ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<unboxString>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rcx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <hExtractTag@plt>
  xor    %edi,%edi
  cmp    $0x40000000,%eax
  sete   %dil
  call   <assert@plt>
  mov    (%rbx),%r12d
  xor    %edi,%edi
  and    $0x1fffffff,%r12d
  cmp    %r14d,%r12d
  setl   %dil
  call   <assert@plt>
  test   %r12d,%r12d
  je     L17
  lea    0x1(%r12),%esi
  mov    $0x1,%eax
  nopl   0x0(%rax)
L0:
  mov    (%rbx,%rax,4),%rdx
  mov    0x4(%rdx),%edx
  mov    %dl,-0x1(%rbp,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L0
L17:
  pop    %rbx
  pop    %rbp
  mov    %r12d,0x0(%r13)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void unboxString(uint32_t *hptr, char *buf, uint32_t buf_len, uint32_t *str_len) {
  uint32_t i;
  uint32_t string_len;
  uint32_t *boxed_char;
  assert(hExtractTag(hptr) == 0x40000000);
  string_len = ((*hptr) & 0x1fffffff);
  assert(string_len <= (buf_len - 1));
  for (i = 0; i < string_len; i++) {
    boxed_char = (((*((uint32_t **)((hptr) + 1 + i)))));
    buf[i] = (char)((*((boxed_char) + 1)));
  }
  *str_len = string_len;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<unboxString>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rcx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <hExtractTag@plt>
  xor    %edi,%edi
  cmp    $0x40000000,%eax
  sete   %dil
  call   <assert@plt>
  mov    (%rbx),%r12d
  xor    %edi,%edi
  and    $0x1fffffff,%r12d
  cmp    %r14d,%r12d
  setl   %dil
  call   <assert@plt>
  test   %r12d,%r12d
  je     L5
  lea    0x1(%r12),%esi
  mov    $0x1,%eax
  nopl   0x0(%rax)
L0:
  mov    (%rbx,%rax,4),%rdx
  mov    0x4(%rdx),%edx
  mov    %dl,-0x1(%rbp,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L0
L5:
  pop    %rbx
  pop    %rbp
  mov    %r12d,0x0(%r13)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void unboxString(uint32_t *hptr, char *buf, uint32_t buf_len, uint32_t *str_len) {
  uint32_t i;
  uint32_t string_len;
  uint32_t *boxed_char;
  assert(hExtractTag(hptr) == 0x40000000);
  string_len = ((*hptr) & 0x1fffffff);
  assert(string_len <= (buf_len - 1));
  for (i = 0; i < string_len; i++) {
    boxed_char = (((*((uint32_t **)((hptr) + 1 + i)))));
    buf[i] = (char)((*((boxed_char) + 1)));
  }
  *str_len = string_len;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<readInput>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movl   $0x400,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
L33:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x11(%rbp)
  cmpb   $0xa,-0x11(%rbp)
  je     L25
  mov    D2(%rip),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x11(%rbp)
  jne    L38
L25:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L37
L38:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x11(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jne    L33
  addl   $0x400,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L33
L37:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *readInput() {
  int buffersize = 1024;
  char *input = malloc(sizeof(char) * buffersize);
  char c;
  int i = 0;
  while (1) {
    c = getchar();
    if (c == '\n' || c == EOF) {
      input[i] = '\0';
      break;
    } else {
      input[i] = c;
    }
    i = i + 1;
    if (i == buffersize) {
      buffersize += 1024;
      input = realloc(input, buffersize);
      input[i] = '\0';
    }
  }
  return input;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readInput>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x400,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x1,%ebx
  mov    $0x400,%r12d
  jmp    L48
L42:
  add    $0x1,%rbx
L48:
  lea    -0x1(%rbx),%r13d
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  je     L16
  mov    D2(%rip),%rdx
  cmp    %al,(%rdx)
  je     L16
  mov    %al,-0x1(%rbp,%rbx,1)
  cmp    %ebx,%r12d
  jne    L42
  add    $0x400,%r12d
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
  movb   $0x0,(%rax,%rbx,1)
  jmp    L42
L16:
  movslq %r13d,%r13
  movb   $0x0,0x0(%rbp,%r13,1)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *readInput() {
  int buffersize = 1024;
  char *input = malloc(sizeof(char) * buffersize);
  char c;
  int i = 0;
  while (1) {
    c = getchar();
    if (c == '\n' || c == EOF) {
      input[i] = '\0';
      break;
    } else {
      input[i] = c;
    }
    i = i + 1;
    if (i == buffersize) {
      buffersize += 1024;
      input = realloc(input, buffersize);
      input[i] = '\0';
    }
  }
  return input;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<readInput>:
  endbr64
  push   %r12
  mov    $0x400,%edi
  mov    $0x400,%r12d
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <malloc@plt>
  mov    %rax,%rbp
  jmp    L38
  nopw   0x0(%rax,%rax,1)
L24:
  add    $0x1,%rbx
L38:
  xor    %eax,%eax
  call   <getchar@plt>
  lea    -0x1(%rbp,%rbx,1),%rdx
  cmp    $0xa,%al
  je     L29
  mov    D8(%rip),%rcx
  cmp    %al,(%rcx)
  je     L29
  mov    %al,(%rdx)
  cmp    %ebx,%r12d
  jne    L24
  add    $0x400,%r12d
  mov    %rbp,%rdi
  mov    %r12d,%esi
  call   <realloc@plt>
  movb   $0x0,(%rax,%rbx,1)
  mov    %rax,%rbp
  jmp    L24
  nopl   0x0(%rax)
L29:
  movb   $0x0,(%rdx)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *readInput() {
  int buffersize = 1024;
  char *input = malloc(sizeof(char) * buffersize);
  char c;
  int i = 0;
  while (1) {
    c = getchar();
    if (c == '\n' || c == EOF) {
      input[i] = '\0';
      break;
    } else {
      input[i] = c;
    }
    i = i + 1;
    if (i == buffersize) {
      buffersize += 1024;
      input = realloc(input, buffersize);
      input[i] = '\0';
    }
  }
  return input;
}
```
user: Translate the below assembly code into C programming language:

```asm
<readInput>:
  endbr64
  push   %r12
  mov    $0x400,%edi
  mov    $0x400,%r12d
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <malloc@plt>
  mov    %rax,%rbp
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L10:
  add    $0x1,%rbx
L15:
  xor    %eax,%eax
  call   <getchar@plt>
  lea    -0x1(%rbp,%rbx,1),%rdx
  cmp    $0xa,%al
  je     L0
  mov    D6(%rip),%rcx
  cmp    %al,(%rcx)
  je     L0
  mov    %al,(%rdx)
  cmp    %ebx,%r12d
  jne    L10
  add    $0x400,%r12d
  mov    %rbp,%rdi
  mov    %r12d,%esi
  call   <realloc@plt>
  movb   $0x0,(%rax,%rbx,1)
  mov    %rax,%rbp
  jmp    L10
  nopl   0x0(%rax)
L0:
  movb   $0x0,(%rdx)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *readInput() {
  int buffersize = 1024;
  char *input = malloc(sizeof(char) * buffersize);
  char c;
  int i = 0;
  while (1) {
    c = getchar();
    if (c == '\n' || c == EOF) {
      input[i] = '\0';
      break;
    } else {
      input[i] = c;
    }
    i = i + 1;
    if (i == buffersize) {
      buffersize += 1024;
      input = realloc(input, buffersize);
      input[i] = '\0';
    }
  }
  return input;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<createMatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L19
L1:
  mov    -0x28(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  addl   $0x1,-0x1c(%rbp)
L19:
  mov    -0x1c(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L1
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int **createMatrix(int r, int c) {
  int i = 0, **pmat;
  pmat = malloc(r * sizeof(int *));
  for (i = 0; i < r; i++) {
    *(pmat + i) = (int *)malloc(c * sizeof(int));
  }
  return pmat;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<createMatrix>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %esi,%ebp
  lea    0x0(,%rdi,8),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L7
  shl    $0x2,%ebp
  mov    %rax,%rbx
  mov    %r12d,%r12d
  lea    (%rax,%r12,8),%r12
L14:
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L14
L7:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int **createMatrix(int r, int c) {
  int i = 0, **pmat;
  pmat = malloc(r * sizeof(int *));
  for (i = 0; i < r; i++) {
    *(pmat + i) = (int *)malloc(c * sizeof(int));
  }
  return pmat;
}
```
user: Translate the below assembly code into C programming language:

```asm
<createMatrix>:
  endbr64
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  lea    0x0(,%r12,8),%edi
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L18
  shl    $0x2,%ebp
  mov    %rax,%rbx
  lea    (%rax,%r12,8),%r12
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %ebp,%edi
  add    $0x8,%rbx
  call   <malloc@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %r12,%rbx
  jne    L5
L18:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int **createMatrix(int r, int c) {
  int i = 0, **pmat;
  pmat = malloc(r * sizeof(int *));
  for (i = 0; i < r; i++) {
    *(pmat + i) = (int *)malloc(c * sizeof(int));
  }
  return pmat;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<createMatrix>:
  endbr64
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  lea    0x0(,%r12,8),%edi
  mov    %esi,%ebp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L11
  shl    $0x2,%ebp
  mov    %rax,%rbx
  lea    (%rax,%r12,8),%r12
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %ebp,%edi
  add    $0x8,%rbx
  call   <malloc@plt>
  mov    %rax,-0x8(%rbx)
  cmp    %r12,%rbx
  jne    L3
L11:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int **createMatrix(int r, int c) {
  int i = 0, **pmat;
  pmat = malloc(r * sizeof(int *));
  for (i = 0; i < r; i++) {
    *(pmat + i) = (int *)malloc(c * sizeof(int));
  }
  return pmat;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fn2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L36
  mov    $0x1,%eax
  jmp    L58
L36:
  subl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L29
  mov    $0x2,%eax
  jmp    L58
L29:
  cmpl   $0x0,-0x4(%rbp)
  jns    L4
  mov    $0x3,%eax
  jmp    L58
L4:
  mov    $0x0,%eax
L58:
  pop    %rbp
  ret
```

assistant: ```c
int fn2(int c) {
  if (c < 0)
    return 1;
  else if (--c == 0)
    return 2;
  else if (c < 0)
    return 3;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<fn2>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  js     L6
  mov    $0x2,%eax
  sub    $0x1,%edi
  je     L6
  mov    %edi,%eax
  sar    $0x1f,%eax
  and    $0x3,%eax
L6:
  ret
```

assistant: ```c
int fn2(int c) {
  if (c < 0)
    return 1;
  else if (--c == 0)
    return 2;
  else if (c < 0)
    return 3;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fn2>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  js     L4
  mov    $0x2,%eax
  sub    $0x1,%edi
  je     L4
  xor    %eax,%eax
  cmp    $0xffffffff,%edi
  sete   %al
  lea    (%rax,%rax,2),%eax
L4:
  ret
```

assistant: ```c
int fn2(int c) {
  if (c < 0)
    return 1;
  else if (--c == 0)
    return 2;
  else if (c < 0)
    return 3;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fn2>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  js     L11
  mov    $0x2,%eax
  sub    $0x1,%edi
  je     L11
  xor    %eax,%eax
  cmp    $0xffffffff,%edi
  sete   %al
  lea    (%rax,%rax,2),%eax
L11:
  ret
```

