ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L5
  lea    D20(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L5:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x10(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  mov    %rax,0x8(%rsp)
  call   <RandomFunc@plt>
  mov    0x10(%rsp),%rsi
  movabs $0x26cf0000569b5dbd,%rax
  cmp    %rax,%rsi
  je     L38
L19:
  xor    %eax,%eax
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L38:
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    0x10(%rsp),%rsi
  jmp    L19
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D20
type_infered: string
name: unpack_data
arguments: data_label: D15
type_infered: string
name: unpack_data
arguments: data_label: D0
type_infered: string

tool: D20: Call this program with %i arguments


tool: D15: %lu


tool: D0: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long input[1];
  unsigned long output[1];
  int randomFuns_i5;
  unsigned long randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 2796453895073390013UL) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%lu\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strspn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L3:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x12(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L10
L20:
  movzbl -0x11(%rbp),%eax
  cmp    -0x12(%rbp),%al
  jne    L10
  jmp    L3
L10:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  cmpb   $0x0,-0x11(%rbp)
  jne    L20
  mov    -0x10(%rbp),%rax
  sub    $0x1,%rax
  sub    -0x28(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strspn(const char *s1, const char *s2) {
  const char *p = s1, *spanp;
  char c, sc;
cont:
  c = *p++;
  for (spanp = s2; (sc = *spanp++) != 0;)
    if (sc == c)
      goto cont;
  return (char *)(p - 1 - s1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strspn>:
  endbr64
  mov    %rdi,%rax
L21:
  lea    0x1(%rax),%r9
  movzbl -0x1(%r9),%r8d
  mov    %rsi,%rdx
L27:
  add    $0x1,%rdx
  movzbl -0x1(%rdx),%ecx
  test   %cl,%cl
  je     L7
  cmp    %cl,%r8b
  jne    L27
  mov    %r9,%rax
  jmp    L21
L7:
  sub    %rdi,%rax
  ret
```

assistant: ```c
char *strspn(const char *s1, const char *s2) {
  const char *p = s1, *spanp;
  char c, sc;
cont:
  c = *p++;
  for (spanp = s2; (sc = *spanp++) != 0;)
    if (sc == c)
      goto cont;
  return (char *)(p - 1 - s1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<strspn>:
  endbr64
  mov    %rdi,%rax
L31:
  lea    0x1(%rax),%r9
  mov    %rsi,%rdx
  movzbl -0x1(%r9),%r8d
  jmp    L12
  nopl   (%rax)
L34:
  cmp    %cl,%r8b
  je     L30
L12:
  movzbl (%rdx),%ecx
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L34
  sub    %rdi,%rax
  ret
  nopl   0x0(%rax)
L30:
  mov    %r9,%rax
  jmp    L31
```

assistant: ```c
char *strspn(const char *s1, const char *s2) {
  const char *p = s1, *spanp;
  char c, sc;
cont:
  c = *p++;
  for (spanp = s2; (sc = *spanp++) != 0;)
    if (sc == c)
      goto cont;
  return (char *)(p - 1 - s1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strspn>:
  endbr64
  mov    %rdi,%rax
L37:
  lea    0x1(%rax),%r9
  mov    %rsi,%rdx
  movzbl -0x1(%r9),%r8d
  jmp    L22
  nopl   (%rax)
L38:
  cmp    %cl,%r8b
  je     L26
L22:
  movzbl (%rdx),%ecx
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L38
  sub    %rdi,%rax
  ret
  nopl   0x0(%rax)
L26:
  mov    %r9,%rax
  jmp    L37
```

assistant: ```c
char *strspn(const char *s1, const char *s2) {
  const char *p = s1, *spanp;
  char c, sc;
cont:
  c = *p++;
  for (spanp = s2; (sc = *spanp++) != 0;)
    if (sc == c)
      goto cont;
