ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<gen_sha256_nonce>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  lea    0x8(%rsi),%rdi
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %r8,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  test   %rax,%rax
  je     L2
  mov    %rax,%rbp
  mov    %rbx,%rdx
  mov    %r15,%rsi
  mov    %rax,%rdi
  xor    %eax,%eax
  call   <memcpy@plt>
  mov    %r14d,%edi
  lea    0x4(%rsp),%r14
  call   <htonl@plt>
  movslq %ebx,%rdi
  mov    $0x4,%edx
  mov    %r14,%rsi
  add    %rbp,%rdi
  mov    %eax,0x4(%rsp)
  xor    %eax,%eax
  call   <memcpy@plt>
  xor    %eax,%eax
  call   <get_random@plt>
  mov    %eax,%edi
  call   <htonl@plt>
  lea    0x4(%rbx),%edi
  mov    $0x4,%edx
  mov    %r14,%rsi
  movslq %edi,%rdi
  mov    %eax,0x4(%rsp)
  xor    %eax,%eax
  add    %rbp,%rdi
  call   <memcpy@plt>
  lea    0x8(%rbx),%esi
  mov    %r13,%rdx
  mov    %rbp,%rdi
  call   <SHA256@plt>
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,(%r12)
  xor    %eax,%eax
L23:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  or     $0xffffffff,%eax
  jmp    L23
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int gen_sha256_nonce(const unsigned char *mac, size_t mac_len, int boardId, unsigned char *nonce, int *nonce_len) {
  unsigned char *d = NULL;
  int value, len;
  if (!(d = malloc(mac_len + sizeof(int) * 2))) {
    return -1;
  }
  len = 0;
  memcpy(d, mac, mac_len);
  len += mac_len;
  value = htonl(boardId);
  memcpy(d + len, &value, sizeof(int));
  len += sizeof(int);
  value = htonl(get_random());
  memcpy(d + len, &value, sizeof(int));
  len += sizeof(int);
  SHA256(d, len, nonce);
  *nonce_len = SHA256_DIGEST_LENGTH;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<test_int_vla_location>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x7,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  mov    %eax,%edx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  mov    %eax,%edx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L39:
  cmp    %rdx,%rsp
  je     L12
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L39
L12:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L34
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L34:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movl   $0x1a,(%rax)
  mov    -0x20(%rbp),%rax
  mov    -0x2c(%rbp),%edx
  movl   $0x1,(%rax,%rdx,4)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <sink@plt>
  mov    %rbx,%rsp
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void test_int_vla_location(void) {
  unsigned nelts = 7;
  int vla[nelts];
  vla[0] = 26;
  vla[nelts] = 1;
  sink(vla);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test_int_vla_location>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x1a,(%rsp)
  movl   $0x1,0x1c(%rsp)
  mov    %rsp,%rdi
  call   <sink@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x38,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_int_vla_location(void) {
  unsigned nelts = 7;
  int vla[nelts];
  vla[0] = 26;
  vla[nelts] = 1;
  sink(vla);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test_int_vla_location>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movl   $0x1a,(%rsp)
  movl   $0x1,0x1c(%rsp)
  call   <sink@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x38,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_int_vla_location(void) {
  unsigned nelts = 7;
