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

```asm
<assert_name_valid>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpb   $0x0,(%rbx)
  je     L12
L16:
  mov    $0x1,%edi
  call   <assert@plt>
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L16
L12:
  pop    %rbx
  ret
```

assistant: ```c
void assert_name_valid(const char *name) {
  const char *c;
  assert(name != NULL);
  for (c = name; *c; c++)
    assert((*c >= 'a' || *c <= 'z') || (*c >= '0' && *c <= '9') || *c == '-');
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<assert_name_valid>:
  endbr64
  test   %rdi,%rdi
  push   %rbx
  mov    %rdi,%rbx
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpb   $0x0,(%rbx)
  je     L17
  nopl   (%rax)
L12:
  mov    $0x1,%edi
  add    $0x1,%rbx
  call   <assert@plt>
  cmpb   $0x0,(%rbx)
  jne    L12
L17:
  pop    %rbx
  ret
```

assistant: ```c
void assert_name_valid(const char *name) {
  const char *c;
  assert(name != NULL);
  for (c = name; *c; c++)
    assert((*c >= 'a' || *c <= 'z') || (*c >= '0' && *c <= '9') || *c == '-');
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<assert_name_valid>:
  endbr64
  test   %rdi,%rdi
  push   %rbx
  mov    %rdi,%rbx
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmpb   $0x0,(%rbx)
  je     L13
  nopl   (%rax)
L6:
  mov    $0x1,%edi
  add    $0x1,%rbx
  call   <assert@plt>
  cmpb   $0x0,(%rbx)
  jne    L6
L13:
  pop    %rbx
  ret
```

assistant: ```c
void assert_name_valid(const char *name) {
  const char *c;
  assert(name != NULL);
  for (c = name; *c; c++)
    assert((*c >= 'a' || *c <= 'z') || (*c >= '0' && *c <= '9') || *c == '-');
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<parse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L3:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x14(%rbp)
L4:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int parse(char *_dst, char *_src, size_t len) {
  int ret = 0;
  char *src = _src;
  char *dst = _dst;
  unsigned int ch;
  while (*src) {
    *dst++ = *src++;
    ret++;
  }
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<parse>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L1
  mov    $0x0,%eax
L27:
  mov    %dl,(%rdi,%rax,1)
  lea    0x1(%rax),%ecx
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L27
L16:
  mov    %ecx,%eax
  ret
L1:
  mov    $0x0,%ecx
  jmp    L16
```

assistant: ```c
int parse(char *_dst, char *_src, size_t len) {
  int ret = 0;
  char *src = _src;
  char *dst = _dst;
  unsigned int ch;
  while (*src) {
    *dst++ = *src++;
    ret++;
  }
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<parse>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L2
  xor    %eax,%eax
  nopl   (%rax)
L8:
  mov    %dl,(%rdi,%rax,1)
  lea    0x1(%rax),%r8d
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L8
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int parse(char *_dst, char *_src, size_t len) {
  int ret = 0;
  char *src = _src;
  char *dst = _dst;
  unsigned int ch;
  while (*src) {
    *dst++ = *src++;
    ret++;
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L4
  xor    %eax,%eax
  nopl   (%rax)
L8:
  mov    %dl,(%rdi,%rax,1)
  lea    0x1(%rax),%r8d
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L8
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int parse(char *_dst, char *_src, size_t len) {
  int ret = 0;
  char *src = _src;
  char *dst = _dst;
  unsigned int ch;
