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

```asm
<strCheck>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmpb   $0xa,(%rax)
  sete   %dl
  movzbl %dl,%edx
  sub    %rdx,%rax
  lea    D24(%rip),%rdx
  cmp    %rax,%rbx
  jae    L18
L25:
  movzbl (%rbx),%ecx
  cmp    %cl,(%rax)
  jne    L35
  sub    $0x1,%rax
  add    $0x1,%rbx
  cmp    %rbx,%rax
  ja     L25
  lea    D24(%rip),%rdx
  jmp    L18
L35:
  lea    D19(%rip),%rdx
L18:
  mov    %rdx,%rax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: label: D24
data_type: string
name: read_data
arguments: label: D19
data_type: string

tool: D24: Is a palindrome


tool: D19: Not a palindrome


assistant: ```c
char *strCheck(char *s) {
  char *b = s;
  char *e = s + strlen(s) - 1;
  if (*e == '\n')
    --e;
  for (; e > b; --e, ++b) {
    if (*e ^ *b)
      return "Not a palindrome\n";
  }
  return "Is a palindrome\n";
}
```
user: Translate the below assembly code into C programming language:

```asm
<strCheck>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  xor    %edx,%edx
  lea    D25(%rip),%r8
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmpb   $0xa,(%rax)
  sete   %dl
  sub    %rdx,%rax
  cmp    %rax,%rbx
  jb     L20
  jmp    L5
  nopl   (%rax)
L37:
  sub    $0x1,%rax
  add    $0x1,%rbx
  cmp    %rbx,%rax
  jbe    L16
L20:
  movzbl (%rbx),%ecx
  cmp    %cl,(%rax)
  je     L37
  lea    D7(%rip),%r8
L5:
  mov    %r8,%rax
  pop    %rbx
  ret
L16:
  lea    D25(%rip),%r8
  pop    %rbx
  mov    %r8,%rax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D25
datatype: string
name: disassemble_data
arguments: tag: D7
datatype: string

tool: D25: Is a palindrome


tool: D7: Not a palindrome


assistant: ```c
char *strCheck(char *s) {
  char *b = s;
  char *e = s + strlen(s) - 1;
  if (*e == '\n')
    --e;
  for (; e > b; --e, ++b) {
    if (*e ^ *b)
      return "Not a palindrome\n";
  }
  return "Is a palindrome\n";
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strCheck>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  xor    %edx,%edx
  lea    D4(%rip),%r8
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  cmpb   $0xa,(%rax)
  sete   %dl
  sub    %rdx,%rax
  cmp    %rax,%rbx
  jb     L34
  jmp    L21
  nopl   (%rax)
L1:
  sub    $0x1,%rax
  add    $0x1,%rbx
  cmp    %rbx,%rax
  jbe    L0
L34:
  movzbl (%rbx),%ecx
  cmp    %cl,(%rax)
  je     L1
  lea    D10(%rip),%r8
L21:
  mov    %r8,%rax
  pop    %rbx
  ret
L0:
  lea    D4(%rip),%r8
  pop    %rbx
  mov    %r8,%rax
  ret
```

tool_calls: name: read_data
arguments: tag: D4
data_type: string
name: read_data
arguments: tag: D10
data_type: string

tool: D4: Is a palindrome


tool: D10: Not a palindrome


assistant: ```c
char *strCheck(char *s) {
  char *b = s;
  char *e = s + strlen(s) - 1;
  if (*e == '\n')
    --e;
  for (; e > b; --e, ++b) {
    if (*e ^ *b)
      return "Not a palindrome\n";
  }
  return "Is a palindrome\n";
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<int_to_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x14,%edi
  call   <emalloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%ecx
  lea    D7(%rip),%rdx
  mov    $0x14,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
datatype: string

tool: D7: %i

assistant: ```c
char *int_to_str(int number) {
  char *buff;
  buff = (char *)emalloc(20);
  snprintf(buff, 20, "%i", number);
  return buff;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<int_to_str>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x14,%edi
  call   <emalloc@plt>
  mov    %rax,%rbx
  mov    %ebp,%ecx
  lea    D4(%rip),%rdx
  mov    $0x14,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
data_type: string

