ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  call   <fork@plt>
  cmp    $0xffffffff,%eax
  je     L10
  cmp    $0x1,%eax
  sbb    %ebx,%ebx
  and    $0x2,%ebx
  add    $0x3,%ebx
  test   %eax,%eax
  lea    D36(%rip),%rbp
  lea    D41(%rip),%rax
  cmovne %rax,%rbp
L12:
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  sub    $0x1,%ebx
  jne    L12
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  lea    D12(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x5,%ebx
  lea    D36(%rip),%rbp
  jmp    L12
```

tool_calls: name: read_data
arguments: data_label: D8
type_infered: string
name: read_data
arguments: data_label: D36
type_infered: string
name: read_data
arguments: data_label: D41
type_infered: string
name: read_data
arguments: data_label: D12
type_infered: string

tool: D8: fork programing starting 


tool: D36: This is the child

tool: D41: This is the parent

tool: D12: fork failed


assistant: ```c
int main() {
  pid_t pid;
  char *message;
  int n;
  printf("fork programing starting \n");
  pid = fork();
  switch (pid) {
  case -1:
    perror("fork failed\n");
    exit(1);
  case 0:
    message = "This is the child";
    n = 5;
    break;
  default:
    message = "This is the parent";
    n = 3;
    break;
  }
  for (; n > 0; n--) {
    puts(message);
    sleep(1);
  }
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D32(%rip),%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <printf@plt>
  call   <fork@plt>
  cmp    $0xffffffff,%eax
  je     L0
  cmp    $0x1,%eax
  lea    D1(%rip),%rbp
  sbb    %ebx,%ebx
  and    $0x2,%ebx
  add    $0x3,%ebx
  test   %eax,%eax
  lea    D4(%rip),%rax
  cmovne %rax,%rbp
  nop
L7:
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  sub    $0x1,%ebx
  jne    L7
  xor    %edi,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L0:
  lea    D31(%rip),%rdi
  mov    $0x5,%ebx
  lea    D1(%rip),%rbp
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L7
```

tool_calls: name: read_data
arguments: label: D32
data_type: string
name: read_data
arguments: label: D1
data_type: string
name: read_data
arguments: label: D4
data_type: string
name: read_data
arguments: label: D31
data_type: string

tool: D32: fork programing starting 


tool: D1: This is the child

tool: D4: This is the parent

tool: D31: fork failed


assistant: ```c
int main() {
  pid_t pid;
  char *message;
  int n;
  printf("fork programing starting \n");
  pid = fork();
  switch (pid) {
  case -1:
    perror("fork failed\n");
    exit(1);
  case 0:
    message = "This is the child";
    n = 5;
    break;
  default:
    message = "This is the parent";
    n = 3;
    break;
  }
  for (; n > 0; n--) {
    puts(message);
    sleep(1);
  }
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D43(%rip),%rdi
  push   %rbx
  sub    $0x8,%rsp
  call   <printf@plt>
  call   <fork@plt>
  cmp    $0xffffffff,%eax
  je     L27
  cmp    $0x1,%eax
  lea    D6(%rip),%rbp
  sbb    %ebx,%ebx
  and    $0x2,%ebx
  add    $0x3,%ebx
  test   %eax,%eax
  lea    D46(%rip),%rax
  cmovne %rax,%rbp
  nop
L15:
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  sub    $0x1,%ebx
  jne    L15
  xor    %edi,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L27:
  lea    D34(%rip),%rdi
  mov    $0x5,%ebx
  lea    D6(%rip),%rbp
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L15
```

tool_calls: name: disassemble_data
arguments: data_label: D43
type_infered: string
name: disassemble_data
arguments: data_label: D6
type_infered: string
name: disassemble_data
arguments: data_label: D46
type_infered: string
name: disassemble_data
arguments: data_label: D34
type_infered: string

tool: D43: fork programing starting 


tool: D6: This is the child

tool: D46: This is the parent

tool: D34: fork failed


assistant: ```c
int main() {
  pid_t pid;
  char *message;
  int n;
  printf("fork programing starting \n");
  pid = fork();
  switch (pid) {
  case -1:
    perror("fork failed\n");
    exit(1);
  case 0:
    message = "This is the child";
    n = 5;
    break;
  default:
    message = "This is the parent";
    n = 3;
    break;
  }
  for (; n > 0; n--) {
    puts(message);
    sleep(1);
  }
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<isPrime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L4:
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  je     L7
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  cmp    %eax,-0x14(%rbp)
  jg     L4
L7:
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  cmp    %eax,-0x4(%rbp)
  jne    L42
  mov    $0x1,%eax
  jmp    L44
L42:
  mov    $0x0,%eax
L44:
  pop    %rbp
  ret
```

