *p--;
  *q = 0;
}
```