ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<time_diff>:
  endbr64
  mov    (%rdi),%rax
  mov    0x8(%rdi),%rcx
  sub    (%rsi),%rax
  mov    0x8(%rsi),%rsi
  mov    %rax,(%rdx)
  cmp    %rsi,%rcx
  jge    L7
  sub    $0x1,%rax
  add    $0x3b9aca00,%rcx
  mov    %rax,(%rdx)
  sub    %rsi,%rcx
  not    %rax
  mov    %rcx,0x8(%rdx)
  shr    $0x3f,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  sub    %rsi,%rcx
  not    %rax
  mov    %rcx,0x8(%rdx)
  shr    $0x3f,%rax
  ret
```

assistant: ```c
int time_diff(struct timespec *tv1, struct timespec *tv2, struct timespec *tvdiff) {
  tvdiff->tv_sec = tv1->tv_sec - tv2->tv_sec;
  if (tv1->tv_nsec < tv2->tv_nsec) {
    tvdiff->tv_nsec = 1000000000 + tv1->tv_nsec - tv2->tv_nsec;
    tvdiff->tv_sec -= 1;
  } else {
    tvdiff->tv_nsec = tv1->tv_nsec - tv2->tv_nsec;
  }
  return (tvdiff->tv_sec >= 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<time_diff>:
  endbr64
  mov    (%rdi),%rax
  mov    0x8(%rdi),%rdi
  sub    (%rsi),%rax
  mov    0x8(%rsi),%rsi
  mov    %rdi,%rcx
  mov    %rax,(%rdx)
  sub    %rsi,%rcx
  cmp    %rsi,%rdi
  jge    L4
  sub    $0x1,%rax
  lea    0x3b9aca00(%rdi),%rcx
  mov    %rax,(%rdx)
  sub    %rsi,%rcx
L4:
  not    %rax
  mov    %rcx,0x8(%rdx)
  shr    $0x3f,%rax
  ret
```

assistant: ```c
int time_diff(struct timespec *tv1, struct timespec *tv2, struct timespec *tvdiff) {
  tvdiff->tv_sec = tv1->tv_sec - tv2->tv_sec;
  if (tv1->tv_nsec < tv2->tv_nsec) {
    tvdiff->tv_nsec = 1000000000 + tv1->tv_nsec - tv2->tv_nsec;
    tvdiff->tv_sec -= 1;
  } else {
    tvdiff->tv_nsec = tv1->tv_nsec - tv2->tv_nsec;
  }
  return (tvdiff->tv_sec >= 0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x30(%rbp)
  movl   $0x2,-0x2c(%rbp)
  movl   $0x22,-0x28(%rbp)
  movl   $0x5,-0x24(%rbp)
  movl   $0x43,-0x20(%rbp)
  movl   $0x3,-0x1c(%rbp)
  movl   $0x17,-0x18(%rbp)
  movl   $0xc,-0x14(%rbp)
  movl   $0xd,-0x10(%rbp)
  movl   $0xa,-0xc(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xa,%edi
  call   <sort@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int test[] = {1, 2, 34, 5, 67, 3, 23, 12, 13, 10};
  sort(10, test);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x22,0x8(%rsp)
  movl   $0x5,0xc(%rsp)
  movl   $0x43,0x10(%rsp)
  movl   $0x3,0x14(%rsp)
  movl   $0x17,0x18(%rsp)
  movl   $0xc,0x1c(%rsp)
  movl   $0xd,0x20(%rsp)
  movl   $0xa,0x24(%rsp)
  mov    %rsp,%rsi
  mov    $0xa,%edi
  call   <sort@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x38,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int test[] = {1, 2, 34, 5, 67, 3, 23, 12, 13, 10};
  sort(10, test);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    $0xa,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x200000001,%rax
  mov    %rsp,%rsi
  mov    %rax,(%rsp)
  movabs $0x500000022,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x300000043,%rax
  mov    %rax,0x10(%rsp)
  movabs $0xc00000017,%rax
  mov    %rax,0x18(%rsp)
  movabs $0xa0000000d,%rax
  mov    %rax,0x20(%rsp)
  call   <sort@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int test[] = {1, 2, 34, 5, 67, 3, 23, 12, 13, 10};
  sort(10, test);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  movdqa D15(%rip),%xmm0
  mov    $0xa,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  mov    %rsp,%rsi
  movaps %xmm0,(%rsp)
  movdqa D13(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <sort@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D15
datatype: byte[16]
name: parse_data
arguments: tag: D8
datatype: byte[8]
name: parse_data
arguments: tag: D13
datatype: byte[16]

