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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L1:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L1
```

assistant: ```c
int main(void) {
  while (1) {
    sleep(1);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L3:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L3
```

assistant: ```c
int main(void) {
  while (1) {
    sleep(1);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<comp_decode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  notl   -0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x80,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x4,%eax
  and    $0x7,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  and    $0x1e,%eax
  or     $0x21,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  sub    $0x21,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L8
  mov    -0x4(%rbp),%eax
  neg    %eax
  jmp    L5
L8:
  mov    -0x4(%rbp),%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int16_t comp_decode(uint8_t usample) {
  usample = ~usample;
  int sign = usample & 0x80;
  int exp = (usample >> 4) & 7;
  int16_t sample = ((((usample & 0xF) << 1) | 0x21) << exp) - 33;
  if (sign)
    return -sample;
  else
    return sample;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<comp_decode>:
  endbr64
  not    %edi
  lea    (%rdi,%rdi,1),%edx
  and    $0x1e,%edx
  or     $0x21,%edx
  mov    %edi,%ecx
  sar    $0x4,%ecx
  and    $0x7,%ecx
  shl    %cl,%edx
  lea    -0x21(%rdx),%ecx
  mov    $0x21,%eax
  sub    %edx,%eax
  test   $0x80,%dil
  cmove  %ecx,%eax
  ret
```

assistant: ```c
int16_t comp_decode(uint8_t usample) {
  usample = ~usample;
  int sign = usample & 0x80;
  int exp = (usample >> 4) & 7;
  int16_t sample = ((((usample & 0xF) << 1) | 0x21) << exp) - 33;
  if (sign)
    return -sample;
  else
    return sample;
}
```
user: What is the c source code of the assembly code below?

```asm
<comp_decode>:
  endbr64
  not    %edi
  mov    $0x21,%eax
  lea    (%rdi,%rdi,1),%edx
  mov    %edi,%ecx
  and    $0x1e,%edx
  sar    $0x4,%ecx
  and    $0x7,%ecx
  or     $0x21,%edx
  shl    %cl,%edx
  sub    %edx,%eax
  lea    -0x21(%rdx),%ecx
  and    $0x80,%edi
  cmove  %ecx,%eax
  ret
```

assistant: ```c
int16_t comp_decode(uint8_t usample) {
  usample = ~usample;
  int sign = usample & 0x80;
  int exp = (usample >> 4) & 7;
  int16_t sample = ((((usample & 0xF) << 1) | 0x21) << exp) - 33;
  if (sign)
    return -sample;
  else
    return sample;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<comp_decode>:
  endbr64
  not    %edi
  mov    $0x21,%eax
  lea    (%rdi,%rdi,1),%edx
  mov    %edi,%ecx
  and    $0x1e,%edx
  sar    $0x4,%ecx
  and    $0x7,%ecx
  or     $0x21,%edx
  shl    %cl,%edx
  sub    %edx,%eax
  lea    -0x21(%rdx),%ecx
  and    $0x80,%edi
  cmove  %ecx,%eax
  ret
```

assistant: ```c
int16_t comp_decode(uint8_t usample) {
  usample = ~usample;
  int sign = usample & 0x80;
  int exp = (usample >> 4) & 7;
  int16_t sample = ((((usample & 0xF) << 1) | 0x21) << exp) - 33;
  if (sign)
    return -sample;
  else
    return sample;
}
```
user: What is the c source code of the assembly code below?

```asm
<ParseString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L17
L19:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,(%rdx)
  addl   $0x1,-0x10(%rbp)
  lea    D8(%rip),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L17:
  cmpq   $0x0,-0x8(%rbp)
  jne    L19
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  mov    -0x10(%rbp),%eax
  leave
  ret
```

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

tool: D8:  	


