ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<print_depth>:
  endbr64
  cmp    $0x1,%edi
  je     L23
  cmp    $0x2,%edi
  je     L4
  cmp    $0x3,%edi
  je     L29
  lea    D35(%rip),%rdi
  jmp    1050 <printf@plt>
  nop
L29:
  lea    D15(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L23:
  lea    D14(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax)
L4:
  lea    D30(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D35
data_type: string
name: retrieve_data
arguments: label: D15
data_type: string
name: retrieve_data
arguments: label: D14
data_type: string
name: retrieve_data
arguments: label: D30
data_type: string

tool: D35: 


tool: D15: 
.

tool: D14: 
.   .   

tool: D30: 
.   

assistant: ```c
void print_depth(int depth) {
  if (depth == 1) {
    printf("\n.   .   ");
  } else if (depth == 2) {
    printf("\n.   ");
  } else if (depth == 3) {
    printf("\n.");
  } else {
    printf("\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<swap5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  mov    %rax,-0x20(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap5(int *a, int *b) {
  int k = a;
  a = b;
  b = k;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<swap5>:
  endbr64
  ret
```

assistant: ```c
void swap5(int *a, int *b) {
  int k = a;
  a = b;
  b = k;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<swap5>:
  endbr64
  ret
```

assistant: ```c
void swap5(int *a, int *b) {
  int k = a;
  a = b;
  b = k;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<swap5>:
  endbr64
  ret
```

assistant: ```c
void swap5(int *a, int *b) {
  int k = a;
  a = b;
  b = k;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<swap_byte4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  jmp    L29
L24:
  mov    -0x10(%rbp),%eax
  cltq
  mov    $0x7,%edx
  sub    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  mov    %dl,-0xc(%rbp,%rax,1)
  addl   $0x1,-0x10(%rbp)
L29:
  mov    -0x10(%rbp),%eax
  cmp    $0x7,%eax
  jbe    L24
  lea    -0xc(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void *swap_byte4(long *theInt) {
  char tmp[4];
  int i;
  for (i = 0; i < sizeof(long); i++)
    tmp[i] = *((char *)theInt + (sizeof(long) - 1 - i));
  *theInt = *((long *)tmp);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<swap_byte4>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rdi),%rax
  lea    0x4(%rsp),%rdx
L19:
  movzbl (%rax),%ecx
  mov    %cl,(%rdx)
  mov    %rax,%rcx
  sub    $0x1,%rax
  add    $0x1,%rdx
  cmp    %rdi,%rcx
  jne    L19
  mov    0x4(%rsp),%rax
  mov    %rax,(%rdi)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *swap_byte4(long *theInt) {
  char tmp[4];
  int i;
  for (i = 0; i < sizeof(long); i++)
    tmp[i] = *((char *)theInt + (sizeof(long) - 1 - i));
  *theInt = *((long *)tmp);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<swap_byte4>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  lea    0x7(%rdi),%rax
  nopl   0x0(%rax)
L12:
  movzbl (%rax),%ecx
  add    $0x1,%rdx
  mov    %cl,-0x1(%rdx)
  mov    %rax,%rcx
  sub    $0x1,%rax
  cmp    %rdi,%rcx
  jne    L12
  mov    0x4(%rsp),%rax
  mov    %rax,(%rdi)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void *swap_byte4(long *theInt) {
  char tmp[4];
  int i;
  for (i = 0; i < sizeof(long); i++)
    tmp[i] = *((char *)theInt + (sizeof(long) - 1 - i));
  *theInt = *((long *)tmp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<swap_byte4>:
  endbr64
  mov    (%rdi),%rax
  bswap  %rax
  mov    %rax,(%rdi)
  ret
```

