ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<check>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movzbl (%rdi),%ebp
  movw   $0x4543,0x5(%rsp)
  movb   $0x53,0x7(%rsp)
  call   <strlen@plt>
  mov    $0x0,%edx
  cmp    $0x43,%bpl
  je     L43
  cmp    $0x45,%bpl
  setne  %dl
  movzbl %dl,%edx
  add    $0x1,%edx
L43:
  test   %eax,%eax
  je     L63
  mov    %rbx,%rdi
  mov    $0x1,%esi
  jmp    L17
L44:
  mov    %esi,%ecx
  jmp    L23
L82:
  mov    %esi,%ecx
  mov    %r8d,%edx
  jmp    L23
L76:
  add    $0x2,%edx
  movslq %edx,%r8
  movzbl 0x5(%rsp,%r8,1),%ebp
  cmp    %bpl,%cl
  jne    L25
  mov    %esi,%ecx
L23:
  add    $0x1,%rdi
  add    $0x1,%esi
  cmp    %eax,%ecx
  je     L63
L17:
  movzbl (%rdi),%ecx
  cmp    %bpl,%cl
  je     L44
  cmp    $0x2,%edx
  je     L76
  lea    0x1(%rdx),%r8d
  movslq %r8d,%r9
  movzbl 0x5(%rsp,%r9,1),%ebp
  cmp    %bpl,%cl
  je     L82
  cmp    $0x1,%edx
  jne    L76
L25:
  lea    D7(%rip),%rdi
  call   <printf@plt>
L48:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L99
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L63:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L48
L99:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D7
type_infered: string
name: parse_data
arguments: tag: D6
type_infered: string

tool: D7: no


tool: D6: yes


assistant: ```c
void check(char *str) {
  char curr_char = str[0];
  char arr[3] = {'C', 'E', 'S'};
  int iter = 0, track = 0;
  int len = strlen(str);
  if (curr_char == 'C') {
    track = 0;
  } else if (curr_char == 'E') {
    track = 1;
  } else {
    track = 2;
  }
  while (iter != len) {
    if (str[iter] == curr_char) {
      iter++;
      continue;
    } else if (track != 2 && str[iter] == arr[track + 1]) {
      iter++;
      track++;
      curr_char = arr[track];
    } else if (track != 1 && str[iter] == arr[track + 2]) {
      iter++;
      track += 2;
      curr_char = arr[track];
    } else {
      printf("no\n");
      return;
    }
  }
  printf("yes\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<check>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  movzbl (%rdi),%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x4543,%eax
  movb   $0x53,0x7(%rsp)
  mov    %ax,0x5(%rsp)
  call   <strlen@plt>
  xor    %edx,%edx
  cmp    $0x43,%bpl
  je     L85
  xor    %edx,%edx
  cmp    $0x45,%bpl
  setne  %dl
  add    $0x1,%edx
L85:
  test   %eax,%eax
  je     L50
  mov    %rbx,%rdi
  mov    $0x1,%esi
  jmp    L117
  cs nopw 0x0(%rax,%rax,1)
L90:
  cmp    $0x2,%edx
  je     L6
  lea    0x1(%rdx),%r8d
  movslq %r8d,%r9
  movzbl 0x5(%rsp,%r9,1),%ebp
  cmp    %bpl,%cl
  je     L129
  cmp    $0x1,%edx
  je     L119
L6:
  add    $0x2,%edx
  cmp    $0x53,%cl
  jne    L119
  mov    %esi,%ecx
  mov    $0x53,%ebp
L115:
  add    $0x1,%rdi
  add    $0x1,%esi
  cmp    %ecx,%eax
  je     L50
L117:
  movzbl (%rdi),%ecx
  cmp    %bpl,%cl
  jne    L90
  mov    %esi,%ecx
  add    $0x1,%rdi
  add    $0x1,%esi
  cmp    %ecx,%eax
  jne    L117
  nopl   0x0(%rax,%rax,1)
L50:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x18,%rsp
  lea    D17(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1090 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L129:
  mov    %esi,%ecx
  mov    %r8d,%edx
  jmp    L115
  nop
L119:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x18,%rsp
  lea    D3(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1090 <printf@plt>
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D17
type_guessed: string
name: parse_data
arguments: label: D3
type_guessed: string