  return (char *)(p - 1 - s1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<hash_djb_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x1505,-0xc(%rbp)
  jmp    L4
L1:
  mov    -0xc(%rbp),%eax
  shl    $0x5,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
L4:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L1
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t hash_djb_str(const void *key) {
  const char *str = key;
  uint32_t hash = 5381;
  while (*str++)
    hash = ((hash << 5) + hash) + *str;
  return hash;
}
```
user: What is the c source code of the assembly code below:

```asm
<hash_djb_str>:
  endbr64
  lea    0x1(%rdi),%rdx
  cmpb   $0x0,(%rdi)
  je     L0
  mov    $0x1505,%eax
L13:
  mov    %eax,%ecx
  shl    $0x5,%ecx
  add    %ecx,%eax
  movzbl (%rdx),%ecx
  movsbl %cl,%esi
  add    %esi,%eax
  add    $0x1,%rdx
  test   %cl,%cl
  jne    L13
  ret
L0:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
uint32_t hash_djb_str(const void *key) {
  const char *str = key;
  uint32_t hash = 5381;
  while (*str++)
    hash = ((hash << 5) + hash) + *str;
  return hash;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<hash_djb_str>:
  endbr64
  cmpb   $0x0,(%rdi)
  lea    0x1(%rdi),%rdx
  mov    $0x1505,%eax
  je     L8
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %eax,%ecx
  movsbl (%rdx),%esi
  add    $0x1,%rdx
  shl    $0x5,%ecx
  add    %ecx,%eax
  add    %esi,%eax
  test   %sil,%sil
  jne    L6
  ret
  xchg   %ax,%ax
L8:
  ret
```

assistant: ```c
uint32_t hash_djb_str(const void *key) {
  const char *str = key;
  uint32_t hash = 5381;
  while (*str++)
    hash = ((hash << 5) + hash) + *str;
  return hash;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hash_djb_str>:
  endbr64
  cmpb   $0x0,(%rdi)
  lea    0x1(%rdi),%rdx
  mov    $0x1505,%eax
  je     L3
  nopw   0x0(%rax,%rax,1)
L7:
  mov    %eax,%ecx
  movsbl (%rdx),%esi
  add    $0x1,%rdx
  shl    $0x5,%ecx
  add    %ecx,%eax
  add    %esi,%eax
  test   %sil,%sil
  jne    L7
  ret
  xchg   %ax,%ax
L3:
  ret
```

assistant: ```c
uint32_t hash_djb_str(const void *key) {
  const char *str = key;
  uint32_t hash = 5381;
  while (*str++)
    hash = ((hash << 5) + hash) + *str;
  return hash;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  add    $0x51e22cdb,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L1
  lea    -0xc(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  mov    %ax,-0xe(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rdx),%edx
  mov    %dx,(%rax)
  lea    -0xc(%rbp),%rax
  movzwl -0xe(%rbp),%edx
  mov    %dx,(%rax)
  jmp    L18
L1:
  mov    -0xc(%rbp),%esi
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  shr    $0x2,%eax
  and    $0x6,%eax
  or     $0x1,%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  and    $0x3c,%eax
  or     %esi,%eax
  mov    %eax,-0xc(%rbp)
L18:
  mov    -0xc(%rbp),%eax
  rol    $0x3,%eax
  imul   $0xddc71a10,%eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned short copy11;
  {
    state[0UL] = input[0UL] + 1373777115U;
    if (state[0UL] & 1U) {
      copy11 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = copy11;
    } else {
      state[0UL] |= ((state[0UL] << (((state[0UL] >> 2U) & 7U) | 1UL)) & 15U) << 2UL;
    }
    output[0UL] = ((state[0UL] << 3U) | (state[0UL] >> 29U)) * 3720813072U;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  lea    0x51e22cdb(%rax),%edx
  test   $0x1,%al
  jne    L5
  mov    %edx,%eax
  shl    $0x10,%eax
  shr    $0x10,%edx
  mov    %dx,%ax
L6:
  rol    $0x3,%eax
  imul   $0xddc71a10,%eax,%eax
  mov    %eax,(%rsi)
  ret
L5:
  mov    %edx,%ecx
  shr    $0x2,%ecx
  and    $0x6,%ecx
