ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<tty_getc_irq>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  nopw   0x0(%rax,%rax,1)
L0:
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  mov    $0x1,%edx
  mov    %ebx,%esi
  mov    $0xa,%edi
  call   <sys_call@plt>
  test   %eax,%eax
  je     L0
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned int tty_getc_irq(char *byte) {
  unsigned int ret = 0;
  while (ret == 0) {
    ret = sys_call(0x0A, (unsigned int)byte, 1, 0, 0);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<readWords>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %ecx,-0x44(%rbp)
  mov    %r8,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1f4,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x50(%rbp),%rax
  movl   $0x0,(%rax)
  jmp    L1
L67:
  lea    -0x20(%rbp),%rdi
  lea    -0x24(%rbp),%rcx
  mov    -0x40(%rbp),%edx
  mov    -0x3c(%rbp),%esi
  mov    -0x38(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <readlnwht@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L18
  mov    -0x50(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x50(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x50(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jge    L61
  shll   -0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x18(%rbp)
L61:
  mov    -0x50(%rbp),%rax
  mov    (%rax),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,(%rdx)
L18:
  mov    -0x24(%rbp),%edx
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  je     L9
L1:
  mov    -0x50(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x44(%rbp)
  jg     L67
  jmp    L8
L9:
  nop
L8:
  mov    -0x50(%rbp),%rax
  mov    (%rax),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
char **readWords(FILE *f, int min_word_chars, int max_word_chars, int max_words, int *nof_words) {
  int buf_siz = 500;
  char **buf = malloc(buf_siz * sizeof(char *));
  int c;
  *nof_words = 0;
  char *str;
  int chars_read;
  while (*nof_words < max_words) {
    str = readlnwht(f, min_word_chars, max_word_chars, &c, &chars_read);
    if (str) {
      (*nof_words)++;
      if (buf_siz < *nof_words) {
        buf = realloc(buf, (buf_siz *= 2) * sizeof(char *));
      }
      buf[*nof_words - 1] = str;
    }
    if (c == EOF) {
      break;
    }
  }
  return realloc(buf, *nof_words * sizeof(char *));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<readWords>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,(%rsp)
  mov    %esi,0x8(%rsp)
  mov    %edx,0xc(%rsp)
  mov    %ecx,%r14d
  mov    %r8,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0xfa0,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  movl   $0x0,0x0(%rbp)
  mov    $0x1f4,%r13d
  lea    0x14(%rsp),%r15
  jmp    L48
L23:
  movslq 0x0(%rbp),%rax
  mov    %rbx,-0x8(%r12,%rax,8)
L29:
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,0x10(%rsp)
  je     L63
L48:
  cmp    %r14d,0x0(%rbp)
  jge    L63
  lea    0x10(%rsp),%rcx
  mov    %r15,%r8
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  mov    (%rsp),%rdi
  call   <readlnwht@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L29
  mov    0x0(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,0x0(%rbp)
  cmp    %r13d,%eax
  jle    L23
  add    %r13d,%r13d
  lea    0x0(,%r13,8),%esi
  mov    %r12,%rdi
  call   <realloc@plt>
  mov    %rax,%r12
  jmp    L23
L63:
  mov    0x0(%rbp),%esi
  shl    $0x3,%esi
  mov    %r12,%rdi
  call   <realloc@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L52
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L52:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **readWords(FILE *f, int min_word_chars, int max_word_chars, int max_words, int *nof_words) {
  int buf_siz = 500;
  char **buf = malloc(buf_siz * sizeof(char *));