assistant: ```c
int ParseString(char *ps, char *fld[]) {
  int field_count = 0;
  char *tok;
  int i;
  field_count = 0;
  for (i = 0, tok = strtok(ps, " \t\n\r"); tok != NULL; tok = strtok(NULL, " \t\n\r"), i++) {
    fld[i] = tok;
    field_count++;
  }
  fld[field_count] = NULL;
  return (field_count);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ParseString>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  lea    D18(%rip),%rsi
  call   <strtok@plt>
  test   %rax,%rax
  je     L29
  mov    $0x1,%ebx
  lea    D18(%rip),%r13
L23:
  mov    %rax,-0x8(%rbp,%rbx,8)
  mov    %ebx,%r12d
  mov    %r13,%rsi
  mov    $0x0,%edi
  call   <strtok@plt>
  add    $0x1,%rbx
  test   %rax,%rax
  jne    L23
L14:
  movslq %r12d,%rax
  movq   $0x0,0x0(%rbp,%rax,8)
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  mov    $0x0,%r12d
  jmp    L14
```

tool_calls: name: read_data
arguments: tag: D18
type_guessed: string

tool: D18:  	


assistant: ```c
int ParseString(char *ps, char *fld[]) {
  int field_count = 0;
  char *tok;
  int i;
  field_count = 0;
  for (i = 0, tok = strtok(ps, " \t\n\r"); tok != NULL; tok = strtok(NULL, " \t\n\r"), i++) {
    fld[i] = tok;
    field_count++;
  }
  fld[field_count] = NULL;
  return (field_count);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ParseString>:
  endbr64
  push   %r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  mov    %rsi,%rbp
  mov    %r12,%rsi
  push   %rbx
  sub    $0x8,%rsp
  call   <strtok@plt>
  test   %rax,%rax
  je     L4
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %rax,-0x8(%rbp,%rbx,8)
  mov    %r12,%rsi
  xor    %edi,%edi
  mov    %ebx,%r13d
  add    $0x1,%rbx
  call   <strtok@plt>
  test   %rax,%rax
  jne    L17
  movslq %r13d,%rax
  lea    0x0(%rbp,%rax,8),%rbp
L14:
  movq   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L4:
  xor    %r13d,%r13d
  jmp    L14
```

tool_calls: name: parse_data
arguments: data_tag: D0
type: string

tool: D0:  	


assistant: ```c
int ParseString(char *ps, char *fld[]) {
  int field_count = 0;
  char *tok;
  int i;
  field_count = 0;
  for (i = 0, tok = strtok(ps, " \t\n\r"); tok != NULL; tok = strtok(NULL, " \t\n\r"), i++) {
    fld[i] = tok;
    field_count++;
  }
  fld[field_count] = NULL;
  return (field_count);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ParseString>:
  endbr64
  push   %r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  mov    %rsi,%rbp
  mov    %r12,%rsi
  push   %rbx
  sub    $0x8,%rsp
  call   <strtok@plt>
  test   %rax,%rax
  je     L15
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %rax,-0x8(%rbp,%rbx,8)
  mov    %r12,%rsi
  xor    %edi,%edi
  mov    %ebx,%r13d
  add    $0x1,%rbx
  call   <strtok@plt>
  test   %rax,%rax
  jne    L28
  movslq %r13d,%rax
  lea    0x0(%rbp,%rax,8),%rbp
L24:
  movq   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L15:
  xor    %r13d,%r13d
  jmp    L24
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_guessed: string

tool: D0:  	


assistant: ```c
int ParseString(char *ps, char *fld[]) {
  int field_count = 0;
  char *tok;
  int i;
  field_count = 0;
  for (i = 0, tok = strtok(ps, " \t\n\r"); tok != NULL; tok = strtok(NULL, " \t\n\r"), i++) {
    fld[i] = tok;
    field_count++;
  }
  fld[field_count] = NULL;
  return (field_count);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<position_cursor_first_blank>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L47
L51:
  movl   $0x0,-0x4(%rbp)
  jmp    L26
L50:
  mov    -0x18(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    %rcx,%rax
  shl    $0x3,%rax
  add    %rcx,%rax
  add    %rsi,%rax
  mov    (%rdx,%rax,4),%eax
  test   %eax,%eax
  jne    L53
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    $0x0,%ecx
  mov    %rax,%rdi
  call   <position_cursor@plt>
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  jmp    L5
L53:
  addl   $0x1,-0x4(%rbp)
L26:
  cmpl   $0x8,-0x4(%rbp)
  jle    L50
  addl   $0x1,-0x8(%rbp)
L47:
  cmpl   $0x8,-0x8(%rbp)
  jle    L51
L5:
  leave
  ret
```

assistant: ```c
void position_cursor_first_blank(uint8_t (*board)[9][9], int *cursor_row, int *cursor_col) {
  int row, col;
  for (row = 0; row < 9; row++) {
    for (col = 0; col < 9; col++) {
      if ((0) == (*board)[row][col]) {
        position_cursor(board, row, col, (0));
        *cursor_row = row;
        *cursor_col = col;
        return;
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<position_cursor_first_blank>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r13
  mov    %rdx,%r12
  mov    %rdi,%rdx
  mov    $0x0,%ebp
  jmp    L20
L39:
  mov    $0x0,%ecx
  mov    %eax,%edx
  mov    %ebp,%esi
  call   <position_cursor@plt>
  mov    %ebp,0x0(%r13)
  mov    %ebx,(%r12)
L42:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  add    $0x1,%ebp
  add    $0x24,%rdx
  cmp    $0x9,%ebp
  je     L42
L20:
  mov    $0x0,%eax
L21:
  mov    %eax,%ebx
  cmpl   $0x0,(%rdx,%rax,4)
  je     L39
  add    $0x1,%rax
  cmp    $0x9,%rax
  jne    L21
  jmp    L7
```

