ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<aes_decrypt>:
  endbr64
  test   %rdx,%rdx
  je     L7
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  nopl   (%rax)
L14:
  mov    $0xffff,%ebx
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  cmp    %rbx,%rbp
  cmovbe %rbp,%rbx
  mov    %rbx,%rcx
  call   <_decrypt@plt>
  mov    %rbx,%rax
  shl    $0x4,%rax
  add    %rax,%r14
  add    %rax,%r12
  sub    %rbx,%rbp
  jne    L14
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  ret
```

assistant: ```c
void aes_decrypt(void *inbuf, void *outbuf, size_t size, uint32_t mode) {
  uint32_t in = (uint32_t)inbuf;
  uint32_t out = (uint32_t)outbuf;
  size_t block_count = size;
  size_t blocks;
  while (block_count != 0) {
    blocks = (block_count >= 0xFFFF) ? 0xFFFF : block_count;
    _decrypt(mode, (void *)in, (void *)out, blocks);
    in += blocks * 0x10;
    out += blocks * 0x10;
    block_count -= blocks;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<aes_decrypt>:
  endbr64
  test   %rdx,%rdx
  je     L2
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  nopl   (%rax)
L16:
  mov    $0xffff,%ebx
  mov    %r12,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  cmp    %rbx,%rbp
  cmovbe %rbp,%rbx
  mov    %rbx,%rcx
  call   <_decrypt@plt>
  mov    %rbx,%rax
  shl    $0x4,%rax
  add    %rax,%r14
  add    %rax,%r12
  sub    %rbx,%rbp
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  ret
```

assistant: ```c
void aes_decrypt(void *inbuf, void *outbuf, size_t size, uint32_t mode) {
  uint32_t in = (uint32_t)inbuf;
  uint32_t out = (uint32_t)outbuf;
  size_t block_count = size;
  size_t blocks;
  while (block_count != 0) {
    blocks = (block_count >= 0xFFFF) ? 0xFFFF : block_count;
    _decrypt(mode, (void *)in, (void *)out, blocks);
    in += blocks * 0x10;
    out += blocks * 0x10;
    block_count -= blocks;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<email_sendmail_command>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x830,%rsp
  mov    %rdi,-0x818(%rbp)
  mov    %rsi,-0x820(%rbp)
  mov    %rdx,-0x828(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x828(%rbp)
  je     L41
  mov    -0x828(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L41
  mov    -0x828(%rbp),%rax
  mov    %rax,%rdi
  call   <timlib_file_exists@plt>
  test   %rax,%rax
  je     L41
  mov    -0x818(%rbp),%rsi
  mov    -0x828(%rbp),%rcx
  mov    -0x820(%rbp),%rdx
  lea    -0x810(%rbp),%rax
  mov    %rsi,%r8
  lea    D5(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L4
L41:
  mov    -0x818(%rbp),%rcx
  mov    -0x820(%rbp),%rdx
  lea    -0x810(%rbp),%rax
  lea    D15(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
L4:
  lea    -0x810(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D5
type: string
name: read_data
arguments: tag: D15
type: string

tool: D5: mutt -s "%s" -a %s -- %s

tool: D15: mutt -s "%s" %s

assistant: ```c
char *email_sendmail_command(char *to_address, char *subject, char *attachment_filename) {
  char sendmail_command[2048];
  if (attachment_filename && *attachment_filename && timlib_file_exists(attachment_filename)) {
    sprintf(sendmail_command, "mutt -s \"%s\" -a %s -- %s", subject, attachment_filename, to_address);
  } else {
    sprintf(sendmail_command, "mutt -s \"%s\" %s", subject, to_address);
  }
  return strdup(sendmail_command);
}
```
user: What is the c source code of the assembly code below:

```asm
<email_sendmail_command>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x810,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L29
  mov    %rdx,%rbx
  cmpb   $0x0,(%rdx)
  jne    L26
L29:
  mov    %rsp,%rdi
  mov    %r12,%rcx
  mov    %rbp,%rdx
  lea    D3(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
L10:
  mov    %rsp,%rdi
  call   <strdup@plt>
  mov    0x808(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L47
  add    $0x810,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  mov    %rdx,%rdi
  call   <timlib_file_exists@plt>
  test   %rax,%rax
  je     L29
  mov    %rsp,%rdi
  mov    %r12,%r8
  mov    %rbx,%rcx
  mov    %rbp,%rdx
  lea    D18(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L10
L47:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: string
name: unpack_data
arguments: data_tag: D18
data_type: string

tool: D3: mutt -s "%s" %s

tool: D18: mutt -s "%s" -a %s -- %s

assistant: ```c
char *email_sendmail_command(char *to_address, char *subject, char *attachment_filename) {
  char sendmail_command[2048];
  if (attachment_filename && *attachment_filename && timlib_file_exists(attachment_filename)) {
    sprintf(sendmail_command, "mutt -s \"%s\" -a %s -- %s", subject, attachment_filename, to_address);
  } else {
    sprintf(sendmail_command, "mutt -s \"%s\" %s", subject, to_address);
  }
  return strdup(sendmail_command);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<email_sendmail_command>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L29
  cmpb   $0x0,(%rdx)
  mov    %rdx,%r12
  jne    L16
L29:
  mov    %rsp,%rbp
  mov    %r14,%rcx
  mov    %r13,%rdx
  xor    %eax,%eax
  lea    D7(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
L22:
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    0x808(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x818,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L16:
  mov    %rdx,%rdi
  call   <timlib_file_exists@plt>
  test   %rax,%rax
  je     L29
  mov    %rsp,%rbp
  mov    %r14,%r8
  mov    %r12,%rcx
  mov    %r13,%rdx
  lea    D19(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <sprintf@plt>
  jmp    L22
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type_guessed: string
name: retrieve_data
arguments: data_label: D19
type_guessed: string

tool: D7: mutt -s "%s" %s

tool: D19: mutt -s "%s" -a %s -- %s

assistant: ```c
char *email_sendmail_command(char *to_address, char *subject, char *attachment_filename) {
  char sendmail_command[2048];
  if (attachment_filename && *attachment_filename && timlib_file_exists(attachment_filename)) {
    sprintf(sendmail_command, "mutt -s \"%s\" -a %s -- %s", subject, attachment_filename, to_address);
  } else {
    sprintf(sendmail_command, "mutt -s \"%s\" %s", subject, to_address);
  }
  return strdup(sendmail_command);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<email_sendmail_command>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L36
  cmpb   $0x0,(%rdx)
  mov    %rdx,%r12
  jne    L35
L36:
  mov    %rsp,%rbp
  mov    %r14,%rcx
  mov    %r13,%rdx
  xor    %eax,%eax
  lea    D6(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
L4:
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    0x808(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L31
  add    $0x818,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L35:
  mov    %rdx,%rdi
  call   <timlib_file_exists@plt>
  test   %rax,%rax
  je     L36
  mov    %rsp,%rbp
  mov    %r14,%r8
  mov    %r12,%rcx
  mov    %r13,%rdx
  lea    D9(%rip),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <sprintf@plt>
  jmp    L4
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D6
data_type: string
name: read_data
arguments: data_label: D9
data_type: string

tool: D6: mutt -s "%s" %s

tool: D9: mutt -s "%s" -a %s -- %s

assistant: ```c
char *email_sendmail_command(char *to_address, char *subject, char *attachment_filename) {
  char sendmail_command[2048];
  if (attachment_filename && *attachment_filename && timlib_file_exists(attachment_filename)) {
    sprintf(sendmail_command, "mutt -s \"%s\" -a %s -- %s", subject, attachment_filename, to_address);
  } else {
    sprintf(sendmail_command, "mutt -s \"%s\" %s", subject, to_address);
  }
  return strdup(sendmail_command);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <prime@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x1,-0xc(%rbp)
  jne    L7
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L29
L7:
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L29:
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L35
L58:
  addl   $0x1,-0x10(%rbp)
L35:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L58
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L44
L16:
  subl   $0x1,-0x10(%rbp)
L44:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  je     L26
  cmpl   $0x0,-0x10(%rbp)
  jg     L16
L26:
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L55
  call   <__stack_chk_fail@plt>
L55:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D27
data_type: string
name: unpack_data
arguments: data_tag: D18
data_type: string
name: unpack_data
arguments: data_tag: D56
data_type: string
name: unpack_data
arguments: data_tag: D44
data_type: string
name: unpack_data
arguments: data_tag: D51
data_type: string
name: unpack_data
arguments: data_tag: D37
data_type: string