  int vla[nelts];
  vla[0] = 26;
  vla[nelts] = 1;
  sink(vla);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test_int_vla_location>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movl   $0x1a,(%rsp)
  movl   $0x1,0x1c(%rsp)
  call   <sink@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x38,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_int_vla_location(void) {
  unsigned nelts = 7;
  int vla[nelts];
  vla[0] = 26;
  vla[nelts] = 1;
  sink(vla);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<game_over>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    $0x0,%eax
  call   <endwin@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
int game_over() {
  sleep(1);
  endwin();
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<game_over>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <sleep@plt>
  mov    $0x0,%eax
  call   <endwin@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int game_over() {
  sleep(1);
  endwin();
  exit(0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<game_over>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <sleep@plt>
  xor    %eax,%eax
  call   <endwin@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1080 <exit@plt>
```

assistant: ```c
int game_over() {
  sleep(1);
  endwin();
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<game_over>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <sleep@plt>
  xor    %eax,%eax
  call   <endwin@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1080 <exit@plt>
```

assistant: ```c
int game_over() {
  sleep(1);
  endwin();
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ms_fread>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <fread@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L28
  cmpl   $0x0,-0x1c(%rbp)
  je     L28
  cmpl   $0x0,-0x20(%rbp)
  je     L28
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ferror@plt>
  test   %rax,%rax
  je     L22
  lea    D6(%rip),%rax
  mov    %rax,%rsi
  mov    $0x2,%edi
  call   <ms_log@plt>
  jmp    L28
L22:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L28
  lea    D13(%rip),%rax
  mov    %rax,%rsi
  mov    $0x2,%edi
  call   <ms_log@plt>
L28:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
datatype: string
name: extract_symbolic_data
arguments: data_tag: D13
datatype: string

tool: D6: ms_fread(): Cannot read input file


tool: D13: ms_fread(): Unknown return from fread()


assistant: ```c
int ms_fread(char *buf, int size, int num, FILE *stream) {
  int read = 0;
  read = (int)fread(buf, size, num, stream);
  if (read <= 0 && size && num) {
    if (ferror(stream))
      ms_log(2, "ms_fread(): Cannot read input file\n");
    else if (!feof(stream))
      ms_log(2, "ms_fread(): Unknown return from fread()\n");
  }
  return read;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ms_fread>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebp
  mov    %edx,%ebx
  mov    %rcx,%r12
  call   <fread@plt>
  mov    %eax,%r13d
  test   %ebp,%ebp
  setne  %cl
  test   %ebx,%ebx
  setne  %dl
  test   %dl,%cl
  je     L26
  test   %eax,%eax
  jle    L44
L26:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L44:
  mov    %r12,%rdi
  call   <ferror@plt>
  test   %rax,%rax
  je     L39
  lea    D7(%rip),%rsi
  mov    $0x2,%edi
  call   <ms_log@plt>
  jmp    L26
L39:
  mov    %r12,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L26
  lea    D17(%rip),%rsi
