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
<local_domain>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  movslq %esi,%rax
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L12:
  cmp    %rdx,%rsp
  je     L2
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L12
L2:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L8
  orq    $0x0,-0x8(%rsp,%rax,1)
L8:
  movb   $0x0,(%rsp)
  mov    %rsp,%rdi
  call   <gethostname@plt>
  mov    %rsp,%rdi
  call   <topdomain@plt>
  mov    %rax,%r12
  mov    %rbx,%rdi
  call   <topdomain@plt>
  mov    %rax,%rsi
  test   %r12,%r12
  je     L26
  test   %rax,%rax
  je     L26
  mov    %r12,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L21:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
L26:
  mov    $0x1,%eax
  jmp    L21
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int local_domain(const char *h) {
  char localhost[MAXHOSTNAMELEN];
  const char *p1, *p2;
  localhost[0] = 0;
  (void)gethostname(localhost, sizeof(localhost));
  p1 = topdomain(localhost);
  p2 = topdomain(h);
  if (p1 == NULL || p2 == NULL || !strcasecmp(p1, p2))
    return (1);
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<local_domain>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  mov    %rdi,%r12
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  mov    D6(%rip),%rax
  mov    %rsp,%rcx
  movslq (%rax),%rax
  mov    %rax,%rsi
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L7
L34:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L34
L7:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L5
L14:
  mov    %rsp,%rdi
  movb   $0x0,(%rsp)
  call   <gethostname@plt>
  mov    %rsp,%rdi
  call   <topdomain@plt>
  mov    %r12,%rdi
  mov    %rax,%r13
  call   <topdomain@plt>
  mov    %rax,%rsi
  test   %r13,%r13
  je     L18
  test   %rax,%rax
  je     L18
  mov    %r13,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L57:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L74
  lea    -0x10(%rbp),%rsp
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
  nopl   0x0(%rax)
L18:
  mov    $0x1,%eax
  jmp    L57
  nop
L5:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L14
L74:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int local_domain(const char *h) {
  char localhost[MAXHOSTNAMELEN];
  const char *p1, *p2;
  localhost[0] = 0;
  (void)gethostname(localhost, sizeof(localhost));
  p1 = topdomain(localhost);
  p2 = topdomain(h);
  if (p1 == NULL || p2 == NULL || !strcasecmp(p1, p2))
    return (1);
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<local_domain>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  mov    %rdi,%r12
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  mov    D5(%rip),%rax
  mov    %rsp,%rcx
  movslq (%rax),%rax
  mov    %rax,%rsi
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L1
L75:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L75
L1:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L44
L49:
  mov    %rsp,%rdi
  movb   $0x0,(%rsp)
  call   <gethostname@plt>
  mov    %rsp,%rdi
  call   <topdomain@plt>
  mov    %r12,%rdi
  mov    %rax,%r13
  call   <topdomain@plt>
  mov    %rax,%rsi
  test   %r13,%r13
  je     L23
  test   %rax,%rax
  je     L23
  mov    %r13,%rdi
  call   <strcasecmp@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
L40:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L63
  lea    -0x10(%rbp),%rsp
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
  nopl   0x0(%rax)
L23:
  mov    $0x1,%eax
  jmp    L40
  nop
L44:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L49
L63:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int local_domain(const char *h) {
  char localhost[MAXHOSTNAMELEN];
  const char *p1, *p2;
  localhost[0] = 0;
  (void)gethostname(localhost, sizeof(localhost));
  p1 = topdomain(localhost);
  p2 = topdomain(h);
  if (p1 == NULL || p2 == NULL || !strcasecmp(p1, p2))
    return (1);
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<badvpn_write_le32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0xc,%rax
  mov    -0x14(%rbp),%edx