  while (*src) {
    *dst++ = *src++;
    ret++;
  }
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  mov    %eax,-0x18(%rbp)
  mov    $0x6,%eax
  mov    %eax,-0x20(%rbp)
  mov    $0x7,%eax
  mov    %eax,-0x1c(%rbp)
  mov    $0x8,%eax
  mov    %eax,-0x14(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,%r10
  call   <simple.0>
  lea    -0x20(%rbp),%rax
  mov    %rax,%r10
  call   <collapse.1>
  lea    -0x20(%rbp),%rax
  mov    %rax,%r10
  call   <reduction.2>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int main(void) {
  int j = 0, k = 6, l = 7, m = 8;
  void simple(void) {
    int i;
#pragma acc parallel
    {
#pragma acc loop
      for (i = 0; i < m; i += k)
        j = (m + i - j) * l;
    }
  }
  void collapse(void) {
    int x, y, z;
#pragma acc parallel
    {
#pragma acc loop collapse(3)
      for (x = 0; x < k; x++)
        for (y = -5; y < l; y++)
          for (z = 0; z < m; z++)
            j += x + y + z;
    }
  }
  void reduction(void) {
    int x, y, z;
#pragma acc parallel reduction(+ : j)
    {
#pragma acc loop reduction(+ : j) collapse(3)
      for (x = 0; x < k; x++)
        for (y = -5; y < l; y++)
          for (z = 0; z < m; z++)
            j += x + y + z;
    }
  }
  simple();
  collapse();
  reduction();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int j = 0, k = 6, l = 7, m = 8;
  void simple(void) {
    int i;
#pragma acc parallel
    {
#pragma acc loop
      for (i = 0; i < m; i += k)
        j = (m + i - j) * l;
    }
  }
  void collapse(void) {
    int x, y, z;
#pragma acc parallel
    {
#pragma acc loop collapse(3)
      for (x = 0; x < k; x++)
        for (y = -5; y < l; y++)
          for (z = 0; z < m; z++)
            j += x + y + z;
    }
  }
  void reduction(void) {
    int x, y, z;
#pragma acc parallel reduction(+ : j)
    {
#pragma acc loop reduction(+ : j) collapse(3)
      for (x = 0; x < k; x++)
        for (y = -5; y < l; y++)
          for (z = 0; z < m; z++)
            j += x + y + z;
    }
  }
  simple();
  collapse();
  reduction();
  return 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   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rax
  mov    %rsp,%rbx
  mov    %rax,0x10(%rsp)
  mov    %rbx,%r10
  movabs $0x700000006,%rax
  mov    %rax,(%rsp)
  movabs $0x8fffffeda,%rax
  mov    %rax,0x8(%rsp)
  call   <collapse.1>
  mov    %rbx,%r10
  call   <collapse.1>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int j = 0, k = 6, l = 7, m = 8;
  void simple(void) {
    int i;
#pragma acc parallel
    {
#pragma acc loop
      for (i = 0; i < m; i += k)
        j = (m + i - j) * l;
    }
  }
  void collapse(void) {
    int x, y, z;
#pragma acc parallel
    {
#pragma acc loop collapse(3)
      for (x = 0; x < k; x++)
        for (y = -5; y < l; y++)
          for (z = 0; z < m; z++)
            j += x + y + z;
    }
  }
  void reduction(void) {
    int x, y, z;
#pragma acc parallel reduction(+ : j)
    {
#pragma acc loop reduction(+ : j) collapse(3)
      for (x = 0; x < k; x++)
        for (y = -5; y < l; y++)
          for (z = 0; z < m; z++)
            j += x + y + z;
    }
  }
  simple();
  collapse();
  reduction();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int j = 0, k = 6, l = 7, m = 8;
  void simple(void) {
    int i;
#pragma acc parallel
    {
#pragma acc loop
      for (i = 0; i < m; i += k)
        j = (m + i - j) * l;
    }
  }
  void collapse(void) {
    int x, y, z;
