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

```asm
<asprintf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rdi,%r12
  mov    %r13d,%edi
  push   %rbp
  mov    %rsi,%rbp
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <vasprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int asprintf(char **restrict strp, const char *restrict format, ...) {
  va_list ap;
  va_start(ap, format);
  int result = vasprintf(strp, format, ap);
  va_end(ap);
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<asprintf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rdi,%r12
  mov    %r13d,%edi
  push   %rbp
  mov    %rsi,%rbp
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <vasprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int asprintf(char **restrict strp, const char *restrict format, ...) {
  va_list ap;
  va_start(ap, format);
  int result = vasprintf(strp, format, ap);
  va_end(ap);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<job_status>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <job_list@plt>
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
int job_status(int argc, char *argv[]) {
  job_list(1, NULL);
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<job_status>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <job_list@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int job_status(int argc, char *argv[]) {
  job_list(1, NULL);
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<job_status>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x1,%edi
  call   <job_list@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int job_status(int argc, char *argv[]) {
  job_list(1, NULL);
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<job_status>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  mov    $0x1,%edi
  call   <job_list@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int job_status(int argc, char *argv[]) {
  job_list(1, NULL);
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fibonacci>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jle    L0
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fibonacci>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <fibonacci>
  add    %ebx,%eax
  jmp    L13
L0:
  cmpl   $0x1,-0x14(%rbp)
  jne    L1
  mov    $0x1,%eax
  jmp    L13
L1:
  cmpl   $0x0,-0x14(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L13
L13:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int fibonacci(int ceil) {
  if (1 < ceil)
    return fibonacci(ceil - 1) + fibonacci(ceil - 2);
  else if (1 == ceil)
    return 1;
  else if (0 == ceil)
    return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fibonacci>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  cmp    $0x1,%edi
  jg     L27
  je     L2
  test   %edi,%edi
  jne    L16
  jmp    L2
L27:
  lea    -0x1(%rdi),%edi
  call   <fibonacci>
  mov    %eax,%ebp
  lea    -0x2(%rbx),%edi
  call   <fibonacci>
  lea    0x0(%rbp,%rax,1),%ebx
L2:
  mov    %ebx,%eax
L16:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fibonacci(int ceil) {
  if (1 < ceil)
    return fibonacci(ceil - 1) + fibonacci(ceil - 2);
  else if (1 == ceil)
    return 1;
  else if (0 == ceil)
    return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<a2u>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,%eax
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %al,-0x14(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L38:
  mov    -0xc(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jg     L14
  mov    -0x18(%rbp),%eax