tool: D17: yes


tool: D3: no


assistant: ```c
void check(char *str) {
  char curr_char = str[0];
  char arr[3] = {'C', 'E', 'S'};
  int iter = 0, track = 0;
  int len = strlen(str);
  if (curr_char == 'C') {
    track = 0;
  } else if (curr_char == 'E') {
    track = 1;
  } else {
    track = 2;
  }
  while (iter != len) {
    if (str[iter] == curr_char) {
      iter++;
      continue;
    } else if (track != 2 && str[iter] == arr[track + 1]) {
      iter++;
      track++;
      curr_char = arr[track];
    } else if (track != 1 && str[iter] == arr[track + 2]) {
      iter++;
      track += 2;
      curr_char = arr[track];
    } else {
      printf("no\n");
      return;
    }
  }
  printf("yes\n");
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<check>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  movzbl (%rdi),%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x4543,%eax
  movb   $0x53,0x7(%rsp)
  mov    %ax,0x5(%rsp)
  call   <strlen@plt>
  xor    %edx,%edx
  cmp    $0x43,%bpl
  je     L50
  xor    %edx,%edx
  cmp    $0x45,%bpl
  setne  %dl
  add    $0x1,%edx
L50:
  test   %eax,%eax
  je     L91
  mov    %rbx,%rdi
  mov    $0x1,%esi
  jmp    L120
  cs nopw 0x0(%rax,%rax,1)
L46:
  cmp    $0x2,%edx
  je     L19
  lea    0x1(%rdx),%r8d
  movslq %r8d,%r9
  movzbl 0x5(%rsp,%r9,1),%ebp
  cmp    %bpl,%cl
  je     L20
  cmp    $0x1,%edx
  je     L25
L19:
  add    $0x2,%edx
  cmp    $0x53,%cl
  jne    L25
  mov    %esi,%ecx
  mov    $0x53,%ebp
L68:
  add    $0x1,%rdi
  add    $0x1,%esi
  cmp    %ecx,%eax
  je     L91
L120:
  movzbl (%rdi),%ecx
  cmp    %bpl,%cl
  jne    L46
  mov    %esi,%ecx
  add    $0x1,%rdi
  add    $0x1,%esi
  cmp    %ecx,%eax
  jne    L120
  nopl   0x0(%rax,%rax,1)
L91:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x18,%rsp
  lea    D9(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1090 <printf@plt>
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %esi,%ecx
  mov    %r8d,%edx
  jmp    L68
  nop
L25:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x18,%rsp
  lea    D1(%rip),%rdi
  pop    %rbx
  pop    %rbp
  jmp    1090 <printf@plt>
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D9
datatype: string
name: retrieve_data
arguments: data_label: D1
datatype: string