#pragma acc parallel
    {
#pragma acc loop collapse(3)
      for (x = 0; x < k; x++)
        for (y = -5; y < l; y++)
          for (z = 0; z < m; z++)
            j += x + y + z;
    }
  }
  void reduction(void) {
    int x, y, z;
#pragma acc parallel reduction(+ : j)
    {
#pragma acc loop reduction(+ : j) collapse(3)
      for (x = 0; x < k; x++)
        for (y = -5; y < l; y++)
          for (z = 0; z < m; z++)
            j += x + y + z;
    }
  }
  simple();
  collapse();
  reduction();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ifaceutil_getip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    D42(%rip),%rax
  mov    (%rax),%rax
  lea    -0x1(%rax),%rdx
  mov    -0xc(%rbp),%eax
  mov    -0x28(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <strncpy@plt>
  mov    D33(%rip),%rax
  mov    (%rax),%ecx
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <socket@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0xffffffff,-0x14(%rbp)
  jne    L29
  mov    $0x0,%eax
  jmp    L37
L29:
  mov    D28(%rip),%rax
  mov    (%rax),%ecx
  lea    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <ioctl@plt>
  cmp    $0xffffffff,%eax
  jne    L18
  mov    $0x0,%eax
  jmp    L37
L18:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  lea    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <inet_ntoa@plt>
  mov    %eax,%edi
  call   <strdup@plt>
L37:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

assistant: ```c
char *ifaceutil_getip(char *iface_name) {
  int fd;
  struct ifreq ifr;
  ifr.ifr_addr.sa_family = AF_INET;
  strncpy(ifr.ifr_name, iface_name, IFNAMSIZ - 1);
  if ((fd = socket(AF_INET, SOCK_DGRAM, 0)) == -1) {
    return NULL;
  }
  if (ioctl(fd, SIOCGIFADDR, &ifr) == -1) {
    return NULL;
  }
  close(fd);
  return strdup(inet_ntoa(((struct sockaddr_in *)&ifr.ifr_addr)->sin_addr));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ifaceutil_getip>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    D3(%rip),%rbx
  mov    (%rbx),%eax
  mov    %eax,(%rsp)
  mov    D21(%rip),%rax
  mov    (%rax),%rdx
  sub    $0x1,%rdx
  mov    0x4(%rsp),%edi
  call   <strncpy@plt>
  mov    $0x0,%edx
  mov    D39(%rip),%rax
  mov    (%rax),%esi
  mov    (%rbx),%edi
  call   <socket@plt>
  cmp    $0xffffffff,%eax
  je     L38
  mov    %eax,%ebx
  mov    %rsp,%rdx
  mov    D10(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <ioctl@plt>
  cmp    $0xffffffff,%eax
  je     L3
  mov    %ebx,%edi
  call   <close@plt>
  mov    (%rsp),%edi
  call   <inet_ntoa@plt>
  mov    %eax,%edi
  call   <strdup@plt>
L49:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x10,%rsp
  pop    %rbx
  ret
L38:
  mov    $0x0,%eax
  jmp    L49
L3:
  mov    $0x0,%eax
  jmp    L49
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ifaceutil_getip(char *iface_name) {
  int fd;
  struct ifreq ifr;
  ifr.ifr_addr.sa_family = AF_INET;
  strncpy(ifr.ifr_name, iface_name, IFNAMSIZ - 1);
  if ((fd = socket(AF_INET, SOCK_DGRAM, 0)) == -1) {
    return NULL;
  }
  if (ioctl(fd, SIOCGIFADDR, &ifr) == -1) {
    return NULL;
  }
  close(fd);
  return strdup(inet_ntoa(((struct sockaddr_in *)&ifr.ifr_addr)->sin_addr));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ifaceutil_getip>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  sub    $0x10,%rsp
  mov    D28(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    0x4(%rsp),%edi
  mov    0x0(%rbp),%eax
