ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<menu>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D19
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D36
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D52
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D39
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D50
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D57
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D5
type_guessed: string

tool: D2: 


tool: D19: 1 - digitar a e b


tool: D36: 2 - somar


tool: D52: 3 - subtrair


tool: D39: 4 - multiplicar


tool: D50: 5 - dividir


tool: D57: 6 - sair


tool: D5: digite sua opcao: 

assistant: ```c
void menu() {
  printf("\n");
  printf("1 - digitar a e b\n");
  printf("2 - somar\n");
  printf("3 - subtrair\n");
  printf("4 - multiplicar\n");
  printf("5 - dividir\n");
  printf("6 - sair\n");
  printf("digite sua opcao: ");
}
```
user: Translate the below assembly code into C programming language:

```asm
<menu>:
  endbr64
  sub    $0x8,%rsp
  lea    D32(%rip),%rdi
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    D52(%rip),%rdi
  call   <printf@plt>
  lea    D53(%rip),%rdi
  call   <printf@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D78(%rip),%rdi
  call   <printf@plt>
  lea    D36(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D32
data_type: string
name: unpack_data
arguments: data_label: D3
data_type: string
name: unpack_data
arguments: data_label: D52
data_type: string
name: unpack_data
arguments: data_label: D53
data_type: string
name: unpack_data
arguments: data_label: D30
data_type: string
name: unpack_data
arguments: data_label: D24
data_type: string
name: unpack_data
arguments: data_label: D78
data_type: string
name: unpack_data
arguments: data_label: D36
data_type: string

tool: D32: 


tool: D3: 1 - digitar a e b


tool: D52: 2 - somar


tool: D53: 3 - subtrair


tool: D30: 4 - multiplicar


tool: D24: 5 - dividir


tool: D78: 6 - sair


tool: D36: digite sua opcao: 

assistant: ```c
void menu() {
  printf("\n");
  printf("1 - digitar a e b\n");
  printf("2 - somar\n");
  printf("3 - subtrair\n");
  printf("4 - multiplicar\n");
  printf("5 - dividir\n");
  printf("6 - sair\n");
  printf("digite sua opcao: ");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<menu>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D60(%rip),%rdi
  call   <printf@plt>
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    D64(%rip),%rdi
  call   <printf@plt>
  lea    D76(%rip),%rdi
  call   <printf@plt>
  lea    D51(%rip),%rdi
  call   <printf@plt>
  lea    D68(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: label: D8
data_type: string
name: parse_data
arguments: label: D60
data_type: string
name: parse_data
arguments: label: D28
data_type: string
name: parse_data
arguments: label: D64
data_type: string
name: parse_data
arguments: label: D76
data_type: string
name: parse_data
arguments: label: D51
data_type: string
name: parse_data
arguments: label: D68
data_type: string
name: parse_data
arguments: label: D16
data_type: string

tool: D8: 


tool: D60: 1 - digitar a e b


tool: D28: 2 - somar


tool: D64: 3 - subtrair


tool: D76: 4 - multiplicar


tool: D51: 5 - dividir


tool: D68: 6 - sair


tool: D16: digite sua opcao: 

assistant: ```c
void menu() {
  printf("\n");
  printf("1 - digitar a e b\n");
  printf("2 - somar\n");
  printf("3 - subtrair\n");
  printf("4 - multiplicar\n");
  printf("5 - dividir\n");
  printf("6 - sair\n");
  printf("digite sua opcao: ");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<menu>:
  endbr64
  sub    $0x8,%rsp
  lea    D43(%rip),%rdi
  call   <printf@plt>
  lea    D52(%rip),%rdi
  call   <printf@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  lea    D77(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D69(%rip),%rdi
  call   <printf@plt>
  lea    D37(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D43
type: string
name: extract_symbolic_data
arguments: tag: D52
type: string
name: extract_symbolic_data
arguments: tag: D9
type: string
name: extract_symbolic_data
arguments: tag: D35
type: string
name: extract_symbolic_data
arguments: tag: D77
type: string
name: extract_symbolic_data
arguments: tag: D5
type: string
name: extract_symbolic_data
arguments: tag: D69
type: string
name: extract_symbolic_data
arguments: tag: D37
type: string