tool: {"D15": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x22,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: {"D8": [0x0d,0x00,0x00,0x00,0x0a,0x00,0x00,0x00]}

tool: {"D13": [0x43,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x17,0x00,0x00,0x00,0x0c,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, char **argv) {
  int test[] = {1, 2, 34, 5, 67, 3, 23, 12, 13, 10};
  sort(10, test);
}
```
user: What is the c source code of the assembly code below:

```asm
<is_bracket>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movsbl -0x4(%rbp),%eax
  cmp    $0x28,%eax
  je     L18
  cmp    $0x29,%eax
  je     L26
  jmp    L19
L18:
  mov    $0x1,%eax
  jmp    L36
L26:
  mov    $0x2,%eax
  jmp    L36
L19:
  mov    $0x0,%eax
L36:
  pop    %rbp
  ret
```

assistant: ```c
int is_bracket(char c) {
  switch (c) {
  case '(':
    return 1;
  case ')':
    return 2;
  default:
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_bracket>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x28,%dil
  je     L0
  cmp    $0x29,%dil
  sete   %al
  movzbl %al,%eax
  add    %eax,%eax
L0:
  ret
```

assistant: ```c
int is_bracket(char c) {
  switch (c) {
  case '(':
    return 1;
  case ')':
    return 2;
  default:
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<is_bracket>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x28,%dil
  je     L7
  xor    %eax,%eax
  cmp    $0x29,%dil
  sete   %al
  add    %eax,%eax
L7:
  ret
```

assistant: ```c
int is_bracket(char c) {
  switch (c) {
  case '(':
    return 1;
  case ')':
    return 2;
  default:
    return 0;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_bracket>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x28,%dil
  je     L3
  xor    %eax,%eax
  cmp    $0x29,%dil
  sete   %al
  add    %eax,%eax
L3:
  ret
```

assistant: ```c
int is_bracket(char c) {
  switch (c) {
  case '(':
    return 1;
  case ')':
    return 2;
  default:
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strerror>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  lea    D0(%rip),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L0
L9:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jne    L40
  mov    -0x10(%rbp),%rax
  mov    0x8(%rax),%rax
  jmp    L13
L40:
  addq   $0x10,-0x10(%rbp)
L0:
  lea    D8(%rip),%rax
  cmp    %rax,-0x10(%rbp)
  jb     L9
  lea    D5(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x14(%rbp)
L27:
  mov    -0x14(%rbp),%ecx
  mov    %ecx,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  mov    %eax,%edx
  shr    $0x3,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%eax
  add    $0x30,%eax
  subq   $0x1,-0x8(%rbp)
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L27
  lea    D33(%rip),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  lea    D37(%rip),%rax
L13:
  leave
  ret
```

assistant: ```c
char *strerror(int err) {
  static struct errlist {
    int err;
    char *what;
  } errlist[] = {
      {0, "No error"}, {1, "Function number invalid"}, {2, "File not found"}, {3, "Path not found"}, {4, "Too many open files"}, {5, "Access denied"}, {6, "Invalid handle"}, {12, "Access code invalid"}, {39, "Insufficient disk space"},
  };
  struct errlist *ep;
  static char unknown[] = "Error 65535";
  unsigned e;
  char *p;
  for (ep = errlist; ep < errlist + sizeof(errlist) / sizeof(errlist[0]); ep++) {
    if (ep->err == err)
      return ep->what;
  }
  p = unknown + sizeof(unknown) - 1;
  e = err;
  do
    *--p = '0' + (e % 10);
  while ((e /= 10) > 0);
  strcpy(unknown + 6, p);
  return unknown;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strerror>:
  endbr64
  lea    D25(%rip),%rax
  lea    0x90(%rax),%rdx
L13:
  cmp    %edi,(%rax)
  je     L20
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L13
  sub    $0x8,%rsp
  lea    D16(%rip),%rsi
  mov    $0xcccccccd,%r8d
L23:
  sub    $0x1,%rsi
  mov    %edi,%eax
  imul   %r8,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edi,%edx
  sub    %ecx,%edx
  add    $0x30,%edx
  mov    %dl,(%rsi)
  mov    %edi,%edx
  mov    %eax,%edi
  cmp    $0x9,%edx
  ja     L23
  lea    D11(%rip),%rdi
  call   <strcpy@plt>
  lea    D18(%rip),%rax
  add    $0x8,%rsp
  ret
L20:
  mov    0x8(%rax),%rax
  ret
```