  mov    $0x2,%edi
  call   <ms_log@plt>
  jmp    L26
```

tool_calls: name: read_data
arguments: tag: D7
type_infered: string
name: read_data
arguments: tag: D17
type_infered: string

tool: D7: ms_fread(): Cannot read input file


tool: D17: ms_fread(): Unknown return from fread()


assistant: ```c
int ms_fread(char *buf, int size, int num, FILE *stream) {
  int read = 0;
  read = (int)fread(buf, size, num, stream);
  if (read <= 0 && size && num) {
    if (ferror(stream))
      ms_log(2, "ms_fread(): Cannot read input file\n");
    else if (!feof(stream))
      ms_log(2, "ms_fread(): Unknown return from fread()\n");
  }
  return read;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ms_fread>:
  endbr64
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <fread@plt>
  test   %ebp,%ebp
  setne  %cl
  test   %ebx,%ebx
  mov    %eax,%r13d
  setne  %dl
  test   %dl,%cl
  je     L1
  test   %eax,%eax
  jle    L41
L1:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L41:
  mov    %r12,%rdi
  call   <ferror@plt>
  test   %rax,%rax
  je     L54
  lea    D6(%rip),%rsi
  mov    $0x2,%edi
  call   <ms_log@plt>
  jmp    L1
L54:
  mov    %r12,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L1
  lea    D0(%rip),%rsi
  mov    $0x2,%edi
  call   <ms_log@plt>
  jmp    L1
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type_infered: string
name: extract_symbolic_data
arguments: data_label: D0
type_infered: string

tool: D6: ms_fread(): Cannot read input file


tool: D0: ms_fread(): Unknown return from fread()


assistant: ```c
int ms_fread(char *buf, int size, int num, FILE *stream) {
  int read = 0;
  read = (int)fread(buf, size, num, stream);
  if (read <= 0 && size && num) {
    if (ferror(stream))
      ms_log(2, "ms_fread(): Cannot read input file\n");
    else if (!feof(stream))
      ms_log(2, "ms_fread(): Unknown return from fread()\n");
  }
  return read;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ms_fread>:
  endbr64
  push   %r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <fread@plt>
  test   %ebp,%ebp
  setne  %cl
  test   %ebx,%ebx
  mov    %eax,%r13d
  setne  %dl
  test   %dl,%cl
  je     L57
  test   %eax,%eax
  jle    L36
L57:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L36:
  mov    %r12,%rdi
  call   <ferror@plt>
  test   %rax,%rax
  je     L13
  lea    D2(%rip),%rsi
  mov    $0x2,%edi
  call   <ms_log@plt>
  jmp    L57
L13:
  mov    %r12,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L57
  lea    D15(%rip),%rsi
  mov    $0x2,%edi
  call   <ms_log@plt>
  jmp    L57
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_guessed: string
name: disassemble_data
arguments: tag: D15
type_guessed: string

tool: D2: ms_fread(): Cannot read input file


tool: D15: ms_fread(): Unknown return from fread()


assistant: ```c
int ms_fread(char *buf, int size, int num, FILE *stream) {
  int read = 0;
  read = (int)fread(buf, size, num, stream);
  if (read <= 0 && size && num) {
    if (ferror(stream))
      ms_log(2, "ms_fread(): Cannot read input file\n");
    else if (!feof(stream))
      ms_log(2, "ms_fread(): Unknown return from fread()\n");
  }
  return read;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bijective_hexavigesimal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L47
L9:
  addl   $0x1,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  sub    %eax,-0x14(%rbp)
  mov    -0x8(%rbp),%eax
  imul   $0x1a,%eax,%eax