  sar    $0x18,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    -0x14(%rbp),%edx
  sar    $0x10,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    -0x14(%rbp),%edx
  sar    $0x8,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    -0x14(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void badvpn_write_le32(uint32_t x, char *c_ptr) {
  uint8_t *ptr = (uint8_t *)c_ptr;
  ptr[3] = x >> 24;
  ptr[2] = x >> 16;
  ptr[1] = x >> 8;
  ptr[0] = x >> 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<badvpn_write_le32>:
  endbr64
  mov    %edi,%eax
  sar    $0x18,%eax
  mov    %eax,0xc(%rsi)
  mov    %edi,%eax
  sar    $0x10,%eax
  mov    %eax,0x8(%rsi)
  mov    %edi,%eax
  sar    $0x8,%eax
  mov    %eax,0x4(%rsi)
  mov    %edi,(%rsi)
  ret
```

assistant: ```c
void badvpn_write_le32(uint32_t x, char *c_ptr) {
  uint8_t *ptr = (uint8_t *)c_ptr;
  ptr[3] = x >> 24;
  ptr[2] = x >> 16;
  ptr[1] = x >> 8;
  ptr[0] = x >> 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<badvpn_write_le32>:
  endbr64
  mov    %edi,%eax
  mov    %edi,(%rsi)
  sar    $0x18,%eax
  mov    %eax,0xc(%rsi)
  mov    %edi,%eax
  sar    $0x10,%eax
  mov    %eax,0x8(%rsi)
  mov    %edi,%eax
  sar    $0x8,%eax
  mov    %eax,0x4(%rsi)
  ret
```

assistant: ```c
void badvpn_write_le32(uint32_t x, char *c_ptr) {
  uint8_t *ptr = (uint8_t *)c_ptr;
  ptr[3] = x >> 24;
  ptr[2] = x >> 16;
  ptr[1] = x >> 8;
  ptr[0] = x >> 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<badvpn_write_le32>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%ecx
  movd   %edi,%xmm0
  mov    %edi,%edx
  sar    $0x8,%eax
  sar    $0x10,%ecx
  sar    $0x18,%edx
  movd   %ecx,%xmm1
  movd   %eax,%xmm3
  movd   %edx,%xmm2
  punpckldq %xmm3,%xmm0
  punpckldq %xmm2,%xmm1
  punpcklqdq %xmm1,%xmm0
  movups %xmm0,(%rsi)
  ret
```

assistant: ```c
void badvpn_write_le32(uint32_t x, char *c_ptr) {
  uint8_t *ptr = (uint8_t *)c_ptr;
  ptr[3] = x >> 24;
  ptr[2] = x >> 16;
  ptr[1] = x >> 8;
  ptr[0] = x >> 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ranged_hist_uint8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %ecx,-0x44(%rbp)
  mov    %r8d,-0x48(%rbp)
  mov    %r9,-0x50(%rbp)
  movq   $0xff,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L48
L17:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L47
L52:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    0x10(%rbp),%rax
  jl     L13
  mov    -0x8(%rbp),%rax
  cmp    0x18(%rbp),%rax
  jg     L13
  mov    -0x8(%rbp),%rax
  sub    0x10(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x50(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
L13:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jge    L12
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  jmp    L80
L12:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jle    L80
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L80:
  mov    -0x48(%rbp),%eax
  cltq
  add    %rax,-0x10(%rbp)
L47:
  mov    -0x40(%rbp),%eax
  imul   -0x48(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x10(%rbp)
  jne    L52
  mov    -0x44(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
L48:
  mov    -0x3c(%rbp),%eax
  imul   -0x44(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x18(%rbp)
  jne    L17
  mov    0x20(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    0x28(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ranged_hist_uint8(const char *image, uint16_t rows, uint16_t cols, uint32_t r_stride, uint32_t c_stride, uint32_t *histogram, uint8_t hist_min, uint8_t hist_max, uint8_t *min, uint8_t *max) {
  uint8_t working_min = 255;
  uint8_t working_max = 0;
  const char *row_start, *pixel;
  for (row_start = image; row_start != image + rows * r_stride; row_start += r_stride) {
    for (pixel = row_start; pixel != row_start + cols * c_stride; pixel += c_stride) {
      uint8_t val = *(uint8_t *)pixel;
      if (val >= hist_min && val <= hist_max)
        histogram[val - hist_min]++;
      if (val < working_min)
        working_min = val;
      else if (val > working_max)
        working_max = val;
