ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<filecopy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    D32(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  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
L23:
  cmp    %rdx,%rsp
  je     L20
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L23
L20:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L80
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L80:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
  jmp    L101
L0:
  mov    -0x2c(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  cmp    %eax,-0x2c(%rbp)
  je     L101
  mov    $0xfffffffd,%eax
  jmp    L82
L101:
  mov    D32(%rip),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x34(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x2c(%rbp)
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x2c(%rbp)
  je     L70
  cmpl   $0x0,-0x2c(%rbp)
  jne    L0
L70:
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x2c(%rbp)
  jne    L74
  mov    $0xfffffffe,%eax
  jmp    L82
L74:
  mov    $0x0,%eax
L82:
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L96
  call   <__stack_chk_fail@plt>
L96:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int filecopy(int fd1) {
  int status;
  char buf[BUFSIZ];
  while (((status = read(fd1, buf, BUFSIZ)) != EOF) && status != 0) {
    if (write(1, buf, status) != status) {
      return -3;
    }
  }
  if (status == EOF) {
    return -2;
  } else {
    return 0;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<filecopy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  mov    D5(%rip),%rax
  movslq (%rax),%rax
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L23:
  cmp    %rdx,%rsp
  je     L58
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L23
L58:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L77
  orq    $0x0,-0x8(%rsp,%rax,1)
L77:
  mov    %rsp,%r12
  mov    D5(%rip),%r14
L29:
  mov    (%r14),%edx
  mov    %r12,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %eax,%ebx
  mov    D21(%rip),%rax
  mov    (%rax),%eax
  test   %ebx,%ebx
  je     L18
  cmp    %ebx,%eax
  je     L18
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  cmp    %ebx,%eax
  je     L29
  mov    $0xfffffffd,%eax
  jmp    L73
L18:
  cmp    %ebx,%eax
  setne  %al
  movzbl %al,%eax
  lea    -0x2(%rax,%rax,1),%eax
L73:
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L70
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L70:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int filecopy(int fd1) {
  int status;
  char buf[BUFSIZ];
  while (((status = read(fd1, buf, BUFSIZ)) != EOF) && status != 0) {
    if (write(1, buf, status) != status) {
      return -3;
    }
  }
  if (status == EOF) {
    return -2;
  } else {
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<filecopy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    D6(%rip),%r14
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movslq (%r14),%rax
  mov    %rax,%rdx
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L83
L39:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L39
L83:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L45
L30:
  mov    D7(%rip),%r15
  mov    %rsp,%rbx
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L96:
  test   %r12d,%r12d
  je     L75
  mov    %r12d,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  cmp    %r12d,%eax
  jne    L52
  mov    (%r14),%edx
L33:
  mov    %rbx,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %eax,%r12d
  mov    (%r15),%eax
  cmp    %r12d,%eax
  jne    L96
  mov    $0xfffffffe,%r12d
L90:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  lea    -0x28(%rbp),%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L75:
  test   %eax,%eax
  mov    $0xfffffffe,%eax
  cmove  %eax,%r12d
  jmp    L90
  nopl   (%rax)
L52:
  mov    $0xfffffffd,%r12d
  jmp    L90
L45:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L30
L40:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int filecopy(int fd1) {
  int status;
  char buf[BUFSIZ];
  while (((status = read(fd1, buf, BUFSIZ)) != EOF) && status != 0) {
    if (write(1, buf, status) != status) {
      return -3;
    }
  }
  if (status == EOF) {
    return -2;
  } else {
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<filecopy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    D8(%rip),%r14
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movslq (%r14),%rax
  mov    %rax,%rdx
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L28
L98:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L98
L28:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L21
L52:
  mov    D16(%rip),%r15
  mov    %rsp,%rbx
  jmp    L103
  nopl   0x0(%rax,%rax,1)
L66:
  test   %r12d,%r12d
  je     L11
  mov    %r12d,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  cmp    %r12d,%eax
  jne    L33
  mov    (%r14),%edx
L103:
  mov    %rbx,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  mov    %eax,%r12d
  mov    (%r15),%eax
  cmp    %r12d,%eax
  jne    L66
  mov    $0xfffffffe,%r12d
L106:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L89
  lea    -0x28(%rbp),%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  test   %eax,%eax
  mov    $0xfffffffe,%eax
  cmove  %eax,%r12d
  jmp    L106
  nopl   (%rax)
L33:
  mov    $0xfffffffd,%r12d
  jmp    L106
L21:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L52
L89:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int filecopy(int fd1) {
  int status;
  char buf[BUFSIZ];
  while (((status = read(fd1, buf, BUFSIZ)) != EOF) && status != 0) {
    if (write(1, buf, status) != status) {
      return -3;
    }
  }
  if (status == EOF) {
    return -2;
  } else {
    return 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<BetSumXByHorse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D15(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L3
  mov    $0x0,%eax
  jmp    L37
L3:
  mov    -0x8(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L13
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <GetValueLength@plt>
  mov    %eax,%edx
  mov    $0x280,%eax
  sub    %edx,%eax
  jmp    L37
L13:
  mov    $0xffffffff,%eax
L37:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D15
type: string
name: unpack_data
arguments: data_tag: D0
type: string