  int c;
  *nof_words = 0;
  char *str;
  int chars_read;
  while (*nof_words < max_words) {
    str = readlnwht(f, min_word_chars, max_word_chars, &c, &chars_read);
    if (str) {
      (*nof_words)++;
      if (buf_siz < *nof_words) {
        buf = realloc(buf, (buf_siz *= 2) * sizeof(char *));
      }
      buf[*nof_words - 1] = str;
    }
    if (c == EOF) {
      break;
    }
  }
  return realloc(buf, *nof_words * sizeof(char *));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<readWords>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  mov    $0xfa0,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x28,%rsp
  mov    %ecx,0x4(%rsp)
  lea    0x14(%rsp),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  movl   $0x0,(%rbx)
  xor    %esi,%esi
  movl   $0x1f4,(%rsp)
  mov    %rax,%rbp
  jmp    L63
  nopl   (%rax)
L23:
  mov    (%rbx),%eax
  lea    0x1(%rax),%esi
  mov    %esi,(%rbx)
  cmp    (%rsp),%esi
  jg     L67
L25:
  movslq %esi,%rax
  mov    %rdx,-0x8(%rbp,%rax,8)
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,0x10(%rsp)
  je     L39
L63:
  cmp    %esi,0x4(%rsp)
  jle    L39
  mov    %r12d,%edx
  lea    0x10(%rsp),%rcx
  mov    %r15,%r8
  mov    %r13d,%esi
  mov    %r14,%rdi
  call   <readlnwht@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L23
  mov    D18(%rip),%rax
  mov    (%rbx),%esi
  mov    (%rax),%eax
  cmp    %eax,0x10(%rsp)
  jne    L63
L39:
  shl    $0x3,%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L61
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L67:
  shll   (%rsp)
  mov    (%rsp),%eax
  mov    %rbp,%rdi
  mov    %rdx,0x8(%rsp)
  lea    0x0(,%rax,8),%esi
  call   <realloc@plt>
  mov    (%rbx),%esi
  mov    0x8(%rsp),%rdx
  mov    %rax,%rbp
  jmp    L25
L61:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **readWords(FILE *f, int min_word_chars, int max_word_chars, int max_words, int *nof_words) {
  int buf_siz = 500;
  char **buf = malloc(buf_siz * sizeof(char *));
  int c;
  *nof_words = 0;
  char *str;
  int chars_read;
  while (*nof_words < max_words) {
    str = readlnwht(f, min_word_chars, max_word_chars, &c, &chars_read);
    if (str) {
      (*nof_words)++;
      if (buf_siz < *nof_words) {
        buf = realloc(buf, (buf_siz *= 2) * sizeof(char *));
      }
      buf[*nof_words - 1] = str;
    }
    if (c == EOF) {
      break;
    }
  }
  return realloc(buf, *nof_words * sizeof(char *));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<readWords>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  mov    $0xfa0,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x28,%rsp
  mov    %ecx,0x4(%rsp)
  lea    0x14(%rsp),%r15
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  movl   $0x0,(%rbx)
  xor    %esi,%esi
  movl   $0x1f4,(%rsp)
  mov    %rax,%rbp
  jmp    L11
  nopl   (%rax)
L62:
  mov    (%rbx),%eax
  lea    0x1(%rax),%esi
  mov    %esi,(%rbx)
  cmp    (%rsp),%esi
  jg     L19
L56:
  movslq %esi,%rax
  mov    %rdx,-0x8(%rbp,%rax,8)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,0x10(%rsp)
  je     L21
L11:
  cmp    %esi,0x4(%rsp)
  jle    L21
  mov    %r12d,%edx
  lea    0x10(%rsp),%rcx
  mov    %r15,%r8
  mov    %r13d,%esi
  mov    %r14,%rdi
  call   <readlnwht@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L62
  mov    D2(%rip),%rax
  mov    (%rbx),%esi
  mov    (%rax),%eax
  cmp    %eax,0x10(%rsp)
  jne    L11
L21:
  shl    $0x3,%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L73
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  shll   (%rsp)
  mov    (%rsp),%eax
  mov    %rbp,%rdi
  mov    %rdx,0x8(%rsp)
  lea    0x0(,%rax,8),%esi
  call   <realloc@plt>
  mov    (%rbx),%esi
  mov    0x8(%rsp),%rdx
  mov    %rax,%rbp
  jmp    L56
