ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x24(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L17
L18:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <str_len@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x20(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jge    L26
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x20(%rbp)
L26:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x24(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <str_copy@plt>
  addl   $0x1,-0x24(%rbp)
L17:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L18
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  addl   $0x2,-0x20(%rbp)
  mov    $0x50,%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %eax,-0x18(%rbp)
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <print_array@plt>
  mov    -0x1c(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <selection_sort_asc@plt>
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x18(%rbp),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <print_array@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D19
type_guessed: string
name: parse_data
arguments: label: D4
type_guessed: string
name: parse_data
arguments: label: D27
type_guessed: string

tool: D19: %s

tool: D4: [Before sorting]


tool: D27: [After sorting]


assistant: ```c
int main() {
  char **data;
  char *s = (char *)malloc(sizeof(char) * 256);
  int i = 0, n = 0, col_width = 0, col_num = 0, l = 0;
  while (scanf("%s", s) == 1) {
    l = str_len(s);
    if (col_width < l)
      col_width = l;
    data[i] = (char *)malloc(sizeof(char) * l);
    str_copy(data[i], s);
    i++;
  }
  n = i;
  col_width += 2;
  col_num = 80 / col_width;
  printf("[Before sorting]\n");
  print_array(data, n, col_width, col_num);
  selection_sort_asc((void **)data, n);
  printf("[After sorting]\n");
  print_array(data, n, col_width, col_num);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  lea    D19(%rip),%r13
  jmp    L17
L3:
  mov    %r12,%rdi
  call   <str_len@plt>
  mov    %eax,%edi
  cmp    %eax,%ebp
  cmovl  %eax,%ebp
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,0x0(,%rbx,8)
  mov    %r12,%rsi
  call   <str_copy@plt>
  add    $0x1,%rbx
L17:
  mov    %ebx,%r14d
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L3
  add    $0x2,%ebp
  mov    $0x50,%eax
  cltd
  idiv   %ebp
  mov    %eax,%ebx
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%ecx
  mov    %ebp,%edx
  mov    %r14d,%esi
  mov    $0x0,%r12d
  mov    %r12,%rdi
  call   <print_array@plt>
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <selection_sort_asc@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%ecx
  mov    %ebp,%edx
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <print_array@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D19
data_type: string
name: parse_data
arguments: data_tag: D11
data_type: string
name: parse_data
arguments: data_tag: D0
data_type: string

tool: D19: %s

tool: D11: [Before sorting]


tool: D0: [After sorting]


assistant: ```c
int main() {
  char **data;
  char *s = (char *)malloc(sizeof(char) * 256);
  int i = 0, n = 0, col_width = 0, col_num = 0, l = 0;
  while (scanf("%s", s) == 1) {
    l = str_len(s);
    if (col_width < l)
      col_width = l;
    data[i] = (char *)malloc(sizeof(char) * l);
    str_copy(data[i], s);
    i++;
  }
  n = i;
  col_width += 2;
  col_num = 80 / col_width;
  printf("[Before sorting]\n");
  print_array(data, n, col_width, col_num);
  selection_sort_asc((void **)data, n);
  printf("[After sorting]\n");
  print_array(data, n, col_width, col_num);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x100,%edi
  push   %r13
  lea    D18(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L9
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %r12,%rdi
  call   <str_len@plt>
  cmp    %eax,%ebp
  mov    %eax,%edi
  cmovl  %eax,%ebp
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,0x0(,%rbx,8)
  mov    %rax,%rdi
  add    $0x1,%rbx
  call   <str_copy@plt>
L9:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L14
  mov    $0x50,%eax
  lea    0x2(%rbp),%r12d
  lea    D13(%rip),%rdi
  xor    %ebp,%ebp
  cltd
  idiv   %r12d
  mov    %eax,%r13d
  call   <printf@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <print_array@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <selection_sort_asc@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <print_array@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D18
type_infered: string
name: unpack_data
arguments: data_label: D13
type_infered: string
name: unpack_data
arguments: data_label: D11
type_infered: string

tool: D18: %s

tool: D13: [Before sorting]


tool: D11: [After sorting]