  mov    %eax,(%rsp)
  mov    D37(%rip),%rax
  mov    (%rax),%rdx
  sub    $0x1,%rdx
  call   <strncpy@plt>
  mov    D26(%rip),%rax
  mov    0x0(%rbp),%edi
  xor    %edx,%edx
  mov    (%rax),%esi
  call   <socket@plt>
  cmp    $0xffffffff,%eax
  je     L21
  mov    %eax,%ebp
  mov    D35(%rip),%rax
  mov    %rsp,%rdx
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <ioctl@plt>
  cmp    $0xffffffff,%eax
  je     L21
  mov    %ebp,%edi
  call   <close@plt>
  mov    (%rsp),%edi
  call   <inet_ntoa@plt>
  mov    %eax,%edi
  call   <strdup@plt>
L7:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L14
  add    $0x10,%rsp
  pop    %rbp
  ret
  xchg   %ax,%ax
L21:
  xor    %eax,%eax
  jmp    L7
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ifaceutil_getip(char *iface_name) {
  int fd;
  struct ifreq ifr;
  ifr.ifr_addr.sa_family = AF_INET;
  strncpy(ifr.ifr_name, iface_name, IFNAMSIZ - 1);
  if ((fd = socket(AF_INET, SOCK_DGRAM, 0)) == -1) {
    return NULL;
  }
  if (ioctl(fd, SIOCGIFADDR, &ifr) == -1) {
    return NULL;
  }
  close(fd);
  return strdup(inet_ntoa(((struct sockaddr_in *)&ifr.ifr_addr)->sin_addr));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ifaceutil_getip>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  sub    $0x10,%rsp
  mov    D29(%rip),%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    0x4(%rsp),%edi
  mov    0x0(%rbp),%eax
  mov    %eax,(%rsp)
  mov    D39(%rip),%rax
  mov    (%rax),%rdx
  sub    $0x1,%rdx
  call   <strncpy@plt>
  mov    D38(%rip),%rax
  mov    0x0(%rbp),%edi
  xor    %edx,%edx
  mov    (%rax),%esi
  call   <socket@plt>
  cmp    $0xffffffff,%eax
  je     L18
  mov    %eax,%ebp
  mov    D20(%rip),%rax
  mov    %rsp,%rdx
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <ioctl@plt>
  cmp    $0xffffffff,%eax
  je     L18
  mov    %ebp,%edi
  call   <close@plt>
  mov    (%rsp),%edi
  call   <inet_ntoa@plt>
  mov    %eax,%edi
  call   <strdup@plt>
L7:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x10,%rsp
  pop    %rbp
  ret
  xchg   %ax,%ax
L18:
  xor    %eax,%eax
  jmp    L7
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ifaceutil_getip(char *iface_name) {
  int fd;
  struct ifreq ifr;
  ifr.ifr_addr.sa_family = AF_INET;
  strncpy(ifr.ifr_name, iface_name, IFNAMSIZ - 1);
  if ((fd = socket(AF_INET, SOCK_DGRAM, 0)) == -1) {
    return NULL;
  }
  if (ioctl(fd, SIOCGIFADDR, &ifr) == -1) {
    return NULL;
  }
  close(fd);
  return strdup(inet_ntoa(((struct sockaddr_in *)&ifr.ifr_addr)->sin_addr));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x2,-0xa0(%rbp)
  movl   $0x5,-0x9c(%rbp)
  movl   $0x1d,-0x98(%rbp)
  movl   $0x3,-0x94(%rbp)
  movl   $0x1a,-0x90(%rbp)
  movl   $0x15,-0x8c(%rbp)
  movl   $0xe,-0x88(%rbp)
  movl   $0xa,-0x84(%rbp)
  movl   $0x9,-0x80(%rbp)
  movl   $0x1,-0x7c(%rbp)
  movl   $0x1e,-0x78(%rbp)
  movl   $0x11,-0x74(%rbp)
  movl   $0x16,-0x70(%rbp)
  movl   $0x19,-0x6c(%rbp)
  movl   $0x4,-0x68(%rbp)
  movl   $0x1b,-0x64(%rbp)
  movl   $0x8,-0x60(%rbp)
  movl   $0x13,-0x5c(%rbp)
  movl   $0x6,-0x58(%rbp)
  movl   $0x10,-0x54(%rbp)
  movl   $0x12,-0x50(%rbp)
  movl   $0x7,-0x4c(%rbp)
  movl   $0xb,-0x48(%rbp)
  movl   $0xc,-0x44(%rbp)
  movl   $0x14,-0x40(%rbp)
  movl   $0x17,-0x3c(%rbp)
  movl   $0xf,-0x38(%rbp)
  movl   $0xd,-0x34(%rbp)
  movl   $0x1c,-0x30(%rbp)
  movl   $0x18,-0x2c(%rbp)
  movl   $0x5,-0x28(%rbp)
  movl   $0xa,-0x24(%rbp)
  movl   $0xf,-0x20(%rbp)