assistant: ```c
void position_cursor_first_blank(uint8_t (*board)[9][9], int *cursor_row, int *cursor_col) {
  int row, col;
  for (row = 0; row < 9; row++) {
    for (col = 0; col < 9; col++) {
      if ((0) == (*board)[row][col]) {
        position_cursor(board, row, col, (0));
        *cursor_row = row;
        *cursor_col = col;
        return;
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<position_cursor_first_blank>:
  endbr64
  push   %r13
  mov    %rdi,%rax
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x8,%rsp
L16:
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L19:
  mov    (%rax,%rbx,4),%ecx
  mov    %ebx,%edx
  test   %ecx,%ecx
  je     L27
  add    $0x1,%rbx
  cmp    $0x9,%rbx
  jne    L19
  add    $0x1,%ebp
  add    $0x24,%rax
  cmp    $0x9,%ebp
  jne    L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %ebp,%esi
  xor    %ecx,%ecx
  call   <position_cursor@plt>
  mov    %ebp,0x0(%r13)
  mov    %ebx,(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void position_cursor_first_blank(uint8_t (*board)[9][9], int *cursor_row, int *cursor_col) {
  int row, col;
  for (row = 0; row < 9; row++) {
    for (col = 0; col < 9; col++) {
      if ((0) == (*board)[row][col]) {
        position_cursor(board, row, col, (0));
        *cursor_row = row;
        *cursor_col = col;
        return;
      }
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<position_cursor_first_blank>:
  endbr64
  push   %r13
  mov    %rdi,%rax
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
L131:
  mov    (%rax),%ebp
  test   %ebp,%ebp
  je     L114
  mov    0x4(%rax),%ebp
  test   %ebp,%ebp
  je     L42
  mov    0x8(%rax),%r11d
  test   %r11d,%r11d
  je     L121
  mov    0xc(%rax),%r10d
  test   %r10d,%r10d
  je     L135
  mov    0x10(%rax),%r9d
  test   %r9d,%r9d
  je     L47
  mov    0x14(%rax),%r8d
  test   %r8d,%r8d
  je     L162
  mov    0x18(%rax),%esi
  test   %esi,%esi
  je     L133
  mov    0x1c(%rax),%ecx
  test   %ecx,%ecx
  je     L157
  mov    0x20(%rax),%edx
  test   %edx,%edx
  je     L11
  add    $0x1,%ebx
  add    $0x24,%rax
  cmp    $0x9,%ebx
  jne    L131
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L42:
  mov    $0x1,%ebp
L114:
  mov    %ebp,%edx
  mov    %ebx,%esi
  xor    %ecx,%ecx
  call   <position_cursor@plt>
  mov    %ebx,(%r12)
  mov    %ebp,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L121:
  mov    $0x2,%ebp
  jmp    L114
L135:
  mov    $0x3,%ebp
  jmp    L114
L47:
  mov    $0x4,%ebp
  jmp    L114
L162:
  mov    $0x5,%ebp
  jmp    L114
L133:
  mov    $0x6,%ebp
  jmp    L114
L157:
  mov    $0x7,%ebp
  jmp    L114
L11:
  mov    $0x8,%ebp
  jmp    L114
```

assistant: ```c
void position_cursor_first_blank(uint8_t (*board)[9][9], int *cursor_row, int *cursor_col) {
  int row, col;
  for (row = 0; row < 9; row++) {
    for (col = 0; col < 9; col++) {
      if ((0) == (*board)[row][col]) {
        position_cursor(board, row, col, (0));
        *cursor_row = row;
        *cursor_col = col;
        return;
      }
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<tripcode_pass>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L15
L42:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  jne    L0
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  test   %rax,%rax
  je     L0
  cmpl   $0x0,-0x10(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L76
L0:
  subl   $0x1,-0x14(%rbp)
L15:
  cmpl   $0x0,-0x14(%rbp)
  jns    L42
L76:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  jne    L34
  cmpl   $0x0,-0x10(%rbp)
  je     L34
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
L34:
  cmpl   $0x0,-0x10(%rbp)
  je     L64
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  jmp    L54
L64:
  mov    $0x0,%eax
L54:
  leave
  ret
```