tool: D43: 


tool: D52: 1 - digitar a e b


tool: D9: 2 - somar


tool: D35: 3 - subtrair


tool: D77: 4 - multiplicar


tool: D5: 5 - dividir


tool: D69: 6 - sair


tool: D37: digite sua opcao: 

assistant: ```c
void menu() {
  printf("\n");
  printf("1 - digitar a e b\n");
  printf("2 - somar\n");
  printf("3 - subtrair\n");
  printf("4 - multiplicar\n");
  printf("5 - dividir\n");
  printf("6 - sair\n");
  printf("digite sua opcao: ");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x7f,%edx
  mov    %rax,%rsi
  mov    $0xffffffff,%edi
  call   <read@plt>
  mov    -0x8(%rbp),%rax
  mov    $0x7f,%edx
  mov    %rax,%rsi
  mov    $0xfffffffe,%edi
  call   <write@plt>
  mov    $0x0,%eax
  call   <newline@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv) {
  puts(argv);
  char *a = malloc(128);
  read(0xffffffff, a, 127);
  write(0xfffffffe, a, 127);
  newline();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    %rsi,%rdi
  call   <puts@plt>
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x7f,%edx
  mov    %rax,%rsi
  mov    $0xffffffff,%edi
  call   <read@plt>
  mov    $0x7f,%edx
  mov    %rbx,%rsi
  mov    $0xfffffffe,%edi
  call   <write@plt>
  mov    $0x0,%eax
  call   <newline@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv) {
  puts(argv);
  char *a = malloc(128);
  read(0xffffffff, a, 127);
  write(0xfffffffe, a, 127);
  newline();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsi,%rdi
  call   <puts@plt>
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    $0x7f,%edx
  mov    $0xffffffff,%edi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <read@plt>
  mov    %rbp,%rsi
  mov    $0x7f,%edx
  mov    $0xfffffffe,%edi
  call   <write@plt>
  xor    %eax,%eax
  call   <newline@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv) {
  puts(argv);
  char *a = malloc(128);
  read(0xffffffff, a, 127);
  write(0xfffffffe, a, 127);
  newline();
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsi,%rdi
  call   <puts@plt>
  mov    $0x80,%edi
  call   <malloc@plt>
  mov    $0x7f,%edx
  mov    $0xffffffff,%edi
  mov    %rax,%rbp
  mov    %rax,%rsi
  call   <read@plt>
  mov    %rbp,%rsi
  mov    $0x7f,%edx
  mov    $0xfffffffe,%edi
  call   <write@plt>
  xor    %eax,%eax
  call   <newline@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char *argv) {
  puts(argv);
  char *a = malloc(128);
  read(0xffffffff, a, 127);
  write(0xfffffffe, a, 127);
  newline();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gpsd_hexpack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jle    L66
  mov    -0x8(%rbp),%eax
  cltq
  cmp    %rax,-0x28(%rbp)
  jae    L2
L66:
  mov    $0xfffffffe,%eax
  jmp    L9
L2:
  movl   $0x0,-0xc(%rbp)
  jmp    L54
L63:
  mov    -0xc(%rbp),%eax
  add    %eax,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <hex2bin@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  je     L31
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  jmp    L42
L31:
  mov    $0xffffffff,%eax
  jmp    L9
L42:
  addl   $0x1,-0xc(%rbp)
L54:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L63
  mov    -0xc(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  sub    %rcx,%rax
  mov    %rax,%rdx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%eax
L9:
  leave
  ret
```

assistant: ```c
int gpsd_hexpack(const char *src, char *dst, size_t len) {
  int i, k, l;
  l = (int)(strlen(src) / 2);
  if ((l < 1) || ((size_t)l > len))
    return -2;
  for (i = 0; i < l; i++)
    if ((k = hex2bin(src + i * 2)) != -1)
      dst[i] = (char)(k & 0xff);
    else
      return -1;
  (void)memset(dst + i, '\0', (size_t)(len - i));
  return l;
}
```
user: Translate the below assembly code into C programming language:

```asm
<gpsd_hexpack>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r15
  mov    %rdx,%r14
  call   <strlen@plt>
  mov    %eax,%r13d
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  cmp    $0x1,%eax
  jle    L49
  movslq %r13d,%rax
  cmp    %r14,%rax
  ja     L42
  mov    %r15,%r12
  mov    $0x0,%ebp
L53:
  mov    %rbx,%rdi
  call   <hex2bin@plt>
  cmp    $0xffffffff,%eax
  je     L1
  mov    %al,(%r12)
  add    $0x1,%ebp
  add    $0x2,%rbx
  add    $0x1,%r12
  cmp    %ebp,%r13d
  jg     L53
  movslq %ebp,%rbp
  mov    %r14,%rdx
  sub    %rbp,%rdx
  lea    (%r15,%rbp,1),%rdi
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %r13d,%eax
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L49:
  mov    $0xfffffffe,%eax
  jmp    L1
L42:
  mov    $0xfffffffe,%eax
  jmp    L1
```