  mov    %eax,-0x8(%rbp)
L47:
  mov    -0x14(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jae    L9
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L46:
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  imul   $0x4ec4ec4f,%rax,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  imul   $0x1a,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  lea    0x61(%rax),%edx
  mov    -0xc(%rbp),%eax
  sub    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  lea    D8(%rip),%rax
  mov    %cl,(%rdx,%rax,1)
  mov    -0x14(%rbp),%eax
  mov    %eax,%eax
  imul   $0x4ec4ec4f,%rax,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jb     L46
  mov    -0xc(%rbp),%eax
  lea    D8(%rip),%rdx
  movb   $0x0,(%rax,%rdx,1)
  movzbl D11(%rip),%eax
  test   %al,%al
  je     L36
  movzbl D8(%rip),%eax
  sub    $0x20,%eax
  mov    %al,D52(%rip)
L36:
  lea    D52(%rip),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *bijective_hexavigesimal(unsigned int a) {
  unsigned int c = 0, x = 1, i;
  static char buf[sizeof(int) * 171 / 100 + 3];
  while (a >= x) {
    c++;
    a -= x;
    x *= 26;
  }
  for (i = 0; i < c; i++) {
    buf[c - i - 1] = 'a' + (a % 26);
    a = a / 26;
  }
  buf[c] = 0;
  if (buf[1])
    buf[0] = buf[0] + ('A' - 'a');
  return buf;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bijective_hexavigesimal>:
  endbr64
  test   %edi,%edi
  je     L31
  mov    $0x1,%eax
  mov    $0x0,%ecx
L27:
  mov    %ecx,%edx
  add    $0x1,%ecx
  sub    %eax,%edi
  imul   $0x1a,%eax,%eax
  cmp    %eax,%edi
  jae    L27
  test   %ecx,%ecx
  je     L39
  lea    D46(%rip),%r9
L1:
  mov    %edx,%esi
  mov    %edi,%eax
  imul   $0x4ec4ec4f,%rax,%rax
  shr    $0x23,%rax
  imul   $0x1a,%eax,%r8d
  sub    %r8d,%edi
  add    $0x61,%edi
  mov    %dil,(%r9,%rsi,1)
  mov    %eax,%edi
  sub    $0x1,%edx
  cmp    $0xffffffff,%edx
  jne    L1
L39:
  mov    %ecx,%ecx
  lea    D46(%rip),%rax
  movb   $0x0,(%rax,%rcx,1)
  cmpb   $0x0,D30(%rip)
  je     L46
  subb   $0x20,D46(%rip)
L46:
  lea    D46(%rip),%rax
  ret
L31:
  mov    %edi,%ecx
  jmp    L39
```

assistant: ```c
char *bijective_hexavigesimal(unsigned int a) {
  unsigned int c = 0, x = 1, i;
  static char buf[sizeof(int) * 171 / 100 + 3];
  while (a >= x) {
    c++;
    a -= x;
    x *= 26;
  }
  for (i = 0; i < c; i++) {
    buf[c - i - 1] = 'a' + (a % 26);
    a = a / 26;
  }
  buf[c] = 0;
  if (buf[1])
    buf[0] = buf[0] + ('A' - 'a');
  return buf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<bijective_hexavigesimal>:
  endbr64
  test   %edi,%edi
  je     L30
  mov    $0x1,%eax
  xor    %ecx,%ecx
  nop
L43:
  sub    %eax,%edi
  imul   $0x1a,%eax,%eax
  mov    %ecx,%edx
  add    $0x1,%ecx
  cmp    %eax,%edi
  jae    L43
  lea    D32(%rip),%r9
  test   %ecx,%ecx
  je     L34
  nopl   0x0(%rax)
L38:
  mov    %edi,%eax
  mov    %edx,%esi
  imul   $0x4ec4ec4f,%rax,%rax
  shr    $0x23,%rax
  imul   $0x1a,%eax,%r8d
  sub    %r8d,%edi
  add    $0x61,%edi
  mov    %dil,(%r9,%rsi,1)
  mov    %eax,%edi
  sub    $0x1,%edx
  jae    L38
L34:
  movb   $0x0,(%r9,%rcx,1)
  cmpb   $0x0,D1(%rip)
  je     L45
  subb   $0x20,D32(%rip)
L45:
  mov    %r9,%rax
  ret
  nopl   (%rax)
L30:
  xor    %ecx,%ecx
  lea    D32(%rip),%r9
  jmp    L34
```

assistant: ```c
char *bijective_hexavigesimal(unsigned int a) {
  unsigned int c = 0, x = 1, i;
  static char buf[sizeof(int) * 171 / 100 + 3];
  while (a >= x) {