assistant: ```c
char *tripcode_pass(char **nameptr) {
  char *name = *nameptr;
  int len = strlen(name);
  int i, found = 0;
  for (i = len - 1; i >= 0; i--) {
    if (name[i] == '#' && &name[i + 1] != NULL) {
      found = !found;
      name[i] = '\0';
      break;
    }
  }
  if (name[0] == '#' && found)
    *nameptr = NULL;
  return (found) ? &name[i + 1] : NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tripcode_pass>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    (%rdi),%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  js     L46
  movslq %eax,%rdx
  add    %rbx,%rdx
  jmp    L39
L50:
  sub    $0x1,%eax
  sub    $0x1,%rdx
  cmp    $0xffffffff,%eax
  je     L63
L39:
  cmpb   $0x23,(%rdx)
  jne    L50
  cmp    $0xffffffffffffffff,%rdx
  je     L50
  movb   $0x0,(%rdx)
  cmpb   $0x23,(%rbx)
  jne    L58
  movq   $0x0,0x0(%rbp)
L58:
  cltq
  lea    0x1(%rbx,%rax,1),%rax
L65:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L63:
  mov    $0x0,%eax
  jmp    L65
L46:
  mov    $0x0,%eax
  jmp    L65
```

assistant: ```c
char *tripcode_pass(char **nameptr) {
  char *name = *nameptr;
  int len = strlen(name);
  int i, found = 0;
  for (i = len - 1; i >= 0; i--) {
    if (name[i] == '#' && &name[i + 1] != NULL) {
      found = !found;
      name[i] = '\0';
      break;
    }
  }
  if (name[0] == '#' && found)
    *nameptr = NULL;
  return (found) ? &name[i + 1] : NULL;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<tripcode_pass>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  js     L32
  movslq %eax,%rdx
  add    %rbx,%rdx
  jmp    L21
  nopl   (%rax)
L29:
  sub    $0x1,%eax
  sub    $0x1,%rdx
  cmp    $0xffffffff,%eax
  je     L32
L21:
  cmpb   $0x23,(%rdx)
  jne    L29
  movb   $0x0,(%rdx)
  cmpb   $0x23,(%rbx)
  jne    L22
  movq   $0x0,0x0(%rbp)
L22:
  add    $0x1,%eax
  add    $0x8,%rsp
  cltq
  add    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *tripcode_pass(char **nameptr) {
  char *name = *nameptr;
  int len = strlen(name);
  int i, found = 0;
  for (i = len - 1; i >= 0; i--) {
    if (name[i] == '#' && &name[i + 1] != NULL) {
      found = !found;
      name[i] = '\0';
      break;
    }
  }
  if (name[0] == '#' && found)
    *nameptr = NULL;
  return (found) ? &name[i + 1] : NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tripcode_pass>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  js     L41
  movslq %eax,%rdx
  add    %rbx,%rdx
  jmp    L18
  nopl   (%rax)
L31:
  sub    $0x1,%eax
  sub    $0x1,%rdx
  cmp    $0xffffffff,%eax
  je     L41
L18:
  cmpb   $0x23,(%rdx)
  jne    L31
  movb   $0x0,(%rdx)
  cmpb   $0x23,(%rbx)
  jne    L42
  movq   $0x0,0x0(%rbp)
L42:
  add    $0x1,%eax
  add    $0x8,%rsp
  cltq
  add    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *tripcode_pass(char **nameptr) {
  char *name = *nameptr;
  int len = strlen(name);
  int i, found = 0;
  for (i = len - 1; i >= 0; i--) {
    if (name[i] == '#' && &name[i + 1] != NULL) {
      found = !found;
      name[i] = '\0';
      break;
    }
  }
  if (name[0] == '#' && found)
    *nameptr = NULL;
  return (found) ? &name[i + 1] : NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L37:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L37
  sub    $0xc70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x9c64(%rbp)
  jmp    L18
L17:
  mov    -0x9c64(%rbp),%eax
  add    $0xc,%eax
  mov    %eax,%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  cmpl   $0x28,-0x9c64(%rbp)
  jne    L2
  mov    -0x9c64(%rbp),%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  jmp    L8
L2:
  mov    -0x9c64(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
L8:
  mov    -0x9c64(%rbp),%eax
  add    $0xc,%eax
  mov    %eax,%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  mov    -0x9c64(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%eax