tool: D9: yes


tool: D1: no


assistant: ```c
void check(char *str) {
  char curr_char = str[0];
  char arr[3] = {'C', 'E', 'S'};
  int iter = 0, track = 0;
  int len = strlen(str);
  if (curr_char == 'C') {
    track = 0;
  } else if (curr_char == 'E') {
    track = 1;
  } else {
    track = 2;
  }
  while (iter != len) {
    if (str[iter] == curr_char) {
      iter++;
      continue;
    } else if (track != 2 && str[iter] == arr[track + 1]) {
      iter++;
      track++;
      curr_char = arr[track];
    } else if (track != 1 && str[iter] == arr[track + 2]) {
      iter++;
      track += 2;
      curr_char = arr[track];
    } else {
      printf("no\n");
      return;
    }
  }
  printf("yes\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<colour_to_mono>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    -0x24(%rbp),%eax
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x3,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L16
L7:
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L37
L23:
  mov    -0x10(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L37:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L23
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %eax,%ecx
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x10(%rbp)
L16:
  mov    -0x1c(%rbp),%eax
  imul   -0x20(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L7
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void colour_to_mono(unsigned char *img, int width, int height, int bitsperpixel, unsigned char *mono) {
  int i, j, v;
  int bytesperpixel = bitsperpixel / 8;
  for (i = 0; i < width * height; i++) {
    v = 0;
    for (j = 0; j < bytesperpixel; j++) {
      v += img[i * bytesperpixel + j];
    }
    mono[i + j] = v / bytesperpixel;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<colour_to_mono>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  lea    0x7(%rcx),%edi
  test   %ecx,%ecx
  cmovns %ecx,%edi
  sar    $0x3,%edi
  imul   %edx,%esi
  test   %esi,%esi
  jle    L59
  mov    %ecx,%ebx
  mov    %r8,%r13
  mov    %edi,%r11d
  mov    %esi,%r12d
  test   %edi,%edi
  mov    $0x1,%r8d
  cmovg  %edi,%r8d
  cmp    $0x7,%ecx
  mov    $0x0,%eax
  cmovle %eax,%r8d
  mov    $0x0,%r10d
  mov    $0x0,%r9d
  jmp    L39
L0:
  movzbl (%rsi,%rdx,1),%ecx
  add    %ecx,%eax
  add    $0x1,%rdx
  cmp    %edx,%edi
  jg     L0
L23:
  lea    (%r9,%r8,1),%ecx
  movslq %ecx,%rcx
  cltd
  idiv   %r11d
  mov    %al,0x0(%r13,%rcx,1)
  add    $0x1,%r9d
  add    %edi,%r10d
  cmp    %r12d,%r9d
  je     L59
L39:
  mov    $0x0,%edx
  mov    $0x0,%eax
  movslq %r10d,%rsi
  add    %rbp,%rsi
  cmp    $0x7,%ebx
  jg     L0
  jmp    L23
L59:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void colour_to_mono(unsigned char *img, int width, int height, int bitsperpixel, unsigned char *mono) {
  int i, j, v;
  int bytesperpixel = bitsperpixel / 8;
  for (i = 0; i < width * height; i++) {
    v = 0;
    for (j = 0; j < bytesperpixel; j++) {
      v += img[i * bytesperpixel + j];
    }
    mono[i + j] = v / bytesperpixel;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<colour_to_mono>:
  endbr64
  mov    %esi,%r9d
  test   %ecx,%ecx
  push   %r12
  lea    0x7(%rcx),%esi
  push   %rbp
  mov    %r8,%rbp
  mov    %r9d,%r8d
  cmovns %ecx,%esi
  imul   %edx,%r8d
  push   %rbx
  sar    $0x3,%esi
  test   %r8d,%r8d
  jle    L38
  xor    %r10d,%r10d
  xor    %r9d,%r9d
  mov    $0x1,%r12d
  test   %esi,%esi
  cmovg  %esi,%r12d
  mov    %rdi,%rbx
  mov    %ecx,%r11d
  nopl   (%rax)
L12:
  movslq %r10d,%rdi
  xor    %edx,%edx
  xor    %eax,%eax
  add    %rbx,%rdi
  mov    %r9d,%ecx
  cmp    $0x7,%r11d
  jle    L7
  nopl   0x0(%rax,%rax,1)
L2:
  movzbl (%rdi,%rdx,1),%ecx
  add    $0x1,%rdx
  add    %ecx,%eax
  cmp    %edx,%esi
  jg     L2
  cltd
  lea    (%r12,%r9,1),%ecx
  idiv   %esi
L7:
  movslq %ecx,%rdx
  add    $0x1,%r9d
  add    %esi,%r10d
  mov    %al,0x0(%rbp,%rdx,1)
  cmp    %r8d,%r9d
  jne    L12
L38:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void colour_to_mono(unsigned char *img, int width, int height, int bitsperpixel, unsigned char *mono) {
  int i, j, v;
  int bytesperpixel = bitsperpixel / 8;
  for (i = 0; i < width * height; i++) {
    v = 0;
    for (j = 0; j < bytesperpixel; j++) {
      v += img[i * bytesperpixel + j];
    }
    mono[i + j] = v / bytesperpixel;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L30
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L63
L30:
  cmpl   $0x1,-0x14(%rbp)
  jle    L51
  cmpl   $0x3,-0x14(%rbp)
  jg     L51
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L22
L46:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L22:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L46
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L63
L51:
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L63:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D18
data_type: string
name: unpack_data
arguments: data_label: D20
data_type: string
name: unpack_data
arguments: data_label: D37
data_type: string
name: unpack_data
arguments: data_label: D10
data_type: string

tool: D18: You only have one argument. You suck.