L73:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char **readWords(FILE *f, int min_word_chars, int max_word_chars, int max_words, int *nof_words) {
  int buf_siz = 500;
  char **buf = malloc(buf_siz * sizeof(char *));
  int c;
  *nof_words = 0;
  char *str;
  int chars_read;
  while (*nof_words < max_words) {
    str = readlnwht(f, min_word_chars, max_word_chars, &c, &chars_read);
    if (str) {
      (*nof_words)++;
      if (buf_siz < *nof_words) {
        buf = realloc(buf, (buf_siz *= 2) * sizeof(char *));
      }
      buf[*nof_words - 1] = str;
    }
    if (c == EOF) {
      break;
    }
  }
  return realloc(buf, *nof_words * sizeof(char *));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<array_freq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L10
L22:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x28(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <str_cmp@plt>
  test   %rax,%rax
  jne    L5
  addl   $0x1,-0x4(%rbp)
L5:
  addl   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L22
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
extern int array_freq(char *keys[], int len, const char *key) {
  int i, cnt = 0;
  for (i = 0; i < len; ++i)
    if (str_cmp(keys[i], key) == 0)
      cnt++;
  return cnt;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<array_freq>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L9
  mov    %rdx,%r12
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%r13
  mov    $0x0,%ebp
L26:
  mov    %r12,%rsi
  mov    (%rbx),%rdi
  call   <str_cmp@plt>
  cmp    $0x1,%rax
  adc    $0x0,%ebp
  add    $0x8,%rbx
  cmp    %r13,%rbx
  jne    L26
L12:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  mov    $0x0,%ebp
  jmp    L12
```

assistant: ```c
extern int array_freq(char *keys[], int len, const char *key) {
  int i, cnt = 0;
  for (i = 0; i < len; ++i)
    if (str_cmp(keys[i], key) == 0)
      cnt++;
  return cnt;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<array_freq>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L7
  lea    -0x1(%rsi),%eax
  mov    %rdx,%rbp
  mov    %rdi,%rbx
  xor    %r12d,%r12d
  lea    0x8(%rdi,%rax,8),%r13
  nopl   0x0(%rax,%rax,1)
L12:
  mov    (%rbx),%rdi
  mov    %rbp,%rsi
  call   <str_cmp@plt>
  cmp    $0x1,%rax
  adc    $0x0,%r12d
  add    $0x8,%rbx
  cmp    %r13,%rbx
  jne    L12
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
extern int array_freq(char *keys[], int len, const char *key) {
  int i, cnt = 0;
  for (i = 0; i < len; ++i)
    if (str_cmp(keys[i], key) == 0)
      cnt++;
  return cnt;
}
```
user: Translate the below assembly code into C programming language:

```asm
<array_freq>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L12
  lea    -0x1(%rsi),%eax
  mov    %rdx,%rbp
  mov    %rdi,%rbx
  xor    %r12d,%r12d
  lea    0x8(%rdi,%rax,8),%r13
  nopl   0x0(%rax,%rax,1)
L19:
  mov    (%rbx),%rdi
  mov    %rbp,%rsi
  call   <str_cmp@plt>
  cmp    $0x1,%rax
  adc    $0x0,%r12d
  add    $0x8,%rbx
  cmp    %r13,%rbx
  jne    L19
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
extern int array_freq(char *keys[], int len, const char *key) {
  int i, cnt = 0;
  for (i = 0; i < len; ++i)
    if (str_cmp(keys[i], key) == 0)
      cnt++;
  return cnt;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x20(%rbp)
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x14(%rbp)
  movl   $0x1,-0x20(%rbp)
  jmp    L20
L42:
  movl   $0x1,-0x1c(%rbp)
  jmp    L22
L48:
  movl   $0x1,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L22:
  mov    -0x1c(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L48
  mov    -0x18(%rbp),%eax
  imul   -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x20(%rbp)
L20:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x20(%rbp)
  jle    L42
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D13