assistant: ```c
int main() {
  char **data;
  char *s = (char *)malloc(sizeof(char) * 256);
  int i = 0, n = 0, col_width = 0, col_num = 0, l = 0;
  while (scanf("%s", s) == 1) {
    l = str_len(s);
    if (col_width < l)
      col_width = l;
    data[i] = (char *)malloc(sizeof(char) * l);
    str_copy(data[i], s);
    i++;
  }
  n = i;
  col_width += 2;
  col_num = 80 / col_width;
  printf("[Before sorting]\n");
  print_array(data, n, col_width, col_num);
  selection_sort_asc((void **)data, n);
  printf("[After sorting]\n");
  print_array(data, n, col_width, col_num);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x100,%edi
  push   %r13
  lea    D22(%rip),%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    %r12,%rdi
  call   <str_len@plt>
  cmp    %eax,%ebp
  mov    %eax,%edi
  cmovl  %eax,%ebp
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,0x0(,%rbx,8)
  mov    %rax,%rdi
  add    $0x1,%rbx
  call   <str_copy@plt>
L14:
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L16
  mov    $0x50,%eax
  lea    0x2(%rbp),%r12d
  lea    D1(%rip),%rdi
  xor    %ebp,%ebp
  cltd
  idiv   %r12d
  mov    %eax,%r13d
  call   <printf@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <print_array@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <selection_sort_asc@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <print_array@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: label: D22
type_guessed: string
name: retrieve_data
arguments: label: D1
type_guessed: string
name: retrieve_data
arguments: label: D4
type_guessed: string

tool: D22: %s

tool: D1: [Before sorting]


tool: D4: [After sorting]


