ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<should_add_conjunction>:
  endbr64
  sub    $0x1,%esi
  js     L28
  movslq %esi,%rsi
  jmp    L18
  xchg   %ax,%ax
L7:
  sub    $0x1,%rsi
  test   %esi,%esi
  js     L3
L18:
  mov    (%rdi,%rsi,4),%eax
  test   %eax,%eax
  je     L7
  mov    $0x1,%eax
  ret
  nopl   (%rax)
L3:
  ret
L28:
  xor    %eax,%eax
  ret
```

assistant: ```c
bool should_add_conjunction(int *arr, int j) {
  bool flag = false;
  for (int x = j - 1; x >= 0; x--) {
    if (arr[x] != 0) {
      flag = true;
      break;
    }
  }
  return flag;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<should_add_conjunction>:
  endbr64
  sub    $0x1,%esi
  js     L26
  movslq %esi,%rsi
  jmp    L3
  xchg   %ax,%ax
L33:
  sub    $0x1,%rsi
  test   %esi,%esi
  js     L9
L3:
  mov    (%rdi,%rsi,4),%eax
  test   %eax,%eax
  je     L33
  mov    $0x1,%eax
  ret
  nopl   (%rax)
L9:
  ret
L26:
  xor    %eax,%eax
  ret
```

assistant: ```c
bool should_add_conjunction(int *arr, int j) {
  bool flag = false;
  for (int x = j - 1; x >= 0; x--) {
    if (arr[x] != 0) {
      flag = true;
      break;
    }
  }
  return flag;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<setup_ich9_pm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xf8,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    $0x601,%edx
  mov    $0x40,%esi
  mov    %eax,%edi
  call   <pci_config_writel@plt>
  mov    -0x4(%rbp),%eax
  mov    $0x80,%edx
  mov    $0x44,%esi
  mov    %eax,%edi
  call   <pci_config_writeb@plt>
  nop
  leave
  ret
```

assistant: ```c
void setup_ich9_pm(void) {
  const int bdf = 0x1f << 3;
  pci_config_writel(bdf, 0x40, 0x601);
  pci_config_writeb(bdf, 0x44, 0x80);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<setup_ich9_pm>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x601,%edx
  mov    $0x40,%esi
  mov    $0xf8,%edi
  call   <pci_config_writel@plt>
  mov    $0x80,%edx
  mov    $0x44,%esi
  mov    $0xf8,%edi
  call   <pci_config_writeb@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void setup_ich9_pm(void) {
  const int bdf = 0x1f << 3;
  pci_config_writel(bdf, 0x40, 0x601);
  pci_config_writeb(bdf, 0x44, 0x80);
}
```
user: What is the c source code of the assembly code below:

```asm
<setup_ich9_pm>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x601,%edx
  mov    $0x40,%esi
  mov    $0xf8,%edi
  call   <pci_config_writel@plt>
  mov    $0x80,%edx
  mov    $0x44,%esi
  mov    $0xf8,%edi
  add    $0x8,%rsp
  jmp    1060 <pci_config_writeb@plt>
```

assistant: ```c
void setup_ich9_pm(void) {
  const int bdf = 0x1f << 3;
  pci_config_writel(bdf, 0x40, 0x601);
  pci_config_writeb(bdf, 0x44, 0x80);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<setup_ich9_pm>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x601,%edx
  mov    $0x40,%esi
  mov    $0xf8,%edi
  call   <pci_config_writel@plt>
  mov    $0x80,%edx
  mov    $0x44,%esi
  mov    $0xf8,%edi
  add    $0x8,%rsp
  jmp    1060 <pci_config_writeb@plt>
```

assistant: ```c
void setup_ich9_pm(void) {
  const int bdf = 0x1f << 3;
  pci_config_writel(bdf, 0x40, 0x601);
  pci_config_writeb(bdf, 0x44, 0x80);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<unpack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  shr    $0x8,%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  shr    $0x10,%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    $0x2,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  shr    $0x18,%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    $0x3,%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void unpack(unsigned long a, unsigned char *b) {
  b[0] = (unsigned char)a;