tool: D15: red

tool: D0: blue

assistant: ```c
int BetSumXByHorse(const char *horse, int value) {
  if (!strcmp(horse, "red")) {
    return 0;
  } else if (!strcmp(horse, "blue")) {
    return 640 - GetValueLength(value);
  } else {
    return -1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<BetSumXByHorse>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  lea    D11(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L14
  lea    D3(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L4
  mov    %ebp,%edi
  call   <GetValueLength@plt>
  mov    %eax,%edx
  mov    $0x280,%eax
  sub    %edx,%eax
L14:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  mov    $0xffffffff,%eax
  jmp    L14
```

tool_calls: name: extract_symbolic_data
arguments: tag: D11
type_guessed: string
name: extract_symbolic_data
arguments: tag: D3
type_guessed: string

tool: D11: red

tool: D3: blue

assistant: ```c
int BetSumXByHorse(const char *horse, int value) {
  if (!strcmp(horse, "red")) {
    return 0;
  } else if (!strcmp(horse, "blue")) {
    return 640 - GetValueLength(value);
  } else {
    return -1;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<BetSumXByHorse>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  lea    D18(%rip),%rsi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %eax,%eax
  je     L20
  lea    D12(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L9
  mov    %r12d,%edi
  call   <GetValueLength@plt>
  mov    %eax,%r8d
  mov    $0x280,%eax
  sub    %r8d,%eax
L20:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nop
L9:
  mov    $0xffffffff,%eax
  jmp    L20
```

tool_calls: name: read_data
arguments: data_tag: D18
type_guessed: string
name: read_data
arguments: data_tag: D12
type_guessed: string

tool: D18: red

tool: D12: blue

assistant: ```c
int BetSumXByHorse(const char *horse, int value) {
  if (!strcmp(horse, "red")) {
    return 0;
  } else if (!strcmp(horse, "blue")) {
    return 640 - GetValueLength(value);
  } else {
    return -1;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<BetSumXByHorse>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  lea    D6(%rip),%rsi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %eax,%eax
  je     L6
  lea    D3(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L4
  mov    %r12d,%edi
  call   <GetValueLength@plt>
  mov    %eax,%r8d
  mov    $0x280,%eax
  sub    %r8d,%eax
L6:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nop
L4:
  mov    $0xffffffff,%eax
  jmp    L6
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type: string
name: extract_symbolic_data
arguments: tag: D3
type: string