tool: D20: Here's your argument:

tool: D37: argv %d = %s


tool: D10: You have too many arguments. You suck

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 1) {
    printf("You only have one argument. You suck.\n");
  } else if (argc > 1 && argc < 4) {
    printf("Here's your argument:\n");
    for (i = 0; i < argc; i++) {
      printf("argv %d = %s\n", i, argv[i]);
    }
    printf("\n");
  } else {
    printf("You have too many arguments. You suck\n");
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  je     L37
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    -0x2(%rdi),%eax
  cmp    $0x1,%eax
  jbe    L20
  lea    D17(%rip),%rdi
  call   <puts@plt>
L2:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L37:
  lea    D30(%rip),%rdi
  call   <puts@plt>
  jmp    L2
L20:
  lea    D20(%rip),%rdi
  call   <puts@plt>
  mov    %ebp,%ebp
  mov    $0x0,%ebx
  lea    D14(%rip),%r13
L23:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L23
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L2
```

tool_calls: name: disassemble_data
arguments: label: D17
type_infered: string
name: disassemble_data
arguments: label: D30
type_infered: string
name: disassemble_data
arguments: label: D20
type_infered: string
name: disassemble_data
arguments: label: D14
type_infered: string

tool: D17: You have too many arguments. You suck

tool: D30: You only have one argument. You suck.

tool: D20: Here's your argument:

tool: D14: argv %d = %s


assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 1) {
    printf("You only have one argument. You suck.\n");
  } else if (argc > 1 && argc < 4) {
    printf("Here's your argument:\n");
    for (i = 0; i < argc; i++) {
      printf("argv %d = %s\n", i, argv[i]);
    }
    printf("\n");
  } else {
    printf("You have too many arguments. You suck\n");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  cmp    $0x1,%ebx
  je     L32
  lea    -0x2(%rbx),%eax
  cmp    $0x1,%eax
  jbe    L39
  lea    D10(%rip),%rdi
  call   <puts@plt>
L1:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  lea    D23(%rip),%rdi
  mov    %rsi,%r12
  xor    %ebp,%ebp
  call   <puts@plt>
  lea    D21(%rip),%r13
L35:
  mov    (%r12,%rbp,8),%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L35
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L1
L32:
  lea    D4(%rip),%rdi
  call   <puts@plt>
  jmp    L1
```

tool_calls: name: unpack_data
arguments: label: D10
datatype: string
name: unpack_data
arguments: label: D23
datatype: string
name: unpack_data
arguments: label: D21
datatype: string
name: unpack_data
arguments: label: D4
datatype: string

tool: D10: You have too many arguments. You suck

tool: D23: Here's your argument:

tool: D21: argv %d = %s


tool: D4: You only have one argument. You suck.