    }
  }
  *min = working_min;
  *max = working_max;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ranged_hist_uint8>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    0x30(%rsp),%rdi
  mov    0x38(%rsp),%rbp
  imul   %ecx,%esi
  movslq %esi,%r14
  add    %rbx,%r14
  cmp    %r14,%rbx
  je     L98
  mov    %edx,%eax
  movslq %ecx,%r13
  imul   %r8d,%eax
  movslq %eax,%r12
  add    %rbx,%r12
  mov    $0x0,%ecx
  mov    $0xff,%esi
  movslq %r8d,%r8
  jmp    L1
L67:
  mov    %rax,%rsi
L46:
  add    %r8,%rdx
  cmp    %rdx,%r11
  je     L55
L86:
  mov    (%rdx),%rax
  cmp    %rdi,%rax
  jl     L82
  cmp    %rbp,%rax
  jg     L82
  mov    %rax,%r10
  sub    %rdi,%r10
  addl   $0x1,(%r9,%r10,4)
L82:
  cmp    %rsi,%rax
  jl     L67
  cmp    %rax,%rcx
  cmovl  %rax,%rcx
  jmp    L46
L55:
  add    %r13,%rbx
  add    %r13,%r12
  cmp    %r14,%rbx
  je     L69
L1:
  mov    %r12,%r11
  mov    %rbx,%rdx
  cmp    %rbx,%r12
  jne    L86
  jmp    L55
L98:
  mov    $0x0,%ecx
  mov    $0xff,%esi
L69:
  mov    0x40(%rsp),%rax
  mov    %rsi,(%rax)
  mov    0x48(%rsp),%rax
  mov    %rcx,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void ranged_hist_uint8(const char *image, uint16_t rows, uint16_t cols, uint32_t r_stride, uint32_t c_stride, uint32_t *histogram, uint8_t hist_min, uint8_t hist_max, uint8_t *min, uint8_t *max) {
  uint8_t working_min = 255;
  uint8_t working_max = 0;
  const char *row_start, *pixel;
  for (row_start = image; row_start != image + rows * r_stride; row_start += r_stride) {
    for (pixel = row_start; pixel != row_start + cols * c_stride; pixel += c_stride) {
      uint8_t val = *(uint8_t *)pixel;
      if (val >= hist_min && val <= hist_max)
        histogram[val - hist_min]++;
      if (val < working_min)
        working_min = val;
      else if (val > working_max)
        working_max = val;
    }
  }
  *min = working_min;
  *max = working_max;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ranged_hist_uint8>:
  endbr64
  imul   %ecx,%esi
  push   %r13
  movslq %r8d,%r13
  push   %r12
  push   %rbp
  movslq %esi,%r12
  push   %rbx
  mov    %rdi,%rbx
  mov    0x30(%rsp),%r8
  add    %rbx,%r12
  mov    0x28(%rsp),%rdi
  cmp    %r12,%rbx
  je     L7
  mov    %edx,%eax
  movslq %ecx,%rbp
  mov    $0xff,%esi
  xor    %ecx,%ecx
  imul   %r13d,%eax
  movslq %eax,%r11
  add    %rbx,%r11
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %rbx,%rdx
  cmp    %rbx,%r11
  jne    L38
  jmp    L52
  nopw   0x0(%rax,%rax,1)
L44:
  cmp    %rax,%rcx
  cmovl  %rax,%rcx
  add    %r13,%rdx
  cmp    %r11,%rdx
  je     L52
L38:
  mov    (%rdx),%rax
  cmp    %rdi,%rax
  jl     L15
  cmp    %r8,%rax
  jg     L15
  mov    %rax,%r10
  sub    %rdi,%r10
  addl   $0x1,(%r9,%r10,4)
L15:
  cmp    %rsi,%rax
  jge    L44
  add    %r13,%rdx
  mov    %rax,%rsi
  cmp    %r11,%rdx
  jne    L38
L52:
  add    %rbp,%rbx
  add    %rbp,%r11
  cmp    %r12,%rbx
  jne    L26
L40:
  mov    0x38(%rsp),%rax
  mov    %rsi,(%rax)
  mov    0x40(%rsp),%rax
  mov    %rcx,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  xor    %ecx,%ecx
  mov    $0xff,%esi
  jmp    L40
```

assistant: ```c
void ranged_hist_uint8(const char *image, uint16_t rows, uint16_t cols, uint32_t r_stride, uint32_t c_stride, uint32_t *histogram, uint8_t hist_min, uint8_t hist_max, uint8_t *min, uint8_t *max) {
  uint8_t working_min = 255;
  uint8_t working_max = 0;
  const char *row_start, *pixel;
  for (row_start = image; row_start != image + rows * r_stride; row_start += r_stride) {
    for (pixel = row_start; pixel != row_start + cols * c_stride; pixel += c_stride) {
      uint8_t val = *(uint8_t *)pixel;
      if (val >= hist_min && val <= hist_max)
        histogram[val - hist_min]++;
      if (val < working_min)
        working_min = val;
      else if (val > working_max)
        working_max = val;
    }
  }
  *min = working_min;
  *max = working_max;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ranged_hist_uint8>:
  endbr64
  imul   %ecx,%esi
  push   %r13
  movslq %r8d,%r13
  push   %r12
  push   %rbp
  movslq %esi,%r12
  push   %rbx