assistant: ```c
int isPrime(int n) {
  int i = 2, j = 0;
  while (n % i != 0 && i++ < n)
    j++;
  if (j == (n - 2))
    return 1;
  else
    return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<isPrime>:
  endbr64
  test   $0x1,%dil
  je     L9
  cmp    $0x2,%edi
  jle    L2
  lea    -0x2(%rdi),%r8d
  mov    $0x0,%ecx
L53:
  lea    0x3(%rcx),%esi
  add    $0x1,%ecx
  mov    %edi,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L20
  cmp    %r8d,%ecx
  jne    L53
  mov    %r8d,%ecx
L20:
  sub    $0x2,%edi
  cmp    %ecx,%edi
  sete   %al
  movzbl %al,%eax
  ret
L9:
  mov    $0x0,%ecx
  jmp    L20
L2:
  mov    $0x0,%ecx
  jmp    L20
```

assistant: ```c
int isPrime(int n) {
  int i = 2, j = 0;
  while (n % i != 0 && i++ < n)
    j++;
  if (j == (n - 2))
    return 1;
  else
    return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<isPrime>:
  endbr64
  test   $0x1,%dil
  je     L3
  cmp    $0x2,%edi
  jle    L3
  lea    -0x2(%rdi),%r8d
  xor    %ecx,%ecx
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L43:
  cmp    %r8d,%ecx
  je     L29
L11:
  mov    %edi,%eax
  lea    0x3(%rcx),%esi
  add    $0x1,%ecx
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L43
L29:
  xor    %eax,%eax
  cmp    %ecx,%r8d
  sete   %al
  ret
  nopl   (%rax)
L3:
  xor    %ecx,%ecx
  lea    -0x2(%rdi),%r8d
  xor    %eax,%eax
  cmp    %ecx,%r8d
  sete   %al
  ret
```

assistant: ```c
int isPrime(int n) {
  int i = 2, j = 0;
  while (n % i != 0 && i++ < n)
    j++;
  if (j == (n - 2))
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<isPrime>:
  endbr64
  test   $0x1,%dil
  je     L39
  cmp    $0x2,%edi
  jle    L39
  lea    -0x2(%rdi),%r8d
  xor    %ecx,%ecx
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L14:
  cmp    %r8d,%ecx
  je     L8
L7:
  mov    %edi,%eax
  lea    0x3(%rcx),%esi
  add    $0x1,%ecx
  cltd
  idiv   %esi
  test   %edx,%edx
  jne    L14
L8:
  xor    %eax,%eax
  cmp    %ecx,%r8d
  sete   %al
  ret
  nopl   (%rax)
L39:
  xor    %ecx,%ecx
  lea    -0x2(%rdi),%r8d
  xor    %eax,%eax
  cmp    %ecx,%r8d
  sete   %al
  ret
```

assistant: ```c
int isPrime(int n) {
  int i = 2, j = 0;
  while (n % i != 0 && i++ < n)
    j++;
  if (j == (n - 2))
    return 1;
  else
    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
<pl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L2
L6:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
L2:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
  nop
  nop
  leave
  ret
```