assistant: ```c
int fn2(int c) {
  if (c < 0)
    return 1;
  else if (--c == 0)
    return 2;
  else if (c < 0)
    return 3;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gui_chat_string_real_pos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  cmpl   $0x0,-0x2c(%rbp)
  jg     L46
  mov    $0x0,%eax
  jmp    L72
L46:
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L16
L38:
  mov    -0x8(%rbp),%rax
  mov    $0x0,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <gui_chat_string_next_char@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L16
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x1f,%al
  jbe    L84
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <utf8_char_size_screen@plt>
  jmp    L9
L84:
  mov    $0x1,%eax
L9:
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L75
  mov    -0x14(%rbp),%eax
  sub    %eax,-0x2c(%rbp)
L75:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <utf8_next_char@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L16:
  cmpq   $0x0,-0x8(%rbp)
  je     L50
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L50
  cmpl   $0x0,-0x2c(%rbp)
  jg     L38
L50:
  mov    -0x10(%rbp),%rax
  sub    -0x28(%rbp),%rax
L72:
  leave
  ret
```

assistant: ```c
int gui_chat_string_real_pos(const char *string, int pos) {
  const char *real_pos, *ptr_string;
  int size_on_screen;
  if (pos <= 0)
    return 0;
  real_pos = string;
  ptr_string = string;
  while (ptr_string && ptr_string[0] && (pos > 0)) {
    ptr_string = gui_chat_string_next_char(NULL, (unsigned char *)ptr_string, 0);
    if (ptr_string) {
      size_on_screen = (((unsigned char)ptr_string[0]) < 32) ? 1 : utf8_char_size_screen(ptr_string);
      if (size_on_screen > 0)
        pos -= size_on_screen;
      ptr_string = utf8_next_char(ptr_string);
      real_pos = ptr_string;
    }
  }
  return 0 + (real_pos - string);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<gui_chat_string_real_pos>:
  endbr64
  mov    $0x0,%eax
  test   %esi,%esi
  jle    L54
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %esi,%ebp
  mov    %rdi,%r12
  mov    $0x1,%r14d
L40:
  mov    %r12,%rbx
L41:
  test   %rbx,%rbx
  je     L77
  cmpb   $0x0,(%rbx)
  je     L77
  test   %ebp,%ebp
  jle    L77
  mov    $0x0,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <gui_chat_string_next_char@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L41
  mov    %r14d,%eax
  cmpb   $0x1f,(%rbx)
  ja     L61
L73:
  sub    %eax,%ebp
L58:
  mov    %rbx,%rdi
  call   <utf8_next_char@plt>
  mov    %rax,%r12
  jmp    L40
L61:
  mov    %rbx,%rdi
  call   <utf8_char_size_screen@plt>
  test   %eax,%eax
  jg     L73
  jmp    L58
L77:
  mov    %r12d,%eax
  sub    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L54:
  ret
```