tool: D4: %i

assistant: ```c
char *int_to_str(int number) {
  char *buff;
  buff = (char *)emalloc(20);
  snprintf(buff, 20, "%i", number);
  return buff;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<int_to_str>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  mov    $0x14,%edi
  push   %r12
  sub    $0x8,%rsp
  call   <emalloc@plt>
  mov    %r13d,%ecx
  mov    $0x14,%esi
  lea    D7(%rip),%rdx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <snprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: tag: D7
type_guessed: string

tool: D7: %i

assistant: ```c
char *int_to_str(int number) {
  char *buff;
  buff = (char *)emalloc(20);
  snprintf(buff, 20, "%i", number);
  return buff;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<int_to_str>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  mov    $0x14,%edi
  push   %r12
  sub    $0x8,%rsp
  call   <emalloc@plt>
  mov    %r13d,%ecx
  mov    $0x14,%esi
  lea    D3(%rip),%rdx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <snprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D3
type: string

tool: D3: %i

assistant: ```c
char *int_to_str(int number) {
  char *buff;
  buff = (char *)emalloc(20);
  snprintf(buff, 20, "%i", number);
  return buff;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<possibilities_naive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L23
  mov    $0x1,%eax
  jmp    L12
L23:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <possibilities_naive>
  mov    %rax,-0x8(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jbe    L28
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <possibilities_naive>
  add    %rax,-0x8(%rbp)
L28:
  cmpl   $0x2,-0x14(%rbp)
  jbe    L17
  mov    -0x14(%rbp),%eax
  sub    $0x3,%eax
  mov    %eax,%edi
  call   <possibilities_naive>
  add    %rax,-0x8(%rbp)
L17:
  mov    -0x8(%rbp),%rax
L12:
  leave
  ret
```

assistant: ```c
unsigned long long possibilities_naive(unsigned n) {
  if (n == 0)
    return 1;
  unsigned long long res = possibilities_naive(n - 1);
  if (n > 1)
    res += possibilities_naive(n - 2);
  if (n > 2)
    res += possibilities_naive(n - 3);
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<possibilities_naive>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x1,%ebp
  test   %edi,%edi
  jne    L47
L37:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L47:
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <possibilities_naive>
  mov    %rax,%rbp
  cmp    $0x1,%ebx
  ja     L10
L42:
  cmp    $0x2,%ebx
  jbe    L37
  lea    -0x3(%rbx),%edi
  call   <possibilities_naive>
  add    %rax,%rbp
  jmp    L37
L10:
  lea    -0x2(%rbx),%edi
  call   <possibilities_naive>
  add    %rax,%rbp
  jmp    L42
```