assistant: ```c
void *swap_byte4(long *theInt) {
  char tmp[4];
  int i;
  for (i = 0; i < sizeof(long); i++)
    tmp[i] = *((char *)theInt + (sizeof(long) - 1 - i));
  *theInt = *((long *)tmp);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<searchy_word_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <word_hash@plt>
  mov    %rax,-0x8(%rbp)
  movabs $0x7fffffffffffffff,%rax
  and    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x1f,%al
  je     L6
  movabs $0x8000000000000000,%rax
  or     %rax,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
unsigned long long searchy_word_hash(const char *str, int len) {
  unsigned long long h = word_hash(str, len);
  h &= 0x7fffffffffffffffULL;
  if (*str != 0x1f) {
    h |= 0x8000000000000000ULL;
  }
  return h;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<searchy_word_hash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <word_hash@plt>
  btr    $0x3f,%rax
  cmpb   $0x1f,(%rbx)
  je     L7
  bts    $0x3f,%rax
L7:
  pop    %rbx
  ret
```

assistant: ```c
unsigned long long searchy_word_hash(const char *str, int len) {
  unsigned long long h = word_hash(str, len);
  h &= 0x7fffffffffffffffULL;
  if (*str != 0x1f) {
    h |= 0x8000000000000000ULL;
  }
  return h;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<searchy_word_hash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <word_hash@plt>
  btr    $0x3f,%rax
  cmpb   $0x1f,(%rbx)
  je     L0
  bts    $0x3f,%rax
L0:
  pop    %rbx
  ret
```

assistant: ```c
unsigned long long searchy_word_hash(const char *str, int len) {
  unsigned long long h = word_hash(str, len);
  h &= 0x7fffffffffffffffULL;
  if (*str != 0x1f) {
    h |= 0x8000000000000000ULL;
  }
  return h;
}
```
user: What is the c source code of the assembly code below?