assistant: ```c
int gui_chat_string_real_pos(const char *string, int pos) {
  const char *real_pos, *ptr_string;
  int size_on_screen;
  if (pos <= 0)
    return 0;
  real_pos = string;
  ptr_string = string;
  while (ptr_string && ptr_string[0] && (pos > 0)) {
    ptr_string = gui_chat_string_next_char(NULL, (unsigned char *)ptr_string, 0);
    if (ptr_string) {
      size_on_screen = (((unsigned char)ptr_string[0]) < 32) ? 1 : utf8_char_size_screen(ptr_string);
      if (size_on_screen > 0)
        pos -= size_on_screen;
      ptr_string = utf8_next_char(ptr_string);
      real_pos = ptr_string;
    }
  }
  return 0 + (real_pos - string);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gui_chat_string_real_pos>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L45
  mov    %rdi,%r13
  test   %rdi,%rdi
  je     L43
  cmpb   $0x0,(%rdi)
  je     L45
  mov    %esi,%r12d
  mov    %rdi,%rbx
  jmp    L67
  nopl   0x0(%rax)
L29:
  cmpb   $0x1f,0x0(%rbp)
  mov    $0x1,%eax
  ja     L38
L74:
  sub    %eax,%r12d
L109:
  mov    %rbp,%rdi
  call   <utf8_next_char@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L43
  cmpb   $0x0,(%rax)
  je     L2
  test   %r12d,%r12d
  jle    L2
L67:
  xor    %edx,%edx
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <gui_chat_string_next_char@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L29
L2:
  sub    %r13d,%ebx
L45:
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L38:
  mov    %rbp,%rdi
  call   <utf8_char_size_screen@plt>
  test   %eax,%eax
  jg     L74
  jmp    L109
L43:
  mov    %r13d,%ebx
  neg    %ebx
  jmp    L45
```

assistant: ```c
int gui_chat_string_real_pos(const char *string, int pos) {
  const char *real_pos, *ptr_string;
  int size_on_screen;
  if (pos <= 0)
    return 0;
  real_pos = string;
  ptr_string = string;
  while (ptr_string && ptr_string[0] && (pos > 0)) {
    ptr_string = gui_chat_string_next_char(NULL, (unsigned char *)ptr_string, 0);
    if (ptr_string) {
      size_on_screen = (((unsigned char)ptr_string[0]) < 32) ? 1 : utf8_char_size_screen(ptr_string);
      if (size_on_screen > 0)
        pos -= size_on_screen;
      ptr_string = utf8_next_char(ptr_string);
      real_pos = ptr_string;
    }
  }
  return 0 + (real_pos - string);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gui_chat_string_real_pos>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L84
  mov    %rdi,%r13
  test   %rdi,%rdi
  je     L94
  cmpb   $0x0,(%rdi)
  je     L84
  mov    %esi,%r12d
  mov    %rdi,%rbx
  jmp    L57
  nopl   0x0(%rax)
L37:
  cmpb   $0x1f,0x0(%rbp)
  mov    $0x1,%eax
  ja     L62
L17:
  sub    %eax,%r12d
L52:
  mov    %rbp,%rdi
  call   <utf8_next_char@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L94
  cmpb   $0x0,(%rax)
  je     L27
  test   %r12d,%r12d
  jle    L27
L57:
  xor    %edx,%edx
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <gui_chat_string_next_char@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L37
L27:
  sub    %r13d,%ebx
L84:
  add    $0x8,%rsp
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L62:
  mov    %rbp,%rdi
  call   <utf8_char_size_screen@plt>
  test   %eax,%eax
  jg     L17
  jmp    L52
L94:
  mov    %r13d,%ebx
  neg    %ebx
  jmp    L84
```

assistant: ```c
int gui_chat_string_real_pos(const char *string, int pos) {
  const char *real_pos, *ptr_string;
  int size_on_screen;
  if (pos <= 0)
    return 0;
  real_pos = string;
  ptr_string = string;
  while (ptr_string && ptr_string[0] && (pos > 0)) {
    ptr_string = gui_chat_string_next_char(NULL, (unsigned char *)ptr_string, 0);
    if (ptr_string) {
      size_on_screen = (((unsigned char)ptr_string[0]) < 32) ? 1 : utf8_char_size_screen(ptr_string);
      if (size_on_screen > 0)
        pos -= size_on_screen;
      ptr_string = utf8_next_char(ptr_string);
      real_pos = ptr_string;
    }
  }
  return 0 + (real_pos - string);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printArray>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L9:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L9
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
type_infered: string

tool: D7: %d 

assistant: ```c
void printArray(int arr[], int size) {
  int i;
  for (i = 0; i < size; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<printArray>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L12
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%r12
  lea    D0(%rip),%rbp
L6:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L6
L12:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D0: %d 

assistant: ```c
void printArray(int arr[], int size) {
  int i;
  for (i = 0; i < size; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<printArray>:
  endbr64
  test   %esi,%esi
  jle    L5
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L2:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L2
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L5:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D2
type_guessed: string

tool: D2: %d 

assistant: ```c
void printArray(int arr[], int size) {
  int i;
  for (i = 0; i < size; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<printArray>:
  endbr64
  test   %esi,%esi
  jle    L14
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L7:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L7
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L14:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
type: string

tool: D4: %d 

assistant: ```c
void printArray(int arr[], int size) {
  int i;
  for (i = 0; i < size; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<broken>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%eax
  call   <clear_screen@plt>
  mov    $0x0,%eax
  call   <reset_term@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type: string

tool: D3: Internal error: %s


assistant: ```c
broken(thing) char *thing;
{
  clear_screen();
  reset_term();
  printf("Internal error: %s\n", thing);
  exit(1);
}
```