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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xfb8(%rbp)
  lea    -0xfb0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x1f4,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xfbc(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L42
L9:
  mov    -0xfbc(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    -0xfb8(%rbp),%eax
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
  mov    -0xfbc(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xfbc(%rbp)
  addl   $0x1,-0xfb8(%rbp)
L42:
  mov    -0xfbc(%rbp),%eax
  test   %eax,%eax
  jne    L9
  mov    -0xfb8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xfb4(%rbp)
  jmp    L21
L34:
  mov    -0xfb4(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  subl   $0x1,-0xfb4(%rbp)
L21:
  cmpl   $0x0,-0xfb4(%rbp)
  jns    L34
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D22
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D8
type_guessed: string

tool: D22: input a number:

tool: D8: %d

assistant: ```c
int main() {
  int num, i = 0, j;
  int arr[1000] = {0};
  printf("input a number:");
  scanf("%d", &num);
  while (num != 0) {
    arr[i] = num % 2;
    num = num / 2;
    i++;
  }
  for (j = i - 1; j >= 0; j--)
    printf("%d", arr[j]);
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfb8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    $0x1f4,%ecx
  rep stos %rax,%es:(%rdi)
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  je     L11
  mov    $0x0,%ecx
L10:
  mov    %ecx,%edi
  mov    %edx,%esi
  shr    $0x1f,%esi
  lea    (%rdx,%rsi,1),%eax
  and    $0x1,%eax
  sub    %esi,%eax
  mov    %eax,0x10(%rsp,%rcx,4)
  mov    %esi,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    %rcx,%rsi
  add    $0x1,%rcx
  test   %eax,%eax
  jne    L10
  movl   $0x0,0xc(%rsp)
  test   %edi,%edi
  js     L11
  movslq %edi,%rdi
  shl    $0x2,%rdi
  lea    0x10(%rsp,%rdi,1),%rbx
  lea    0xc(%rsp,%rdi,1),%rbp
  mov    %esi,%esi
  shl    $0x2,%rsi
  sub    %rsi,%rbp
  lea    D0(%rip),%r12
L2:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L2
L11:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xfb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  mov    $0x0,%eax
  add    $0xfc0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D25
type: string
name: retrieve_data
arguments: data_tag: D0
type: string

tool: D25: input a number:

tool: D0: %d

assistant: ```c
int main() {
  int num, i = 0, j;
  int arr[1000] = {0};
  printf("input a number:");
  scanf("%d", &num);
  while (num != 0) {
    arr[i] = num % 2;
    num = num / 2;
    i++;
  }
  for (j = i - 1; j >= 0; j--)
    printf("%d", arr[j]);
  printf("\n");
  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
  mov    $0x1f4,%ecx
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  sub    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfb8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  xor    %edx,%edx
  test   %eax,%eax
  je     L6
  nopl   0x0(%rax)
L35:
  mov    %eax,%ecx
  mov    %edx,%r12d
  shr    $0x1f,%ecx
  lea    (%rax,%rcx,1),%esi
  and    $0x1,%esi
  sub    %ecx,%esi
  mov    %eax,%ecx
  shr    $0x1f,%ecx
  mov    %esi,(%rbx,%rdx,4)
  add    $0x1,%rdx
  add    %ecx,%eax
  sar    %eax
  jne    L35
  sub    $0x1,%r12d
  movl   $0x0,0xc(%rsp)
  movslq %r12d,%r12
  jmp    L14
  nopl   0x0(%rax,%rax,1)
L18:
  mov    (%rbx,%r12,4),%esi
  sub    $0x1,%r12
L14:
  xor    %eax,%eax
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0xffffffff,%r12d
  jne    L18
L6:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xfb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0xfc0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D5
type_guessed: string
name: read_data
arguments: data_label: D8
type_guessed: string

tool: D5: %d

tool: D8: input a number:

assistant: ```c
int main() {
  int num, i = 0, j;
  int arr[1000] = {0};
  printf("input a number:");
  scanf("%d", &num);
  while (num != 0) {
    arr[i] = num % 2;
    num = num / 2;
    i++;
  }
  for (j = i - 1; j >= 0; j--)
    printf("%d", arr[j]);
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x1f4,%ecx
  push   %rbp
  push   %rbx
  lea    D7(%rip),%rbx
  sub    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfb8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  lea    0xc(%rsp),%r12
  mov    %rbp,%rdi
  rep stos %rax,%es:(%rdi)
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  xor    %ecx,%ecx
  test   %eax,%eax
  je     L38
  nop
L1:
  mov    %eax,%esi
  movslq %ecx,%rdi
  shr    $0x1f,%esi
  lea    (%rax,%rsi,1),%edx
  add    %esi,%eax
  and    $0x1,%edx
  sub    %esi,%edx
  mov    %edx,0x0(%rbp,%rcx,4)
  mov    %rcx,%rdx
  add    $0x1,%rcx
  sar    %eax
  jne    L1
  shl    $0x2,%rdi
  movl   $0x0,0xc(%rsp)
  shl    $0x2,%rdx
  add    %rdi,%r12
  add    %rdi,%rbp
  sub    %rdx,%r12
  nopw   0x0(%rax,%rax,1)
L3:
  mov    0x0(%rbp),%esi
  mov    %rbx,%rdi
  xor    %eax,%eax
  sub    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L3
L38:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xfb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0xfc0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type: string
name: disassemble_data
arguments: data_label: D8
type: string

tool: D7: %d

tool: D8: input a number:

assistant: ```c
int main() {
  int num, i = 0, j;
  int arr[1000] = {0};
  printf("input a number:");
  scanf("%d", &num);
  while (num != 0) {
    arr[i] = num % 2;
    num = num / 2;
    i++;
  }
  for (j = i - 1; j >= 0; j--)
    printf("%d", arr[j]);
  printf("\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<arr_ops>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L23
  mov    %rdi,%r15
  mov    %rdx,%r14
  mov    %esi,%eax
  mov    %rax,0x8(%rsp)
  mov    $0x0,%r13d
  lea    D27(%rip),%r12
L33:
  mov    %r14,%rbp
  mov    (%r14),%rdx
  mov    %r13d,%esi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%r14),%rax
  movzbl (%rax),%esi
  test   %sil,%sil
  je     L16
  mov    $0x1,%ebx
L8:
  movsbl %sil,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x0(%rbp),%rax
  movzbl (%rax,%rbx,1),%esi
  add    $0x1,%rbx
  test   %sil,%sil
  jne    L8
L16:
  lea    0x10(%rsp),%rbx
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <strcpy_argv@plt>
  mov    %rbx,%rsi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%r15,%r13,8),%rsi
  lea    D45(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    (%r15,%r13,8),%rdi
  mov    %rbx,%rsi
  call   <strcpy_my@plt>
  lea    D22(%rip),%rdi
  call   <puts@plt>
  add    $0x1,%r13
  add    $0x8,%r14
  cmp    0x8(%rsp),%r13
  jne    L33
L23:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D27
datatype: string
name: extract_symbolic_data
arguments: data_tag: D15
datatype: string
name: extract_symbolic_data
arguments: data_tag: D28
datatype: string
name: extract_symbolic_data
arguments: data_tag: D45
datatype: string
name: extract_symbolic_data
arguments: data_tag: D22
datatype: string

tool: D27: * ( * ( argv + i ) + j ) = '%c';


tool: D15: * ( argv + ( i = %d ) ) = '%s' _whole string_;



tool: D28: getted string tmp:'%s'


tool: D45: string * ( arr + i ):'%s'