```asm
<searchy_word_hash>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <word_hash@plt>
  btr    $0x3f,%rax
  cmpb   $0x1f,(%rbx)
  je     L7
  bts    $0x3f,%rax
L7:
  pop    %rbx
  ret
```

assistant: ```c
unsigned long long searchy_word_hash(const char *str, int len) {
  unsigned long long h = word_hash(str, len);
  h &= 0x7fffffffffffffffULL;
  if (*str != 0x1f) {
    h |= 0x8000000000000000ULL;
  }
  return h;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L2
L13:
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <box@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L2:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  test   %eax,%eax
  jne    L13
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D12
datatype: string
name: read_data
arguments: data_tag: D28
datatype: string
name: read_data
arguments: data_tag: D29
datatype: string

tool: D12: %d

tool: D28: %d %d

tool: D29: %.10f


assistant: ```c
int main() {
  int n = 0;
  scanf("%d", &n);
  while (n--) {
    int x = 0, y = 0;
    scanf("%d %d", &x, &y);
    printf("%.10f\n", box(x, y));
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L18
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D13(%rip),%rbp
  lea    D2(%rip),%rbx
L3:
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%edi
  call   <box@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L3
L18:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D3
type_guessed: string
name: unpack_data
arguments: tag: D13
type_guessed: string
name: unpack_data
arguments: tag: D2
type_guessed: string

tool: D3: %d

tool: D13: %d %d

tool: D2: %.10f


assistant: ```c
int main() {
  int n = 0;
  scanf("%d", &n);
  while (n--) {
    int x = 0, y = 0;
    scanf("%d %d", &x, &y);
    printf("%.10f\n", box(x, y));
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D29(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  movl   $0x0,0xc(%rsp)
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L14
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D20(%rip),%rbp
  lea    D15(%rip),%rbx
  xchg   %ax,%ax
L8:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  call   <scanf@plt>
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%edi
  call   <box@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L8
L14:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D29
datatype: string
name: unpack_data
arguments: tag: D20
datatype: string
name: unpack_data
arguments: tag: D15
datatype: string

tool: D29: %d

tool: D20: %d %d

tool: D15: %.10f


assistant: ```c
int main() {
  int n = 0;
  scanf("%d", &n);
  while (n--) {
    int x = 0, y = 0;
    scanf("%d %d", &x, &y);
    printf("%.10f\n", box(x, y));
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D20(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  movl   $0x0,0xc(%rsp)
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L10
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D9(%rip),%rbp
  lea    D24(%rip),%rbx
  xchg   %ax,%ax
L16:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  call   <scanf@plt>
  mov    0x14(%rsp),%esi
  mov    0x10(%rsp),%edi
  call   <box@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L16
L10:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D20
type_guessed: string
name: extract_symbolic_data
arguments: label: D9
type_guessed: string
name: extract_symbolic_data
arguments: label: D24
type_guessed: string

tool: D20: %d

tool: D9: %d %d

tool: D24: %.10f


assistant: ```c
int main() {
  int n = 0;
  scanf("%d", &n);
  while (n--) {
    int x = 0, y = 0;
    scanf("%d %d", &x, &y);
    printf("%.10f\n", box(x, y));
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<send_char_async>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x140,%rsp
  mov    %edi,-0x124(%rbp)
  mov    %esi,-0x128(%rbp)
  mov    %edx,%eax
  mov    %rcx,-0x138(%rbp)
  mov    %al,-0x12c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x110(%rbp)
  movq   $0x0,-0x108(%rbp)
  movq   $0x0,-0x100(%rbp)
  movq   $0x0,-0xf8(%rbp)
  movq   $0x0,-0xf0(%rbp)
  movq   $0x0,-0xe8(%rbp)
  movq   $0x0,-0xe0(%rbp)
  movq   $0x0,-0xd8(%rbp)
  movq   $0x0,-0xd0(%rbp)
  movq   $0x0,-0xc8(%rbp)
  movq   $0x0,-0xc0(%rbp)
  movq   $0x0,-0xb8(%rbp)