assistant: ```c
int main() {
  char **data;
  char *s = (char *)malloc(sizeof(char) * 256);
  int i = 0, n = 0, col_width = 0, col_num = 0, l = 0;
  while (scanf("%s", s) == 1) {
    l = str_len(s);
    if (col_width < l)
      col_width = l;
    data[i] = (char *)malloc(sizeof(char) * l);
    str_copy(data[i], s);
    i++;
  }
  n = i;
  col_width += 2;
  col_num = 80 / col_width;
  printf("[Before sorting]\n");
  print_array(data, n, col_width, col_num);
  selection_sort_asc((void **)data, n);
  printf("[After sorting]\n");
  print_array(data, n, col_width, col_num);
}
```
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
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x64,-0x28(%rbp)
  movss  D26(%rip),%xmm0
  movss  %xmm0,-0x24(%rbp)
  movsd  D61(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movb   $0x53,-0x29(%rbp)
  movl   $0x69726843,-0x15(%rbp)
  movw   $0x73,-0x11(%rbp)
  movl   $0x7472614d,-0xf(%rbp)
  movw   $0x6e69,-0xb(%rbp)
  movb   $0x0,-0x9(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D65(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  cvtss2sd -0x24(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  lea    D80(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%eax
  mov    %eax,%esi
  lea    D78(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x15(%rbp),%rax
  mov    %rax,%rsi
  lea    D76(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xf(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%edx
  lea    -0xf(%rbp),%rcx
  lea    -0x15(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D26
type_infered: byte[4]
name: read_data
arguments: label: D61
type_infered: f64
name: read_data
arguments: label: D65
type_infered: string
name: read_data
arguments: label: D17
type_infered: string
name: read_data
arguments: label: D80
type_infered: string
name: read_data
arguments: label: D78
type_infered: string
name: read_data
arguments: label: D76
type_infered: string
name: read_data
arguments: label: D2
type_infered: string
name: read_data
arguments: label: D9
type_infered: string

tool: {"D26": [0x7b,0x14,0x16,0x40]}

tool: D61: 56789.4532

tool: D65: You are %d miles away.


tool: D17: You have %f levels of power,


tool: D80: You have %f awesome super powers


tool: D78: I have an initial %c.


tool: D76: I have a first name %s


tool: D2: I have a last name %s


tool: D9: My whole name is %s %c. %s


assistant: ```c
int main(int argc, const char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'S';
  char first_name[] = "Chris";
  char last_name[] = "Martin";
  printf("You are %d miles away.\n", distance);
  printf("You have %f levels of power,\n", power);
  printf("You have %f awesome super powers\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("I have a first name %s\n", first_name);
  printf("I have a last name %s\n", last_name);
  printf("My whole name is %s %c. %s\n", first_name, initial, last_name);
  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
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x69726843,0xb(%rsp)
  movw   $0x73,0xf(%rsp)
  movl   $0x7472614d,0x11(%rsp)
  movw   $0x6e69,0x15(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    $0x64,%esi
  lea    D29(%rip),%rdi
  call   <printf@plt>
  movsd  D68(%rip),%xmm0
  lea    D25(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  D6(%rip),%xmm0
  lea    D3(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x53,%esi
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xb(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x11(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D89(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rcx
  mov    $0x53,%edx
  mov    %rbx,%rsi
  lea    D86(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D29
data_type: string
name: read_data
arguments: data_tag: D68
data_type: byte[8]
name: read_data
arguments: data_tag: D25
data_type: string
name: read_data
arguments: data_tag: D6
data_type: f64
name: read_data
arguments: data_tag: D3
data_type: string
name: read_data
arguments: data_tag: D24
data_type: string
name: read_data
arguments: data_tag: D8
data_type: string
name: read_data
arguments: data_tag: D89
data_type: string
name: read_data
arguments: data_tag: D86
data_type: string

tool: D29: You are %d miles away.


tool: {"D68": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D25: You have %f levels of power,


tool: D6: 56789.4532

tool: D3: You have %f awesome super powers


tool: D24: I have an initial %c.


tool: D8: I have a first name %s


tool: D89: I have a last name %s


tool: D86: My whole name is %s %c. %s


assistant: ```c
int main(int argc, const char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'S';
  char first_name[] = "Chris";
  char last_name[] = "Martin";
  printf("You are %d miles away.\n", distance);
  printf("You have %f levels of power,\n", power);
  printf("You have %f awesome super powers\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("I have a first name %s\n", first_name);
  printf("I have a last name %s\n", last_name);
  printf("My whole name is %s %c. %s\n", first_name, initial, last_name);
  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    $0x73,%edx
  mov    $0x6e69,%ecx
  mov    $0x64,%esi
  push   %rbp
  lea    D37(%rip),%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %dx,0xf(%rsp)
  lea    0xb(%rsp),%rbp
  lea    0x11(%rsp),%r12
  mov    %cx,0x15(%rsp)
  movl   $0x69726843,0xb(%rsp)
  movl   $0x7472614d,0x11(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D32(%rip),%xmm0
  mov    $0x1,%eax
  lea    D56(%rip),%rdi
  call   <printf@plt>
  movsd  D18(%rip),%xmm0
  mov    $0x1,%eax
  lea    D51(%rip),%rdi
  call   <printf@plt>
  mov    $0x53,%esi
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D64(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x53,%edx
  mov    %rbp,%rsi
  lea    D85(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D37
type_guessed: string
name: unpack_data
arguments: data_label: D32
type_guessed: byte[8]
name: unpack_data
arguments: data_label: D56
type_guessed: string
name: unpack_data
arguments: data_label: D18
type_guessed: f64
name: unpack_data
arguments: data_label: D51
type_guessed: string
name: unpack_data
arguments: data_label: D26
type_guessed: string
name: unpack_data
arguments: data_label: D41
type_guessed: string
name: unpack_data
arguments: data_label: D64
type_guessed: string
name: unpack_data
arguments: data_label: D85
type_guessed: string

tool: D37: You are %d miles away.


tool: {"D32": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D56: You have %f levels of power,


tool: D18: 56789.4532

tool: D51: You have %f awesome super powers


tool: D26: I have an initial %c.


tool: D41: I have a first name %s


tool: D64: I have a last name %s


tool: D85: My whole name is %s %c. %s


assistant: ```c
int main(int argc, const char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'S';
  char first_name[] = "Chris";
  char last_name[] = "Martin";
  printf("You are %d miles away.\n", distance);
  printf("You have %f levels of power,\n", power);
  printf("You have %f awesome super powers\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("I have a first name %s\n", first_name);
  printf("I have a last name %s\n", last_name);
  printf("My whole name is %s %c. %s\n", first_name, initial, last_name);
  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
  push   %r12
  mov    $0x73,%edx
  mov    $0x6e69,%ecx
  mov    $0x64,%esi
  push   %rbp
  lea    D14(%rip),%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %dx,0xf(%rsp)
  lea    0xb(%rsp),%rbp
  lea    0x11(%rsp),%r12
  mov    %cx,0x15(%rsp)
  movl   $0x69726843,0xb(%rsp)
  movl   $0x7472614d,0x11(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D44(%rip),%xmm0
  mov    $0x1,%eax
  lea    D74(%rip),%rdi
  call   <printf@plt>
  movsd  D65(%rip),%xmm0
  mov    $0x1,%eax
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    $0x53,%esi
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D68(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D63(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x53,%edx
  mov    %rbp,%rsi
  lea    D50(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D14
type_guessed: string
name: read_data
arguments: data_tag: D44
type_guessed: byte[8]
name: read_data
arguments: data_tag: D74
type_guessed: string
name: read_data
arguments: data_tag: D65
type_guessed: f64
name: read_data
arguments: data_tag: D30
type_guessed: string
name: read_data
arguments: data_tag: D28
type_guessed: string
name: read_data
arguments: data_tag: D68
type_guessed: string
name: read_data
arguments: data_tag: D63
type_guessed: string
name: read_data
arguments: data_tag: D50
type_guessed: string

tool: D14: You are %d miles away.


tool: {"D44": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D74: You have %f levels of power,


tool: D65: 56789.4532

tool: D30: You have %f awesome super powers


tool: D28: I have an initial %c.


tool: D68: I have a first name %s


tool: D63: I have a last name %s


tool: D50: My whole name is %s %c. %s


assistant: ```c
int main(int argc, const char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'S';
  char first_name[] = "Chris";
  char last_name[] = "Martin";
  printf("You are %d miles away.\n", distance);
  printf("You have %f levels of power,\n", power);
  printf("You have %f awesome super powers\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("I have a first name %s\n", first_name);
  printf("I have a last name %s\n", last_name);
  printf("My whole name is %s %c. %s\n", first_name, initial, last_name);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<isPrime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x4(%rbp)
  jmp    L36
L10:
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  divl   -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L20
  mov    $0x0,%eax
  jmp    L5
L20:
  addl   $0x1,-0x4(%rbp)
L36:
  mov    -0x14(%rbp),%eax
  shr    %eax
  cmp    %eax,-0x4(%rbp)
  jbe    L10
  mov    $0x1,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
bool isPrime(unsigned int i) {
  unsigned int n;
  for (n = 2; n <= i / 2; n++) {
    if (i % n == 0)
      return false;
  }
  return true;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isPrime>:
  endbr64
  mov    %edi,%esi
  shr    %esi
  cmp    $0x3,%edi
  jbe    L4
  test   $0x1,%dil
  je     L25
  mov    $0x2,%ecx
L18:
  add    $0x1,%ecx
  cmp    %esi,%ecx
  ja     L33
  mov    %edi,%eax
  mov    $0x0,%edx
  div    %ecx
  test   %edx,%edx
  jne    L18
  mov    $0x0,%eax
  ret
L33:
  mov    $0x1,%eax
  ret
L4:
  mov    $0x1,%eax
  ret
L25:
  mov    $0x0,%eax
  ret
```

