ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  mov    $0xb2d05e00,%esi
  sub    $0x8,%rsp
  lea    D16(%rip),%rdi
  mov    %rsi,%rdx
  call   <printf@plt>
  mov    $0xc8,%edx
  mov    $0xc8,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x10001,%edx
  mov    $0x10001,%esi
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  movabs $0x2ce37e5bf8d54c02,%rsi
  mov    %rsi,%rdx
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D16
type_infered: string
name: disassemble_data
arguments: label: D2
type_infered: string
name: disassemble_data
arguments: label: D24
type_infered: string
name: disassemble_data
arguments: label: D12
type_infered: string

tool: D16: un = %u and not %d


tool: D2: end = %hd and %d


tool: D24: big = %ld and not %hd


tool: D12: verybig = %lld and not %ld


assistant: ```c
int main(void) {
  unsigned int un = 3000000000;
  short end = 200;
  long big = 65537;
  long long verybig = 3234567890864000002;
  printf("un = %u and not %d\n", un, un);
  printf("end = %hd and %d\n", end, end);
  printf("big = %ld and not %hd\n", big, big);
  printf("verybig = %lld and not %ld\n", verybig, verybig);
  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
  mov    $0xb2d05e00,%esi
  sub    $0x8,%rsp
  lea    D26(%rip),%rdi
  mov    %rsi,%rdx
  call   <printf@plt>
  mov    $0xc8,%edx
  mov    $0xc8,%esi
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    $0x10001,%edx
  mov    $0x10001,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  movabs $0x2ce37e5bf8d54c02,%rsi
  mov    %rsi,%rdx
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D26
datatype: string
name: unpack_data
arguments: label: D24
datatype: string
name: unpack_data
arguments: label: D5
datatype: string
name: unpack_data
arguments: label: D20
datatype: string

tool: D26: un = %u and not %d


tool: D24: end = %hd and %d


tool: D5: big = %ld and not %hd


tool: D20: verybig = %lld and not %ld


assistant: ```c
int main(void) {
  unsigned int un = 3000000000;
  short end = 200;
  long big = 65537;
  long long verybig = 3234567890864000002;
  printf("un = %u and not %d\n", un, un);
  printf("end = %hd and %d\n", end, end);
  printf("big = %ld and not %hd\n", big, big);
  printf("verybig = %lld and not %ld\n", verybig, verybig);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x10,-0x30(%rbp)
  mov    -0x30(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L32:
  cmp    %rdx,%rsp
  je     L74
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L32
L74:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L64
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L64:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x4,-0x2c(%rbp)
  lea    -0x3c(%rbp),%rax
  mov    %rax,%rdi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L77
L27:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x64,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    -0x38(%rbp),%ecx
  movslq %ecx,%rcx
  mov    %edx,(%rax,%rcx,4)
  addl   $0x1,-0x38(%rbp)
L77:
  mov    -0x38(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L27
  mov    -0x2c(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x30(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <callerBP@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L39
L71:
  mov    -0x20(%rbp),%rax
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L39:
  mov    -0x34(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L71
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbx,%rsp
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L79
  call   <__stack_chk_fail@plt>
L79:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D5
data_type: string

tool: D5: %d 

assistant: ```c
int main() {
  int N = 16;
  int d[N];
  int threads = 4;
  time_t t;
  srand((unsigned)time(&t));
  int j;
  for (j = 0; j < N; ++j) {
    d[j] = (rand() % 100);
  }
  callerBP(N, d, threads);
  int i;
  for (i = 0; i < N; ++i) {
    printf("%d ", d[i]);
  }
  printf("\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  lea    0x10(%rsp),%rbx
  lea    0x50(%rsp),%r12
  mov    %rbx,%rbp
L1:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,0x0(%rbp)
  add    $0x4,%rbp
  cmp    %r12,%rbp
  jne    L1
  lea    0x10(%rsp),%rsi
  mov    $0x4,%edx
  mov    $0x10,%edi
  call   <callerBP@plt>
  lea    D3(%rip),%rbp
L4:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L4
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D3
datatype: string