  movq   $0x0,-0xb0(%rbp)
  movq   $0x0,-0xa8(%rbp)
  movq   $0x0,-0xa0(%rbp)
  movq   $0x0,-0x98(%rbp)
  movq   $0x0,-0x90(%rbp)
  movq   $0x0,-0x88(%rbp)
  movq   $0x0,-0x80(%rbp)
  movq   $0x0,-0x78(%rbp)
  movq   $0x0,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movzbl -0x12c(%rbp),%eax
  mov    %al,-0x110(%rbp)
  movl   $0x1,-0x120(%rbp)
  mov    $0x28,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,-0x118(%rbp)
  mov    -0x118(%rbp),%rax
  mov    $0x28,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x118(%rbp),%rax
  movl   $0xffffffff,(%rax)
  mov    -0x118(%rbp),%rax
  lea    -0x110(%rbp),%rdx
  mov    %rdx,0x8(%rax)
  mov    -0x118(%rbp),%rax
  mov    -0x120(%rbp),%edx
  mov    %edx,0x10(%rax)
  mov    -0x118(%rbp),%rax
  mov    -0x128(%rbp),%edx
  mov    %edx,0x14(%rax)
  mov    D27(%rip),%rax
  mov    (%rax),%edx
  mov    -0x118(%rbp),%rax
  mov    %edx,0x20(%rax)
  mov    -0x118(%rbp),%rax
  mov    -0x138(%rbp),%rdx
  mov    %rdx,0x18(%rax)
L7:
  mov    D20(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x118(%rbp),%rdx
  mov    -0x124(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <ioctl@plt>
  mov    %eax,-0x11c(%rbp)
  cmpl   $0x0,-0x11c(%rbp)
  jns    L10
  mov    D0(%rip),%rax
  mov    (%rax),%rdx
  mov    D24(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L7
L10:
  mov    -0x11c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

assistant: ```c
int send_char_async(int fd, int endpoint, char msg, char *label) {
  char buffer[256] = {0};
  buffer[0] = msg;
  int length = 1;
  struct usbdevfs_urb *urb = calloc(1, sizeof(struct usbdevfs_urb));
  memset(urb, 0, sizeof(struct usbdevfs_urb));
  int res;
  urb->status = -1;
  urb->buffer = buffer;
  urb->buffer_length = length;
  urb->endpoint = endpoint;
  urb->type = USBDEVFS_URB_TYPE_BULK;
  urb->usercontext = label;
  do {
    res = ioctl(fd, USBDEVFS_SUBMITURB, urb);
  } while ((res < 0) && (errno == EINTR));
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<send_char_async>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %edi,%ebp
  mov    %esi,%r13d
  mov    %rcx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movq   $0x0,0x60(%rsp)
  movq   $0x0,0x68(%rsp)
  movq   $0x0,0x70(%rsp)
  movq   $0x0,0x78(%rsp)
  movq   $0x0,0x80(%rsp)
  movq   $0x0,0x88(%rsp)
  movq   $0x0,0x90(%rsp)
  movq   $0x0,0x98(%rsp)
  movq   $0x0,0xa0(%rsp)
  movq   $0x0,0xa8(%rsp)
  movq   $0x0,0xb0(%rsp)
  movq   $0x0,0xb8(%rsp)
  movq   $0x0,0xc0(%rsp)
  movq   $0x0,0xc8(%rsp)
  movq   $0x0,0xd0(%rsp)
  movq   $0x0,0xd8(%rsp)
  movq   $0x0,0xe0(%rsp)
  movq   $0x0,0xe8(%rsp)
  movq   $0x0,0xf0(%rsp)
  movq   $0x0,0xf8(%rsp)
  mov    %dl,(%rsp)
  mov    $0x28,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,%rbx
  mov    $0x28,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0xffffffff,(%rbx)
  mov    %rsp,%rax
  mov    %rax,0x8(%rbx)
  movl   $0x1,0x10(%rbx)
  mov    %r13d,0x14(%rbx)
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x20(%rbx)
  mov    %r12,0x18(%rbx)
  mov    D12(%rip),%r12
L2:
  mov    %rbx,%rdx
  mov    (%r12),%esi
  mov    %ebp,%edi
  call   <ioctl@plt>
  test   %eax,%eax
  jns    L18
  mov    D25(%rip),%rcx
  mov    D15(%rip),%rdx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L2
L18:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int send_char_async(int fd, int endpoint, char msg, char *label) {
  char buffer[256] = {0};
  buffer[0] = msg;
  int length = 1;
  struct usbdevfs_urb *urb = calloc(1, sizeof(struct usbdevfs_urb));
  memset(urb, 0, sizeof(struct usbdevfs_urb));
  int res;
  urb->status = -1;
  urb->buffer = buffer;
  urb->buffer_length = length;
  urb->endpoint = endpoint;
  urb->type = USBDEVFS_URB_TYPE_BULK;
  urb->usercontext = label;
  do {
    res = ioctl(fd, USBDEVFS_SUBMITURB, urb);
  } while ((res < 0) && (errno == EINTR));
  return res;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<send_char_async>:
  endbr64
  push   %r13
  pxor   %xmm0,%xmm0
  mov    %esi,%r13d
  mov    $0x28,%esi
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %edi,%ebp
  mov    $0x1,%edi
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  mov    %dl,(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  movaps %xmm0,0x80(%rsp)