    c++;
    a -= x;
    x *= 26;
  }
  for (i = 0; i < c; i++) {
    buf[c - i - 1] = 'a' + (a % 26);
    a = a / 26;
  }
  buf[c] = 0;
  if (buf[1])
    buf[0] = buf[0] + ('A' - 'a');
  return buf;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bijective_hexavigesimal>:
  endbr64
  test   %edi,%edi
  je     L65
  mov    $0x1,%edx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L113:
  sub    %edx,%edi
  imul   $0x1a,%edx,%edx
  mov    %eax,%ecx
  add    $0x1,%eax
  cmp    %edx,%edi
  jae    L113
  lea    D30(%rip),%r8
  test   %eax,%eax
  je     L121
  mov    %edi,%edx
  mov    %ecx,%esi
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x23,%rdx
  imul   $0x1a,%edx,%r9d
  sub    %r9d,%edi
  add    $0x61,%edi
  mov    %dil,(%r8,%rsi,1)
  cmp    $0x1,%eax
  je     L121
  mov    %edx,%esi
  lea    -0x1(%rcx),%edi
  imul   $0x4ec4ec4f,%rsi,%rsi
  shr    $0x23,%rsi
  imul   $0x1a,%esi,%r9d
  sub    %r9d,%edx
  add    $0x61,%edx
  mov    %dl,(%r8,%rdi,1)
  cmp    $0x2,%eax
  je     L121
  mov    %esi,%edx
  lea    -0x2(%rcx),%edi
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x23,%rdx
  imul   $0x1a,%edx,%r9d
  sub    %r9d,%esi
  add    $0x61,%esi
  mov    %sil,(%r8,%rdi,1)
  cmp    $0x3,%eax
  je     L121
  mov    %edx,%esi
  lea    -0x3(%rcx),%edi
  imul   $0x4ec4ec4f,%rsi,%rsi
  shr    $0x23,%rsi
  imul   $0x1a,%esi,%r9d
  sub    %r9d,%edx
  add    $0x61,%edx
  mov    %dl,(%r8,%rdi,1)
  cmp    $0x4,%eax
  je     L121
  mov    %esi,%edx
  lea    -0x4(%rcx),%edi
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x23,%rdx
  imul   $0x1a,%edx,%r9d
  sub    %r9d,%esi
  add    $0x61,%esi
  mov    %sil,(%r8,%rdi,1)
  cmp    $0x5,%eax
  je     L121
  mov    %edx,%esi
  lea    -0x5(%rcx),%edi
  imul   $0x4ec4ec4f,%rsi,%rsi
  shr    $0x23,%rsi
  imul   $0x1a,%esi,%r9d
  sub    %r9d,%edx
  add    $0x61,%edx
  mov    %dl,(%r8,%rdi,1)
  cmp    $0x6,%eax
  je     L121
  lea    -0x6(%rcx),%edx
  add    $0x61,%esi
  mov    %sil,(%r8,%rdx,1)
  cmp    $0x7,%eax
  je     L121
  lea    -0x7(%rcx),%edx
  movb   $0x61,(%r8,%rdx,1)
  cmp    $0x8,%eax
  je     L121
  lea    -0x8(%rcx),%edx
  movb   $0x61,(%r8,%rdx,1)
L121:
  movb   $0x0,(%r8,%rax,1)
  cmpb   $0x0,D14(%rip)
  je     L1
  subb   $0x20,D30(%rip)
L1:
  mov    %r8,%rax
  ret
  nop
L65:
  xor    %eax,%eax
  lea    D30(%rip),%r8
  jmp    L121
```

assistant: ```c
char *bijective_hexavigesimal(unsigned int a) {
  unsigned int c = 0, x = 1, i;
  static char buf[sizeof(int) * 171 / 100 + 3];
  while (a >= x) {
    c++;
    a -= x;
    x *= 26;
  }
  for (i = 0; i < c; i++) {
    buf[c - i - 1] = 'a' + (a % 26);
    a = a / 26;
  }
  buf[c] = 0;
  if (buf[1])
    buf[0] = buf[0] + ('A' - 'a');
  return buf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<heapsort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L22
L6:
  mov    -0xc(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <heapify@plt>
  subl   $0x1,-0xc(%rbp)
L22:
  cmpl   $0x0,-0xc(%rbp)
  jns    L6
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L8
L14:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <heapify@plt>
L8:
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  test   %eax,%eax
  jne    L14
  nop
  nop
  leave
  ret
```

assistant: ```c
void heapsort(int *values, int size) {
  for (int i = size / 2 - 1; i >= 0; i--) {
    heapify(values, size, i);
  }
  int len = size;
  while (len--) {
    int tmp = values[0];