assistant: ```c
int gpsd_hexpack(const char *src, char *dst, size_t len) {
  int i, k, l;
  l = (int)(strlen(src) / 2);
  if ((l < 1) || ((size_t)l > len))
    return -2;
  for (i = 0; i < l; i++)
    if ((k = hex2bin(src + i * 2)) != -1)
      dst[i] = (char)(k & 0xff);
    else
      return -1;
  (void)memset(dst + i, '\0', (size_t)(len - i));
  return l;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<gpsd_hexpack>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%ebp
  shr    $0x1f,%ebp
  add    %eax,%ebp
  sar    %ebp
  cmp    $0x1,%eax
  jle    L4
  movslq %ebp,%rax
  cmp    %r12,%rax
  ja     L4
  mov    %r13,%r15
  xor    %r14d,%r14d
  jmp    L54
  xchg   %ax,%ax
L7:
  add    $0x1,%r14d
  mov    %al,(%r15)
  add    $0x2,%rbx
  add    $0x1,%r15
  cmp    %r14d,%ebp
  jle    L49
L54:
  mov    %rbx,%rdi
  call   <hex2bin@plt>
  cmp    $0xffffffff,%eax
  jne    L7
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L49:
  movslq %r14d,%r14
  mov    %r12,%rdx
  xor    %esi,%esi
  lea    0x0(%r13,%r14,1),%rdi
  sub    %r14,%rdx
  call   <memset@plt>
  add    $0x8,%rsp
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  mov    $0xfffffffe,%eax
  jmp    L13
```

assistant: ```c
int gpsd_hexpack(const char *src, char *dst, size_t len) {
  int i, k, l;
  l = (int)(strlen(src) / 2);
  if ((l < 1) || ((size_t)l > len))
    return -2;
  for (i = 0; i < l; i++)
    if ((k = hex2bin(src + i * 2)) != -1)
      dst[i] = (char)(k & 0xff);
    else
      return -1;
  (void)memset(dst + i, '\0', (size_t)(len - i));
  return l;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gpsd_hexpack>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%ebp
  shr    $0x1f,%ebp
  add    %eax,%ebp
  sar    %ebp
  cmp    $0x1,%eax
  jle    L50
  movslq %ebp,%rax
  cmp    %r12,%rax
  ja     L50
  mov    %r13,%r15
  xor    %r14d,%r14d
  jmp    L48
  xchg   %ax,%ax
L59:
  add    $0x1,%r14d
  mov    %al,(%r15)
  add    $0x2,%rbx
  add    $0x1,%r15
  cmp    %r14d,%ebp
  jle    L6
L48:
  mov    %rbx,%rdi
  call   <hex2bin@plt>
  cmp    $0xffffffff,%eax
  jne    L59
L10:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L6:
  movslq %r14d,%r14
  mov    %r12,%rdx
  xor    %esi,%esi
  lea    0x0(%r13,%r14,1),%rdi
  sub    %r14,%rdx
  call   <memset@plt>
  add    $0x8,%rsp
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L50:
  mov    $0xfffffffe,%eax
  jmp    L10
```