  movaps %xmm0,0x90(%rsp)
  movaps %xmm0,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  movaps %xmm0,0xc0(%rsp)
  movaps %xmm0,0xd0(%rsp)
  movaps %xmm0,0xe0(%rsp)
  movaps %xmm0,0xf0(%rsp)
  call   <calloc@plt>
  mov    $0x28,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <memset@plt>
  mov    %rsp,%rax
  mov    %r12,0x18(%rbx)
  mov    D4(%rip),%r12
  mov    %rax,0x8(%rbx)
  mov    D22(%rip),%rax
  movl   $0xffffffff,(%rbx)
  mov    (%rax),%eax
  movl   $0x1,0x10(%rbx)
  mov    %r13d,0x14(%rbx)
  mov    %eax,0x20(%rbx)
  jmp    L34
  nopl   0x0(%rax)
L37:
  mov    D14(%rip),%rdx
  mov    D18(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  jne    L7
L34:
  mov    (%r12),%esi
  mov    %rbx,%rdx
  mov    %ebp,%edi
  call   <ioctl@plt>
  test   %eax,%eax
  js     L37
L7:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int send_char_async(int fd, int endpoint, char msg, char *label) {
  char buffer[256] = {0};
  buffer[0] = msg;
  int length = 1;
  struct usbdevfs_urb *urb = calloc(1, sizeof(struct usbdevfs_urb));
  memset(urb, 0, sizeof(struct usbdevfs_urb));
  int res;
  urb->status = -1;
  urb->buffer = buffer;
  urb->buffer_length = length;
  urb->endpoint = endpoint;
  urb->type = USBDEVFS_URB_TYPE_BULK;
  urb->usercontext = label;
  do {
    res = ioctl(fd, USBDEVFS_SUBMITURB, urb);
  } while ((res < 0) && (errno == EINTR));
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<send_char_async>:
  endbr64
  push   %r13
  pxor   %xmm0,%xmm0
  mov    %esi,%r13d
  mov    $0x28,%esi
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  mov    %edi,%ebp
  mov    $0x1,%edi
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  mov    %dl,(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movaps %xmm0,0x70(%rsp)
  movaps %xmm0,0x80(%rsp)
  movaps %xmm0,0x90(%rsp)
  movaps %xmm0,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  movaps %xmm0,0xc0(%rsp)
  movaps %xmm0,0xd0(%rsp)
  movaps %xmm0,0xe0(%rsp)
  movaps %xmm0,0xf0(%rsp)
  call   <calloc@plt>
  mov    $0x28,%edx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <memset@plt>
  mov    %rsp,%rax
  mov    %r12,0x18(%rbx)
  mov    D35(%rip),%r12
  mov    %rax,0x8(%rbx)
  mov    D2(%rip),%rax
  movl   $0xffffffff,(%rbx)
  mov    (%rax),%eax
  movl   $0x1,0x10(%rbx)
  mov    %r13d,0x14(%rbx)
  mov    %eax,0x20(%rbx)
  jmp    L13
  nopl   0x0(%rax)
L23:
  mov    D9(%rip),%rdx
  mov    D24(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  jne    L21
L13:
  mov    (%r12),%esi
  mov    %rbx,%rdx
  mov    %ebp,%edi
  call   <ioctl@plt>
  test   %eax,%eax
  js     L23
L21:
  mov    0x108(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int send_char_async(int fd, int endpoint, char msg, char *label) {
  char buffer[256] = {0};
  buffer[0] = msg;
  int length = 1;
  struct usbdevfs_urb *urb = calloc(1, sizeof(struct usbdevfs_urb));
  memset(urb, 0, sizeof(struct usbdevfs_urb));
  int res;
  urb->status = -1;
  urb->buffer = buffer;
  urb->buffer_length = length;
  urb->endpoint = endpoint;
  urb->type = USBDEVFS_URB_TYPE_BULK;
  urb->usercontext = label;
  do {
    res = ioctl(fd, USBDEVFS_SUBMITURB, urb);
  } while ((res < 0) && (errno == EINTR));
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hpack_encode_str_raw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L0
  mov    -0x18(%rbp),%rax
  movq   $0x0,(%rax)
L0:
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    $0x7,%ecx
  mov    %rax,%rdi
  call   <hpack_encode_int@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L17
  mov    -0x8(%rbp),%rax
  shl    $0x3,%rax
  add    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  sub    %rax,-0x20(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  cmp    %rax,%rdx
  cmova  %rax,%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
L17:
  mov    -0x10(%rbp),%rax
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
size_t hpack_encode_str_raw(uint8_t *restrict buf, size_t size, const char *str) {
  size_t len = strlen(str);
  if (size > 0)
    *buf = 0;
  size_t ret = hpack_encode_int(buf, size, len, 7);
  if (ret < size) {
    buf += ret;
    size -= ret;
    memcpy(buf, str, (len <= size) ? len : size);
  }
  ret += len;
  return ret;
}
```