tool: D3: %d 

assistant: ```c
int main() {
  int N = 16;
  int d[N];
  int threads = 4;
  time_t t;
  srand((unsigned)time(&t));
  int j;
  for (j = 0; j < N; ++j) {
    d[j] = (rand() % 100);
  }
  callerBP(N, d, threads);
  int i;
  for (i = 0; i < N; ++i) {
    printf("%d ", d[i]);
  }
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  lea    0x10(%rsp),%r12
  call   <time@plt>
  mov    %r12,%rbx
  lea    0x50(%rsp),%rbp
  mov    %r12,%r13
  mov    %eax,%edi
  call   <srand@plt>
  nop
L15:
  xor    %eax,%eax
  add    $0x4,%r13
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x25,%rdx
  sub    %ecx,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%r13)
  cmp    %rbp,%r13
  jne    L15
  mov    %r12,%rsi
  mov    $0x4,%edx
  mov    $0x10,%edi
  call   <callerBP@plt>
  lea    D1(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L17:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L17
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_guessed: string

tool: D1: %d 

assistant: ```c
int main() {
  int N = 16;
  int d[N];
  int threads = 4;
  time_t t;
  srand((unsigned)time(&t));
  int j;
  for (j = 0; j < N; ++j) {
    d[j] = (rand() % 100);
  }
  callerBP(N, d, threads);
  int i;
  for (i = 0; i < N; ++i) {
    printf("%d ", d[i]);
  }
  printf("\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  lea    0x10(%rsp),%r12
  call   <time@plt>
  mov    %r12,%rbx
  lea    0x50(%rsp),%rbp
  mov    %r12,%r13
  mov    %eax,%edi
  call   <srand@plt>
  nop
L24:
  xor    %eax,%eax
  add    $0x4,%r13
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x25,%rdx
  sub    %ecx,%edx
  imul   $0x64,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%r13)
  cmp    %rbp,%r13
  jne    L24
  mov    %r12,%rsi
  mov    $0x4,%edx
  mov    $0x10,%edi
  call   <callerBP@plt>
  lea    D6(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L22:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L22
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x68,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type: string

tool: D6: %d 

assistant: ```c
int main() {
  int N = 16;
  int d[N];
  int threads = 4;
  time_t t;
  srand((unsigned)time(&t));
  int j;
  for (j = 0; j < N; ++j) {
    d[j] = (rand() % 100);
  }
  callerBP(N, d, threads);
  int i;
  for (i = 0; i < N; ++i) {
    printf("%d ", d[i]);
  }
  printf("\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<md_write_e_type>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  je     L75
  cmpl   $0x3,-0x4(%rbp)
  jg     L38
  cmpl   $0x2,-0x4(%rbp)
  je     L108
  cmpl   $0x2,-0x4(%rbp)
  jg     L38
  cmpl   $0x0,-0x4(%rbp)
  je     L19
  cmpl   $0x1,-0x4(%rbp)
  je     L49
  jmp    L38
L19:
  lea    D3(%rip),%rax
  jmp    L45
L49:
  lea    D48(%rip),%rax
  jmp    L45
L108:
  lea    D48(%rip),%rax
  jmp    L45
L75:
  lea    D45(%rip),%rax
  jmp    L45
L38:
  lea    D3(%rip),%rax
L45:
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
datatype: string
name: disassemble_data
arguments: data_label: D48
datatype: string
name: disassemble_data
arguments: data_label: D45
datatype: string

tool: D3: |--|

tool: D48: o--|

tool: D45: o--o

assistant: ```c
char *md_write_e_type(int type) {
  switch (type) {
  case 0:
    return ("|--|");
    break;
  case 1:
    return ("o--|");
    break;
  case 2:
    return ("o--|");
    break;
  case 3:
    return ("o--o");
    break;
  }
  return ("|--|");
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<md_write_e_type>:
  endbr64
  cmp    $0x2,%edi
  jg     L1
  test   %edi,%edi
  lea    D28(%rip),%rax
  lea    D10(%rip),%rdx
  cmovg  %rdx,%rax
  ret
L1:
  cmp    $0x3,%edi
  lea    D28(%rip),%rax
  lea    D4(%rip),%rdx
  cmove  %rdx,%rax
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D28
data_type: string
name: retrieve_data
arguments: data_label: D10
data_type: string
name: retrieve_data
arguments: data_label: D4
data_type: string

