ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x1a,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x30(%rbp),%rax
  and    $0xff,%eax
  mov    %rax,-0x28(%rbp)
  mov    -0x30(%rbp),%rax
  and    $0xff00,%eax
  mov    %rax,-0x20(%rbp)
  shrq   $0x8,-0x20(%rbp)
  mov    -0x30(%rbp),%rax
  and    $0xff0000,%eax
  mov    %rax,-0x18(%rbp)
  shrq   $0x10,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  and    $0xff000000,%eax
  mov    %rax,-0x10(%rbp)
  shrq   $0x18,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    -0x28(%rbp),%rdi
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    %rdi,%r9
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L6
L14:
  nop
L6:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L14
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D7
datatype: string
name: retrieve_data
arguments: label: D12
datatype: string
name: retrieve_data
arguments: label: D14
datatype: string

tool: D7: Enter a eight-digit hex value %c 


tool: D12: %x

tool: D14: 0x%04X is composed of %2X, %2X, %2X and %02X


assistant: ```c
int main(void) {
  unsigned long int hex, upper, lower, middle_1, middle_2;
  printf("Enter a eight-digit hex value %c \n", 26);
  scanf("%x", &hex);
  lower = hex & 0x000000FF;
  middle_2 = hex & 0x0000FF00;
  middle_2 >>= 8;
  middle_1 = hex & 0x00FF0000;
  middle_1 >>= 16;
  upper = hex & 0xFF000000;
  upper >>= 24;
  printf("0x%04X is composed of %2X, %2X, %2X and %02X\n", hex, upper, middle_1, middle_2, lower);
  while (getchar() != '\n')
    continue;
  getchar();
  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
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x1a,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rsi
  mov    %rsi,%rax
  movzbl %ah,%eax
  mov    %rax,%r8
  mov    %rsi,%rcx
  shr    $0x10,%rcx
  movzbl %cl,%ecx
  mov    %rsi,%rdx
  shr    $0x18,%rdx
  movzbl %dl,%edx
  movzbl %sil,%r9d
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L2:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L2
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D4
datatype: string
name: retrieve_data
arguments: tag: D2
datatype: string
name: retrieve_data
arguments: tag: D24
datatype: string

tool: D4: Enter a eight-digit hex value %c 


tool: D2: %x

tool: D24: 0x%04X is composed of %2X, %2X, %2X and %02X


assistant: ```c
int main(void) {
  unsigned long int hex, upper, lower, middle_1, middle_2;
  printf("Enter a eight-digit hex value %c \n", 26);
  scanf("%x", &hex);
  lower = hex & 0x000000FF;
  middle_2 = hex & 0x0000FF00;
  middle_2 >>= 8;
  middle_1 = hex & 0x00FF0000;
  middle_1 >>= 16;
  upper = hex & 0xFF000000;
  upper >>= 24;
  printf("0x%04X is composed of %2X, %2X, %2X and %02X\n", hex, upper, middle_1, middle_2, lower);
  while (getchar() != '\n')
    continue;
  getchar();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x1a,%esi
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D0(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rsi
  lea    D3(%rip),%rdi
  mov    %rsi,%rax
  mov    %rsi,%rcx
  mov    %rsi,%rdx
  movzbl %sil,%r9d
  shr    $0x10,%rcx
  movzbl %ah,%eax
  shr    $0x18,%rdx
  mov    %rax,%r8
  movzbl %cl,%ecx
  movzbl %dl,%edx
  xor    %eax,%eax
  call   <printf@plt>
L14:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L14
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type_infered: string
name: retrieve_data
arguments: data_label: D0
type_infered: string
name: retrieve_data
arguments: data_label: D3
type_infered: string

