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   %r13
  lea    D38(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rsi
  call   <scanf@plt>
  movzbl 0x2(%rsp),%eax
  cmp    $0x64,%al
  je     L16
  cmp    $0x68,%al
  je     L8
L17:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  lea    0x9(%rsp),%rbp
  lea    D25(%rip),%rdi
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <get_numbers@plt>
  mov    0x4(%rax),%r12d
  mov    (%rax),%edi
  mov    0x8(%rax),%r13d
  call   <dec_to_hex@plt>
  mov    %r12d,%edi
  mov    %rax,%rbp
  call   <dec_to_hex@plt>
  mov    %r13d,%edi
  mov    %rax,%r12
  call   <dec_to_hex@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D6(%rip),%rdi
  mov    %rax,%rcx
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L17
L8:
  lea    0x9(%rsp),%rsi
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  movzwl 0xa(%rsp),%eax
  lea    0x3(%rsp),%rdi
  xor    %esi,%esi
  mov    $0x10,%edx
  mov    %ax,0x3(%rsp)
  movzwl 0xc(%rsp),%eax
  mov    %ax,0x5(%rsp)
  movzwl 0xe(%rsp),%eax
  mov    %ax,0x7(%rsp)
  call   <strtol@plt>
  lea    0x5(%rsp),%rdi
  xor    %esi,%esi
  mov    $0x10,%edx
  mov    %rax,%rbx
  call   <strtol@plt>
  lea    0x7(%rsp),%rdi
  xor    %esi,%esi
  mov    $0x10,%edx
  mov    %rax,%rbp
  call   <strtol@plt>
  mov    %ebp,%edx
  mov    %ebx,%esi
  lea    D13(%rip),%rdi
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L17
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D38
type: string
name: disassemble_data
arguments: tag: D25
type: string
name: disassemble_data
arguments: tag: D6
type: string
name: disassemble_data
arguments: tag: D13
type: string

tool: D38: %c

tool: D25:  %[^
]s

tool: D6: #%s%s%s

tool: D13: rgb(%d, %d, %d)

assistant: ```c
int main() {
  char number_type;
  scanf("%c", &number_type);
  if (number_type == 'd') {
    char dec_code_color[15];
    scanf(" %[^\n]s", dec_code_color);
    int *numbers = get_numbers(dec_code_color);
    int red_dec;
    red_dec = numbers[0];
    int green_dec;
    green_dec = numbers[1];
    int blue_dec;
    blue_dec = numbers[2];
    char *red_hex;
    red_hex = dec_to_hex(red_dec);
    char *green_hex;
    green_hex = dec_to_hex(green_dec);
    char *blue_hex;
    blue_hex = dec_to_hex(blue_dec);
    printf("#%s%s%s", red_hex, green_hex, blue_hex);
  } else if (number_type == 'h') {
    char hex_code_color[9];
    scanf(" %[^\n]s", hex_code_color);
    char red_hex[] = {hex_code_color[1], hex_code_color[2]};
    char green_hex[] = {hex_code_color[3], hex_code_color[4]};
    char blue_hex[] = {hex_code_color[5], hex_code_color[6]};
    int red_dec;
    red_dec = (int)strtol(red_hex, NULL, 16);
    int green_dec;
    green_dec = (int)strtol(green_hex, NULL, 16);
    int blue_dec;
    blue_dec = (int)strtol(blue_hex, NULL, 16);
    printf("rgb(%d, %d, %d)", red_dec, green_dec, blue_dec);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<StripCR>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L7
L11:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L6
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L0
L6:
  subl   $0x1,-0x4(%rbp)
L7:
  cmpl   $0x0,-0x4(%rbp)
  jg     L11
  nop