tool: D6: red

tool: D3: blue

assistant: ```c
int BetSumXByHorse(const char *horse, int value) {
  if (!strcmp(horse, "red")) {
    return 0;
  } else if (!strcmp(horse, "blue")) {
    return 640 - GetValueLength(value);
  } else {
    return -1;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ce_run>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L2
L14:
  mov    $0x1,%edi
  call   <ce_poller_poll@plt>
  test   %rax,%rax
  je     L63
  mov    $0xffffffff,%eax
  jmp    L42
L63:
  mov    $0x0,%eax
  call   <ce_poller_react@plt>
  test   %rax,%rax
  je     L88
  mov    $0xffffffff,%eax
  jmp    L42
L88:
  movl   $0x0,-0x8(%rbp)
  jmp    L97
L3:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <ce_get_coroutine_status@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  je     L50
  cmpl   $0x3,-0x4(%rbp)
  jne    L34
L50:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <ce_coroutine_resume@plt>
L34:
  addl   $0x1,-0x8(%rbp)
L97:
  mov    $0x0,%eax
  call   <ce_coroutine_cnt@plt>
  cmp    %eax,-0x8(%rbp)
  jl     L3
L2:
  mov    $0x0,%eax
  call   <ce_coroutine_cnt@plt>
  test   %eax,%eax
  jg     L14
  mov    $0x0,%eax
  call   <ce_close_scheduler@plt>
L42:
  leave
  ret
```

assistant: ```c
int ce_run() {
  while (ce_coroutine_cnt() > 0) {
    int crtn_id;
    if (ce_poller_poll(1) != 0) {
      return -1;
    }
    if (ce_poller_react() != 0) {
      return -1;
    }
    for (crtn_id = 0; crtn_id < ce_coroutine_cnt(); crtn_id++) {
      int status = ce_get_coroutine_status(crtn_id);
      if (status == 1 || status == 3) {
        ce_coroutine_resume(crtn_id);
      }
    }
  }
  return ce_close_scheduler();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ce_run>:
  endbr64
  push   %rbx
L22:
  mov    $0x0,%eax
  call   <ce_coroutine_cnt@plt>
  test   %eax,%eax
  jle    L20
  mov    $0x1,%edi
  call   <ce_poller_poll@plt>
  test   %rax,%rax
  jne    L59
  call   <ce_poller_react@plt>
  test   %rax,%rax
  jne    L4
  mov    $0x0,%ebx
  jmp    L24
L70:
  add    $0x1,%ebx
L24:
  mov    $0x0,%eax
  call   <ce_coroutine_cnt@plt>
  cmp    %ebx,%eax
  jle    L22
  mov    %ebx,%edi
  call   <ce_get_coroutine_status@plt>
  and    $0xfffffffd,%eax
  cmp    $0x1,%eax
  jne    L70
  mov    %ebx,%edi
  call   <ce_coroutine_resume@plt>
  jmp    L70
L20:
  mov    $0x0,%eax
  call   <ce_close_scheduler@plt>
L10:
  pop    %rbx
  ret
L59:
  mov    $0xffffffff,%eax
  jmp    L10
L4:
  mov    $0xffffffff,%eax
  jmp    L10
```