  imul   -0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x14(%rbp)
L3:
  movsbl -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <a2d@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jns    L38
  jmp    L12
L14:
  nop
L12:
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    -0x10(%rbp),%edx
  mov    %edx,(%rax)
  movzbl -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
char a2u(char ch, const char **src, int base, unsigned int *nump) {
  const char *p = *src;
  unsigned int num = 0;
  int digit = 0;
  while ((digit = a2d(ch)) >= 0) {
    if (digit > base) {
      break;
    }
    num = num * base + digit;
    ch = *p++;
  }
  *src = p;
  *nump = num;
  return ch;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<a2u>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r14
  mov    %edx,%r13d
  mov    %rcx,%r15
  mov    %edi,%r12d
  mov    (%rsi),%rbp
  mov    $0x0,%ebx
L20:
  movsbl %r12b,%edi
  call   <a2d@plt>
  cmp    %r13d,%eax
  jg     L24
  test   %eax,%eax
  js     L24
  imul   %r13d,%ebx
  add    %eax,%ebx
  add    $0x1,%rbp
  movzbl -0x1(%rbp),%r12d
  jmp    L20
L24:
  mov    %rbp,(%r14)
  mov    %ebx,(%r15)
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char a2u(char ch, const char **src, int base, unsigned int *nump) {
  const char *p = *src;
  unsigned int num = 0;
  int digit = 0;
  while ((digit = a2d(ch)) >= 0) {
    if (digit > base) {
      break;
    }
    num = num * base + digit;
    ch = *p++;
  }
  *src = p;
  *nump = num;
  return ch;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<a2u>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  mov    (%rsi),%r14
  jmp    L10
  nopl   0x0(%rax)
L9:
  test   %eax,%eax
  js     L20
  imul   %ebp,%ebx
  movzbl (%r14),%r15d
  add    $0x1,%r14
  add    %eax,%ebx
L10:
  movsbl %r15b,%edi
  call   <a2d@plt>
  cmp    %ebp,%eax
  jle    L9
L20:
  mov    %r14,(%r12)
  mov    %r15d,%eax
  mov    %ebx,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char a2u(char ch, const char **src, int base, unsigned int *nump) {
  const char *p = *src;
  unsigned int num = 0;
  int digit = 0;
  while ((digit = a2d(ch)) >= 0) {
    if (digit > base) {
      break;
    }
    num = num * base + digit;
    ch = *p++;
  }
  *src = p;
  *nump = num;
  return ch;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<a2u>:
  endbr64
  push   %r15
  mov    %edi,%r15d
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  mov    (%rsi),%r14
  jmp    L11
  nopl   0x0(%rax)
L4:
  test   %eax,%eax
  js     L29
  imul   %ebp,%ebx
  movzbl (%r14),%r15d
  add    $0x1,%r14
  add    %eax,%ebx
L11:
  movsbl %r15b,%edi
  call   <a2d@plt>
  cmp    %ebp,%eax
  jle    L4
L29:
  mov    %r14,(%r12)
  mov    %r15d,%eax
  mov    %ebx,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char a2u(char ch, const char **src, int base, unsigned int *nump) {
  const char *p = *src;
  unsigned int num = 0;
  int digit = 0;
  while ((digit = a2d(ch)) >= 0) {
    if (digit > base) {
      break;
    }
    num = num * base + digit;
    ch = *p++;
  }
  *src = p;
  *nump = num;
  return ch;
}
```
user: What is the c source code of the assembly code below?

```asm
<get_stat_birthtime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0xffffffff,-0x8(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
struct timespec get_stat_birthtime(struct stat const *st) {
  struct timespec t;
  t.tv_sec = -1;
  t.tv_nsec = -1;
  (void)st;
  return t;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_stat_birthtime>:
  endbr64
  mov    $0xffffffffffffffff,%rax
  ret
```