tool: D27: PRIMES: ENTER A NUMBER: 

tool: D18: %d

tool: D56: PRIME!

tool: D44: NOT A PRIME!

tool: D51: Next Prime: %d


tool: D37: Prev Prime: %d


assistant: ```c
int main() {
  printf("PRIMES: ENTER A NUMBER: ");
  int c;
  scanf("%d", &c);
  int v = prime(c);
  if (v == 1)
    printf("PRIME!\n");
  else
    printf("NOT A PRIME!\n");
  int y = c + 1;
  while (prime(y) != 1) {
    y += 1;
  }
  printf("Next Prime: %d\n", y);
  y = c - 1;
  while (prime(y) != 1 && y > 0) {
    y -= 1;
  }
  printf("Prev Prime: %d\n", y);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D37(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  je     L11
  lea    D25(%rip),%rdi
  call   <puts@plt>
L68:
  mov    0x4(%rsp),%eax
  lea    0x1(%rax),%ebx
  jmp    L44
L11:
  lea    D21(%rip),%rdi
  call   <puts@plt>
  jmp    L68
L19:
  add    $0x1,%ebx
L44:
  mov    %ebx,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L19
  mov    %ebx,%esi
  lea    D49(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%ebx
L9:
  mov    %ebx,%edi
  call   <prime@plt>
  test   %ebx,%ebx
  jle    L55
  cmp    $0x1,%eax
  je     L55
  sub    $0x1,%ebx
  jmp    L9
L55:
  mov    %ebx,%esi
  lea    D30(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D42
data_type: string
name: disassemble_data
arguments: label: D37
data_type: string
name: disassemble_data
arguments: label: D25
data_type: string
name: disassemble_data
arguments: label: D21
data_type: string
name: disassemble_data
arguments: label: D49
data_type: string
name: disassemble_data
arguments: label: D30
data_type: string

tool: D42: PRIMES: ENTER A NUMBER: 

tool: D37: %d

tool: D25: NOT A PRIME!

tool: D21: PRIME!

tool: D49: Next Prime: %d


tool: D30: Prev Prime: %d


assistant: ```c
int main() {
  printf("PRIMES: ENTER A NUMBER: ");
  int c;
  scanf("%d", &c);
  int v = prime(c);
  if (v == 1)
    printf("PRIME!\n");
  else
    printf("NOT A PRIME!\n");
  int y = c + 1;
  while (prime(y) != 1) {
    y += 1;
  }
  printf("Next Prime: %d\n", y);
  y = c - 1;
  while (prime(y) != 1 && y > 0) {
    y -= 1;
  }
  printf("Prev Prime: %d\n", y);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D22(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D18(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  je     L6
  lea    D51(%rip),%rdi
  call   <puts@plt>
L55:
  mov    0x4(%rsp),%eax
  lea    0x1(%rax),%ebp
  jmp    L29
  nopl   (%rax)
L36:
  add    $0x1,%ebp
L29:
  mov    %ebp,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L36
  mov    %ebp,%esi
  lea    D54(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%ebp
  jmp    L16
  nopl   0x0(%rax,%rax,1)
L31:
  cmp    $0x1,%eax
  je     L44
  sub    $0x1,%ebp
L16:
  mov    %ebp,%edi
  call   <prime@plt>
  test   %ebp,%ebp
  jg     L31
L44:
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L6:
  lea    D25(%rip),%rdi
  call   <puts@plt>
  jmp    L55
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D22
type_infered: string
name: parse_data
arguments: tag: D18
type_infered: string
name: parse_data
arguments: tag: D51
type_infered: string
name: parse_data
arguments: tag: D54
type_infered: string
name: parse_data
arguments: tag: D23
type_infered: string
name: parse_data
arguments: tag: D25
type_infered: string

tool: D22: PRIMES: ENTER A NUMBER: 

tool: D18: %d

tool: D51: NOT A PRIME!