  movl   $0x14,-0x1c(%rbp)
  movl   $0x19,-0x18(%rbp)
  lea    -0xa0(%rbp),%rax
  mov    $0x22,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <qsort@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0xa4(%rbp)
  jmp    L15
L6:
  mov    -0xa4(%rbp),%eax
  cltq
  mov    -0xa0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xa4(%rbp)
L15:
  cmpl   $0x22,-0xa4(%rbp)
  jle    L6
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
type_infered: string

tool: D4: %d 

assistant: ```c
main() {
  int v[35] = {2, 5, 29, 3, 26, 21, 14, 10, 9, 1, 30, 17, 22, 25, 4, 27, 8, 19, 6, 16, 18, 7, 11, 12, 20, 23, 15, 13, 28, 24, 5, 10, 15, 20, 25};
  int i;
  qsort(v, 0, 34);
  printf("\n");
  for (i = 0; i < 35; i++)
    printf("%d ", v[i]);
  printf("\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  movl   $0x2,(%rsp)
  movl   $0x5,0x4(%rsp)
  movl   $0x1d,0x8(%rsp)
  movl   $0x3,0xc(%rsp)
  movl   $0x1a,0x10(%rsp)
  movl   $0x15,0x14(%rsp)
  movl   $0xe,0x18(%rsp)
  movl   $0xa,0x1c(%rsp)
  movl   $0x9,0x20(%rsp)
  movl   $0x1,0x24(%rsp)
  movl   $0x1e,0x28(%rsp)
  movl   $0x11,0x2c(%rsp)
  movl   $0x16,0x30(%rsp)
  movl   $0x19,0x34(%rsp)
  movl   $0x4,0x38(%rsp)
  movl   $0x1b,0x3c(%rsp)
  movl   $0x8,0x40(%rsp)
  movl   $0x13,0x44(%rsp)
  movl   $0x6,0x48(%rsp)
  movl   $0x10,0x4c(%rsp)
  movl   $0x12,0x50(%rsp)
  movl   $0x7,0x54(%rsp)
  movl   $0xb,0x58(%rsp)
  movl   $0xc,0x5c(%rsp)
  movl   $0x14,0x60(%rsp)
  movl   $0x17,0x64(%rsp)
  movl   $0xf,0x68(%rsp)
  movl   $0xd,0x6c(%rsp)
  movl   $0x1c,0x70(%rsp)
  movl   $0x18,0x74(%rsp)
  movl   $0x5,0x78(%rsp)
  movl   $0xa,0x7c(%rsp)
  movl   $0xf,0x80(%rsp)
  movl   $0x14,0x84(%rsp)
  movl   $0x19,0x88(%rsp)
  mov    %rsp,%rbx
  mov    $0x22,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <qsort@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    0x8c(%rsp),%r12
  lea    D7(%rip),%rbp
L6:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L6
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0xa0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D7
datatype: string

tool: D7: %d 

assistant: ```c
main() {
  int v[35] = {2, 5, 29, 3, 26, 21, 14, 10, 9, 1, 30, 17, 22, 25, 4, 27, 8, 19, 6, 16, 18, 7, 11, 12, 20, 23, 15, 13, 28, 24, 5, 10, 15, 20, 25};
  int i;
  qsort(v, 0, 34);
  printf("\n");
  for (i = 0; i < 35; i++)
    printf("%d ", v[i]);
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x22,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  movabs $0x500000002,%rax
  mov    %rsp,%rbx
  lea    0x8c(%rsp),%r12
  movl   $0x19,0x88(%rsp)
  mov    %rbx,%rdi
  mov    %rax,(%rsp)
  movabs $0x30000001d,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x150000001a,%rax
  mov    %rax,0x10(%rsp)
  movabs $0xa0000000e,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x100000009,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x110000001e,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x1900000016,%rax
  mov    %rax,0x30(%rsp)
  movabs $0x1b00000004,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x1300000008,%rax
  mov    %rax,0x40(%rsp)
  movabs $0x1000000006,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x700000012,%rax
  mov    %rax,0x50(%rsp)
  movabs $0xc0000000b,%rax
  mov    %rax,0x58(%rsp)
  movabs $0x1700000014,%rax
  mov    %rax,0x60(%rsp)
  movabs $0xd0000000f,%rax
  mov    %rax,0x68(%rsp)
  movabs $0x180000001c,%rax
  mov    %rax,0x70(%rsp)