L0:
  nop
  leave
  ret
```

assistant: ```c
void StripCR(char *string) {
  int i;
  for (i = strlen(string) - 1; i > 0; i--)
    if (string[i] == '\n') {
      string[i] = '\0';
      break;
    }
}
```
user: Translate the below assembly code into C programming language:

```asm
<StripCR>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  test   %edx,%edx
  jle    L15
  mov    %eax,%ecx
  movslq %edx,%rdx
  add    %rbx,%rdx
  cltq
  lea    -0x2(%rbx,%rax,1),%rax
  lea    -0x2(%rcx),%ecx
  sub    %rcx,%rax
L4:
  cmpb   $0xa,(%rdx)
  je     L22
  sub    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L4
  jmp    L15
L22:
  movb   $0x0,(%rdx)
L15:
  pop    %rbx
  ret
```

assistant: ```c
void StripCR(char *string) {
  int i;
  for (i = strlen(string) - 1; i > 0; i--)
    if (string[i] == '\n') {
      string[i] = '\0';
      break;
    }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<StripCR>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  test   %edx,%edx
  jle    L21
  mov    %eax,%ecx
  cltq
  movslq %edx,%rdx
  lea    -0x2(%rbx,%rax,1),%rax
  sub    $0x2,%ecx
  add    %rbx,%rdx
  sub    %rcx,%rax
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L6:
  sub    $0x1,%rdx
  cmp    %rax,%rdx
  je     L21
L9:
  cmpb   $0xa,(%rdx)
  jne    L6
  movb   $0x0,(%rdx)
L21:
  pop    %rbx
  ret
```

assistant: ```c
void StripCR(char *string) {
  int i;
  for (i = strlen(string) - 1; i > 0; i--)
    if (string[i] == '\n') {
      string[i] = '\0';
      break;
    }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<StripCR>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  test   %edx,%edx
  jle    L38
  mov    %eax,%ecx
  cltq
  movslq %edx,%rdx
  lea    -0x2(%rbx,%rax,1),%rax
  sub    $0x2,%ecx
  add    %rbx,%rdx
  sub    %rcx,%rax
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L22:
  sub    $0x1,%rdx
  cmp    %rax,%rdx
  je     L38
L18:
  cmpb   $0xa,(%rdx)
  jne    L22
  movb   $0x0,(%rdx)
L38:
  pop    %rbx
  ret
```

assistant: ```c
void StripCR(char *string) {
  int i;
  for (i = strlen(string) - 1; i > 0; i--)
    if (string[i] == '\n') {
      string[i] = '\0';
      break;
    }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_system_time_in_microseconds>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x20(%rbp),%eax
  imul   $0xf4240,%eax,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
long get_system_time_in_microseconds(void) {
  struct timeval tempo;
  gettimeofday(&tempo, NULL);
  return tempo.tv_sec * 1000000 + tempo.tv_usec;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_system_time_in_microseconds>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  imul   $0xf4240,(%rsp),%eax
  cltq
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long get_system_time_in_microseconds(void) {
  struct timeval tempo;
  gettimeofday(&tempo, NULL);
  return tempo.tv_sec * 1000000 + tempo.tv_usec;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_system_time_in_microseconds>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,(%rsp),%eax
  cltq
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long get_system_time_in_microseconds(void) {
  struct timeval tempo;