tool: D54: Next Prime: %d


tool: D23: Prev Prime: %d


tool: D25: PRIME!

assistant: ```c
int main() {
  printf("PRIMES: ENTER A NUMBER: ");
  int c;
  scanf("%d", &c);
  int v = prime(c);
  if (v == 1)
    printf("PRIME!\n");
  else
    printf("NOT A PRIME!\n");
  int y = c + 1;
  while (prime(y) != 1) {
    y += 1;
  }
  printf("Next Prime: %d\n", y);
  y = c - 1;
  while (prime(y) != 1 && y > 0) {
    y -= 1;
  }
  printf("Prev Prime: %d\n", y);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D3(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D21(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  je     L58
  lea    D25(%rip),%rdi
  call   <puts@plt>
L71:
  mov    0x4(%rsp),%eax
  lea    0x1(%rax),%ebp
  jmp    L47
  nopl   (%rax)
L65:
  add    $0x1,%ebp
L47:
  mov    %ebp,%edi
  call   <prime@plt>
  cmp    $0x1,%eax
  jne    L65
  mov    %ebp,%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%ebp
  jmp    L68
  nopl   0x0(%rax,%rax,1)
L48:
  cmp    $0x1,%eax
  je     L74
  sub    $0x1,%ebp
L68:
  mov    %ebp,%edi
  call   <prime@plt>
  test   %ebp,%ebp
  jg     L48
L74:
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L78
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L58:
  lea    D52(%rip),%rdi
  call   <puts@plt>
  jmp    L71
L78:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D3
type: string
name: parse_data
arguments: data_tag: D21
type: string
name: parse_data
arguments: data_tag: D25
type: string
name: parse_data
arguments: data_tag: D6
type: string
name: parse_data
arguments: data_tag: D28
type: string
name: parse_data
arguments: data_tag: D52
type: string

tool: D3: PRIMES: ENTER A NUMBER: 

tool: D21: %d

tool: D25: NOT A PRIME!

tool: D6: Next Prime: %d


tool: D28: Prev Prime: %d


tool: D52: PRIME!

assistant: ```c
int main() {
  printf("PRIMES: ENTER A NUMBER: ");
  int c;
  scanf("%d", &c);
  int v = prime(c);
  if (v == 1)
    printf("PRIME!\n");
  else
    printf("NOT A PRIME!\n");
  int y = c + 1;
  while (prime(y) != 1) {
    y += 1;
  }
  printf("Next Prime: %d\n", y);
  y = c - 1;
  while (prime(y) != 1 && y > 0) {
    y -= 1;
  }
  printf("Prev Prime: %d\n", y);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<AlarmFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D0
data_type: string

tool: D0: No correct answer from the mote. 
Check the firware is XEE-compatible. Or try again.

assistant: ```c
void AlarmFunc(int signo) {
  printf("No correct answer from the mote. \n"
         "Check the firware is XEE-compatible. Or try again.");
  exit(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<AlarmFunc>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D4
type_infered: string

tool: D4: No correct answer from the mote. 
Check the firware is XEE-compatible. Or try again.

assistant: ```c
void AlarmFunc(int signo) {
  printf("No correct answer from the mote. \n"
         "Check the firware is XEE-compatible. Or try again.");
  exit(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<AlarmFunc>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type_guessed: string

tool: D0: No correct answer from the mote. 
Check the firware is XEE-compatible. Or try again.

assistant: ```c
void AlarmFunc(int signo) {
  printf("No correct answer from the mote. \n"
         "Check the firware is XEE-compatible. Or try again.");
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<AlarmFunc>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D1
type_guessed: string

tool: D1: No correct answer from the mote. 
Check the firware is XEE-compatible. Or try again.