tool: D22: -------------------

assistant: ```c
void arr_ops(char **array, int argc, char **argv) {
  int i = 0;
  int j = 0;
  char tmp[100];
  for (i = 0; i < argc; i++) {
    printf("* ( argv + ( i = %d ) ) = '%s' _whole string_;\n\n", i, *(argv + i));
    j = 0;
    while (*(*(argv + i) + j) != '\0') {
      printf("* ( * ( argv + i ) + j ) = '%c';\n", *(*(argv + i) + j));
      j++;
    }
    strcpy_argv(tmp, *(argv + i));
    printf("getted string tmp:'%s'\n", tmp);
    printf("string * ( arr + i ):'%s'\n", *(array + i));
    strcpy_my(*(array + i), tmp);
    printf("-------------------\n");
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<arr_ops>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L5
  movslq %esi,%rax
  mov    %rdi,%r14
  mov    %rdx,%rbx
  xor    %r12d,%r12d
  mov    %rax,0x8(%rsp)
  lea    0x10(%rsp),%r13
  lea    D40(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L30:
  mov    (%rbx),%rdx
  mov    %r12d,%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rbx),%r8
  movsbl (%r8),%esi
  test   %sil,%sil
  je     L13
  mov    $0x1,%r15d
  cs nopw 0x0(%rax,%rax,1)
L23:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rbx),%r8
  movsbl (%r8,%r15,1),%esi
  add    $0x1,%r15
  test   %sil,%sil
  jne    L23
L13:
  mov    %r8,%rsi
  mov    %r13,%rdi
  add    $0x8,%rbx
  call   <strcpy_argv@plt>
  mov    %r13,%rsi
  lea    D34(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%r14,%r12,8),%rsi
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%r14,%r12,8),%rdi
  mov    %r13,%rsi
  add    $0x1,%r12
  call   <strcpy_my@plt>
  lea    D28(%rip),%rdi
  call   <puts@plt>
  cmp    0x8(%rsp),%r12
  jne    L30
L5:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D40
data_type: string
name: read_data
arguments: label: D6
data_type: string
name: read_data
arguments: label: D34
data_type: string
name: read_data
arguments: label: D3
data_type: string
name: read_data
arguments: label: D28
data_type: string

tool: D40: * ( * ( argv + i ) + j ) = '%c';


tool: D6: * ( argv + ( i = %d ) ) = '%s' _whole string_;



tool: D34: getted string tmp:'%s'


tool: D3: string * ( arr + i ):'%s'


tool: D28: -------------------

assistant: ```c
void arr_ops(char **array, int argc, char **argv) {
  int i = 0;
  int j = 0;
  char tmp[100];
  for (i = 0; i < argc; i++) {
    printf("* ( argv + ( i = %d ) ) = '%s' _whole string_;\n\n", i, *(argv + i));
    j = 0;
    while (*(*(argv + i) + j) != '\0') {
      printf("* ( * ( argv + i ) + j ) = '%c';\n", *(*(argv + i) + j));
      j++;
    }
    strcpy_argv(tmp, *(argv + i));
    printf("getted string tmp:'%s'\n", tmp);
    printf("string * ( arr + i ):'%s'\n", *(array + i));
    strcpy_my(*(array + i), tmp);
    printf("-------------------\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<arr_ops>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L24
  movslq %esi,%rax
  mov    %rdi,%r14
  mov    %rdx,%rbx
  xor    %r12d,%r12d
  mov    %rax,0x8(%rsp)
  lea    0x10(%rsp),%r13
  lea    D47(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L49:
  mov    (%rbx),%rdx
  mov    %r12d,%esi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rbx),%r8
  movsbl (%r8),%esi
  test   %sil,%sil
  je     L5
  mov    $0x1,%r15d
  cs nopw 0x0(%rax,%rax,1)
L12:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rbx),%r8
  movsbl (%r8,%r15,1),%esi
  add    $0x1,%r15
  test   %sil,%sil
  jne    L12
L5:
  mov    %r8,%rsi
  mov    %r13,%rdi
  add    $0x8,%rbx
  call   <strcpy_argv@plt>
  mov    %r13,%rsi
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%r14,%r12,8),%rsi
  lea    D21(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%r14,%r12,8),%rdi
  mov    %r13,%rsi
  add    $0x1,%r12
  call   <strcpy_my@plt>
  lea    D35(%rip),%rdi
  call   <puts@plt>
  cmp    0x8(%rsp),%r12
  jne    L49
L24:
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D47
data_type: string
name: unpack_data
arguments: label: D29
data_type: string
name: unpack_data
arguments: label: D2
data_type: string
name: unpack_data
arguments: label: D21
data_type: string
name: unpack_data
arguments: label: D35
data_type: string