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 1) {
    printf("You only have one argument. You suck.\n");
  } else if (argc > 1 && argc < 4) {
    printf("Here's your argument:\n");
    for (i = 0; i < argc; i++) {
      printf("argv %d = %s\n", i, argv[i]);
    }
    printf("\n");
  } else {
    printf("You have too many arguments. You suck\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  cmp    $0x1,%ebx
  je     L21
  lea    -0x2(%rbx),%eax
  cmp    $0x1,%eax
  jbe    L12
  lea    D32(%rip),%rdi
  call   <puts@plt>
L33:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  lea    D35(%rip),%rdi
  mov    %rsi,%r12
  xor    %ebp,%ebp
  call   <puts@plt>
  lea    D38(%rip),%r13
L15:
  mov    (%r12,%rbp,8),%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L15
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L33
L21:
  lea    D10(%rip),%rdi
  call   <puts@plt>
  jmp    L33
```

tool_calls: name: parse_data
arguments: data_label: D32
type_guessed: string
name: parse_data
arguments: data_label: D35
type_guessed: string
name: parse_data
arguments: data_label: D38
type_guessed: string
name: parse_data
arguments: data_label: D10
type_guessed: string

tool: D32: You have too many arguments. You suck

tool: D35: Here's your argument:

tool: D38: argv %d = %s


tool: D10: You only have one argument. You suck.

assistant: ```c
int main(int argc, char *argv[]) {
  int i = 0;
  if (argc == 1) {
    printf("You only have one argument. You suck.\n");
  } else if (argc > 1 && argc < 4) {
    printf("Here's your argument:\n");
    for (i = 0; i < argc; i++) {
      printf("argv %d = %s\n", i, argv[i]);
    }
    printf("\n");
  } else {
    printf("You have too many arguments. You suck\n");
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<linear_root>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  neg    %eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x14(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
data_type: string
name: extract_symbolic_data
arguments: label: D17
data_type: string

tool: D9: The line has a zero at x = %.3lf

tool: D17: 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-


assistant: ```c
void linear_root(int CoeffB, int CoeffC) {
  double linear_root;
  linear_root = ((double)-(CoeffC)) / (double)CoeffB;
  printf("The line has a zero at x = %.3lf", linear_root);
  printf("\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-\n\n");
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<linear_root>:
  endbr64
  sub    $0x8,%rsp
  neg    %esi
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %edi,%xmm1
  divsd  %xmm1,%xmm0
  lea    D14(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <puts@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D14
type_guessed: string
name: read_data
arguments: data_tag: D4
type_guessed: string

tool: D14: The line has a zero at x = %.3lf

tool: D4: 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-


assistant: ```c
void linear_root(int CoeffB, int CoeffC) {
  double linear_root;
  linear_root = ((double)-(CoeffC)) / (double)CoeffB;
  printf("The line has a zero at x = %.3lf", linear_root);
  printf("\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-\n\n");
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<linear_root>:
  endbr64
  neg    %esi
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  sub    $0x8,%rsp
  cvtsi2sd %edi,%xmm1
  mov    $0x1,%eax
  lea    D1(%rip),%rdi
  cvtsi2sd %esi,%xmm0
  divsd  %xmm1,%xmm0
  call   <printf@plt>
  lea    D14(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <puts@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D1
type_infered: string
name: unpack_data
arguments: data_tag: D14
type_infered: string

tool: D1: The line has a zero at x = %.3lf

tool: D14: 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-


assistant: ```c
void linear_root(int CoeffB, int CoeffC) {
  double linear_root;
  linear_root = ((double)-(CoeffC)) / (double)CoeffB;
  printf("The line has a zero at x = %.3lf", linear_root);
  printf("\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-\n\n");
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<linear_root>:
  endbr64
  neg    %esi
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  sub    $0x8,%rsp
  cvtsi2sd %edi,%xmm1
  mov    $0x1,%eax
  lea    D12(%rip),%rdi
  cvtsi2sd %esi,%xmm0
  divsd  %xmm1,%xmm0
  call   <printf@plt>
  lea    D10(%rip),%rdi
  add    $0x8,%rsp
  jmp    1060 <puts@plt>
```

tool_calls: name: read_data
arguments: tag: D12
data_type: string
name: read_data
arguments: tag: D10
data_type: string

tool: D12: The line has a zero at x = %.3lf

tool: D10: 
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-


assistant: ```c
void linear_root(int CoeffB, int CoeffC) {
  double linear_root;
  linear_root = ((double)-(CoeffC)) / (double)CoeffB;
  printf("The line has a zero at x = %.3lf", linear_root);
  printf("\n-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-\n\n");
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<parse_varlist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  lea    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  lea    D11(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtok_r@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L58
  lea    -0x20(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <find_variable@plt>
  mov    -0x20(%rbp),%eax
  test   %eax,%eax
  jle    L28
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  sub    $0x1,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x1c(%rbp)
  jmp    L53
L54:
  lea    -0x20(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <find_variable@plt>
  mov    -0x20(%rbp),%eax
  test   %eax,%eax
  jle    L34
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  sub    $0x1,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x1c(%rbp)
L53:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdx
  lea    D11(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok_r@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L54
  mov    -0x38(%rbp),%rax
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
  jmp    L50
L58:
  nop
  jmp    L50
L28:
  nop
  jmp    L50
L34:
  nop
L50:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L60
  call   <__stack_chk_fail@plt>
L60:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D11
type_guessed: string

tool: D11:  ,

assistant: ```c
void parse_varlist(char *varlist, int *ivars, int *n) {
  char *item;
  char *toksave;
  int v, i = 0;
  item = strtok_r(varlist, " ,", &toksave);
  if (item[0] == 0)
    return;
  find_variable(item, &v);
  if (v <= 0)
    return;
  ivars[i] = v - 1;
  i++;
  while ((item = strtok_r(NULL, " ,", &toksave)) != NULL) {
    find_variable(item, &v);
    if (v <= 0)
      return;
    ivars[i] = v - 1;
    i++;
  }
  *n = i;
}
```