  b[1] = (unsigned char)(a >> 8);
  b[2] = (unsigned char)(a >> 16);
  b[3] = (unsigned char)(a >> 24);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<unpack>:
  endbr64
  mov    %rdi,%rax
  mov    %dil,(%rsi)
  movzbl %ah,%ecx
  mov    %cl,0x1(%rsi)
  mov    %rdi,%rdx
  shr    $0x10,%rdx
  mov    %dl,0x2(%rsi)
  shr    $0x18,%rax
  mov    %al,0x3(%rsi)
  ret
```

assistant: ```c
void unpack(unsigned long a, unsigned char *b) {
  b[0] = (unsigned char)a;
  b[1] = (unsigned char)(a >> 8);
  b[2] = (unsigned char)(a >> 16);
  b[3] = (unsigned char)(a >> 24);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<unpack>:
  endbr64
  mov    %edi,(%rsi)
  ret
```

assistant: ```c
void unpack(unsigned long a, unsigned char *b) {
  b[0] = (unsigned char)a;
  b[1] = (unsigned char)(a >> 8);
  b[2] = (unsigned char)(a >> 16);
  b[3] = (unsigned char)(a >> 24);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<unpack>:
  endbr64
  mov    %edi,(%rsi)
  ret
```

assistant: ```c
void unpack(unsigned long a, unsigned char *b) {
  b[0] = (unsigned char)a;
  b[1] = (unsigned char)(a >> 8);
  b[2] = (unsigned char)(a >> 16);
  b[3] = (unsigned char)(a >> 24);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stonith_signal_set_simple_handler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x24(%rbp),%rax
  mov    %rax,%rdi
  call   <sigemptyset@plt>
  test   %rax,%rax
  jns    L21
  mov    $0xffffffff,%eax
  jmp    L42
L21:
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x48(%rbp),%rdx
  lea    -0x20(%rbp),%rcx
  mov    -0x34(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <sigaction@plt>
  test   %rax,%rax
  jns    L8
  mov    $0xffffffff,%eax
  jmp    L42
L8:
  mov    $0x0,%eax
L42:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

assistant: ```c
int stonith_signal_set_simple_handler(int sig, void (*handler)(int), struct sigaction *oldact) {
  struct sigaction sa;
  sigset_t mask;
  (void)stonith_signal_set_simple_handler;
  if (sigemptyset(&mask) < 0) {
    return (-1);
  }
  sa.sa_handler = handler;
  sa.sa_mask = mask;
  sa.sa_flags = 0;
  if (sigaction(sig, &sa, oldact) < 0) {
    return (-1);
  }
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<stonith_signal_set_simple_handler>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  call   <sigemptyset@plt>
  test   %rax,%rax
  js     L12
  mov    %r12,0x10(%rsp)
  mov    0xc(%rsp),%eax
  mov    %eax,0x20(%rsp)
  movq   $0x0,0x18(%rsp)
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdx
  mov    %ebx,%edi
  call   <sigaction@plt>
  sar    $0x3f,%rax
L8:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    $0xffffffff,%eax
  jmp    L8
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int stonith_signal_set_simple_handler(int sig, void (*handler)(int), struct sigaction *oldact) {
  struct sigaction sa;
  sigset_t mask;
  (void)stonith_signal_set_simple_handler;
  if (sigemptyset(&mask) < 0) {
    return (-1);
  }
  sa.sa_handler = handler;
  sa.sa_mask = mask;
  sa.sa_flags = 0;
  if (sigaction(sig, &sa, oldact) < 0) {
    return (-1);
  }
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<stonith_signal_set_simple_handler>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  call   <sigemptyset@plt>
  test   %rax,%rax
  js     L3
  mov    0xc(%rsp),%eax
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdx
  mov    %ebp,%edi
  mov    %rbx,0x10(%rsp)
  mov    %eax,0x20(%rsp)
  movq   $0x0,0x18(%rsp)
  call   <sigaction@plt>
  sar    $0x3f,%rax
L17:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L3:
  mov    $0xffffffff,%eax
  jmp    L17