assistant: ```c
char *strerror(int err) {
  static struct errlist {
    int err;
    char *what;
  } errlist[] = {
      {0, "No error"}, {1, "Function number invalid"}, {2, "File not found"}, {3, "Path not found"}, {4, "Too many open files"}, {5, "Access denied"}, {6, "Invalid handle"}, {12, "Access code invalid"}, {39, "Insufficient disk space"},
  };
  struct errlist *ep;
  static char unknown[] = "Error 65535";
  unsigned e;
  char *p;
  for (ep = errlist; ep < errlist + sizeof(errlist) / sizeof(errlist[0]); ep++) {
    if (ep->err == err)
      return ep->what;
  }
  p = unknown + sizeof(unknown) - 1;
  e = err;
  do
    *--p = '0' + (e % 10);
  while ((e /= 10) > 0);
  strcpy(unknown + 6, p);
  return unknown;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strerror>:
  endbr64
  lea    D23(%rip),%rax
  lea    0x90(%rax),%rdx
  nopw   0x0(%rax,%rax,1)
L5:
  cmp    %edi,(%rax)
  je     L20
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L5
  sub    $0x8,%rsp
  lea    D14(%rip),%rsi
  mov    $0xcccccccd,%r8d
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %edi,%eax
  mov    %edi,%edx
  sub    $0x1,%rsi
  imul   %r8,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    $0x30,%edx
  mov    %dl,(%rsi)
  mov    %edi,%edx
  mov    %eax,%edi
  cmp    $0x9,%edx
  ja     L14
  lea    D29(%rip),%rdi
  call   <strcpy@plt>
  lea    D10(%rip),%rax
  add    $0x8,%rsp
  ret
  nopl   (%rax)
L20:
  mov    0x8(%rax),%rax
  ret
```

assistant: ```c
char *strerror(int err) {
  static struct errlist {
    int err;
    char *what;
  } errlist[] = {
      {0, "No error"}, {1, "Function number invalid"}, {2, "File not found"}, {3, "Path not found"}, {4, "Too many open files"}, {5, "Access denied"}, {6, "Invalid handle"}, {12, "Access code invalid"}, {39, "Insufficient disk space"},
  };
  struct errlist *ep;
  static char unknown[] = "Error 65535";
  unsigned e;
  char *p;
  for (ep = errlist; ep < errlist + sizeof(errlist) / sizeof(errlist[0]); ep++) {
    if (ep->err == err)
      return ep->what;
  }
  p = unknown + sizeof(unknown) - 1;
  e = err;
  do
    *--p = '0' + (e % 10);
  while ((e /= 10) > 0);
  strcpy(unknown + 6, p);
  return unknown;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strerror>:
  endbr64
  cmp    D104(%rip),%edi
  je     L77
  cmp    D125(%rip),%edi
  je     L13
  cmp    D76(%rip),%edi
  je     L113
  cmp    D24(%rip),%edi
  je     L47
  cmp    D106(%rip),%edi
  je     L100
  cmp    D197(%rip),%edi
  je     L56
  cmp    D118(%rip),%edi
  je     L19
  cmp    D29(%rip),%edi
  je     L45
  cmp    %edi,D81(%rip)
  lea    D89(%rip),%rsi
  mov    $0xcccccccd,%r8d
  lea    D81(%rip),%rax
  je     L44
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L2:
  mov    %edi,%eax
  mov    %edi,%edx
  sub    $0x1,%rsi
  imul   %r8,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    $0x30,%edx
  mov    %dl,(%rsi)
  mov    %edi,%edx
  mov    %eax,%edi
  cmp    $0x9,%edx
  ja     L2
  lea    D138(%rip),%rdi
  call   <strcpy@plt>
  lea    D19(%rip),%rax
  add    $0x8,%rsp
  ret
L77:
  lea    D104(%rip),%rax
L44:
  mov    0x8(%rax),%rax
  ret
L13:
  lea    D125(%rip),%rax
  jmp    L44
L113:
  lea    D76(%rip),%rax
  jmp    L44
L47:
  lea    D24(%rip),%rax
  jmp    L44
L100:
  lea    D106(%rip),%rax
  jmp    L44
L56:
  lea    D197(%rip),%rax
  jmp    L44
L19:
  lea    D118(%rip),%rax
  jmp    L44
L45:
  lea    D29(%rip),%rax
  jmp    L44
```