assistant: ```c
struct timespec get_stat_birthtime(struct stat const *st) {
  struct timespec t;
  t.tv_sec = -1;
  t.tv_nsec = -1;
  (void)st;
  return t;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<pattern_whiteSquares>:
  endbr64
  movabs $0x44eff0004517f800,%rax
  mov    %rax,(%rdi)
  movabs $0x448ff0004517f800,%rax
  mov    %rax,0x8(%rdi)
  movabs $0x43bfc0004517f800,%rax
  mov    %rax,0x10(%rdi)
  movabs $0x42ff000044eff000,%rax
  mov    %rax,0x18(%rdi)
  sub    $0x600000,%rax
  mov    %rax,0x20(%rdi)
  sub    $0xd03000,%rax
  mov    %rax,0x28(%rdi)
  movabs $0x43bfc00042ff0000,%rax
  mov    %rax,0x30(%rdi)
  movabs $0x448ff00042ff0000,%rax
  mov    %rax,0x38(%rdi)
  movabs $0x44eff00042ff0000,%rax
  mov    %rax,0x40(%rdi)
  movabs $0x4517f80043bfc000,%rax
  mov    %rax,0x48(%rdi)
  add    $0xd03000,%rax
  mov    %rax,0x50(%rdi)
  add    $0x600000,%rax
  mov    %rax,0x58(%rdi)
  ret
```

assistant: ```c
void pattern_whiteSquares(float *p) {
  float up_res = 8.0;
  int up_res_pattern_block_size = up_res * 32.0;
  int up_res_pattern_block_size_2 = up_res_pattern_block_size / 2;
  p[0] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 9;
  p[1] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 7;
  p[2] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 9;
  p[3] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 4;
  p[4] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 9;
  p[5] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 1;
  p[6] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 7;
  p[7] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 0;
  p[8] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 4;
  p[9] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 0;
  p[10] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 1;
  p[11] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 0;
  p[12] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 0;
  p[13] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 1;
  p[14] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 0;
  p[15] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 4;
  p[16] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 0;
  p[17] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 7;
  p[18] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 1;
  p[19] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 9;
  p[20] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 4;
  p[21] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 9;
  p[22] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 7;
  p[23] = up_res_pattern_block_size_2 - 0.5 + up_res_pattern_block_size * 9;
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<init_modem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  lea    D54(%rip),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <send_modem_command@plt>
  test   %rax,%rax
  je     L32
  mov    $0xffffffff,%eax
  jmp    L22
L32:
  mov    $0x1,%edi
  call   <sleep@plt>
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  lea    D25(%rip),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <send_modem_command@plt>
  test   %rax,%rax
  je     L0
  mov    $0xffffffff,%eax
  jmp    L22
L0:
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  lea    D24(%rip),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <send_modem_command@plt>
  mov    $0x0,%eax
L22:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
datatype: string
name: retrieve_data
arguments: data_label: D54
datatype: string
name: retrieve_data
arguments: data_label: D44
datatype: string
name: retrieve_data
arguments: data_label: D25
datatype: string
name: retrieve_data
arguments: data_label: D39
datatype: string
name: retrieve_data
arguments: data_label: D24
datatype: string

tool: D4: sending ATZ command...


tool: D54: ATZ

tool: D44: sending caller ID command...


tool: D25: AT+VCID=1

tool: D39: sending FAX mode command...


tool: D24: AT+FCLASS=2

assistant: ```c
int init_modem(int fd) {
  printf("sending ATZ command...\n");
  if (send_modem_command(fd, "ATZ\r") != 0) {
    return (-1);
  }
  sleep(1);
  printf("sending caller ID command...\n");
  if (send_modem_command(fd, "AT+VCID=1\r") != 0) {
    return (-1);
  }
  printf("sending FAX mode command...\n");
  send_modem_command(fd, "AT+FCLASS=2\r");
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<init_modem>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    D32(%rip),%rdi
  call   <printf@plt>
  lea    D43(%rip),%rsi
  mov    %ebx,%edi
  call   <send_modem_command@plt>
  test   %rax,%rax
  jne    L22
  mov    $0x1,%edi
  call   <sleep@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D31(%rip),%rsi
  mov    %ebx,%edi
  call   <send_modem_command@plt>
  test   %rax,%rax
  jne    L28
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    D36(%rip),%rsi
  mov    %ebx,%edi
  call   <send_modem_command@plt>
  mov    $0x0,%eax
L13:
  pop    %rbx
  ret
L22:
  mov    $0xffffffff,%eax
  jmp    L13
L28:
  mov    $0xffffffff,%eax
  jmp    L13
```