tool: D47: * ( * ( argv + i ) + j ) = '%c';


tool: D29: * ( argv + ( i = %d ) ) = '%s' _whole string_;



tool: D2: getted string tmp:'%s'


tool: D21: string * ( arr + i ):'%s'


tool: D35: -------------------

assistant: ```c
void arr_ops(char **array, int argc, char **argv) {
  int i = 0;
  int j = 0;
  char tmp[100];
  for (i = 0; i < argc; i++) {
    printf("* ( argv + ( i = %d ) ) = '%s' _whole string_;\n\n", i, *(argv + i));
    j = 0;
    while (*(*(argv + i) + j) != '\0') {
      printf("* ( * ( argv + i ) + j ) = '%c';\n", *(*(argv + i) + j));
      j++;
    }
    strcpy_argv(tmp, *(argv + i));
    printf("getted string tmp:'%s'\n", tmp);
    printf("string * ( arr + i ):'%s'\n", *(array + i));
    strcpy_my(*(array + i), tmp);
    printf("-------------------\n");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<rop3_159_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  xor    -0x1c(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x20(%rbp),%eax
  and    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  notb   -0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  and    -0x1c(%rbp),%al
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  and    %edx,%eax
  or     %eax,%ecx
  mov    %ecx,%edx
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  and    %eax,%ecx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    %ecx,%eax
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_159_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = *D ^ S;
  stk1 = T & stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rop3_159_0_1>:
  endbr64
  mov    %rdi,%rcx
  mov    %esi,%edi
  mov    %esi,%eax
  xor    (%rcx),%al
  and    %edx,%eax
  not    %esi
  and    %edx,%esi
  and    (%rcx),%sil
  not    %edx
  or     %edx,%esi
  not    %eax
  and    %edi,%eax
  or     %eax,%esi
  mov    %sil,(%rcx)
  ret
```

assistant: ```c
void rop3_159_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = *D ^ S;
  stk1 = T & stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rop3_159_0_1>:
  endbr64
  mov    %rdi,%rcx
  mov    %esi,%edi
  movzbl (%rcx),%eax
  xor    %esi,%eax
  not    %esi
  and    %edx,%eax
  and    %edx,%esi
  not    %edx
  and    (%rcx),%sil
  not    %eax
  or     %edx,%esi
  and    %edi,%eax
  or     %eax,%esi
  mov    %sil,(%rcx)
  ret
```