  gettimeofday(&tempo, NULL);
  return tempo.tv_sec * 1000000 + tempo.tv_usec;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_system_time_in_microseconds>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,(%rsp),%eax
  cltq
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long get_system_time_in_microseconds(void) {
  struct timeval tempo;
  gettimeofday(&tempo, NULL);
  return tempo.tv_sec * 1000000 + tempo.tv_usec;
}
```
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    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L1
L22:
  mov    -0x10(%rbp),%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  shl    $0x3,%ecx
  sub    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  je     L38
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <have7@plt>
  test   %rax,%rax
  je     L40
L38:
  addl   $0x1,-0xc(%rbp)
L40:
  addl   $0x1,-0x10(%rbp)
L1:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L22
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_guessed: string
name: retrieve_data
arguments: tag: D19
type_guessed: string

tool: D3: %d

tool: D19: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int n, i, sum = 0, now, t;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    if (i % 7 == 0 || have7(i)) {
      sum++;
    }
  }
  printf("%d\n", sum);
  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    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L17
  mov    $0x0,%ebp
  mov    $0x1,%ebx
  jmp    L2
L49:
  add    $0x1,%ebp
L15:
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jle    L0
L2:
  movslq %ebx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %ebx,%eax
  sar    $0x2,%eax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    0x0(,%rax,8),%edx
  sub    %eax,%edx
  cmp    %edx,%ebx
  je     L49
  mov    %ebx,%edi
  call   <have7@plt>
  test   %rax,%rax
  je     L15
  jmp    L49
L17:
  mov    $0x0,%ebp
L0:
  mov    %ebp,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
type_infered: string
name: read_data
arguments: tag: D2
type_infered: string

tool: D7: %d

tool: D2: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int n, i, sum = 0, now, t;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    if (i % 7 == 0 || have7(i)) {
      sum++;
    }
  }
  printf("%d\n", sum);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rdi
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L52
  mov    $0x1,%ebx
  jmp    L63
  nopl   0x0(%rax,%rax,1)
L65:
  add    $0x1,%ebp
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jle    L52
L63:
  imul   $0xb6db6db7,%ebx,%eax
  cmp    $0x24924924,%eax
  jbe    L65
  mov    %ebx,%edi
  call   <have7@plt>
  test   %rax,%rax
  jne    L65
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jg     L63
  xchg   %ax,%ax
L52:
  mov    %ebp,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L60
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L60:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D14
type_guessed: string
name: retrieve_data
arguments: tag: D4
type_guessed: string

tool: D14: %d

tool: D4: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int n, i, sum = 0, now, t;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    if (i % 7 == 0 || have7(i)) {