type_infered: string
name: unpack_data
arguments: data_label: D3
type_infered: string

tool: D13: Enter any number

tool: D3: %d

assistant: ```c
int main() {
  int a, b, i = 1, j, p, m, sum;
  printf("Enter any number");
  scanf("%d", &a);
  m = 1;
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      sum = 1;
    }
    p = sum * m;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L11
  add    $0x1,%edx
  mov    $0x1,%eax
L14:
  add    $0x1,%eax
  cmp    %edx,%eax
  jne    L14
L11:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D16
type_guessed: string
name: unpack_data
arguments: label: D11
type_guessed: string

tool: D16: Enter any number

tool: D11: %d

assistant: ```c
int main() {
  int a, b, i = 1, j, p, m, sum;
  printf("Enter any number");
  scanf("%d", &a);
  m = 1;
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      sum = 1;
    }
    p = sum * m;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D12(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D12
data_type: string
name: extract_symbolic_data
arguments: data_label: D10
data_type: string

tool: D12: Enter any number

tool: D10: %d

assistant: ```c
int main() {
  int a, b, i = 1, j, p, m, sum;
  printf("Enter any number");
  scanf("%d", &a);
  m = 1;
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      sum = 1;
    }
    p = sum * m;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D15(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D15
type: string
name: parse_data
arguments: data_label: D8
type: string

tool: D15: Enter any number

tool: D8: %d

assistant: ```c
int main() {
  int a, b, i = 1, j, p, m, sum;
  printf("Enter any number");
  scanf("%d", &a);
  m = 1;
  for (i = 1; i <= a; i++) {
    for (j = 1; j <= b; j++) {
      sum = 1;
    }
    p = sum * m;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x14(%rbp)
  movl   $0x2,-0x10(%rbp)
  movl   $0x3,-0xc(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x2,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_infered: string
name: unpack_data
arguments: data_label: D0
type_infered: string

tool: D5: I'm going to pick contestant number %i


tool: D0: %i


assistant: ```c
int main() {
  int contestants[] = {1, 2, 3};
  int *choice = contestants;
  contestants[0] = 2;
  contestants[1] = contestants[2];
  contestants[2] = *choice;
  printf("I'm going to pick contestant number %i\n", contestants[2]);
  printf("%i\n", contestants[1]);
  printf("%i\n", contestants[0]);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2,%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  lea    D17(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x2,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: label: D12
datatype: string
name: disassemble_data
arguments: label: D17
datatype: string

tool: D12: I'm going to pick contestant number %i


tool: D17: %i


assistant: ```c
int main() {
  int contestants[] = {1, 2, 3};
  int *choice = contestants;
  contestants[0] = 2;
  contestants[1] = contestants[2];
  contestants[2] = *choice;
  printf("I'm going to pick contestant number %i\n", contestants[2]);
  printf("%i\n", contestants[1]);
  printf("%i\n", contestants[0]);
  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
  mov    $0x2,%esi
  lea    D12(%rip),%rbp
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
type: string
name: disassemble_data
arguments: data_tag: D17
type: string

tool: D12: %i


tool: D17: I'm going to pick contestant number %i


assistant: ```c
int main() {
  int contestants[] = {1, 2, 3};
  int *choice = contestants;
  contestants[0] = 2;
  contestants[1] = contestants[2];
  contestants[2] = *choice;
  printf("I'm going to pick contestant number %i\n", contestants[2]);
  printf("%i\n", contestants[1]);