assistant: ```c
void AlarmFunc(int signo) {
  printf("No correct answer from the mote. \n"
         "Check the firware is XEE-compatible. Or try again.");
  exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fastfib>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L8
L59:
  cmpl   $0x1,-0x24(%rbp)
  ja     L50
  mov    -0x20(%rbp),%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,(%rax)
  jmp    L62
L50:
  lea    -0x14(%rbp),%rax
  cmp    %rax,-0x20(%rbp)
  jne    L16
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L62
L16:
  lea    -0x14(%rbp),%rax
  add    $0x4,%rax
  cmp    %rax,-0x20(%rbp)
  jne    L92
  mov    -0x14(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L62
L92:
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
L62:
  addq   $0x4,-0x20(%rbp)
  lea    -0x14(%rbp),%rax
  add    $0x8,%rax
  cmp    %rax,-0x20(%rbp)
  jbe    L106
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L106:
  addl   $0x1,-0x24(%rbp)
L8:
  mov    -0x24(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jbe    L59
  lea    -0x14(%rbp),%rax
  cmp    %rax,-0x20(%rbp)
  jne    L86
  mov    -0x20(%rbp),%rax
  mov    0x8(%rax),%eax
  jmp    L102
L86:
  mov    -0x20(%rbp),%rax
  mov    -0x4(%rax),%eax
L102:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L97
  call   <__stack_chk_fail@plt>
L97:
  leave
  ret
```

assistant: ```c
unsigned int fastfib(unsigned int n) {
  unsigned int a[3];
  unsigned int *p = a;
  unsigned int i;
  for (i = 0; i <= n; ++i) {
    if (i < 2)
      *p = i;
    else {
      if (p == a)
        *p = *(a + 1) + *(a + 2);
      else if (p == a + 1)
        *p = *a + *(a + 2);
      else
        *p = *a + *(a + 1);
    }
    if (++p > a + 2)
      p = a;
  }
  return p == a ? *(p + 2) : *(p - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fastfib>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x0,%edx
  lea    0xc(%rsp),%rax
  lea    0x14(%rsp),%r8
  mov    %rax,%rsi
  lea    0x10(%rsp),%r9
  jmp    L106
L82:
  mov    0x14(%rsp),%ecx
  add    0x10(%rsp),%ecx
L76:
  mov    %ecx,(%rax)
  add    $0x4,%rax
  jmp    L36
L37:
  mov    0x14(%rsp),%ecx
  add    0xc(%rsp),%ecx
  jmp    L76
L56:
  mov    %edx,%ecx
L27:
  mov    %ecx,(%rax)
  add    $0x4,%rax
  cmp    %r8,%rax
  cmova  %rsi,%rax
L36:
  add    $0x1,%edx
  cmp    %edx,%edi
  jb     L41
L106:
  cmp    $0x1,%edx
  jbe    L56
  cmp    %rsi,%rax
  je     L82
  cmp    %r9,%rax
  je     L37
  mov    0x10(%rsp),%ecx
  add    0xc(%rsp),%ecx
  jmp    L27
L41:
  lea    0xc(%rsp),%rdx
  cmp    %rdx,%rax
  je     L51
  mov    -0x4(%rax),%eax
L26:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L89
  add    $0x28,%rsp
  ret
L51:
  mov    0x14(%rsp),%eax
  jmp    L26
L89:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int fastfib(unsigned int n) {
  unsigned int a[3];
  unsigned int *p = a;
  unsigned int i;
  for (i = 0; i <= n; ++i) {
    if (i < 2)
      *p = i;
    else {
      if (p == a)
        *p = *(a + 1) + *(a + 2);
      else if (p == a + 1)
        *p = *a + *(a + 2);
      else
        *p = *a + *(a + 1);
    }
    if (++p > a + 2)
      p = a;
  }
  return p == a ? *(p + 2) : *(p - 1);
}
```