      sum++;
    }
  }
  printf("%d\n", sum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rdi
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  cmp    $0x1,%edx
  jle    L58
  mov    $0x1,%ebx
  jmp    L18
  nopl   (%rax)
L43:
  add    $0x1,%ebx
  add    $0x1,%ebp
  cmp    %ebx,%edx
  jle    L58
L18:
  imul   $0xb6db6db7,%ebx,%eax
  cmp    $0x24924924,%eax
  jbe    L43
  mov    %ebx,%edi
  call   <have7@plt>
  mov    0x4(%rsp),%edx
  test   %rax,%rax
  jne    L43
  add    $0x1,%ebx
  cmp    %ebx,%edx
  jg     L18
  xchg   %ax,%ax
L58:
  mov    %ebp,%esi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D9
data_type: string
name: unpack_data
arguments: data_label: D17
data_type: string

tool: D9: %d

tool: D17: %d


assistant: ```c
int main(int argc, const char *argv[]) {
  int n, i, sum = 0, now, t;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    if (i % 7 == 0 || have7(i)) {
      sum++;
    }
  }
  printf("%d\n", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<can_make_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L189
L159:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jne    L75
  movl   $0x1,-0x18(%rbp)
L75:
  mov    -0x8(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jne    L134
  movl   $0x1,-0x14(%rbp)
L134:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jne    L157
  movl   $0x1,-0x10(%rbp)
L157:
  mov    -0x4(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jne    L1
  movl   $0x1,-0xc(%rbp)
L1:
  cmpl   $0x6,-0x8(%rbp)
  jne    L35
  cmpl   $0x9,-0x24(%rbp)
  jne    L35
  movl   $0x1,-0x18(%rbp)
L35:
  cmpl   $0x9,-0x8(%rbp)
  jne    L144
  cmpl   $0x6,-0x24(%rbp)
  jne    L144
  movl   $0x1,-0x18(%rbp)
L144:
  cmpl   $0x6,-0x4(%rbp)
  jne    L84
  cmpl   $0x9,-0x24(%rbp)
  jne    L84
  movl   $0x1,-0x10(%rbp)
L84:
  cmpl   $0x9,-0x4(%rbp)
  jne    L4
  cmpl   $0x6,-0x24(%rbp)
  jne    L4
  movl   $0x1,-0x10(%rbp)
L4:
  addl   $0x1,-0x1c(%rbp)
L189:
  cmpl   $0x5,-0x1c(%rbp)
  jle    L159
  cmpl   $0x0,-0x18(%rbp)
  je     L66
  cmpl   $0x0,-0xc(%rbp)
  jne    L165
L66:
  cmpl   $0x0,-0x14(%rbp)
  je     L24
  cmpl   $0x0,-0x10(%rbp)
  je     L24
L165:
  mov    $0x1,%eax
  jmp    L95
L24:
  mov    $0x0,%eax
L95:
  pop    %rbp
  ret
```

assistant: ```c
int can_make_n(int n1, int n2, int d1[], int d2[]) {
  int i = 0;
  int n1_in_d1 = 0;
  int n2_in_d1 = 0;
  int n1_in_d2 = 0;
  int n2_in_d2 = 0;
  int d_1, d_2;
  for (i = 0; i < 6; i++) {
    d_1 = d1[i];
    d_2 = d2[i];
    if (d_1 == n1)
      n1_in_d1 = 1;
    if (d_1 == n2)
      n2_in_d1 = 1;
    if (d_2 == n1)
      n1_in_d2 = 1;
    if (d_2 == n2)
      n2_in_d2 = 1;
    if (d_1 == 6)
      if (n1 == 9)
        n1_in_d1 = 1;
    if (d_1 == 9)
      if (n1 == 6)
        n1_in_d1 = 1;
    if (d_2 == 6)
      if (n1 == 9)
        n1_in_d2 = 1;
    if (d_2 == 9)
      if (n1 == 6)
        n1_in_d2 = 1;
  }
  return (n1_in_d1 && n2_in_d2) || (n2_in_d1 && n1_in_d2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<can_make_n>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r11d
  mov    %rdx,%r14
  mov    %rcx,%r13
  mov    $0x0,%eax