assistant: ```c
void pl(char *str) {
  putchar('\n');
  while (*str)
    putchar(*str++);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pl>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xa,%edi
  call   <putchar@plt>
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L3
L15:
  add    $0x1,%rbx
  movsbl %dil,%edi
  call   <putchar@plt>
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L15
L3:
  pop    %rbx
  ret
```

assistant: ```c
void pl(char *str) {
  putchar('\n');
  while (*str)
    putchar(*str++);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pl>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xa,%edi
  call   <putchar@plt>
  movsbl (%rbx),%edi
  test   %dil,%dil
  je     L14
  nopw   0x0(%rax,%rax,1)
L4:
  call   <putchar@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L4
L14:
  pop    %rbx
  ret
```

assistant: ```c
void pl(char *str) {
  putchar('\n');
  while (*str)
    putchar(*str++);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pl>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xa,%edi
  call   <putchar@plt>
  movsbl (%rbx),%edi
  test   %dil,%dil
  je     L18
  nopw   0x0(%rax,%rax,1)
L19:
  call   <putchar@plt>
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L19
L18:
  pop    %rbx
  ret
```

assistant: ```c
void pl(char *str) {
  putchar('\n');
  while (*str)
    putchar(*str++);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x0,-0x94(%rbp)
  mov    -0x94(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x90(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  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
L64:
  cmp    %rdx,%rsp
  je     L92
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L64
L92:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L41
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L41:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x88(%rbp)
  movl   $0x0,-0x94(%rbp)
  jmp    L53
L85:
  lea    -0x80(%rbp),%rdx
  mov    -0x94(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  nop
  mov    -0x94(%rbp),%eax
  cltq
  movzbl -0x80(%rbp,%rax,1),%eax
  test   %al,%al
  jne    L10
  mov    $0x0,%eax
  jmp    L24
L10:
  mov    -0x94(%rbp),%eax
  cltq
  movzbl -0x80(%rbp,%rax,1),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,%eax
  mov    %eax,%ecx
  mov    -0x88(%rbp),%rdx
  mov    -0x94(%rbp),%eax
  cltq
  mov    %cl,(%rdx,%rax,1)
  mov    -0x88(%rbp),%rdx
  mov    -0x94(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  cmp    $0x7a,%al
  jle    L77
  mov    -0x88(%rbp),%rdx
  mov    -0x94(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  sub    $0x1a,%eax
  mov    %eax,%ecx
  mov    -0x88(%rbp),%rdx
  mov    -0x94(%rbp),%eax
  cltq
  mov    %cl,(%rdx,%rax,1)
  mov    -0x88(%rbp),%rdx
  mov    -0x94(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L18
L77:
  mov    -0x88(%rbp),%rdx
  mov    -0x94(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L18:
  addl   $0x1,-0x94(%rbp)
L53:
  cmpl   $0x64,-0x94(%rbp)
  jle    L85
  mov    $0x0,%eax
L24:
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L52
  call   <__stack_chk_fail@plt>
L52:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D16
type_guessed: string

tool: D16: %c

assistant: ```c
int main() {
  int i = 0;
  char stri[100], stro[i];
  for (i = 0; i <= 100; i++) {
    scanf("%c", &stri[i]);
    while (stri[i] == '\0') {
      return 0;
    }
    stro[i] = stri[i] + stri[i] - 96;
    if (stro[i] > 122) {
      stro[i] = stro[i] - 122 + 96;
      printf("%c", stro[i]);
    } else
      printf("%c", stro[i]);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x65(%rsp),%r12
  lea    D5(%rip),%rbp
  jmp    L50
L45:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
L10:
  add    $0x1,%rbx
  cmp    %r12,%rbx
  je     L43
L50:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  movzbl (%rbx),%esi
  test   %sil,%sil
  je     L43
  lea    -0x60(%rsi,%rsi,1),%esi
  cmp    $0x7a,%sil
  jle    L45
  sub    $0x1a,%esi
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L10
L43:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
datatype: string

tool: D5: %c

assistant: ```c
int main() {
  int i = 0;
  char stri[100], stro[i];
  for (i = 0; i <= 100; i++) {
    scanf("%c", &stri[i]);
    while (stri[i] == '\0') {
      return 0;
    }
    stro[i] = stri[i] + stri[i] - 96;
    if (stro[i] > 122) {
      stro[i] = stro[i] - 122 + 96;
      printf("%c", stro[i]);
    } else
      printf("%c", stro[i]);
  }
  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