assistant: ```c
int gpsd_hexpack(const char *src, char *dst, size_t len) {
  int i, k, l;
  l = (int)(strlen(src) / 2);
  if ((l < 1) || ((size_t)l > len))
    return -2;
  for (i = 0; i < l; i++)
    if ((k = hex2bin(src + i * 2)) != -1)
      dst[i] = (char)(k & 0xff);
    else
      return -1;
  (void)memset(dst + i, '\0', (size_t)(len - i));
  return l;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<h2o_mem_swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x150,%rsp
  mov    %rdi,-0x138(%rbp)
  mov    %rsi,-0x140(%rbp)
  mov    %rdx,-0x148(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x138(%rbp),%rax
  mov    %rax,-0x128(%rbp)
  mov    -0x140(%rbp),%rax
  mov    %rax,-0x120(%rbp)
  jmp    L10
L13:
  mov    -0x148(%rbp),%rax
  mov    $0x100,%edx
  cmp    %rdx,%rax
  cmova  %rdx,%rax
  mov    %rax,-0x118(%rbp)
  mov    -0x118(%rbp),%rdx
  mov    -0x128(%rbp),%rcx
  lea    -0x110(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x118(%rbp),%rdx
  mov    -0x120(%rbp),%rcx
  mov    -0x128(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x118(%rbp),%rdx
  lea    -0x110(%rbp),%rcx
  mov    -0x120(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x118(%rbp),%rax
  sub    %rax,-0x148(%rbp)
  mov    -0x118(%rbp),%rax
  add    %rax,-0x128(%rbp)
  mov    -0x118(%rbp),%rax
  add    %rax,-0x120(%rbp)
L10:
  cmpq   $0x0,-0x148(%rbp)
  jne    L13
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
void h2o_mem_swap(void *_x, void *_y, size_t len) {
  char *x = _x, *y = _y;
  char buf[256];
  while (len != 0) {
    size_t blocksz = len < sizeof(buf) ? len : sizeof(buf);
    memcpy(buf, x, blocksz);
    memcpy(x, y, blocksz);
    memcpy(y, buf, blocksz);
    len -= blocksz;
    x += blocksz;
    y += blocksz;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<h2o_mem_swap>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L7
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbp
  mov    %rsp,%r14
L29:
  mov    $0x100,%ebx
  cmp    %rbx,%rbp
  cmovbe %rbp,%rbx
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  add    %rbx,%r12
  add    %rbx,%r13
  sub    %rbx,%rbp
  jne    L29
L7:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void h2o_mem_swap(void *_x, void *_y, size_t len) {
  char *x = _x, *y = _y;
  char buf[256];
  while (len != 0) {
    size_t blocksz = len < sizeof(buf) ? len : sizeof(buf);
    memcpy(buf, x, blocksz);
    memcpy(x, y, blocksz);
    memcpy(y, buf, blocksz);
    len -= blocksz;
    x += blocksz;
    y += blocksz;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<h2o_mem_swap>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L19
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbp
  mov    %rsp,%r14
  nopw   0x0(%rax,%rax,1)
L28:
  mov    $0x100,%ebx
  mov    %r12,%rsi
  mov    %r14,%rdi
  cmp    %rbx,%rbp
  cmovbe %rbp,%rbx
  mov    %rbx,%rdx
  call   <memcpy@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %rbx,%rdx
  call   <memcpy@plt>
  mov    %r13,%rdi
  mov    %rbx,%rdx
  mov    %r14,%rsi
  call   <memcpy@plt>
  add    %rbx,%r12
  add    %rbx,%r13
  sub    %rbx,%rbp
  jne    L28
L19:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void h2o_mem_swap(void *_x, void *_y, size_t len) {
  char *x = _x, *y = _y;
  char buf[256];
  while (len != 0) {
    size_t blocksz = len < sizeof(buf) ? len : sizeof(buf);
    memcpy(buf, x, blocksz);
    memcpy(x, y, blocksz);
    memcpy(y, buf, blocksz);
    len -= blocksz;
    x += blocksz;
    y += blocksz;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<h2o_mem_swap>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L15
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbp
  mov    %rsp,%r14
  nopw   0x0(%rax,%rax,1)
L24:
  mov    $0x100,%ebx
  mov    %r12,%rsi
  mov    %r14,%rdi
  cmp    %rbx,%rbp
  cmovbe %rbp,%rbx
  mov    %rbx,%rdx
  call   <memcpy@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %rbx,%rdx
  call   <memcpy@plt>
  mov    %r13,%rdi
  mov    %rbx,%rdx
  mov    %r14,%rsi
  call   <memcpy@plt>
  add    %rbx,%r12
  add    %rbx,%r13
  sub    %rbx,%rbp
  jne    L24
L15:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void h2o_mem_swap(void *_x, void *_y, size_t len) {
  char *x = _x, *y = _y;
  char buf[256];
  while (len != 0) {
    size_t blocksz = len < sizeof(buf) ? len : sizeof(buf);
    memcpy(buf, x, blocksz);
    memcpy(x, y, blocksz);
    memcpy(y, buf, blocksz);
    len -= blocksz;
    x += blocksz;
    y += blocksz;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cypher>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L69
L43:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x4d,%al
  jle    L74
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jle    L68
L74:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x6d,%al
  jle    L25
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jg     L25
L68:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    -0xd(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L107
L25:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x4d,%al
  jg     L31
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jg     L29
L31:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x6d,%al
  jg     L107
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L107
L29:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  lea    0xd(%rax),%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L107:
  addl   $0x1,-0x4(%rbp)
L69:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L43
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D8
type: string