  mov    $0x0,%ebx
  mov    $0x0,%r9d
  mov    $0x0,%ebp
  mov    $0x0,%r10d
  mov    $0x1,%esi
  mov    $0x0,%r15d
  jmp    L71
L82:
  mov    %r15d,%r8d
  mov    $0x1,%r10d
  jmp    L26
L55:
  mov    %edx,%r8d
  mov    $0x1,%r10d
L5:
  cmp    $0x9,%ecx
  jne    L97
  test   %r8b,%r8b
  cmovne %esi,%r9d
L97:
  add    $0x4,%rax
  cmp    $0x18,%rax
  je     L91
L71:
  mov    (%r14,%rax,1),%edx
  mov    0x0(%r13,%rax,1),%ecx
  cmp    %edi,%edx
  cmove  %esi,%r10d
  cmp    %r11d,%edx
  cmove  %esi,%ebp
  cmp    %edi,%ecx
  cmove  %esi,%r9d
  cmp    %r11d,%ecx
  cmove  %esi,%ebx
  cmp    $0x9,%edi
  sete   %r12b
  cmp    $0x6,%edx
  jne    L34
  test   %r12b,%r12b
  jne    L82
L34:
  cmp    $0x6,%edi
  sete   %r8b
  cmp    $0x9,%edx
  sete   %dl
  and    %r8b,%dl
  jne    L55
L26:
  cmp    $0x6,%ecx
  jne    L5
  test   %r12b,%r12b
  je     L5
  mov    $0x1,%r9d
  jmp    L97
L91:
  test   %r10d,%r10d
  je     L53
  mov    $0x1,%eax
  test   %ebx,%ebx
  je     L53
L52:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L53:
  test   %ebp,%ebp
  setne  %dl
  test   %r9d,%r9d
  setne  %al
  movzbl %al,%eax
  and    %edx,%eax
  jmp    L52
```

assistant: ```c
int can_make_n(int n1, int n2, int d1[], int d2[]) {
  int i = 0;
  int n1_in_d1 = 0;
  int n2_in_d1 = 0;
  int n1_in_d2 = 0;
  int n2_in_d2 = 0;
  int d_1, d_2;
  for (i = 0; i < 6; i++) {
    d_1 = d1[i];
    d_2 = d2[i];
    if (d_1 == n1)
      n1_in_d1 = 1;
    if (d_1 == n2)
      n2_in_d1 = 1;
    if (d_2 == n1)
      n1_in_d2 = 1;
    if (d_2 == n2)
      n2_in_d2 = 1;
    if (d_1 == 6)
      if (n1 == 9)
        n1_in_d1 = 1;
    if (d_1 == 9)
      if (n1 == 6)
        n1_in_d1 = 1;
    if (d_2 == 6)
      if (n1 == 9)
        n1_in_d2 = 1;
    if (d_2 == 9)
      if (n1 == 6)
        n1_in_d2 = 1;
  }
  return (n1_in_d1 && n2_in_d2) || (n2_in_d1 && n1_in_d2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<can_make_n>:
  endbr64
  push   %r14
  mov    %edi,%r9d
  xor    %r10d,%r10d
  xor    %r11d,%r11d
  push   %r13
  xor    %eax,%eax
  mov    %rdx,%r13
  xor    %edx,%edx
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  mov    $0x1,%esi
  push   %rbx
  xor    %ebx,%ebx
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L73:
  cmp    $0x6,%r9d
  sete   %r12b
  mov    %r12d,%r14d
  cmp    $0x9,%edi
  jne    L97
  test   %r12b,%r12b
  je     L97
  mov    $0x1,%eax
L92:
  cmp    $0x6,%r8d
  je     L76
  cmp    $0x6,%r9d
  sete   %r14b
L4:
  cmp    $0x9,%r8d
  jne    L76
  test   %r14b,%r14b
  cmovne %esi,%r10d
L76:
  add    $0x4,%rdx
  cmp    $0x18,%rdx
  je     L35
L1:
  mov    0x0(%r13,%rdx,1),%edi
  mov    (%rcx,%rdx,1),%r8d
  cmp    %r9d,%edi
  cmove  %esi,%eax
  cmp    %ebp,%edi
  cmove  %esi,%r11d
  cmp    %r9d,%r8d
  cmove  %esi,%r10d
  cmp    %ebp,%r8d
  cmove  %esi,%ebx
  cmp    $0x6,%edi
  jne    L73
  cmp    $0x9,%r9d
  jne    L92
  cmp    $0x6,%r8d
  mov    $0x1,%eax
  cmove  %esi,%r10d
  add    $0x4,%rdx
  cmp    $0x18,%rdx
  jne    L1
L35:
  and    %r10d,%r11d
  and    %ebx,%eax
  pop    %rbx
  pop    %rbp
  cmove  %r11d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L97:
  cmp    $0x6,%r8d
  jne    L4
  cmp    $0x9,%r9d