  push   %rbx
  lea    D3(%rip),%rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  jmp    L6
  cs nopw 0x0(%rax,%rax,1)
L26:
  sub    $0x1a,%esi
  mov    %rbx,%rdi
  movzbl %sil,%esi
  call   <printf@plt>
L41:
  add    $0x1,%rbp
L6:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movzbl 0x0(%rbp),%esi
  test   %sil,%sil
  je     L36
  lea    -0x60(%rsi,%rsi,1),%esi
  cmp    $0x7a,%sil
  jg     L26
  movsbl %sil,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  jmp    L41
  nop
L36:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
data_type: string

tool: D3: %c

assistant: ```c
int main() {
  int i = 0;
  char stri[100], stro[i];
  for (i = 0; i <= 100; i++) {
    scanf("%c", &stri[i]);
    while (stri[i] == '\0') {
      return 0;
    }
    stro[i] = stri[i] + stri[i] - 96;
    if (stro[i] > 122) {
      stro[i] = stro[i] - 122 + 96;
      printf("%c", stro[i]);
    } else
      printf("%c", stro[i]);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  lea    D3(%rip),%rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  jmp    L36
  cs nopw 0x0(%rax,%rax,1)
L49:
  sub    $0x1a,%esi
  mov    %rbx,%rdi
  movzbl %sil,%esi
  call   <printf@plt>
L12:
  add    $0x1,%rbp
L36:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movzbl 0x0(%rbp),%esi
  test   %sil,%sil
  je     L15
  lea    -0x60(%rsi,%rsi,1),%esi
  cmp    $0x7a,%sil
  jg     L49
  movsbl %sil,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  jmp    L12
  nop
L15:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D3
type_guessed: string

tool: D3: %c

assistant: ```c
int main() {
  int i = 0;
  char stri[100], stro[i];
  for (i = 0; i <= 100; i++) {
    scanf("%c", &stri[i]);
    while (stri[i] == '\0') {
      return 0;
    }
    stro[i] = stri[i] + stri[i] - 96;
    if (stro[i] > 122) {
      stro[i] = stro[i] - 122 + 96;
      printf("%c", stro[i]);
    } else
      printf("%c", stro[i]);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ieee_swcr_to_fpcr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  shl    $0x23,%rax
  mov    %rax,%rdx
  movabs $0x3f0000000000000,%rax
  and    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  shl    $0x24,%rax
  mov    %rax,%rdx
  movabs $0x1000000000000,%rax
  and    %rdx,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  and    $0x7e0000,%eax
  test   %rax,%rax
  je     L35
  movabs $0x8000000000000000,%rax
  jmp    L10
L35:
  mov    $0x0,%eax
L10:
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  not    %rax
  shl    $0x30,%rax
  mov    %rax,%rdx
  movabs $0xe000000000000,%rax
  and    %rdx,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  not    %rax
  shl    $0x39,%rax
  mov    %rax,%rdx
  movabs $0x6000000000000000,%rax
  and    %rdx,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  and    $0x2000,%eax
  test   %rax,%rax
  je     L7
  movabs $0x3000000000000000,%rax
  jmp    L18
L7:
  mov    $0x0,%eax
L18:
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  not    %rax
  shl    $0x29,%rax
  mov    %rax,%rdx
  movabs $0x800000000000,%rax
  and    %rdx,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long ieee_swcr_to_fpcr(unsigned long sw) {
  unsigned long fp;
  fp = (sw & ((1UL << 17) | (1UL << 18) | (1UL << 19) | (1UL << 20) | (1UL << 21) | (1UL << 22))) << 35;
  fp |= (sw & (1UL << 12)) << 36;
  fp |= (sw & ((1UL << 17) | (1UL << 18) | (1UL << 19) | (1UL << 20) | (1UL << 21) | (1UL << 22)) ? (1UL << 63) : 0);
  fp |= (~sw & ((1UL << 1) | (1UL << 2) | (1UL << 3))) << 48;
  fp |= (~sw & ((1UL << 4) | (1UL << 5))) << 57;
  fp |= (sw & (1UL << 13) ? (1UL << 60) | (1UL << 61) : 0);
  fp |= (~sw & (1UL << 6)) << 41;
  return fp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ieee_swcr_to_fpcr>:
  endbr64
  mov    %rdi,%rax
  mov    %rdi,%rcx
  shl    $0x23,%rcx
  movabs $0x3f0000000000000,%rdx
  and    %rdx,%rcx
  test   $0x7e0000,%edi
  setne  %dil
  movzbl %dil,%edi
  shl    $0x3f,%rdi
  mov    %rax,%rdx
  not    %rdx
  mov    %rax,%rsi
  shl    $0x24,%rsi
  movabs $0x1000000000000,%r8
  and    %r8,%rsi
  or     %rcx,%rsi
  mov    %rdx,%rcx
  shl    $0x30,%rcx
  movabs $0xe000000000000,%r8
  and    %r8,%rcx
  or     %rcx,%rsi
  mov    %rdx,%rcx
  shl    $0x39,%rcx
  movabs $0x6000000000000000,%r8
  and    %r8,%rcx
  or     %rcx,%rsi
  and    $0x2000,%eax
  mov    %rax,%rcx
  movabs $0x3000000000000000,%rax
  cmovne %rax,%rcx
  mov    %rdx,%rax
  shl    $0x29,%rax
  movabs $0x800000000000,%rdx
  and    %rdx,%rax
  or     %rsi,%rax
  or     %rdi,%rax
  or     %rcx,%rax
  ret
```