tool: D9: Enter a eight-digit hex value %c 


tool: D0: %x

tool: D3: 0x%04X is composed of %2X, %2X, %2X and %02X


assistant: ```c
int main(void) {
  unsigned long int hex, upper, lower, middle_1, middle_2;
  printf("Enter a eight-digit hex value %c \n", 26);
  scanf("%x", &hex);
  lower = hex & 0x000000FF;
  middle_2 = hex & 0x0000FF00;
  middle_2 >>= 8;
  middle_1 = hex & 0x00FF0000;
  middle_1 >>= 16;
  upper = hex & 0xFF000000;
  upper >>= 24;
  printf("0x%04X is composed of %2X, %2X, %2X and %02X\n", hex, upper, middle_1, middle_2, lower);
  while (getchar() != '\n')
    continue;
  getchar();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x1a,%esi
  lea    D10(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rsi
  lea    D14(%rip),%rdi
  mov    %rsi,%rax
  mov    %rsi,%rcx
  mov    %rsi,%rdx
  movzbl %sil,%r9d
  shr    $0x10,%rcx
  movzbl %ah,%eax
  shr    $0x18,%rdx
  mov    %rax,%r8
  movzbl %cl,%ecx
  movzbl %dl,%edx
  xor    %eax,%eax
  call   <printf@plt>
L6:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L6
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D10
datatype: string
name: retrieve_data
arguments: data_tag: D9
datatype: string
name: retrieve_data
arguments: data_tag: D14
datatype: string

tool: D10: Enter a eight-digit hex value %c 


tool: D9: %x

tool: D14: 0x%04X is composed of %2X, %2X, %2X and %02X


assistant: ```c
int main(void) {
  unsigned long int hex, upper, lower, middle_1, middle_2;
  printf("Enter a eight-digit hex value %c \n", 26);
  scanf("%x", &hex);
  lower = hex & 0x000000FF;
  middle_2 = hex & 0x0000FF00;
  middle_2 >>= 8;
  middle_1 = hex & 0x00FF0000;
  middle_1 >>= 16;
  upper = hex & 0xFF000000;
  upper >>= 24;
  printf("0x%04X is composed of %2X, %2X, %2X and %02X\n", hex, upper, middle_1, middle_2, lower);
  while (getchar() != '\n')
    continue;
  getchar();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L41
  mov    $0x0,%eax
  jmp    L40
L41:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L11
  mov    $0x0,%eax
  jmp    L40
L11:
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L12
L39:
  mov    -0x18(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L12:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L39
  mov    -0x8(%rbp),%rax
L40:
  leave
  ret
```

assistant: ```c
char *strdup(const char *org) {
  if (org == NULL)
    return NULL;
  char *newstr = malloc(strlen(org) + 1);
  char *p;
  if (newstr == NULL)
    return NULL;
  p = newstr;
  while (*org)
    *p++ = *org++;
  return newstr;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strdup>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L7
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L7
  mov    $0x0,%edx
L25:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L25
L7:
  pop    %rbx
  ret
```

assistant: ```c
char *strdup(const char *org) {
  if (org == NULL)
    return NULL;
  char *newstr = malloc(strlen(org) + 1);
  char *p;
  if (newstr == NULL)
    return NULL;
  p = newstr;
  while (*org)
    *p++ = *org++;
  return newstr;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strdup>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L5
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L5
  xor    %edx,%edx
  nopl   0x0(%rax)
L7:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L7
L5:
  pop    %rbx
  ret
```

assistant: ```c
char *strdup(const char *org) {
  if (org == NULL)
    return NULL;
  char *newstr = malloc(strlen(org) + 1);
  char *p;
  if (newstr == NULL)
    return NULL;
  p = newstr;
  while (*org)
    *p++ = *org++;
  return newstr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strdup>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  test   %rax,%rax
  je     L13
  movzbl (%rbx),%ecx
  test   %cl,%cl
  je     L13
  xor    %edx,%edx
  nopl   0x0(%rax)
L6:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rbx,%rdx,1),%ecx
  test   %cl,%cl
  jne    L6
L13:
  pop    %rbx
  ret
```