  or     $0x1,%ecx
  mov    %edx,%eax
  shl    %cl,%eax
  shl    $0x2,%eax
  and    $0x3c,%eax
  or     %edx,%eax
  jmp    L6
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned short copy11;
  {
    state[0UL] = input[0UL] + 1373777115U;
    if (state[0UL] & 1U) {
      copy11 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = copy11;
    } else {
      state[0UL] |= ((state[0UL] << (((state[0UL] >> 2U) & 7U) | 1UL)) & 15U) << 2UL;
    }
    output[0UL] = ((state[0UL] << 3U) | (state[0UL] >> 29U)) * 3720813072U;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%edx
  lea    0x51e22cdb(%rdx),%eax
  and    $0x1,%edx
  jne    L5
  rol    $0x10,%eax
  rol    $0x3,%eax
  imul   $0xddc71a10,%eax,%eax
  mov    %eax,(%rsi)
  ret
L5:
  mov    %eax,%ecx
  mov    %eax,%edx
  shr    $0x2,%ecx
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%edx
  shl    $0x2,%edx
  and    $0x3c,%edx
  or     %edx,%eax
  rol    $0x3,%eax
  imul   $0xddc71a10,%eax,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned short copy11;
  {
    state[0UL] = input[0UL] + 1373777115U;
    if (state[0UL] & 1U) {
      copy11 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = copy11;
    } else {
      state[0UL] |= ((state[0UL] << (((state[0UL] >> 2U) & 7U) | 1UL)) & 15U) << 2UL;
    }
    output[0UL] = ((state[0UL] << 3U) | (state[0UL] >> 29U)) * 3720813072U;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%edx
  lea    0x51e22cdb(%rdx),%eax
  and    $0x1,%edx
  jne    L8
  rol    $0x10,%eax
  rol    $0x3,%eax
  imul   $0xddc71a10,%eax,%eax
  mov    %eax,(%rsi)
  ret
L8:
  mov    %eax,%ecx
  mov    %eax,%edx
  shr    $0x2,%ecx
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%edx
  shl    $0x2,%edx
  and    $0x3c,%edx
  or     %edx,%eax
  rol    $0x3,%eax
  imul   $0xddc71a10,%eax,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned short copy11;
  {
    state[0UL] = input[0UL] + 1373777115U;
    if (state[0UL] & 1U) {
      copy11 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = copy11;
    } else {
      state[0UL] |= ((state[0UL] << (((state[0UL] >> 2U) & 7U) | 1UL)) & 15U) << 2UL;
    }
    output[0UL] = ((state[0UL] << 3U) | (state[0UL] >> 29U)) * 3720813072U;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<openstdin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    D20(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <fileno@plt>
  mov    -0x8(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <tcgetattr@plt>
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x10,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rax
  movq   $0x0,0x8(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    D27(%rip),%rax
  mov    (%rax),%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    D34(%rip),%rax
  mov    (%rax),%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  mov    D31(%rip),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <tcflush@plt>
  mov    D46(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x18(%rbp),%rdx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <tcsetattr@plt>
  nop
  leave
  ret
```

assistant: ```c
void openstdin(int *fd, struct termios *oldtio, struct termios *newtio) {
  *fd = fileno(stdin);
  tcgetattr(*fd, oldtio);
  memcpy(newtio, oldtio, sizeof(*oldtio));
  newtio->c_lflag = 0;
  newtio->c_cc[VTIME] = 0;
  newtio->c_cc[VMIN] = 1;
  tcflush(*fd, TCIFLUSH);
  tcsetattr(*fd, TCSANOW, newtio);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<openstdin>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
  mov    D46(%rip),%rax
  mov    (%rax),%edi
  call   <fileno@plt>
  mov    %eax,%edi