tool_calls: name: retrieve_data
arguments: data_tag: D32
datatype: string
name: retrieve_data
arguments: data_tag: D43
datatype: string
name: retrieve_data
arguments: data_tag: D2
datatype: string
name: retrieve_data
arguments: data_tag: D31
datatype: string
name: retrieve_data
arguments: data_tag: D10
datatype: string
name: retrieve_data
arguments: data_tag: D36
datatype: string

tool: D32: sending ATZ command...


tool: D43: ATZ

tool: D2: sending caller ID command...


tool: D31: AT+VCID=1

tool: D10: sending FAX mode command...


tool: D36: AT+FCLASS=2

assistant: ```c
int init_modem(int fd) {
  printf("sending ATZ command...\n");
  if (send_modem_command(fd, "ATZ\r") != 0) {
    return (-1);
  }
  sleep(1);
  printf("sending caller ID command...\n");
  if (send_modem_command(fd, "AT+VCID=1\r") != 0) {
    return (-1);
  }
  printf("sending FAX mode command...\n");
  send_modem_command(fd, "AT+FCLASS=2\r");
  return (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<init_modem>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    D33(%rip),%rsi
  mov    %ebp,%edi
  call   <send_modem_command@plt>
  test   %rax,%rax
  jne    L17
  mov    $0x1,%edi
  call   <sleep@plt>
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    D28(%rip),%rsi
  mov    %ebp,%edi
  call   <send_modem_command@plt>
  test   %rax,%rax
  jne    L17
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  lea    D9(%rip),%rsi
  call   <send_modem_command@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   (%rax)
L17:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D11
data_type: string
name: parse_data
arguments: data_label: D33
data_type: string
name: parse_data
arguments: data_label: D38
data_type: string
name: parse_data
arguments: data_label: D28
data_type: string
name: parse_data
arguments: data_label: D29
data_type: string
name: parse_data
arguments: data_label: D9
data_type: string

tool: D11: sending ATZ command...


tool: D33: ATZ

tool: D38: sending caller ID command...


tool: D28: AT+VCID=1

tool: D29: sending FAX mode command...


tool: D9: AT+FCLASS=2

assistant: ```c
int init_modem(int fd) {
  printf("sending ATZ command...\n");
  if (send_modem_command(fd, "ATZ\r") != 0) {
    return (-1);
  }
  sleep(1);
  printf("sending caller ID command...\n");
  if (send_modem_command(fd, "AT+VCID=1\r") != 0) {
    return (-1);
  }
  printf("sending FAX mode command...\n");
  send_modem_command(fd, "AT+FCLASS=2\r");
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<init_modem>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rsi
  mov    %ebp,%edi
  call   <send_modem_command@plt>
  test   %rax,%rax
  jne    L5
  mov    $0x1,%edi
  call   <sleep@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D48(%rip),%rsi
  mov    %ebp,%edi
  call   <send_modem_command@plt>
  test   %rax,%rax
  jne    L5
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  lea    D57(%rip),%rsi
  call   <send_modem_command@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
  nopl   (%rax)
L5:
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D16
datatype: string
name: disassemble_data
arguments: tag: D15
datatype: string
name: disassemble_data
arguments: tag: D24
datatype: string
name: disassemble_data
arguments: tag: D48
datatype: string
name: disassemble_data
arguments: tag: D13
datatype: string
name: disassemble_data
arguments: tag: D57
datatype: string

tool: D16: sending ATZ command...


tool: D15: ATZ

tool: D24: sending caller ID command...


tool: D48: AT+VCID=1

tool: D13: sending FAX mode command...


tool: D57: AT+FCLASS=2

assistant: ```c
int init_modem(int fd) {
  printf("sending ATZ command...\n");
  if (send_modem_command(fd, "ATZ\r") != 0) {
    return (-1);
  }
  sleep(1);
  printf("sending caller ID command...\n");
  if (send_modem_command(fd, "AT+VCID=1\r") != 0) {
    return (-1);
  }
  printf("sending FAX mode command...\n");
  send_modem_command(fd, "AT+FCLASS=2\r");
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_cp_buff>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <ft_strnew@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  test   %rax,%rax
  jne    L50
  mov    $0xffffffff,%eax
  jmp    L48
L50:
  mov    -0x10(%rbp),%rax
  mov    -0x1c(%rbp),%edx
  mov    -0x30(%rbp),%rsi
  mov    %edx,%ecx
  mov    $0xa,%edx
  mov    %rax,%rdi
  call   <ft_memccpy@plt>
  mov    -0x10(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <ft_strchr@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  je     L40
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
L40:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ft_strjoin@plt>
  mov    -0x28(%rbp),%rdx
  mov    %rax,(%rdx)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strdel@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strdel@plt>
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L30
  mov    $0xffffffff,%eax
  jmp    L48
L30:
  mov    $0x1,%eax
L48:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int ft_cp_buff(char **line, char *buff) {
  char *c;
  char *tmp;
  int len;
  len = ft_strlen(buff);
  tmp = ft_strnew(len);
  if (!tmp)
    return (-1);
  ft_memccpy(tmp, buff, '\n', len);
  c = ft_strchr(tmp, '\n');
  if (c)
    *c = '\0';
  c = *line;
  *line = ft_strjoin(*line, tmp);
  ft_strdel(&c);
  ft_strdel(&tmp);
  if (!*line)
    return (-1);
  return (1);
}
```
user: What is the c source code of the assembly code below?