tool: D28: |--|

tool: D10: o--|

tool: D4: o--o

assistant: ```c
char *md_write_e_type(int type) {
  switch (type) {
  case 0:
    return ("|--|");
    break;
  case 1:
    return ("o--|");
    break;
  case 2:
    return ("o--|");
    break;
  case 3:
    return ("o--o");
    break;
  }
  return ("|--|");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<md_write_e_type>:
  endbr64
  cmp    $0x2,%edi
  jg     L4
  test   %edi,%edi
  lea    D11(%rip),%rdx
  lea    D15(%rip),%rax
  cmovg  %rdx,%rax
  ret
  xchg   %ax,%ax
L4:
  cmp    $0x3,%edi
  lea    D31(%rip),%rdx
  lea    D15(%rip),%rax
  cmove  %rdx,%rax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D11
type_guessed: string
name: disassemble_data
arguments: data_label: D15
type_guessed: string
name: disassemble_data
arguments: data_label: D31
type_guessed: string

tool: D11: o--|

tool: D15: |--|

tool: D31: o--o

assistant: ```c
char *md_write_e_type(int type) {
  switch (type) {
  case 0:
    return ("|--|");
    break;
  case 1:
    return ("o--|");
    break;
  case 2:
    return ("o--|");
    break;
  case 3:
    return ("o--o");
    break;
  }
  return ("|--|");
}
```
user: Translate the below assembly code into C programming language:

```asm
<md_write_e_type>:
  endbr64
  cmp    $0x2,%edi
  jg     L7
  test   %edi,%edi
  lea    D20(%rip),%rdx
  lea    D33(%rip),%rax
  cmovg  %rdx,%rax
  ret
  xchg   %ax,%ax
L7:
  cmp    $0x3,%edi
  lea    D32(%rip),%rdx
  lea    D33(%rip),%rax
  cmove  %rdx,%rax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D20
datatype: string
name: extract_symbolic_data
arguments: data_tag: D33
datatype: string
name: extract_symbolic_data
arguments: data_tag: D32
datatype: string