assistant: ```c
char *strerror(int err) {
  static struct errlist {
    int err;
    char *what;
  } errlist[] = {
      {0, "No error"}, {1, "Function number invalid"}, {2, "File not found"}, {3, "Path not found"}, {4, "Too many open files"}, {5, "Access denied"}, {6, "Invalid handle"}, {12, "Access code invalid"}, {39, "Insufficient disk space"},
  };
  struct errlist *ep;
  static char unknown[] = "Error 65535";
  unsigned e;
  char *p;
  for (ep = errlist; ep < errlist + sizeof(errlist) / sizeof(errlist[0]); ep++) {
    if (ep->err == err)
      return ep->what;
  }
  p = unknown + sizeof(unknown) - 1;
  e = err;
  do
    *--p = '0' + (e % 10);
  while ((e /= 10) > 0);
  strcpy(unknown + 6, p);
  return unknown;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<remove_spaces>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L5
  mov    $0x0,%eax
  jmp    L40
L5:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L32
L33:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L16
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x28(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L16:
  addq   $0x1,-0x28(%rbp)
L32:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L33
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  sub    -0x8(%rbp),%rax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
L40:
  leave
  ret
```

assistant: ```c
char *remove_spaces(const char *o) {
  int len;
  char *m, *no_spaces;
  if (!o)
    return (NULL);
  len = strlen(o);
  no_spaces = (char *)malloc(len + 1);
  m = no_spaces;
  while (*o) {
    if (!isspace(*o)) {
      *m++ = *o;
    }
    o++;
  }
  *m = '\0';
  len = m - no_spaces + 1;
  no_spaces = (char *)(realloc(no_spaces, len));
  return (no_spaces);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<remove_spaces>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L49
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L9
  mov    %rax,%rbp
  jmp    L35
L32:
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  je     L54
L35:
  movsbl %dil,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L32
  movzbl (%rbx),%eax
  mov    %al,0x0(%rbp)
  lea    0x1(%rbp),%rbp
  jmp    L32
L9:
  mov    %rax,%rbp
L54:
  movb   $0x0,0x0(%rbp)
  sub    %r12,%rbp
  lea    0x1(%rbp),%esi
  mov    %r12,%rdi
  call   <realloc@plt>
L51:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L49:
  mov    %rdi,%rax
  jmp    L51
```