assistant: ```c
bool isPrime(unsigned int i) {
  unsigned int n;
  for (n = 2; n <= i / 2; n++) {
    if (i % n == 0)
      return false;
  }
  return true;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<isPrime>:
  endbr64
  mov    %edi,%esi
  shr    %esi
  cmp    $0x3,%edi
  jbe    L45
  test   $0x1,%dil
  je     L43
  mov    $0x2,%ecx
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L49:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  test   %edx,%edx
  je     L43
L1:
  mov    %ecx,%eax
  add    $0x1,%ecx
  cmp    %eax,%esi
  jne    L49
L45:
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L43:
  xor    %eax,%eax
  ret
```

assistant: ```c
bool isPrime(unsigned int i) {
  unsigned int n;
  for (n = 2; n <= i / 2; n++) {
    if (i % n == 0)
      return false;
  }
  return true;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<isPrime>:
  endbr64
  mov    %edi,%esi
  shr    %esi
  cmp    $0x3,%edi
  jbe    L29
  test   $0x1,%dil
  je     L11
  mov    $0x2,%ecx
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L33:
  mov    %edi,%eax
  xor    %edx,%edx
  div    %ecx
  test   %edx,%edx
  je     L11
L40:
  mov    %ecx,%eax
  add    $0x1,%ecx
  cmp    %eax,%esi
  jne    L33
L29:
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax)
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
bool isPrime(unsigned int i) {
  unsigned int n;
  for (n = 2; n <= i / 2; n++) {
    if (i % n == 0)
      return false;
  }
  return true;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <foo@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  foo(0, 0, 0);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <foo@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  foo(0, 0, 0);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edx,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <foo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  foo(0, 0, 0);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edx,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <foo@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  foo(0, 0, 0);
  return 0;
}
```