  mov    %eax,0x0(%rbp)
  mov    %r12,%rsi
  call   <tcgetattr@plt>
  mov    $0x10,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  movq   $0x0,0x8(%rbx)
  mov    D16(%rip),%rax
  mov    (%rax),%rdx
  mov    (%rbx),%rax
  movl   $0x0,(%rax,%rdx,4)
  mov    D17(%rip),%rax
  mov    (%rax),%rdx
  mov    (%rbx),%rax
  movl   $0x1,(%rax,%rdx,4)
  mov    D7(%rip),%rax
  mov    (%rax),%esi
  mov    0x0(%rbp),%edi
  call   <tcflush@plt>
  mov    %rbx,%rdx
  mov    D10(%rip),%rax
  mov    (%rax),%esi
  mov    0x0(%rbp),%edi
  call   <tcsetattr@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void openstdin(int *fd, struct termios *oldtio, struct termios *newtio) {
  *fd = fileno(stdin);
  tcgetattr(*fd, oldtio);
  memcpy(newtio, oldtio, sizeof(*oldtio));
  newtio->c_lflag = 0;
  newtio->c_cc[VTIME] = 0;
  newtio->c_cc[VMIN] = 1;
  tcflush(*fd, TCIFLUSH);
  tcsetattr(*fd, TCSANOW, newtio);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<openstdin>:
  endbr64
  mov    D10(%rip),%rax
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rax),%edi
  call   <fileno@plt>
  mov    %rbp,%rsi
  mov    %eax,(%rbx)
  mov    %eax,%edi
  call   <tcgetattr@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x10,%edx
  call   <memcpy@plt>
  mov    D36(%rip),%rdx
  mov    (%r12),%rax
  movq   $0x0,0x8(%r12)
  mov    (%rdx),%rdx
  movl   $0x0,(%rax,%rdx,4)
  mov    D0(%rip),%rdx
  mov    (%rdx),%rdx
  movl   $0x1,(%rax,%rdx,4)
  mov    D38(%rip),%rax
  mov    (%rbx),%edi
  mov    (%rax),%esi
  call   <tcflush@plt>
  mov    D18(%rip),%rax
  mov    (%rbx),%edi
  mov    %r12,%rdx
  pop    %rbx
  pop    %rbp
  mov    (%rax),%esi
  pop    %r12
  jmp    10d0 <tcsetattr@plt>
```

assistant: ```c
void openstdin(int *fd, struct termios *oldtio, struct termios *newtio) {
  *fd = fileno(stdin);
  tcgetattr(*fd, oldtio);
  memcpy(newtio, oldtio, sizeof(*oldtio));
  newtio->c_lflag = 0;
  newtio->c_cc[VTIME] = 0;
  newtio->c_cc[VMIN] = 1;
  tcflush(*fd, TCIFLUSH);
  tcsetattr(*fd, TCSANOW, newtio);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<openstdin>:
  endbr64
  mov    D48(%rip),%rax
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rax),%edi
  call   <fileno@plt>
  mov    %rbp,%rsi
  mov    %eax,(%rbx)
  mov    %eax,%edi
  call   <tcgetattr@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x10,%edx
  call   <memcpy@plt>
  mov    D38(%rip),%rdx
  mov    (%r12),%rax
  movq   $0x0,0x8(%r12)
  mov    (%rdx),%rdx
  movl   $0x0,(%rax,%rdx,4)
  mov    D9(%rip),%rdx
  mov    (%rdx),%rdx
  movl   $0x1,(%rax,%rdx,4)
  mov    D36(%rip),%rax
  mov    (%rbx),%edi
  mov    (%rax),%esi
  call   <tcflush@plt>
  mov    D11(%rip),%rax
  mov    (%rbx),%edi
  mov    %r12,%rdx
  pop    %rbx
  pop    %rbp
  mov    (%rax),%esi
  pop    %r12
  jmp    10d0 <tcsetattr@plt>
```

assistant: ```c
void openstdin(int *fd, struct termios *oldtio, struct termios *newtio) {
  *fd = fileno(stdin);
  tcgetattr(*fd, oldtio);
  memcpy(newtio, oldtio, sizeof(*oldtio));
  newtio->c_lflag = 0;
  newtio->c_cc[VTIME] = 0;
  newtio->c_cc[VMIN] = 1;
  tcflush(*fd, TCIFLUSH);
  tcsetattr(*fd, TCSANOW, newtio);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<interface_sigwinch_adjust>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    D35(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <resize_term@plt>
  mov    D10(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <wrefresh@plt>
  mov    D48(%rip),%rax
  mov    (%rax),%eax
  mov    D39(%rip),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <signal@plt>
  nop
  leave
  ret
```

assistant: ```c
void interface_sigwinch_adjust(int sig) {
  resize_term(LINES, COLS);