assistant: ```c
char *strdup(const char *org) {
  if (org == NULL)
    return NULL;
  char *newstr = malloc(strlen(org) + 1);
  char *p;
  if (newstr == NULL)
    return NULL;
  p = newstr;
  while (*org)
    *p++ = *org++;
  return newstr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<utl_strcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L70
  cmpq   $0x0,-0x10(%rbp)
  jne    L70
  mov    $0x0,%eax
  jmp    L62
L70:
  cmpq   $0x0,-0x8(%rbp)
  je     L54
  cmpq   $0x0,-0x10(%rbp)
  jne    L79
L54:
  mov    $0x0,%eax
  jmp    L62
L5:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L61
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  jmp    L62
L61:
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L79:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L57
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L5
L57:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
L62:
  pop    %rbp
  ret
```

assistant: ```c
int utl_strcmp(char *first, char *second) {
  if (!first && !second)
    return (0);
  else if (!first || !second)
    return (0);
  while (*first && *second) {
    if (*first != *second)
      return (*first - *second);
    first++;
    second++;
  }
  return (*first - *second);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<utl_strcmp>:
  endbr64
  test   %rdi,%rdi
  je     L32
  test   %rsi,%rsi
  je     L32
  movzbl (%rdi),%eax
  test   %al,%al
  je     L8
L41:
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L8
  cmp    %al,%dl
  jne    L24
  add    $0x1,%rdi
  add    $0x1,%rsi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L41
L8:
  movsbl %al,%eax
  movsbl (%rsi),%edx
  sub    %edx,%eax
  ret
L24:
  movsbl %al,%eax
  movsbl %dl,%edx
  sub    %edx,%eax
  ret
L32:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int utl_strcmp(char *first, char *second) {
  if (!first && !second)
    return (0);
  else if (!first || !second)
    return (0);
  while (*first && *second) {
    if (*first != *second)
      return (*first - *second);
    first++;
    second++;
  }
  return (*first - *second);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<utl_strcmp>:
  endbr64
  test   %rdi,%rdi
  je     L20
  test   %rsi,%rsi
  je     L20
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L49
  jmp    L54
  nopw   0x0(%rax,%rax,1)
L73:
  cmp    %al,%dl
  jne    L27
  movsbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  lea    0x1(%rsi),%rdx
  test   %al,%al
  je     L39
  mov    %rdx,%rsi
L49:
  movsbl (%rsi),%edx
  test   %dl,%dl
  jne    L73
L27:
  sub    %edx,%eax
  ret
  nopl   0x0(%rax)
L20:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L39:
  movsbl 0x1(%rsi),%edx
  xor    %eax,%eax
  sub    %edx,%eax
  ret
L54:
  movsbl (%rsi),%edx
  xor    %eax,%eax
  jmp    L27
```

assistant: ```c
int utl_strcmp(char *first, char *second) {
  if (!first && !second)
    return (0);
  else if (!first || !second)
    return (0);
  while (*first && *second) {
    if (*first != *second)
      return (*first - *second);
    first++;
    second++;
  }
  return (*first - *second);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<utl_strcmp>:
  endbr64
  test   %rdi,%rdi
  je     L67
  test   %rsi,%rsi
  je     L67
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L57
  jmp    L63
  nopw   0x0(%rax,%rax,1)
L69:
  cmp    %al,%dl
  jne    L1
  movsbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  lea    0x1(%rsi),%rdx
  test   %al,%al
  je     L45
  mov    %rdx,%rsi
L57:
  movsbl (%rsi),%edx
  test   %dl,%dl
  jne    L69
L1:
  sub    %edx,%eax
  ret
  nopl   0x0(%rax)
L67:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L45:
  movsbl 0x1(%rsi),%edx
  xor    %eax,%eax
  sub    %edx,%eax
  ret
L63:
  movsbl (%rsi),%edx
  xor    %eax,%eax
  jmp    L1
```

assistant: ```c
int utl_strcmp(char *first, char *second) {
  if (!first && !second)
    return (0);
  else if (!first || !second)
    return (0);
  while (*first && *second) {
    if (*first != *second)
      return (*first - *second);
    first++;
    second++;
  }
  return (*first - *second);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<log_quit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L7
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L7:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  mov    -0xb4(%rbp),%ecx
  mov    -0xc8(%rbp),%rdx
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <log_doit@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    $0x2,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void log_quit(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  log_doit(0, LOG_ERR, fmt, ap);
  va_end(ap);
  exit(2);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<log_quit>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %rbx,%rdx
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  mov    $0x0,%edi
  call   <log_doit@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    $0x2,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void log_quit(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  log_doit(0, LOG_ERR, fmt, ap);
  va_end(ap);
  exit(2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<log_quit>:
  endbr64
  push   %r12
  mov    %rdi,%rsi
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    D2(%rip),%rax
  mov    %ebp,%ecx
  xor    %edi,%edi
  mov    %r12,%rdx
  mov    (%rax),%esi
  call   <log_doit@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    $0x2,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void log_quit(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  log_doit(0, LOG_ERR, fmt, ap);
  va_end(ap);
  exit(2);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<log_quit>:
  endbr64
  push   %r12
  mov    %rdi,%rsi
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    D1(%rip),%rax
  mov    %ebp,%ecx
  xor    %edi,%edi
  mov    %r12,%rdx
  mov    (%rax),%esi
  call   <log_doit@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    $0x2,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void log_quit(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  log_doit(0, LOG_ERR, fmt, ap);
  va_end(ap);
  exit(2);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L1:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
  cmpb   $0x0,-0x9(%rbp)
  jne    L1
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strcpy(char *d, const char *s) {
  char *r = d;
  unsigned char c;
  do {
    c = *d = *s;
    d++;
    s++;
  } while (c != 0);
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  mov    $0x0,%edx
L6:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L6
  ret
```