assistant: ```c
unsigned long long possibilities_naive(unsigned n) {
  if (n == 0)
    return 1;
  unsigned long long res = possibilities_naive(n - 1);
  if (n > 1)
    res += possibilities_naive(n - 2);
  if (n > 2)
    res += possibilities_naive(n - 3);
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<possibilities_naive>:
  endbr64
  test   %edi,%edi
  je     L24
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
L22:
  lea    -0x1(%rbx),%edi
  call   <possibilities_naive>
  mov    %rax,%rbp
  cmp    $0x1,%ebx
  je     L11
  lea    -0x2(%rbx),%edi
  call   <possibilities_naive>
  add    %rbp,%rax
  cmp    $0x2,%ebx
  je     L27
  add    %rax,%r12
  sub    $0x3,%ebx
  jne    L22
  lea    0x1(%r12),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  lea    (%rax,%r12,1),%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L27:
  add    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L24:
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned long long possibilities_naive(unsigned n) {
  if (n == 0)
    return 1;
  unsigned long long res = possibilities_naive(n - 1);
  if (n > 1)
    res += possibilities_naive(n - 2);
  if (n > 2)
    res += possibilities_naive(n - 3);
  return res;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<grwp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jle    L49
  movl   $0x1,-0x8(%rbp)
  jmp    L8
L2:
  mov    -0x18(%rbp),%rax
  lea    0x280(%rax),%rdx
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    (%rdx,%rax,4),%edx
  mov    -0x18(%rbp),%rax
  lea    0x280(%rax),%rcx
  mov    -0x8(%rbp),%eax
  cltq
  mov    (%rcx,%rax,4),%eax
  cmp    %eax,%edx
  jne    L56
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%esi
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%ecx
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  add    %esi,%ecx
  mov    -0x18(%rbp),%rax
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L17
L50:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%esi
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%ecx
  mov    -0x18(%rbp),%rax
  movslq %esi,%rdx
  mov    %ecx,(%rax,%rdx,4)
  mov    -0x18(%rbp),%rax
  lea    0x280(%rax),%rdx
  mov    -0x18(%rbp),%rax
  lea    0x280(%rax),%rcx
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%esi
  mov    -0x4(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%edx
  movslq %esi,%rax
  mov    %edx,(%rcx,%rax,4)
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jl     L50
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  subl   $0x1,-0x8(%rbp)
L56:
  addl   $0x1,-0x8(%rbp)
L8:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L2
L49:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void grwp(int list[2][160], int *indexliste) {
  if (*(indexliste) > 1) {
    int i = 1;
    while (i < *(indexliste)) {
      if (list[1][i - 1] == list[1][i]) {
        int j;
        list[0][i - 1] = list[0][i - 1] + list[0][i];
        j = i + 1;
        while (j < *(indexliste)) {
          list[0][j - 1] = list[0][j];
          list[1][j - 1] = list[1][j];
          j++;
        }
        *(indexliste) = *(indexliste)-1;
        i--;
      }
      i++;
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<grwp>:
  endbr64
  cmpl   $0x1,(%rsi)
  jle    L62
  mov    $0x1,%eax
  jmp    L56
L63:
  sub    $0x1,%edx
  mov    %edx,(%rsi)
  mov    %ecx,%eax
L46:
  add    $0x1,%eax
  cmp    %eax,(%rsi)
  jle    L62
L56:
  lea    -0x1(%rax),%ecx
  movslq %eax,%rdx
  lea    (%rdi,%rdx,4),%rdx
  mov    0x280(%rdx),%r9d
  cmp    %r9d,0x27c(%rdx)
  jne    L46
  mov    (%rdx),%r8d
  add    %r8d,-0x4(%rdx)
  add    $0x1,%eax
  mov    (%rsi),%edx
  cmp    %edx,%eax
  jge    L63
  cltq
L6:
  mov    (%rdi,%rax,4),%edx
  mov    %edx,-0x4(%rdi,%rax,4)
  mov    0x280(%rdi,%rax,4),%edx
  mov    %edx,0x27c(%rdi,%rax,4)
  mov    (%rsi),%edx
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L6
  jmp    L63
L62:
  ret
```

assistant: ```c
void grwp(int list[2][160], int *indexliste) {
  if (*(indexliste) > 1) {
    int i = 1;
    while (i < *(indexliste)) {
      if (list[1][i - 1] == list[1][i]) {
        int j;
        list[0][i - 1] = list[0][i - 1] + list[0][i];
        j = i + 1;
        while (j < *(indexliste)) {
          list[0][j - 1] = list[0][j];
          list[1][j - 1] = list[1][j];
          j++;
        }
        *(indexliste) = *(indexliste)-1;
        i--;
      }
      i++;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<grwp>:
  endbr64
  cmpl   $0x1,(%rsi)
  mov    $0x1,%ecx
  jg     L16
  jmp    L1
L34:
  mov    (%rsi),%edx
  mov    %eax,%ecx
  cmp    %edx,%ecx
  jge    L1
L16:
  movslq %ecx,%rax
  lea    (%rdi,%rax,4),%rdx
  lea    0x1(%rcx),%eax
  mov    0x280(%rdx),%r9d
  cmp    %r9d,0x27c(%rdx)
  jne    L34
  mov    (%rdx),%r8d
  add    %r8d,-0x4(%rdx)
  mov    (%rsi),%edx
  cmp    %edx,%eax
  jge    L43
  cltq
  nopl   0x0(%rax)
L14:
  mov    (%rdi,%rax,4),%edx
  mov    %edx,-0x4(%rdi,%rax,4)
  mov    0x280(%rdi,%rax,4),%edx
  mov    %edx,0x27c(%rdi,%rax,4)
  mov    (%rsi),%edx
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L14
L43:
  sub    $0x1,%edx
  mov    %edx,(%rsi)
  cmp    %edx,%ecx
  jl     L16
L1:
  ret
```