  movabs $0xa00000005,%rax
  mov    %rax,0x78(%rsp)
  movabs $0x140000000f,%rax
  mov    %rax,0x80(%rsp)
  call   <qsort@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  nopw   0x0(%rax,%rax,1)
L15:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L15
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
data_type: string

tool: D5: %d 

assistant: ```c
main() {
  int v[35] = {2, 5, 29, 3, 26, 21, 14, 10, 9, 1, 30, 17, 22, 25, 4, 27, 8, 19, 6, 16, 18, 7, 11, 12, 20, 23, 15, 13, 28, 24, 5, 10, 15, 20, 25};
  int i;
  qsort(v, 0, 34);
  printf("\n");
  for (i = 0; i < 35; i++)
    printf("%d ", v[i]);
  printf("\n");
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x22,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D49(%rip),%rbp
  push   %rbx
  sub    $0xa0,%rsp
  movdqa D67(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    D83(%rip),%rax
  movl   $0x19,0x88(%rsp)
  movaps %xmm0,(%rsp)
  movdqa D74(%rip),%xmm0
  mov    %rbx,%rdi
  lea    0x8c(%rsp),%r12
  mov    %rax,0x80(%rsp)
  movaps %xmm0,0x10(%rsp)
  movdqa D1(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movdqa D34(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  movdqa D31(%rip),%xmm0
  movaps %xmm0,0x40(%rsp)
  movdqa D55(%rip),%xmm0
  movaps %xmm0,0x50(%rsp)
  movdqa D8(%rip),%xmm0
  movaps %xmm0,0x60(%rsp)
  movdqa D3(%rip),%xmm0
  movaps %xmm0,0x70(%rsp)
  call   <qsort@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  xchg   %ax,%ax
L5:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L5
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D49
type: string
name: disassemble_data
arguments: data_label: D67
type: byte[16]
name: disassemble_data
arguments: data_label: D83
type: byte[8]
name: disassemble_data
arguments: data_label: D74
type: byte[16]
name: disassemble_data
arguments: data_label: D1
type: byte[16]
name: disassemble_data
arguments: data_label: D34
type: byte[16]
name: disassemble_data
arguments: data_label: D31
type: byte[16]
name: disassemble_data
arguments: data_label: D55
type: byte[16]
name: disassemble_data
arguments: data_label: D8
type: byte[16]
name: disassemble_data
arguments: data_label: D3
type: byte[16]

tool: D49: %d 

tool: {"D67": [0x02,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x1d,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D83": [0x0f,0x00,0x00,0x00,0x14,0x00,0x00,0x00]}

tool: {"D74": [0x1a,0x00,0x00,0x00,0x15,0x00,0x00,0x00,0x0e,0x00,0x00,0x00,0x0a,0x00,0x00,0x00]}

tool: {"D1": [0x09,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x1e,0x00,0x00,0x00,0x11,0x00,0x00,0x00]}

tool: {"D34": [0x16,0x00,0x00,0x00,0x19,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x1b,0x00,0x00,0x00]}

tool: {"D31": [0x08,0x00,0x00,0x00,0x13,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x10,0x00,0x00,0x00]}

tool: {"D55": [0x12,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x0b,0x00,0x00,0x00,0x0c,0x00,0x00,0x00]}

tool: {"D8": [0x14,0x00,0x00,0x00,0x17,0x00,0x00,0x00,0x0f,0x00,0x00,0x00,0x0d,0x00,0x00,0x00]}

tool: {"D3": [0x1c,0x00,0x00,0x00,0x18,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x0a,0x00,0x00,0x00]}

assistant: ```c
main() {
  int v[35] = {2, 5, 29, 3, 26, 21, 14, 10, 9, 1, 30, 17, 22, 25, 4, 27, 8, 19, 6, 16, 18, 7, 11, 12, 20, 23, 15, 13, 28, 24, 5, 10, 15, 20, 25};
  int i;
  qsort(v, 0, 34);
  printf("\n");
  for (i = 0; i < 35; i++)
    printf("%d ", v[i]);
  printf("\n");
}
```