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int stonith_signal_set_simple_handler(int sig, void (*handler)(int), struct sigaction *oldact) {
  struct sigaction sa;
  sigset_t mask;
  (void)stonith_signal_set_simple_handler;
  if (sigemptyset(&mask) < 0) {
    return (-1);
  }
  sa.sa_handler = handler;
  sa.sa_mask = mask;
  sa.sa_flags = 0;
  if (sigaction(sig, &sa, oldact) < 0) {
    return (-1);
  }
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<stonith_signal_set_simple_handler>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdi
  call   <sigemptyset@plt>
  test   %rax,%rax
  js     L2
  mov    0xc(%rsp),%eax
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdx
  mov    %ebp,%edi
  mov    %rbx,0x10(%rsp)
  mov    %eax,0x20(%rsp)
  movq   $0x0,0x18(%rsp)
  call   <sigaction@plt>
  sar    $0x3f,%rax
L12:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L2:
  mov    $0xffffffff,%eax
  jmp    L12
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int stonith_signal_set_simple_handler(int sig, void (*handler)(int), struct sigaction *oldact) {
  struct sigaction sa;
  sigset_t mask;
  (void)stonith_signal_set_simple_handler;
  if (sigemptyset(&mask) < 0) {
    return (-1);
  }
  sa.sa_handler = handler;
  sa.sa_mask = mask;
  sa.sa_flags = 0;
  if (sigaction(sig, &sa, oldact) < 0) {
    return (-1);
  }
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cs5530_disable_softvga>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x80009444,%edi
  call   <gfx_pci_config_read@plt>
  mov    %rax,-0x8(%rbp)
  orq    $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x80009444,%edi
  call   <gfx_pci_config_write@plt>
  mov    $0x80009444,%edi
  call   <gfx_pci_config_read@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L5
  mov    $0x1,%eax
  jmp    L6
L5:
  mov    $0x0,%eax
L6:
  leave
  ret
```

assistant: ```c
int cs5530_disable_softvga(void) {
  unsigned long reg_val;
  reg_val = gfx_pci_config_read(0x80009444);
  reg_val |= 0x1;
  gfx_pci_config_write(0x80009444, reg_val);
  reg_val = gfx_pci_config_read(0x80009444);
  if ((reg_val & 0x1) == 0x1)
    return (1);
  else
    return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cs5530_disable_softvga>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x80009444,%edi
  call   <gfx_pci_config_read@plt>
  or     $0x1,%rax
  mov    %rax,%rsi
  mov    $0x80009444,%edi
  call   <gfx_pci_config_write@plt>
  mov    $0x80009444,%edi
  call   <gfx_pci_config_read@plt>
  and    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int cs5530_disable_softvga(void) {
  unsigned long reg_val;
  reg_val = gfx_pci_config_read(0x80009444);
  reg_val |= 0x1;
  gfx_pci_config_write(0x80009444, reg_val);
  reg_val = gfx_pci_config_read(0x80009444);
  if ((reg_val & 0x1) == 0x1)
    return (1);
  else
    return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cs5530_disable_softvga>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x80009444,%edi
  call   <gfx_pci_config_read@plt>
  mov    $0x80009444,%edi
  or     $0x1,%rax
  mov    %rax,%rsi
  call   <gfx_pci_config_write@plt>
  mov    $0x80009444,%edi
  call   <gfx_pci_config_read@plt>
  add    $0x8,%rsp
  and    $0x1,%eax
  ret
```

assistant: ```c
int cs5530_disable_softvga(void) {
  unsigned long reg_val;
  reg_val = gfx_pci_config_read(0x80009444);
  reg_val |= 0x1;
  gfx_pci_config_write(0x80009444, reg_val);
  reg_val = gfx_pci_config_read(0x80009444);
  if ((reg_val & 0x1) == 0x1)
    return (1);
  else
    return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cs5530_disable_softvga>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x80009444,%edi
  call   <gfx_pci_config_read@plt>
  mov    $0x80009444,%edi
  or     $0x1,%rax
  mov    %rax,%rsi