assistant: ```c
int ce_run() {
  while (ce_coroutine_cnt() > 0) {
    int crtn_id;
    if (ce_poller_poll(1) != 0) {
      return -1;
    }
    if (ce_poller_react() != 0) {
      return -1;
    }
    for (crtn_id = 0; crtn_id < ce_coroutine_cnt(); crtn_id++) {
      int status = ce_get_coroutine_status(crtn_id);
      if (status == 1 || status == 3) {
        ce_coroutine_resume(crtn_id);
      }
    }
  }
  return ce_close_scheduler();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ce_run>:
  endbr64
  push   %rbx
  nopl   (%rax)
L12:
  xor    %eax,%eax
  call   <ce_coroutine_cnt@plt>
  test   %eax,%eax
  jle    L46
  mov    $0x1,%edi
  call   <ce_poller_poll@plt>
  test   %rax,%rax
  jne    L3
  call   <ce_poller_react@plt>
  test   %rax,%rax
  jne    L3
  xor    %ebx,%ebx
  jmp    L4
L59:
  add    $0x1,%ebx
L4:
  xor    %eax,%eax
  call   <ce_coroutine_cnt@plt>
  cmp    %ebx,%eax
  jle    L12
  mov    %ebx,%edi
  call   <ce_get_coroutine_status@plt>
  and    $0xfffffffd,%eax
  cmp    $0x1,%eax
  jne    L59
  mov    %ebx,%edi
  call   <ce_coroutine_resume@plt>
  jmp    L59
L3:
  mov    $0xffffffff,%eax
  pop    %rbx
  ret
L46:
  xor    %eax,%eax
  pop    %rbx
  jmp    10d0 <ce_close_scheduler@plt>
```

assistant: ```c
int ce_run() {
  while (ce_coroutine_cnt() > 0) {
    int crtn_id;
    if (ce_poller_poll(1) != 0) {
      return -1;
    }
    if (ce_poller_react() != 0) {
      return -1;
    }
    for (crtn_id = 0; crtn_id < ce_coroutine_cnt(); crtn_id++) {
      int status = ce_get_coroutine_status(crtn_id);
      if (status == 1 || status == 3) {
        ce_coroutine_resume(crtn_id);
      }
    }
  }
  return ce_close_scheduler();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ce_run>:
  endbr64
  push   %rbx
  nopl   (%rax)
L51:
  xor    %eax,%eax
  call   <ce_coroutine_cnt@plt>
  test   %eax,%eax
  jle    L5
  mov    $0x1,%edi
  call   <ce_poller_poll@plt>
  test   %rax,%rax
  jne    L60
  call   <ce_poller_react@plt>
  test   %rax,%rax
  jne    L60
  xor    %ebx,%ebx
  jmp    L7
L15:
  add    $0x1,%ebx
L7:
  xor    %eax,%eax
  call   <ce_coroutine_cnt@plt>
  cmp    %ebx,%eax
  jle    L51
  mov    %ebx,%edi
  call   <ce_get_coroutine_status@plt>
  and    $0xfffffffd,%eax
  cmp    $0x1,%eax
  jne    L15
  mov    %ebx,%edi
  call   <ce_coroutine_resume@plt>
  jmp    L15
L60:
  mov    $0xffffffff,%eax
  pop    %rbx
  ret
L5:
  xor    %eax,%eax
  pop    %rbx
  jmp    10d0 <ce_close_scheduler@plt>
```