  mov    -0x9c60(%rbp,%rax,4),%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  mov    -0x9c64(%rbp),%eax
  mov    -0x9c60(%rbp,%rax,4),%edx
  mov    -0x9c64(%rbp),%eax
  add    $0x2,%eax
  mov    %eax,%eax
  mov    -0x9c60(%rbp,%rax,4),%eax
  add    %eax,%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  mov    -0x9c64(%rbp),%eax
  mov    -0x9c60(%rbp,%rax,4),%edx
  mov    -0x9c64(%rbp),%eax
  add    $0x3,%eax
  mov    %eax,%eax
  mov    -0x9c60(%rbp,%rax,4),%eax
  add    %eax,%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  mov    -0x9c64(%rbp),%eax
  mov    -0x9c60(%rbp,%rax,4),%edx
  mov    -0x9c64(%rbp),%eax
  add    $0x4,%eax
  mov    %eax,%eax
  mov    -0x9c60(%rbp,%rax,4),%eax
  add    %eax,%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  mov    -0x9c64(%rbp),%eax
  mov    -0x9c60(%rbp,%rax,4),%edx
  mov    -0x9c64(%rbp),%eax
  add    $0x5,%eax
  mov    %eax,%eax
  mov    -0x9c60(%rbp,%rax,4),%eax
  add    %eax,%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  mov    -0x9c64(%rbp),%eax
  mov    -0x9c60(%rbp,%rax,4),%edx
  mov    -0x9c64(%rbp),%eax
  add    $0x6,%eax
  mov    %eax,%eax
  mov    -0x9c60(%rbp,%rax,4),%eax
  add    %eax,%edx
  mov    -0x9c64(%rbp),%eax
  mov    %edx,-0x9c60(%rbp,%rax,4)
  addl   $0x1,-0x9c64(%rbp)
L18:
  cmpl   $0x270f,-0x9c64(%rbp)
  jbe    L17
  mov    -0x9c10(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

assistant: ```c
int test() {
  int a[10000 + 6];
  unsigned i;
  for (i = 0; i < 10000; i++) {
    a[i] = i + 12;
    if (i == 40)
      a[i] = i;
    else
      a[i] = i + 1;
    a[i] = i + 12;
    a[i] = a[i + 1];
    a[i] += a[i + 2];
    a[i] += a[i + 3];
    a[i] += a[i + 4];
    a[i] += a[i + 5];
    a[i] += a[i + 6];
  }
  return a[20];
}
```
user: Translate the below assembly code into C programming language:

```asm
<test>:
  endbr64
  lea    -0x9000(%rsp),%r11
L17:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L17
  sub    $0xc68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  lea    0x9c40(%rsp),%rcx
L23:
  mov    0x8(%rdx),%eax
  add    0x4(%rdx),%eax
  add    0xc(%rdx),%eax
  add    0x10(%rdx),%eax
  add    0x14(%rdx),%eax
  add    0x18(%rdx),%eax
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L23
  mov    0x50(%rsp),%eax
  mov    0x9c58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x9c68,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test() {
  int a[10000 + 6];
  unsigned i;
  for (i = 0; i < 10000; i++) {
    a[i] = i + 12;
    if (i == 40)
      a[i] = i;
    else
      a[i] = i + 1;
    a[i] = i + 12;
    a[i] = a[i + 1];
    a[i] += a[i + 2];
    a[i] += a[i + 3];
    a[i] += a[i + 4];
    a[i] += a[i + 5];
    a[i] += a[i + 6];
  }
  return a[20];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test>:
  endbr64
  lea    -0x9000(%rsp),%r11
L22:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L22
  sub    $0xc68,%rsp
  xor    %ecx,%ecx
  mov    $0x28,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x9c58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  nopl   0x0(%rax)
L27:
  cmp    $0x28,%ecx
  mov    %esi,%eax
  cmovne 0xa0(%rsp),%eax
  add    $0x1,%ecx
  add    $0x4,%rdx
  mov    %eax,0xa0(%rsp)
  mov    0x4(%rdx),%eax
  add    (%rdx),%eax
  add    0x8(%rdx),%eax
  add    0xc(%rdx),%eax
  add    0x10(%rdx),%eax
  add    0x14(%rdx),%eax
  mov    %eax,-0x4(%rdx)
  cmp    $0x2710,%ecx
  jne    L27
  mov    0x50(%rsp),%eax
  mov    0x9c58(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x9c68,%rsp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test() {
  int a[10000 + 6];
  unsigned i;
  for (i = 0; i < 10000; i++) {
    a[i] = i + 12;
    if (i == 40)
      a[i] = i;
    else
      a[i] = i + 1;
    a[i] = i + 12;
    a[i] = a[i + 1];
    a[i] += a[i + 2];
    a[i] += a[i + 3];
    a[i] += a[i + 4];
    a[i] += a[i + 5];
    a[i] += a[i + 6];
  }
  return a[20];
}
```