assistant: ```c
void rop3_159_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = *D ^ S;
  stk1 = T & stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rop3_159_0_1>:
  endbr64
  mov    %rdi,%rcx
  mov    %esi,%edi
  movzbl (%rcx),%eax
  xor    %esi,%eax
  not    %esi
  and    %edx,%eax
  and    %edx,%esi
  not    %edx
  and    (%rcx),%sil
  not    %eax
  or     %edx,%esi
  and    %edi,%eax
  or     %eax,%esi
  mov    %sil,(%rcx)
  ret
```

assistant: ```c
void rop3_159_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = *D ^ S;
  stk1 = T & stk2;
  stk1 = ~stk1;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x230,%rsp
  mov    %edi,-0x224(%rbp)
  mov    %rsi,-0x230(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x220(%rbp)
  movl   $0x0,-0x21c(%rbp)
  movl   $0x1,-0x218(%rbp)
  movl   $0x0,-0x214(%rbp)
  cmpl   $0x3,-0x224(%rbp)
  jne    L11
  mov    -0x230(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  sub    $0x1,%eax
  mov    %eax,-0x220(%rbp)
  mov    -0x230(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  sub    $0x1,%eax
  mov    %eax,-0x21c(%rbp)
L11:
  lea    -0x110(%rbp),%rax
  mov    %rax,%rdi
  call   <Aones@plt>
  mov    -0x21c(%rbp),%edx
  mov    -0x220(%rbp),%ecx
  lea    -0x210(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <M_init@plt>
  jmp    L35
L66:
  lea    -0x21c(%rbp),%rdi
  lea    -0x220(%rbp),%rcx
  lea    -0x110(%rbp),%rdx
  mov    -0x218(%rbp),%esi
  lea    -0x210(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <move@plt>
  mov    %eax,-0x218(%rbp)
L61:
  cmpl   $0x3f,-0x218(%rbp)
  jle    L66
  lea    -0x210(%rbp),%rax
  mov    %rax,%rdi
  call   <printb@plt>
  addl   $0x1,-0x214(%rbp)
  mov    -0x214(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
L5:
  lea    -0x21c(%rbp),%rdi
  lea    -0x220(%rbp),%rcx
  lea    -0x110(%rbp),%rdx
  mov    -0x218(%rbp),%esi
  lea    -0x210(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <move@plt>
  mov    %eax,-0x218(%rbp)
L65:
  cmpl   $0x40,-0x218(%rbp)
  je     L5
L35:
  cmpl   $0x0,-0x218(%rbp)
  jne    L61
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L67
  call   <__stack_chk_fail@plt>
L67:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
data_type: string

tool: D7: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int m = 0, n = 0;
  int mvnum = 1, i = 0;
  int M[8][8], mvchoice[(8) * (8) + 1];
  if (argc == 3)
    m = atoi(argv[1]) - 1, n = atoi(argv[2]) - 1;
  Aones(mvchoice);
  M_init(M, m, n);
  while (mvnum) {
    while (mvnum < (8) * (8))
      mvnum = move(M, mvnum, mvchoice, &m, &n);
    printb(M);
    printf("%d\n", ++i);
    while (mvnum == (8) * (8))
      mvnum = move(M, mvnum, mvchoice, &m, &n);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x228,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x218(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  cmp    $0x3,%edi
  je     L90
L57:
  lea    0x110(%rsp),%rdi
  call   <Aones@plt>
  lea    0x10(%rsp),%rdi
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  call   <M_init@plt>
  mov    $0x0,%r14d
  mov    $0x1,%ebx
  lea    0xc(%rsp),%r13
  lea    0x8(%rsp),%r12
  lea    0x10(%rsp),%rbp
  lea    D2(%rip),%r15
  jmp    L52
L90:
  mov    %rsi,%rbx
  mov    0x8(%rsi),%rdi
  call   <atoi@plt>
  sub    $0x1,%eax
  mov    %eax,0x8(%rsp)
  mov    0x10(%rbx),%rdi
  call   <atoi@plt>
  sub    $0x1,%eax
  mov    %eax,0xc(%rsp)
  jmp    L57
L2:
  lea    0x110(%rsp),%rdx
  mov    %r13,%r8
  mov    %r12,%rcx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <move@plt>
  mov    %eax,%ebx
  cmp    $0x3f,%eax
  jle    L2
L39:
  mov    %rbp,%rdi
  call   <printb@plt>
  add    $0x1,%r14d
  mov    %r14d,%esi
  mov    %r15,%rdi
  call   <printf@plt>
  cmp    $0x40,%ebx
  jne    L69
L31:
  lea    0x110(%rsp),%rdx
  mov    %r13,%r8
  mov    %r12,%rcx
  mov    $0x40,%esi
  mov    %rbp,%rdi
  call   <move@plt>
  mov    %eax,%ebx
  cmp    $0x40,%eax
  je     L31
L69:
  test   %ebx,%ebx
  je     L36
L52:
  cmp    $0x3f,%ebx
  jle    L2
  jmp    L39
L36:
  mov    0x218(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x228,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D2
data_type: string

tool: D2: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int m = 0, n = 0;
  int mvnum = 1, i = 0;
  int M[8][8], mvchoice[(8) * (8) + 1];
  if (argc == 3)
    m = atoi(argv[1]) - 1, n = atoi(argv[2]) - 1;
  Aones(mvchoice);
  M_init(M, m, n);
  while (mvnum) {
    while (mvnum < (8) * (8))
      mvnum = move(M, mvnum, mvchoice, &m, &n);
    printb(M);
    printf("%d\n", ++i);
    while (mvnum == (8) * (8))
      mvnum = move(M, mvnum, mvchoice, &m, &n);
  }
  return 0;
}
```