    values[0] = values[len];
    values[len] = tmp;
    heapify(values, len, 0);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<heapsort>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %esi,%ebp
  shr    $0x1f,%ebp
  add    %esi,%ebp
  sar    %ebp
  sub    $0x1,%ebp
  js     L10
L8:
  mov    %ebp,%edx
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <heapify@plt>
  sub    $0x1,%ebp
  cmp    $0xffffffff,%ebp
  jne    L8
L10:
  lea    -0x1(%r12),%ebp
  test   %r12d,%r12d
  je     L39
  movslq %ebp,%rbp
  movslq %r12d,%rax
  sub    $0x2,%rax
  lea    -0x1(%r12),%edx
  sub    %rdx,%rax
  mov    %rax,%r12
L26:
  mov    (%rbx),%eax
  mov    (%rbx,%rbp,4),%edx
  mov    %edx,(%rbx)
  mov    %eax,(%rbx,%rbp,4)
  mov    $0x0,%edx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <heapify@plt>
  sub    $0x1,%rbp
  cmp    %r12,%rbp
  jne    L26
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void heapsort(int *values, int size) {
  for (int i = size / 2 - 1; i >= 0; i--) {
    heapify(values, size, i);
  }
  int len = size;
  while (len--) {
    int tmp = values[0];
    values[0] = values[len];
    values[len] = tmp;
    heapify(values, len, 0);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<heapsort>:
  endbr64
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %r12d,%ebp
  shr    $0x1f,%ebp
  push   %rbx
  mov    %rdi,%rbx
  add    %r12d,%ebp
  sar    %ebp
  sub    $0x1,%ebp
  js     L19
  xchg   %ax,%ax
L11:
  mov    %ebp,%edx
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <heapify@plt>
  sub    $0x1,%ebp
  jae    L11
L19:
  lea    -0x1(%r12),%eax
  test   %r12d,%r12d
  je     L13
  movslq %eax,%rbp
  sub    $0x2,%r12
  mov    %eax,%eax
  sub    %rax,%r12
  nopl   0x0(%rax,%rax,1)
L24:
  mov    (%rbx,%rbp,4),%edx
  mov    (%rbx),%eax
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    %edx,(%rbx)
  xor    %edx,%edx
  mov    %eax,(%rbx,%rbp,4)
  sub    $0x1,%rbp
  call   <heapify@plt>
  cmp    %r12,%rbp
  jne    L24
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void heapsort(int *values, int size) {
  for (int i = size / 2 - 1; i >= 0; i--) {
    heapify(values, size, i);
  }
  int len = size;
  while (len--) {
    int tmp = values[0];
    values[0] = values[len];
    values[len] = tmp;
    heapify(values, len, 0);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<heapsort>:
  endbr64
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %r12d,%ebp
  shr    $0x1f,%ebp
  push   %rbx
  mov    %rdi,%rbx
  add    %r12d,%ebp
  sar    %ebp
  sub    $0x1,%ebp
  js     L23
  xchg   %ax,%ax
L27:
  mov    %ebp,%edx
  mov    %r12d,%esi
  mov    %rbx,%rdi
  call   <heapify@plt>
  sub    $0x1,%ebp
  jae    L27
L23:
  lea    -0x1(%r12),%eax
  test   %r12d,%r12d
  je     L7
  movslq %eax,%rbp
  sub    $0x2,%r12
  mov    %eax,%eax
  sub    %rax,%r12
  nopl   0x0(%rax,%rax,1)
L32:
  mov    (%rbx,%rbp,4),%edx
  mov    (%rbx),%eax
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    %edx,(%rbx)
  xor    %edx,%edx
  mov    %eax,(%rbx,%rbp,4)
  sub    $0x1,%rbp
  call   <heapify@plt>
  cmp    %rbp,%r12
  jne    L32
L7:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void heapsort(int *values, int size) {
  for (int i = size / 2 - 1; i >= 0; i--) {
    heapify(values, size, i);
  }
  int len = size;
  while (len--) {
    int tmp = values[0];
    values[0] = values[len];
    values[len] = tmp;
    heapify(values, len, 0);
  }
}
```