```asm
<ft_cp_buff>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsi,%rdi
  call   <ft_strlen@plt>
  mov    %eax,%r12d
  mov    %eax,%edi
  call   <ft_strnew@plt>
  mov    %rax,%rdi
  mov    %rax,0x10(%rsp)
  mov    $0xffffffff,%eax
  test   %rdi,%rdi
  je     L0
  mov    %r12d,%ecx
  mov    $0xa,%edx
  mov    %rbp,%rsi
  call   <ft_memccpy@plt>
  mov    $0xa,%esi
  mov    0x10(%rsp),%rdi
  call   <ft_strchr@plt>
  test   %rax,%rax
  je     L5
  movb   $0x0,(%rax)
L5:
  mov    (%rbx),%rdi
  mov    %rdi,0x8(%rsp)
  mov    0x10(%rsp),%rsi
  call   <ft_strjoin@plt>
  mov    %rax,(%rbx)
  lea    0x8(%rsp),%rdi
  call   <ft_strdel@plt>
  lea    0x10(%rsp),%rdi
  call   <ft_strdel@plt>
  cmpq   $0x1,(%rbx)
  sbb    %eax,%eax
  or     $0x1,%eax
L0:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_cp_buff(char **line, char *buff) {
  char *c;
  char *tmp;
  int len;
  len = ft_strlen(buff);
  tmp = ft_strnew(len);
  if (!tmp)
    return (-1);
  ft_memccpy(tmp, buff, '\n', len);
  c = ft_strchr(tmp, '\n');
  if (c)
    *c = '\0';
  c = *line;
  *line = ft_strjoin(*line, tmp);
  ft_strdel(&c);
  ft_strdel(&tmp);
  if (!*line)
    return (-1);
  return (1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_cp_buff>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <ft_strlen@plt>
  mov    %eax,%edi
  mov    %eax,%r12d
  call   <ft_strnew@plt>
  mov    %rax,0x10(%rsp)
  test   %rax,%rax
  je     L33
  mov    %rax,%rdi
  mov    %rbp,%rsi
  mov    %r12d,%ecx
  mov    $0xa,%edx
  call   <ft_memccpy@plt>
  mov    0x10(%rsp),%rdi
  mov    $0xa,%esi
  call   <ft_strchr@plt>
  test   %rax,%rax
  je     L22
  movb   $0x0,(%rax)
L22:
  mov    (%rbx),%rdi
  mov    0x10(%rsp),%rsi
  mov    %rdi,0x8(%rsp)
  call   <ft_strjoin@plt>
  lea    0x8(%rsp),%rdi
  mov    %rax,(%rbx)
  call   <ft_strdel@plt>
  lea    0x10(%rsp),%rdi
  call   <ft_strdel@plt>
  cmpq   $0x0,(%rbx)
  mov    $0x1,%eax
  je     L33
L14:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L33:
  mov    $0xffffffff,%eax
  jmp    L14
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_cp_buff(char **line, char *buff) {
  char *c;
  char *tmp;
  int len;
  len = ft_strlen(buff);
  tmp = ft_strnew(len);
  if (!tmp)
    return (-1);
  ft_memccpy(tmp, buff, '\n', len);
  c = ft_strchr(tmp, '\n');
  if (c)
    *c = '\0';
  c = *line;
  *line = ft_strjoin(*line, tmp);
  ft_strdel(&c);
  ft_strdel(&tmp);
  if (!*line)
    return (-1);
  return (1);
}
```