tool: D8: %si 

assistant: ```c
void cypher(char arr[100]) {
  int i;
  char ch;
  for (i = 0; arr[i] != '\0'; ++i) {
    if (arr[i] >= 78 && arr[i] <= 90 || arr[i] >= 110 && arr[i] <= 122) {
      arr[i] = arr[i] - 13;
    } else if (arr[i] <= 77 && arr[i] >= 65 || arr[i] <= 109 && arr[i] >= 97) {
      arr[i] = arr[i] + 13;
    }
  }
  printf("%si ", arr);
}
```
user: What is the c source code of the assembly code below:

```asm
<cypher>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L40
  mov    %rdi,%rdx
  jmp    L48
L43:
  sub    $0x41,%ecx
  cmp    $0xc,%cl
  ja     L47
  add    $0xd,%r9d
  mov    %r9b,(%rdx)
L47:
  add    $0x1,%rdx
  movzbl (%rdx),%eax
  test   %al,%al
  je     L40
L48:
  mov    %eax,%r9d
  mov    %eax,%ecx
  and    $0xffffffdf,%ecx
  lea    -0x4e(%rcx),%r8d
  cmp    $0xc,%r8b
  ja     L43
  sub    $0xd,%eax
  mov    %al,(%rdx)
  jmp    L47
L40:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_guessed: string

tool: D6: %si 

assistant: ```c
void cypher(char arr[100]) {
  int i;
  char ch;
  for (i = 0; arr[i] != '\0'; ++i) {
    if (arr[i] >= 78 && arr[i] <= 90 || arr[i] >= 110 && arr[i] <= 122) {
      arr[i] = arr[i] - 13;
    } else if (arr[i] <= 77 && arr[i] >= 65 || arr[i] <= 109 && arr[i] >= 97) {
      arr[i] = arr[i] + 13;
    }
  }
  printf("%si ", arr);
}
```