assistant: ```c
unsigned long ieee_swcr_to_fpcr(unsigned long sw) {
  unsigned long fp;
  fp = (sw & ((1UL << 17) | (1UL << 18) | (1UL << 19) | (1UL << 20) | (1UL << 21) | (1UL << 22))) << 35;
  fp |= (sw & (1UL << 12)) << 36;
  fp |= (sw & ((1UL << 17) | (1UL << 18) | (1UL << 19) | (1UL << 20) | (1UL << 21) | (1UL << 22)) ? (1UL << 63) : 0);
  fp |= (~sw & ((1UL << 1) | (1UL << 2) | (1UL << 3))) << 48;
  fp |= (~sw & ((1UL << 4) | (1UL << 5))) << 57;
  fp |= (sw & (1UL << 13) ? (1UL << 60) | (1UL << 61) : 0);
  fp |= (~sw & (1UL << 6)) << 41;
  return fp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ieee_swcr_to_fpcr>:
  endbr64
  movabs $0x3f0000000000000,%rdx
  mov    %rdi,%rcx
  mov    %rdi,%rax
  xor    %edi,%edi
  shl    $0x23,%rcx
  mov    %rax,%rsi
  movabs $0x1000000000000,%r8
  and    %rdx,%rcx
  test   $0x7e0000,%eax
  mov    %rax,%rdx
  setne  %dil
  shl    $0x24,%rsi
  not    %rdx
  and    %r8,%rsi
  shl    $0x3f,%rdi
  movabs $0xe000000000000,%r8
  or     %rcx,%rsi
  mov    %rdx,%rcx
  shl    $0x30,%rcx
  and    %r8,%rcx
  movabs $0x6000000000000000,%r8
  or     %rcx,%rsi
  mov    %rdx,%rcx
  shl    $0x39,%rcx
  and    %r8,%rcx
  or     %rcx,%rsi
  and    $0x2000,%eax
  mov    %rax,%rcx
  movabs $0x3000000000000000,%rax
  cmovne %rax,%rcx
  mov    %rdx,%rax
  movabs $0x800000000000,%rdx
  shl    $0x29,%rax
  and    %rdx,%rax
  or     %rsi,%rax
  or     %rdi,%rax
  or     %rcx,%rax
  ret
```

assistant: ```c
unsigned long ieee_swcr_to_fpcr(unsigned long sw) {
  unsigned long fp;
  fp = (sw & ((1UL << 17) | (1UL << 18) | (1UL << 19) | (1UL << 20) | (1UL << 21) | (1UL << 22))) << 35;
  fp |= (sw & (1UL << 12)) << 36;
  fp |= (sw & ((1UL << 17) | (1UL << 18) | (1UL << 19) | (1UL << 20) | (1UL << 21) | (1UL << 22)) ? (1UL << 63) : 0);
  fp |= (~sw & ((1UL << 1) | (1UL << 2) | (1UL << 3))) << 48;
  fp |= (~sw & ((1UL << 4) | (1UL << 5))) << 57;
  fp |= (sw & (1UL << 13) ? (1UL << 60) | (1UL << 61) : 0);
  fp |= (~sw & (1UL << 6)) << 41;
  return fp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ieee_swcr_to_fpcr>:
  endbr64
  movabs $0x3f0000000000000,%rdx
  mov    %rdi,%rcx
  mov    %rdi,%rax
  xor    %edi,%edi
  shl    $0x23,%rcx
  mov    %rax,%rsi
  movabs $0x1000000000000,%r8
  and    %rdx,%rcx
  test   $0x7e0000,%eax
  mov    %rax,%rdx
  setne  %dil
  shl    $0x24,%rsi
  not    %rdx
  and    %r8,%rsi
  shl    $0x3f,%rdi
  movabs $0xe000000000000,%r8
  or     %rcx,%rsi
  mov    %rdx,%rcx
  shl    $0x30,%rcx
  and    %r8,%rcx
  movabs $0x6000000000000000,%r8
  or     %rcx,%rsi
  mov    %rdx,%rcx
  shl    $0x39,%rcx
  and    %r8,%rcx
  or     %rcx,%rsi
  and    $0x2000,%eax
  mov    %rax,%rcx
  movabs $0x3000000000000000,%rax
  cmovne %rax,%rcx
  mov    %rdx,%rax
  movabs $0x800000000000,%rdx
  shl    $0x29,%rax
  and    %rdx,%rax
  or     %rsi,%rax
  or     %rdi,%rax
  or     %rcx,%rax
  ret
```