assistant: ```c
char *remove_spaces(const char *o) {
  int len;
  char *m, *no_spaces;
  if (!o)
    return (NULL);
  len = strlen(o);
  no_spaces = (char *)malloc(len + 1);
  m = no_spaces;
  while (*o) {
    if (!isspace(*o)) {
      *m++ = *o;
    }
    o++;
  }
  *m = '\0';
  len = m - no_spaces + 1;
  no_spaces = (char *)(realloc(no_spaces, len));
  return (no_spaces);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<remove_spaces>:
  endbr64
  test   %rdi,%rdi
  je     L30
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  movsbl (%rbx),%edi
  mov    %rax,%r12
  mov    %rax,%rbp
  test   %dil,%dil
  je     L32
  nopl   0x0(%rax,%rax,1)
L16:
  call   <isspace@plt>
  test   %eax,%eax
  jne    L31
  movzbl (%rbx),%eax
  add    $0x1,%rbp
  mov    %al,-0x1(%rbp)
L31:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L16
  mov    %rbp,%rsi
  sub    %r12,%rsi
  add    $0x1,%esi
L7:
  movb   $0x0,0x0(%rbp)
  mov    %r12,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10b0 <realloc@plt>
  nopl   0x0(%rax)
L30:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  mov    $0x1,%esi
  jmp    L7
```

assistant: ```c
char *remove_spaces(const char *o) {
  int len;
  char *m, *no_spaces;
  if (!o)
    return (NULL);
  len = strlen(o);
  no_spaces = (char *)malloc(len + 1);
  m = no_spaces;
  while (*o) {
    if (!isspace(*o)) {
      *m++ = *o;
    }
    o++;
  }
  *m = '\0';
  len = m - no_spaces + 1;
  no_spaces = (char *)(realloc(no_spaces, len));
  return (no_spaces);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<remove_spaces>:
  endbr64
  test   %rdi,%rdi
  je     L52
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  movsbl (%rbx),%edi
  mov    %rax,%r12
  mov    %rax,%rbp
  test   %dil,%dil
  je     L21
  nopl   0x0(%rax,%rax,1)
L26:
  call   <isspace@plt>
  test   %eax,%eax
  jne    L57
  movzbl (%rbx),%edx
  add    $0x1,%rbx
  lea    0x1(%rbp),%rax
  mov    %dl,0x0(%rbp)
  movsbl (%rbx),%edi
  mov    %rax,%rbp
  test   %dil,%dil
  jne    L26
L63:
  mov    %rbp,%rsi
  sub    %r12,%rsi
  add    $0x1,%esi
L4:
  movb   $0x0,0x0(%rbp)
  mov    %r12,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10b0 <realloc@plt>
  nopl   0x0(%rax,%rax,1)
L57:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  jne    L26
  jmp    L63
  nop
L52:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L21:
  mov    $0x1,%esi
  jmp    L4
```

assistant: ```c
char *remove_spaces(const char *o) {
  int len;
  char *m, *no_spaces;
  if (!o)
    return (NULL);
  len = strlen(o);
  no_spaces = (char *)malloc(len + 1);
  m = no_spaces;
  while (*o) {
    if (!isspace(*o)) {
      *m++ = *o;
    }
    o++;
  }
  *m = '\0';
  len = m - no_spaces + 1;
  no_spaces = (char *)(realloc(no_spaces, len));
  return (no_spaces);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L85
  negl   -0x14(%rbp)
L85:
  cmpl   $0x2,-0x14(%rbp)
  jne    L98
  mov    $0x1,%eax
  jmp    L58
L98:
  cmpl   $0x1,-0x14(%rbp)
  je     L2
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L99
L2:
  mov    $0x0,%eax
  jmp    L58
L99:
  movl   $0x3,-0x4(%rbp)
  jmp    L75
L73:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L61
  mov    $0x0,%eax
  jmp    L58
L61:
  addl   $0x1,-0x4(%rbp)
L75:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x4(%rbp)
  jle    L73
  mov    $0x1,%eax
L58:
  pop    %rbp
  ret
```

assistant: ```c
int is_prime(int number) {
  int i;
  if (number < 0) {
    number *= -1;
  }
  if (number == 2) {
    return 1;
  }
  if (number == 1 || number % 2 == 0) {
    return 0;
  }
  for (i = 3; i <= (number / 2); i++) {
    if (number % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_prime>:
  endbr64
  mov    %edi,%esi
  neg    %esi
  cmovs  %edi,%esi
  mov    $0x1,%eax
  cmp    $0x2,%esi
  je     L45
  mov    $0x0,%eax
  cmp    $0x1,%esi
  je     L45
  test   $0x1,%sil
  je     L45
  mov    %esi,%edi
  shr    $0x1f,%edi
  add    %esi,%edi
  sar    %edi
  cmp    $0x5,%esi
  jle    L66
  movslq %esi,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%edx
  mov    %esi,%eax
  sub    %edx,%eax
  je     L45
  mov    $0x3,%ecx
L16:
  add    $0x1,%ecx
  cmp    %edi,%ecx
  jg     L15
  mov    %esi,%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %edx,%edx
  jne    L16
L45:
  ret
L15:
  mov    $0x1,%eax
  ret
L66:
  mov    $0x1,%eax
  jmp    L45
```