assistant: ```c
int ce_run() {
  while (ce_coroutine_cnt() > 0) {
    int crtn_id;
    if (ce_poller_poll(1) != 0) {
      return -1;
    }
    if (ce_poller_react() != 0) {
      return -1;
    }
    for (crtn_id = 0; crtn_id < ce_coroutine_cnt(); crtn_id++) {
      int status = ce_get_coroutine_status(crtn_id);
      if (status == 1 || status == 3) {
        ce_coroutine_resume(crtn_id);
      }
    }
  }
  return ce_close_scheduler();
}
```
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    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x30(%rbp)
  movl   $0x9,-0x2c(%rbp)
  movl   $0x4,-0x28(%rbp)
  movl   $0x2,-0x24(%rbp)
  movl   $0x5,-0x20(%rbp)
  movl   $0x8,-0x1c(%rbp)
  movl   $0x7,-0x18(%rbp)
  movl   $0x6,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x9,-0x34(%rbp)
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <PrintArray@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x34(%rbp),%eax
  lea    -0x1(%rax),%edx
  lea    -0x30(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <QuickSortMiddleBase@plt>
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <PrintArray@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <system@plt>
  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: data_tag: D24
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D9
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D38
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D7
type_guessed: string

tool: D24: output array


tool: D9: begin quick sort


tool: D38: quick sorted


tool: D7: pause

assistant: ```c
int main(int argc, char **argv) {
  int array[] = {3, 9, 4, 2, 5, 8, 7, 6, 1};
  int length = sizeof(array) / sizeof(int);
  printf("output array\n");
  PrintArray(array, length);
  printf("begin quick sort\n");
  QuickSortMiddleBase(array, 0, length - 1);
  printf("quick sorted\n");
  PrintArray(array, length);
  system("pause");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x3,(%rsp)
  movl   $0x9,0x4(%rsp)
  movl   $0x4,0x8(%rsp)
  movl   $0x2,0xc(%rsp)
  movl   $0x5,0x10(%rsp)
  movl   $0x8,0x14(%rsp)
  movl   $0x7,0x18(%rsp)
  movl   $0x6,0x1c(%rsp)
  movl   $0x1,0x20(%rsp)
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    $0x9,%esi
  mov    %rbx,%rdi
  call   <PrintArray@plt>
  lea    D20(%rip),%rdi
  call   <printf@plt>
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <QuickSortMiddleBase@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  mov    $0x9,%esi
  mov    %rbx,%rdi
  call   <PrintArray@plt>
  lea    D31(%rip),%rdi
  call   <system@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D28
type: string
name: read_data
arguments: label: D20
type: string
name: read_data
arguments: label: D39
type: string
name: read_data
arguments: label: D31
type: string

tool: D28: output array


tool: D20: begin quick sort


tool: D39: quick sorted


tool: D31: pause

assistant: ```c
int main(int argc, char **argv) {
  int array[] = {3, 9, 4, 2, 5, 8, 7, 6, 1};
  int length = sizeof(array) / sizeof(int);
  printf("output array\n");
  PrintArray(array, length);
  printf("begin quick sort\n");
  QuickSortMiddleBase(array, 0, length - 1);
  printf("quick sorted\n");
  PrintArray(array, length);
  system("pause");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rdi
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x900000003,%rax
  mov    %rsp,%rbp
  movl   $0x1,0x20(%rsp)
  mov    %rax,(%rsp)
  movabs $0x200000004,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x800000005,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x600000007,%rax
  mov    %rax,0x18(%rsp)
  call   <printf@plt>
  mov    $0x9,%esi
  mov    %rbp,%rdi
  call   <PrintArray@plt>
  lea    D10(%rip),%rdi
  call   <printf@plt>
  xor    %esi,%esi
  mov    $0x8,%edx
  mov    %rbp,%rdi
  call   <QuickSortMiddleBase@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x9,%esi
  call   <PrintArray@plt>
  lea    D32(%rip),%rdi
  call   <system@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
type_guessed: string
name: retrieve_data
arguments: data_tag: D10
type_guessed: string
name: retrieve_data
arguments: data_tag: D29
type_guessed: string
name: retrieve_data
arguments: data_tag: D32
type_guessed: string

tool: D18: output array


tool: D10: begin quick sort


tool: D29: quick sorted


tool: D32: pause

assistant: ```c
int main(int argc, char **argv) {
  int array[] = {3, 9, 4, 2, 5, 8, 7, 6, 1};
  int length = sizeof(array) / sizeof(int);
  printf("output array\n");
  PrintArray(array, length);
  printf("begin quick sort\n");
  QuickSortMiddleBase(array, 0, length - 1);
  printf("quick sorted\n");
  PrintArray(array, length);
  system("pause");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D23(%rip),%rdi
  sub    $0x30,%rsp
  movdqa D19(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  movl   $0x1,0x20(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D1(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  call   <printf@plt>
  mov    $0x9,%esi
  mov    %rbp,%rdi
  call   <PrintArray@plt>
  lea    D58(%rip),%rdi
  call   <printf@plt>
  xor    %esi,%esi
  mov    $0x8,%edx
  mov    %rbp,%rdi
  call   <QuickSortMiddleBase@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x9,%esi
  call   <PrintArray@plt>
  lea    D50(%rip),%rdi
  call   <system@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D23
type_infered: string
name: unpack_data
arguments: data_label: D19
type_infered: byte[16]
name: unpack_data
arguments: data_label: D1
type_infered: byte[16]
name: unpack_data
arguments: data_label: D58
type_infered: string
name: unpack_data
arguments: data_label: D30
type_infered: string
name: unpack_data
arguments: data_label: D50
type_infered: string

tool: D23: output array


tool: {"D19": [0x03,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: {"D1": [0x05,0x00,0x00,0x00,0x08,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: D58: begin quick sort


tool: D30: quick sorted


tool: D50: pause

assistant: ```c
int main(int argc, char **argv) {
  int array[] = {3, 9, 4, 2, 5, 8, 7, 6, 1};
  int length = sizeof(array) / sizeof(int);
  printf("output array\n");
  PrintArray(array, length);
  printf("begin quick sort\n");
  QuickSortMiddleBase(array, 0, length - 1);
  printf("quick sorted\n");
  PrintArray(array, length);
  system("pause");
  return 0;
}
```