assistant: ```c
char *strcpy(char *d, const char *s) {
  char *r = d;
  unsigned char c;
  do {
    c = *d = *s;
    d++;
    s++;
  } while (c != 0);
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L2:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L2
  ret
```

assistant: ```c
char *strcpy(char *d, const char *s) {
  char *r = d;
  unsigned char c;
  do {
    c = *d = *s;
    d++;
    s++;
  } while (c != 0);
  return r;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strcpy>:
  endbr64
  mov    %rdi,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L6:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L6
  ret
```

assistant: ```c
char *strcpy(char *d, const char *s) {
  char *r = d;
  unsigned char c;
  do {
    c = *d = *s;
    d++;
    s++;
  } while (c != 0);
  return r;
}
```
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    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x750,%rsp
  mov    %edi,-0x2744(%rbp)
  mov    %rsi,-0x2750(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x2720(%rbp)
  movq   $0x0,-0x2718(%rbp)
  lea    -0x2710(%rbp),%rax
  mov    $0x2700,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x0,-0x2734(%rbp)
  movl   $0x1,-0x273c(%rbp)
  jmp    L18
L55:
  mov    -0x273c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x2738(%rbp)
  jmp    L12
L27:
  mov    -0x273c(%rbp),%eax
  imul   -0x2738(%rbp),%eax
  mov    %eax,-0x2730(%rbp)
  mov    -0x273c(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <digit_usage@plt>
  mov    %eax,-0x272c(%rbp)
  mov    -0x2738(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <digit_usage@plt>
  mov    %eax,-0x2728(%rbp)
  mov    -0x2730(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <digit_usage@plt>
  mov    %eax,-0x2724(%rbp)
  mov    -0x272c(%rbp),%eax
  and    -0x2728(%rbp),%eax
  and    -0x2724(%rbp),%eax
  test   %eax,%eax
  jne    L36
  mov    -0x272c(%rbp),%eax
  or     -0x2728(%rbp),%eax
  or     -0x2724(%rbp),%eax
  cmp    $0x3fe,%eax
  jne    L36
  mov    -0x2730(%rbp),%eax
  cltq
  movzbl -0x2720(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L5
  mov    -0x2730(%rbp),%eax
  add    %eax,-0x2734(%rbp)
L5:
  mov    -0x2730(%rbp),%eax
  cltq
  movb   $0x1,-0x2720(%rbp,%rax,1)
L36:
  addl   $0x1,-0x2738(%rbp)
L12:
  mov    $0x2710,%eax
  cltd
  idivl  -0x273c(%rbp)
  cmp    %eax,-0x2738(%rbp)
  jle    L27
  addl   $0x1,-0x273c(%rbp)
L18:
  cmpl   $0x270f,-0x273c(%rbp)
  jle    L55
  mov    -0x2734(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L71
  call   <__stack_chk_fail@plt>
L71:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
data_type: string

tool: D9: %d


assistant: ```c
int main(int argc, char **argv) {
  char p[10000] = {0};
  int a, aa, b, bb, c, cc, s = 0;
  for (a = 1; a < 10000; a++) {
    for (b = a + 1; b < 10000 / a + 1; b++) {
      c = a * b;
      aa = digit_usage(a, 0);
      bb = digit_usage(b, 0);
      cc = digit_usage(c, 0);
      if ((aa & bb & cc) == 0 && (aa | bb | cc) == 0x3fe) {
        if (!p[c])
          s += c;
        p[c] = 1;
      }
    }
  }
  printf("%d\n", s);
  return 0;
}
```