  printf("%i\n", contestants[0]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x2,%esi
  lea    D4(%rip),%rbp
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x2,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
type: string
name: unpack_data
arguments: tag: D19
type: string

tool: D4: %i


tool: D19: I'm going to pick contestant number %i


assistant: ```c
int main() {
  int contestants[] = {1, 2, 3};
  int *choice = contestants;
  contestants[0] = 2;
  contestants[1] = contestants[2];
  contestants[2] = *choice;
  printf("I'm going to pick contestant number %i\n", contestants[2]);
  printf("%i\n", contestants[1]);
  printf("%i\n", contestants[0]);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<icbrt_with_guess>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %edi,-0xc(%rbp)
  mov    %esi,-0x10(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jg     L23
  mov    $0x0,%eax
  jmp    L3
L23:
  cmpl   $0x0,-0x10(%rbp)
  jg     L47
  movl   $0x1,-0x10(%rbp)
L47:
  mov    -0x10(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  cltd
  idiv   %ecx
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  sub    %edx,%eax
  movslq %eax,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%eax
  mov    %edx,%ebx
  sub    %eax,%ebx
  sub    %ebx,-0x10(%rbp)
  test   %ebx,%ebx
  jne    L47
  mov    -0x10(%rbp),%eax
  imul   %eax,%eax
  imul   -0x10(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jge    L32
  subl   $0x1,-0x10(%rbp)
L32:
  mov    -0x10(%rbp),%eax
L3:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int icbrt_with_guess(int a, int guess) {
  register int delta;
  if (a <= 0)
    return 0;
  if (guess < 1)
    guess = 1;
  do {
    delta = (guess - a / (guess * guess)) / 3;
    guess -= delta;
  } while (delta != 0);
  if (guess * guess * guess > a)
    guess--;
  return guess;
}
```
user: What is the c source code of the assembly code below:

```asm
<icbrt_with_guess>:
  endbr64
  mov    $0x0,%ecx
  test   %edi,%edi
  jle    L15
  test   %esi,%esi
  mov    $0x1,%ecx
  cmovg  %esi,%ecx
L18:
  mov    %ecx,%esi
  imul   %ecx,%esi
  mov    %edi,%eax
  cltd
  idiv   %esi
  mov    %ecx,%esi
  sub    %eax,%esi
  movslq %esi,%rdx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x1f,%esi
  sub    %esi,%edx
  sub    %edx,%ecx
  test   %edx,%edx
  jne    L18
  mov    %ecx,%eax
  imul   %ecx,%eax
  imul   %ecx,%eax
  cmp    %edi,%eax
  setg   %al
  movzbl %al,%eax
  sub    %eax,%ecx
L15:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int icbrt_with_guess(int a, int guess) {
  register int delta;
  if (a <= 0)
    return 0;
  if (guess < 1)
    guess = 1;
  do {
    delta = (guess - a / (guess * guess)) / 3;
    guess -= delta;
  } while (delta != 0);
  if (guess * guess * guess > a)
    guess--;
  return guess;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<icbrt_with_guess>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  jle    L2
  test   %esi,%esi
  mov    $0x1,%r8d
  cmovg  %esi,%r8d
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %r8d,%ecx
  mov    %edi,%eax
  imul   %r8d,%ecx
  cltd
  idiv   %ecx
  mov    %r8d,%ecx
  sub    %eax,%ecx
  movslq %ecx,%rdx
  sar    $0x1f,%ecx
  imul   $0x55555556,%rdx,%rdx
  shr    $0x20,%rdx
  sub    %ecx,%edx
  sub    %edx,%r8d
  test   %edx,%edx
  jne    L19
  mov    %r8d,%eax
  imul   %r8d,%eax
  imul   %r8d,%eax
  cmp    %edi,%eax
  setg   %al
  movzbl %al,%eax
  sub    %eax,%r8d
L2:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int icbrt_with_guess(int a, int guess) {
  register int delta;
  if (a <= 0)
    return 0;
  if (guess < 1)
    guess = 1;
  do {
    delta = (guess - a / (guess * guess)) / 3;
    guess -= delta;
  } while (delta != 0);
  if (guess * guess * guess > a)
    guess--;
  return guess;
}
```