assistant: ```c
unsigned long ieee_swcr_to_fpcr(unsigned long sw) {
  unsigned long fp;
  fp = (sw & ((1UL << 17) | (1UL << 18) | (1UL << 19) | (1UL << 20) | (1UL << 21) | (1UL << 22))) << 35;
  fp |= (sw & (1UL << 12)) << 36;
  fp |= (sw & ((1UL << 17) | (1UL << 18) | (1UL << 19) | (1UL << 20) | (1UL << 21) | (1UL << 22)) ? (1UL << 63) : 0);
  fp |= (~sw & ((1UL << 1) | (1UL << 2) | (1UL << 3))) << 48;
  fp |= (~sw & ((1UL << 4) | (1UL << 5))) << 57;
  fp |= (sw & (1UL << 13) ? (1UL << 60) | (1UL << 61) : 0);
  fp |= (~sw & (1UL << 6)) << 41;
  return fp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xrop3_233_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x1c(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x18(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x1c(%rbp),%eax
  and    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  notl   -0x10(%rbp)
  mov    -0x14(%rbp),%eax
  and    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  xor    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  xor    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_233_0_0(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  unsigned stk4;
  stk4 = D & S;
  stk4 = ~stk4;
  stk3 = T & stk4;
  stk2 = S ^ stk3;
  stk1 = D ^ stk2;
  stk1 = ~stk1;
  return stk1;
}
```