assistant: ```c
int is_prime(int number) {
  int i;
  if (number < 0) {
    number *= -1;
  }
  if (number == 2) {
    return 1;
  }
  if (number == 1 || number % 2 == 0) {
    return 0;
  }
  for (i = 3; i <= (number / 2); i++) {
    if (number % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_prime>:
  endbr64
  mov    %edi,%esi
  mov    $0x1,%edx
  neg    %esi
  cmovs  %edi,%esi
  cmp    $0x2,%esi
  je     L16
  xor    %edx,%edx
  cmp    $0x1,%esi
  je     L16
  test   $0x1,%sil
  je     L16
  mov    %esi,%edi
  sar    %edi
  cmp    $0x5,%esi
  jle    L0
  mov    $0xaaaaaaab,%eax
  mov    %esi,%edx
  imul   %rax,%rdx
  shr    $0x21,%rdx
  lea    (%rdx,%rdx,2),%eax
  mov    %esi,%edx
  sub    %eax,%edx
  je     L16
  add    $0x1,%edi
  mov    $0x3,%ecx
  jmp    L36
  nopl   (%rax)
L22:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L16
L36:
  add    $0x1,%ecx
  cmp    %edi,%ecx
  jne    L22
L0:
  mov    $0x1,%edx
L16:
  mov    %edx,%eax
  ret
```

assistant: ```c
int is_prime(int number) {
  int i;
  if (number < 0) {
    number *= -1;
  }
  if (number == 2) {
    return 1;
  }
  if (number == 1 || number % 2 == 0) {
    return 0;
  }
  for (i = 3; i <= (number / 2); i++) {
    if (number % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<is_prime>:
  endbr64
  mov    %edi,%esi
  mov    $0x1,%edx
  neg    %esi
  cmovs  %edi,%esi
  cmp    $0x2,%esi
  je     L55
  xor    %edx,%edx
  cmp    $0x1,%esi
  je     L55
  test   $0x1,%sil
  je     L55
  mov    %esi,%edi
  sar    %edi
  cmp    $0x5,%esi
  jle    L20
  mov    $0xaaaaaaab,%eax
  mov    %esi,%edx
  imul   %rax,%rdx
  shr    $0x21,%rdx
  lea    (%rdx,%rdx,2),%eax
  mov    %esi,%edx
  sub    %eax,%edx
  je     L55
  add    $0x1,%edi
  mov    $0x3,%ecx
  jmp    L5
  nopl   (%rax)
L11:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L55
L5:
  add    $0x1,%ecx
  cmp    %edi,%ecx
  jne    L11
L20:
  mov    $0x1,%edx
L55:
  mov    %edx,%eax
  ret
```

assistant: ```c
int is_prime(int number) {
  int i;
  if (number < 0) {
    number *= -1;
  }
  if (number == 2) {
    return 1;
  }
  if (number == 1 || number % 2 == 0) {
    return 0;
  }
  for (i = 3; i <= (number / 2); i++) {
    if (number % i == 0) {
      return 0;
    }
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<htoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  lea    D8(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D8
data_type: string

tool: D8: %x

assistant: ```c
int htoi(char *s) {
  int num;
  sscanf(s, "%x", &num);
  return num;
}
```