  wrefresh(curscr);
  signal(SIGWINCH, interface_sigwinch_adjust);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<interface_sigwinch_adjust>:
  endbr64
  sub    $0x8,%rsp
  mov    D20(%rip),%rax
  mov    (%rax),%esi
  mov    D33(%rip),%rax
  mov    (%rax),%edi
  call   <resize_term@plt>
  mov    D42(%rip),%rax
  mov    (%rax),%edi
  call   <wrefresh@plt>
  mov    D16(%rip),%rsi
  mov    D35(%rip),%rax
  mov    (%rax),%edi
  call   <signal@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void interface_sigwinch_adjust(int sig) {
  resize_term(LINES, COLS);
  wrefresh(curscr);
  signal(SIGWINCH, interface_sigwinch_adjust);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<interface_sigwinch_adjust>:
  endbr64
  sub    $0x8,%rsp
  mov    D26(%rip),%rax
  mov    (%rax),%esi
  mov    D22(%rip),%rax
  mov    (%rax),%edi
  call   <resize_term@plt>
  mov    D27(%rip),%rax
  mov    (%rax),%edi
  call   <wrefresh@plt>
  mov    D16(%rip),%rax
  mov    D4(%rip),%rsi
  mov    (%rax),%edi
  add    $0x8,%rsp
  jmp    1080 <signal@plt>
```

assistant: ```c
void interface_sigwinch_adjust(int sig) {
  resize_term(LINES, COLS);
  wrefresh(curscr);
  signal(SIGWINCH, interface_sigwinch_adjust);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<interface_sigwinch_adjust>:
  endbr64
  sub    $0x8,%rsp
  mov    D7(%rip),%rax
  mov    (%rax),%esi
  mov    D40(%rip),%rax
  mov    (%rax),%edi
  call   <resize_term@plt>
  mov    D34(%rip),%rax
  mov    (%rax),%edi
  call   <wrefresh@plt>
  mov    D22(%rip),%rax
  mov    D6(%rip),%rsi
  mov    (%rax),%edi
  add    $0x8,%rsp
  jmp    1080 <signal@plt>
```

assistant: ```c
void interface_sigwinch_adjust(int sig) {
  resize_term(LINES, COLS);
  wrefresh(curscr);
  signal(SIGWINCH, interface_sigwinch_adjust);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L34:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L44
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  jmp    L8
L44:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jne    L8
  addl   $0x1,-0x10(%rbp)
L8:
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L34
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D27
type_infered: string
name: parse_data
arguments: label: D23
type_infered: string

tool: D27: %d

tool: D23: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int max_value = 0;
  int max_count = 0;
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    if (t > max_value) {
      max_value = t;
      max_count = 1;
    } else if (t == max_value) {
      max_count++;
    }
  }
  printf("%d\n", max_count);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D19(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,(%rsp)
  jle    L56
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  mov    $0x0,%ebp
  lea    D19(%rip),%r13
  jmp    L43
L35:
  mov    %eax,%ebp
  mov    $0x1,%r12d
L31:
  add    $0x1,%ebx
  cmp    %ebx,(%rsp)
  jle    L5
L43:
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebp,%eax
  jg     L35
  sete   %al
  movzbl %al,%eax
  add    %eax,%r12d
  jmp    L31
L56:
  mov    $0x0,%r12d
L5:
  mov    %r12d,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D19
type_infered: string
name: extract_symbolic_data
arguments: data_label: D4
type_infered: string

tool: D19: %d

tool: D4: %d


assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int max_value = 0;
  int max_count = 0;
  for (int i = 0; i < n; i++) {
    int t;
    scanf("%d", &t);
    if (t > max_value) {
      max_value = t;
      max_count = 1;
    } else if (t == max_value) {
      max_count++;
    }
  }
  printf("%d\n", max_count);
  return 0;
}
```