tool: D20: o--|

tool: D33: |--|

tool: D32: o--o

assistant: ```c
char *md_write_e_type(int type) {
  switch (type) {
  case 0:
    return ("|--|");
    break;
  case 1:
    return ("o--|");
    break;
  case 2:
    return ("o--|");
    break;
  case 3:
    return ("o--o");
    break;
  }
  return ("|--|");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  lea    0x20(%rsp),%rbx
  mov    %rbx,%rdi
  call   <print_input@plt>
  mov    %rsp,%rbp
  lea    0x40(%rsp),%r12
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <run@plt>
  mov    $0x0,%ecx
  mov    $0xffffffff,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <fill@plt>
  mov    %rbx,%rdi
  call   <print_input@plt>
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <run@plt>
  mov    $0x0,%ecx
  mov    $0xf0f0f0f0,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <fill@plt>
  mov    %rbx,%rdi
  call   <print_input@plt>
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <run@plt>
  mov    $0x1010101,%ecx
  mov    $0x1010101,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <fill@plt>
  mov    %rbx,%rdi
  call   <print_input@plt>
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <run@plt>
  mov    $0xfefefeff,%ecx
  mov    $0xffffffff,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <fill@plt>
  mov    %rbx,%rdi
  call   <print_input@plt>
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <run@plt>
  mov    $0x1010101,%ecx
  mov    $0x0,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <fill@plt>
  mov    %rbx,%rdi
  call   <print_input@plt>
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <run@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char _input[256 / 8 * 2] = {0};
  char _output[256 / 8] = {0};
  void *input = _input;
  void *output = _output;
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xffffffffffffffffl, 0);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xf0f0f0f0f0f0f0f0l, 0);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0x0101010101010101l, 0x9456010101010101l);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xffffffffffffffffl, -0x0101010101010101l);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0, 0x0101010101010101l);
  print_input(input);
  run(input, input + 32, output);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r13
  pxor   %xmm0,%xmm0
  push   %r12
  push   %rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rbp
  mov    %rsp,%r12
  lea    0x40(%rsp),%r13
  mov    %rbp,%rdi
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  xor    %ecx,%ecx
  mov    $0xffffffff,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  xor    %ecx,%ecx
  mov    $0xf0f0f0f0,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  mov    $0x1010101,%ecx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x1010101,%edx
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  mov    $0xfefefeff,%ecx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0xffffffff,%edx
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  xor    %edx,%edx
  mov    %r12,%rsi
  mov    $0x1010101,%ecx
  mov    %rbp,%rdi
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char _input[256 / 8 * 2] = {0};
  char _output[256 / 8] = {0};
  void *input = _input;
  void *output = _output;
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xffffffffffffffffl, 0);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xf0f0f0f0f0f0f0f0l, 0);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0x0101010101010101l, 0x9456010101010101l);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xffffffffffffffffl, -0x0101010101010101l);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0, 0x0101010101010101l);
  print_input(input);
  run(input, input + 32, output);
  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   %r13
  pxor   %xmm0,%xmm0
  push   %r12
  push   %rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rbp
  mov    %rsp,%r12
  lea    0x40(%rsp),%r13
  mov    %rbp,%rdi
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  xor    %ecx,%ecx
  mov    $0xffffffff,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  xor    %ecx,%ecx
  mov    $0xf0f0f0f0,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  mov    $0x1010101,%ecx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x1010101,%edx
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  mov    $0xfefefeff,%ecx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0xffffffff,%edx
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  xor    %edx,%edx
  mov    %r12,%rsi
  mov    $0x1010101,%ecx
  mov    %rbp,%rdi
  call   <fill@plt>
  mov    %rbp,%rdi
  call   <print_input@plt>
  mov    %r12,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <run@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  char _input[256 / 8 * 2] = {0};
  char _output[256 / 8] = {0};
  void *input = _input;
  void *output = _output;
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xffffffffffffffffl, 0);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xf0f0f0f0f0f0f0f0l, 0);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0x0101010101010101l, 0x9456010101010101l);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0xffffffffffffffffl, -0x0101010101010101l);
  print_input(input);
  run(input, input + 32, output);
  fill(input, output, 0, 0x0101010101010101l);
  print_input(input);
  run(input, input + 32, output);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<inverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jne    L17
  mov    $0xffffffff,%eax
  jmp    L6
L17:
  mov    -0x4(%rbp),%eax
  lea    -0x2(%rax),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <expo@plt>
L6:
  leave
  ret
```

assistant: ```c
int inverse(int base, int n) {
  if (n == 0)
    return -1;
  return expo(base, n, base - 2);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<inverse>:
  endbr64
  test   %esi,%esi
  je     L5
  sub    $0x8,%rsp
  lea    -0x2(%rdi),%edx
  call   <expo@plt>
  add    $0x8,%rsp
  ret
L5:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int inverse(int base, int n) {
  if (n == 0)
    return -1;
  return expo(base, n, base - 2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inverse>:
  endbr64
  test   %esi,%esi
  je     L4
  lea    -0x2(%rdi),%edx
  jmp    1050 <expo@plt>
L4:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int inverse(int base, int n) {
  if (n == 0)
    return -1;
  return expo(base, n, base - 2);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<inverse>:
  endbr64
  test   %esi,%esi
  je     L0
  lea    -0x2(%rdi),%edx
  jmp    1050 <expo@plt>
L0:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int inverse(int base, int n) {
  if (n == 0)
    return -1;
  return expo(base, n, base - 2);
}
```