  mov    %rdi,%rbx
  mov    0x30(%rsp),%r8
  add    %rbx,%r12
  mov    0x28(%rsp),%rdi
  cmp    %r12,%rbx
  je     L89
  mov    %edx,%eax
  movslq %ecx,%rbp
  mov    $0xff,%esi
  xor    %ecx,%ecx
  imul   %r13d,%eax
  movslq %eax,%r11
  add    %rbx,%r11
  nopw   0x0(%rax,%rax,1)
L76:
  mov    %rbx,%rdx
  cmp    %rbx,%r11
  jne    L44
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L66:
  cmp    %rax,%rcx
  cmovl  %rax,%rcx
  add    %r13,%rdx
  cmp    %rdx,%r11
  je     L17
L44:
  mov    (%rdx),%rax
  cmp    %rdi,%rax
  jl     L83
  cmp    %r8,%rax
  jg     L83
  mov    %rax,%r10
  sub    %rdi,%r10
  addl   $0x1,(%r9,%r10,4)
L83:
  cmp    %rsi,%rax
  jge    L66
  add    %r13,%rdx
  mov    %rax,%rsi
  cmp    %rdx,%r11
  jne    L44
L17:
  add    %rbp,%rbx
  add    %rbp,%r11
  cmp    %r12,%rbx
  jne    L76
L42:
  mov    0x38(%rsp),%rax
  mov    %rsi,(%rax)
  mov    0x40(%rsp),%rax
  mov    %rcx,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L89:
  xor    %ecx,%ecx
  mov    $0xff,%esi
  jmp    L42
```

assistant: ```c
void ranged_hist_uint8(const char *image, uint16_t rows, uint16_t cols, uint32_t r_stride, uint32_t c_stride, uint32_t *histogram, uint8_t hist_min, uint8_t hist_max, uint8_t *min, uint8_t *max) {
  uint8_t working_min = 255;
  uint8_t working_max = 0;
  const char *row_start, *pixel;
  for (row_start = image; row_start != image + rows * r_stride; row_start += r_stride) {
    for (pixel = row_start; pixel != row_start + cols * c_stride; pixel += c_stride) {
      uint8_t val = *(uint8_t *)pixel;
      if (val >= hist_min && val <= hist_max)
        histogram[val - hist_min]++;
      if (val < working_min)
        working_min = val;
      else if (val > working_max)
        working_max = val;
    }
  }
  *min = working_min;
  *max = working_max;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L17
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L16
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L16
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L16:
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L44
L17:
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L44:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D35
data_type: string
name: retrieve_data
arguments: tag: D23
data_type: string
name: retrieve_data
arguments: tag: D2
data_type: string

tool: D35: %d

tool: D23: Not Leap Year

tool: D2: Leap Year

assistant: ```c
int main() {
  int year;
  scanf("%d", &year);
  if (year % 4 == 0) {
    if ((year % 100 == 0) && !(year % 400 == 0))
      printf("Not Leap Year");
    printf("Leap Year");
  } else
    printf("Not Leap Year");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D37(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x3,%al
  jne    L43
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  sar    $0x25,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x64,%edx,%edx
  cmp    %edx,%eax
  jne    L46
  mov    $0x190,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L50
L46:
  lea    D35(%rip),%rdi
  call   <printf@plt>
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L50:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  jmp    L46
L43:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  jmp    L21
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D37
type_guessed: string
name: disassemble_data
arguments: data_tag: D35
type_guessed: string
name: disassemble_data
arguments: data_tag: D5
type_guessed: string

tool: D37: %d

tool: D35: Leap Year

tool: D5: Not Leap Year

assistant: ```c
int main() {
  int year;
  scanf("%d", &year);
  if (year % 4 == 0) {
    if ((year % 100 == 0) && !(year % 400 == 0))
      printf("Not Leap Year");
    printf("Leap Year");
  } else
    printf("Not Leap Year");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D39(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x3,%al
  jne    L46
  imul   $0xc28f5c29,%eax,%edx
  add    $0x51eb850,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  jbe    L33
L45:
  lea    D19(%rip),%rdi
  call   <printf@plt>
L59:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L33:
  mov    $0x190,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L45
  lea    D23(%rip),%rdi
  call   <printf@plt>
  jmp    L45
L46:
  lea    D23(%rip),%rdi
  call   <printf@plt>
  jmp    L59
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D39
type: string
name: extract_symbolic_data
arguments: data_tag: D19
type: string
name: extract_symbolic_data
arguments: data_tag: D23
type: string

tool: D39: %d

tool: D19: Leap Year

tool: D23: Not Leap Year

assistant: ```c
int main() {
  int year;
  scanf("%d", &year);
  if (year % 4 == 0) {
    if ((year % 100 == 0) && !(year % 400 == 0))
      printf("Not Leap Year");
    printf("Leap Year");
  } else
    printf("Not Leap Year");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   $0x3,%al
  jne    L59
  imul   $0xc28f5c29,%eax,%edx
  add    $0x51eb850,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  jbe    L4
L28:
  lea    D32(%rip),%rdi
  call   <printf@plt>
L26:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L4:
  mov    $0x190,%ecx
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L28
  lea    D35(%rip),%rdi
  call   <printf@plt>
  jmp    L28
L59:
  lea    D35(%rip),%rdi
  call   <printf@plt>
  jmp    L26
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
datatype: string
name: unpack_data
arguments: data_label: D32
datatype: string
name: unpack_data
arguments: data_label: D35
datatype: string

tool: D9: %d

tool: D32: Leap Year

tool: D35: Not Leap Year

assistant: ```c
int main() {
  int year;
  scanf("%d", &year);
  if (year % 4 == 0) {
    if ((year % 100 == 0) && !(year % 400 == 0))
      printf("Not Leap Year");
    printf("Leap Year");
  } else
    printf("Not Leap Year");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<printbinary>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L1:
  movzbl -0x14(%rbp),%edx
  mov    $0x7,%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  movzbl %al,%eax
  and    $0x1,%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L0:
  cmpl   $0x7,-0x4(%rbp)
  jle    L1
  nop
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
type_infered: string

tool: D9: %hhu

assistant: ```c
void printbinary(unsigned char value) {
  for (int i = 0; i < 8; i++)
    printf("%hhu", (value >> 7 - i) & 0x01);
}
```
user: Translate the below assembly code into C programming language:

```asm
<printbinary>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    $0x7,%ebx
  lea    D6(%rip),%r12
L9:
  movzbl %bpl,%esi
  mov    %ebx,%ecx
  sar    %cl,%esi
  and    $0x1,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  cmp    $0xffffffff,%ebx
  jne    L9
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
type: string

tool: D6: %hhu

assistant: ```c
void printbinary(unsigned char value) {
  for (int i = 0; i < 8; i++)
    printf("%hhu", (value >> 7 - i) & 0x01);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<printbinary>:
  endbr64
  push   %r12
  lea    D5(%rip),%r12
  push   %rbp
  movzbl %dil,%ebp
  push   %rbx
  mov    $0x7,%ebx
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %ebx,%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  sar    %cl,%esi
  and    $0x1,%esi
  call   <printf@plt>
  sub    $0x1,%ebx
  jae    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
type_guessed: string

tool: D5: %hhu

assistant: ```c
void printbinary(unsigned char value) {
  for (int i = 0; i < 8; i++)
    printf("%hhu", (value >> 7 - i) & 0x01);
}
```
user: What is the c source code of the assembly code below:

```asm
<printbinary>:
  endbr64
  push   %r12
  lea    D2(%rip),%r12
  push   %rbp
  movzbl %dil,%ebp
  push   %rbx
  mov    $0x7,%ebx
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %ebx,%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  sar    %cl,%esi
  and    $0x1,%esi
  call   <printf@plt>
  sub    $0x1,%ebx
  jae    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: label: D2
type_infered: string

tool: D2: %hhu

assistant: ```c
void printbinary(unsigned char value) {
  for (int i = 0; i < 8; i++)
    printf("%hhu", (value >> 7 - i) & 0x01);
}
```