assistant: ```c
void grwp(int list[2][160], int *indexliste) {
  if (*(indexliste) > 1) {
    int i = 1;
    while (i < *(indexliste)) {
      if (list[1][i - 1] == list[1][i]) {
        int j;
        list[0][i - 1] = list[0][i - 1] + list[0][i];
        j = i + 1;
        while (j < *(indexliste)) {
          list[0][j - 1] = list[0][j];
          list[1][j - 1] = list[1][j];
          j++;
        }
        *(indexliste) = *(indexliste)-1;
        i--;
      }
      i++;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<grwp>:
  endbr64
  mov    (%rsi),%edx
  mov    $0x1,%ecx
  cmp    $0x1,%edx
  jg     L23
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %eax,%ecx
  cmp    %edx,%ecx
  jge    L18
L23:
  movslq %ecx,%rax
  lea    (%rdi,%rax,4),%r8
  lea    0x1(%rcx),%eax
  mov    0x280(%r8),%r9d
  cmp    %r9d,0x27c(%r8)
  jne    L26
  mov    (%r8),%edx
  add    %edx,-0x4(%r8)
  mov    (%rsi),%edx
  cmp    %eax,%edx
  jle    L64
  cltq
  nopw   0x0(%rax,%rax,1)
L11:
  mov    (%rdi,%rax,4),%edx
  mov    %edx,-0x4(%rdi,%rax,4)
  mov    0x280(%rdi,%rax,4),%edx
  mov    %edx,0x27c(%rdi,%rax,4)
  mov    (%rsi),%edx
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L11
L64:
  sub    $0x1,%edx
  mov    %edx,(%rsi)
  cmp    %edx,%ecx
  jl     L23
L18:
  ret
```

assistant: ```c
void grwp(int list[2][160], int *indexliste) {
  if (*(indexliste) > 1) {
    int i = 1;
    while (i < *(indexliste)) {
      if (list[1][i - 1] == list[1][i]) {
        int j;
        list[0][i - 1] = list[0][i - 1] + list[0][i];
        j = i + 1;
        while (j < *(indexliste)) {
          list[0][j - 1] = list[0][j];
          list[1][j - 1] = list[1][j];
          j++;
        }
        *(indexliste) = *(indexliste)-1;
        i--;
      }
      i++;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<hello>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D9
type_infered: string
name: parse_data
arguments: data_label: D0
type_infered: string

tool: D9: Hello, library world.`


tool: D0: Hello, this is a security fix.`


assistant: ```c
void hello(void) {
  printf("Hello, library world.`\n");
  printf("Hello, this is a security fix.`\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hello>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D10
data_type: string
name: extract_symbolic_data
arguments: label: D2
data_type: string

tool: D10: Hello, library world.`


tool: D2: Hello, this is a security fix.`


assistant: ```c
void hello(void) {
  printf("Hello, library world.`\n");
  printf("Hello, this is a security fix.`\n");
}
```
user: Translate the below assembly code into C programming language:

```asm
<hello>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D2(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type: string
name: retrieve_data
arguments: data_label: D2
type: string

tool: D5: Hello, library world.`


tool: D2: Hello, this is a security fix.`


assistant: ```c
void hello(void) {
  printf("Hello, library world.`\n");
  printf("Hello, this is a security fix.`\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hello>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D8
type: string
name: disassemble_data
arguments: data_label: D14
type: string

tool: D8: Hello, library world.`


tool: D14: Hello, this is a security fix.`


assistant: ```c
void hello(void) {
  printf("Hello, library world.`\n");
  printf("Hello, this is a security fix.`\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  add    $0xb45,%ax
  mov    %ax,-0xa(%rbp)
  movzwl -0xa(%rbp),%eax
  shr    $0x4,%ax
  movzwl %ax,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L53
  movzwl -0xa(%rbp),%eax
  shr    $0x4,%ax
  movzwl %ax,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L21
  movzwl -0xa(%rbp),%eax
  shr    $0x2,%ax