  cmove  %esi,%r10d
  jmp    L76
```

assistant: ```c
int can_make_n(int n1, int n2, int d1[], int d2[]) {
  int i = 0;
  int n1_in_d1 = 0;
  int n2_in_d1 = 0;
  int n1_in_d2 = 0;
  int n2_in_d2 = 0;
  int d_1, d_2;
  for (i = 0; i < 6; i++) {
    d_1 = d1[i];
    d_2 = d2[i];
    if (d_1 == n1)
      n1_in_d1 = 1;
    if (d_1 == n2)
      n2_in_d1 = 1;
    if (d_2 == n1)
      n1_in_d2 = 1;
    if (d_2 == n2)
      n2_in_d2 = 1;
    if (d_1 == 6)
      if (n1 == 9)
        n1_in_d1 = 1;
    if (d_1 == 9)
      if (n1 == 6)
        n1_in_d1 = 1;
    if (d_2 == 6)
      if (n1 == 9)
        n1_in_d2 = 1;
    if (d_2 == 9)
      if (n1 == 6)
        n1_in_d2 = 1;
  }
  return (n1_in_d1 && n2_in_d2) || (n2_in_d1 && n1_in_d2);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<can_make_n>:
  endbr64
  push   %r14
  mov    %esi,%r10d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %rdx,%rbx
  cmp    $0x9,%edi
  jne    L123
  xor    %edx,%edx
  xor    %r9d,%r9d
  xor    %esi,%esi
  xor    %r11d,%r11d
  xor    %eax,%eax
  mov    $0x1,%ecx
  jmp    L87
  nopl   (%rax)
L69:
  cmp    $0x6,%r8d
  cmove  %ecx,%esi
  add    $0x4,%rdx
  cmp    $0x18,%rdx
  je     L121
L87:
  mov    (%rbx,%rdx,1),%edi
  mov    0x0(%rbp,%rdx,1),%r8d
  cmp    $0x9,%edi
  cmove  %ecx,%eax
  cmp    %r10d,%edi
  cmove  %ecx,%r11d
  cmp    $0x9,%r8d
  cmove  %ecx,%esi
  cmp    %r10d,%r8d
  cmove  %ecx,%r9d
  cmp    $0x6,%edi
  jne    L69
  cmp    $0x6,%r8d
  mov    $0x1,%eax
  cmove  %ecx,%esi
  add    $0x4,%rdx
  cmp    $0x18,%rdx
  jne    L87
L121:
  and    %r11d,%esi
  and    %r9d,%eax
  pop    %rbx
  pop    %rbp
  cmove  %esi,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L123:
  xor    %ecx,%ecx
  xor    %r9d,%r9d
  xor    %esi,%esi
  xor    %r11d,%r11d
  xor    %eax,%eax
  mov    $0x1,%edx
L32:
  mov    (%rbx,%rcx,1),%r12d
  mov    0x0(%rbp,%rcx,1),%r8d
  cmp    %r12d,%edi
  cmove  %edx,%eax
  cmp    %r12d,%r10d
  cmove  %edx,%r11d
  cmp    %r8d,%edi
  cmove  %edx,%esi
  cmp    %r8d,%r10d
  cmove  %edx,%r9d
  cmp    $0x6,%r12d
  je     L82
  cmp    $0x6,%edi
  sete   %r14b
  mov    %r14d,%r13d
  cmp    $0x9,%r12d
  jne    L62
  test   %r14b,%r14b
  je     L62
  mov    $0x1,%eax
L82:
  cmp    $0x6,%r8d
  je     L107
  cmp    $0x6,%edi
  sete   %r13b
L6:
  cmp    $0x9,%r8d
  jne    L107
  test   %r13b,%r13b
  cmovne %edx,%esi
L107:
  add    $0x4,%rcx
  cmp    $0x18,%rcx
  jne    L32
  jmp    L121
L62:
  cmp    $0x6,%r8d
  jne    L6
  jmp    L107
```

assistant: ```c
int can_make_n(int n1, int n2, int d1[], int d2[]) {
  int i = 0;
  int n1_in_d1 = 0;
  int n2_in_d1 = 0;
  int n1_in_d2 = 0;
  int n2_in_d2 = 0;
  int d_1, d_2;
  for (i = 0; i < 6; i++) {
    d_1 = d1[i];
    d_2 = d2[i];
    if (d_1 == n1)
      n1_in_d1 = 1;
    if (d_1 == n2)
      n2_in_d1 = 1;
    if (d_2 == n1)
      n1_in_d2 = 1;
    if (d_2 == n2)
      n2_in_d2 = 1;
    if (d_1 == 6)
      if (n1 == 9)
        n1_in_d1 = 1;
    if (d_1 == 9)
      if (n1 == 6)
        n1_in_d1 = 1;
    if (d_2 == 6)
      if (n1 == 9)
        n1_in_d2 = 1;
    if (d_2 == 9)
      if (n1 == 6)
        n1_in_d2 = 1;
  }
  return (n1_in_d1 && n2_in_d2) || (n2_in_d1 && n1_in_d2);
}
```