  movzwl %ax,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L12
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  mov    %al,-0xc(%rbp)
  lea    -0xa(%rbp),%rdx
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  lea    -0xa(%rbp),%rax
  movzbl -0xc(%rbp),%edx
  mov    %dl,(%rax)
  lea    -0xa(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0xb(%rbp)
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  lea    -0xa(%rbp),%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl -0xb(%rbp),%edx
  mov    %dl,(%rax)
  jmp    L53
L12:
  movzwl -0xa(%rbp),%edx
  movzwl -0xa(%rbp),%eax
  imul   %edx,%eax
  mov    %ax,-0xa(%rbp)
  movzwl -0xa(%rbp),%eax
  add    %eax,%eax
  mov    %ax,-0xa(%rbp)
  jmp    L53
L21:
  movzwl -0xa(%rbp),%eax
  add    %eax,%eax
  mov    %ax,-0xa(%rbp)
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  lea    -0xa(%rbp),%rdx
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  lea    -0xa(%rbp),%rax
  movzbl -0xd(%rbp),%edx
  mov    %dl,(%rax)
L53:
  movzwl -0xa(%rbp),%eax
  lea    -0x12bc(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  char copy12;
  char copy13;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
      if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
        if ((state[0UL] >> (unsigned short)2) & (unsigned short)1) {
          copy11 = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy11;
          copy12 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy12;
        } else {
          state[0UL] *= state[0UL];
          state[0UL] += state[0UL];
        }
      } else {
        state[0UL] += state[0UL];
        copy13 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy13;
      }
    }
    output[0UL] = (state[0UL] + 852556293UL) + (unsigned short)62271;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0xb45,%ax
  test   $0x10,%al
  je     L6
  test   $0x4,%al
  jne    L6
  imul   %eax,%eax
  add    %eax,%eax
L6:
  sub    $0x12bc,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  char copy12;
  char copy13;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
      if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
        if ((state[0UL] >> (unsigned short)2) & (unsigned short)1) {
          copy11 = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy11;
          copy12 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy12;
        } else {
          state[0UL] *= state[0UL];
          state[0UL] += state[0UL];
        }
      } else {
        state[0UL] += state[0UL];
        copy13 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy13;
      }
    }
    output[0UL] = (state[0UL] + 852556293UL) + (unsigned short)62271;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  add    $0xb45,%ax
  test   $0x10,%al
  je     L0
  test   $0x4,%al
  je     L3
L0:
  sub    $0x12bc,%ax
  mov    %ax,(%rsi)
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  imul   %eax,%eax
  add    %eax,%eax
  sub    $0x12bc,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  char copy11;
  char copy12;
  char copy13;
  {
    state[0UL] = input[0UL] + (unsigned short)2885;
    if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
      if ((state[0UL] >> (unsigned short)4) & (unsigned short)1) {
        if ((state[0UL] >> (unsigned short)2) & (unsigned short)1) {
          copy11 = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy11;
          copy12 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy12;
        } else {
          state[0UL] *= state[0UL];
          state[0UL] += state[0UL];
        }
      } else {
        state[0UL] += state[0UL];
        copy13 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy13;
      }
    }
    output[0UL] = (state[0UL] + 852556293UL) + (unsigned short)62271;
  }
}
```