  call   <gfx_pci_config_write@plt>
  mov    $0x80009444,%edi
  call   <gfx_pci_config_read@plt>
  add    $0x8,%rsp
  and    $0x1,%eax
  ret
```

assistant: ```c
int cs5530_disable_softvga(void) {
  unsigned long reg_val;
  reg_val = gfx_pci_config_read(0x80009444);
  reg_val |= 0x1;
  gfx_pci_config_write(0x80009444, reg_val);
  reg_val = gfx_pci_config_read(0x80009444);
  if ((reg_val & 0x1) == 0x1)
    return (1);
  else
    return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ft_atoi>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  movzbl (%rdi),%esi
  mov    $0x0,%edi
  test   %sil,%sil
  je     L46
  lea    -0x9(%rsi),%eax
  cmp    $0x5,%al
  setbe  %al
  cmp    $0x20,%sil
  sete   %dl
  or     %dl,%al
  je     L74
  cmp    $0x2d,%sil
  je     L74
  mov    $0x1,%edx
L58:
  mov    %edx,%eax
  movzbl (%rbx,%rdx,1),%esi
  add    $0x1,%rdx
  lea    -0x9(%rsi),%ecx
  cmp    $0x5,%cl
  setbe  %cl
  cmp    $0x20,%sil
  sete   %dil
  or     %dil,%cl
  je     L51
  cmp    $0x2d,%sil
  jne    L58
L51:
  mov    %eax,0xc(%rsp)
L55:
  movq   $0x1,0x10(%rsp)
  lea    0x1(%rax),%edx
  movslq %edx,%rdx
  add    %rbx,%rdx
  movzbl (%rdx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L29
  mov    $0x1,%eax
L40:
  lea    (%rax,%rax,4),%rax
  add    %rax,%rax
  movzbl 0x1(%rdx),%edi
  lea    -0x30(%rdi),%ecx
  add    $0x1,%rdx
  cmp    $0x9,%cl
  jbe    L40
  mov    %rax,0x10(%rsp)
L29:
  lea    0xc(%rsp),%rdx
  movsbl %sil,%esi
  lea    0x10(%rsp),%rdi
  call   <is_negative@plt>
  movslq 0xc(%rsp),%rdx
  movzbl (%rbx,%rdx,1),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L87
  mov    0x10(%rsp),%rcx
  lea    0x1(%rbx,%rdx,1),%rsi
  mov    $0x0,%edi
  movabs $0x6666666666666667,%r8
L21:
  movsbl %al,%eax
  sub    $0x30,%eax
  imul   %ecx,%eax
  lea    (%rax,%rdi,1),%edi
  mov    %rcx,%rax
  imul   %r8
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  mov    %rdx,%rcx
  sub    %rax,%rcx
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L21
L46:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L119
  mov    %edi,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L74:
  mov    $0x0,%eax
  jmp    L55
L87:
  mov    $0x0,%edi
  jmp    L46
L119:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_atoi(char *str) {
  int num_start;
  int mult_it;
  long int mult;
  int resault;
  num_start = 0;
  if (str[num_start] == 0)
    return (0);
  while (((str[num_start] > '\10' && str[num_start] <= '\16') || str[num_start] == '\40') && str[num_start] != '-')
    num_start++;
  mult = 1;
  mult_it = num_start;
  while (str[++mult_it] >= '0' && str[mult_it] <= '9')
    mult *= 10;
  is_negative(&mult, str[num_start], &num_start);
  resault = 0;
  while (str[num_start] >= '0' && str[num_start] <= '9') {
    resault += (str[num_start++] - 48) * mult;
    mult /= 10;
  }
  return (resault);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_atoi>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  movsbl (%rdi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  test   %sil,%sil
  je     L50
  lea    -0x9(%rsi),%eax
  mov    %rdi,%rbx
  cmp    $0x5,%al
  mov    $0x1,%eax
  jbe    L111
  cmp    $0x20,%sil
  jne    L51
  cs nopw 0x0(%rax,%rax,1)
L111:
  movsbl (%rbx,%rax,1),%esi
  mov    %eax,%ecx
  add    $0x1,%rax
  lea    -0x9(%rsi),%edx
  cmp    $0x5,%dl
  jbe    L111
  cmp    $0x20,%sil
  je     L111
  mov    %ecx,0xc(%rsp)
  add    $0x1,%ecx
  movslq %ecx,%rax
L113:
  movq   $0x1,0x10(%rsp)
  movzbl (%rbx,%rax,1),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L119
  movslq %ecx,%rcx
  mov    $0x1,%eax
  lea    0x1(%rbx,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L81:
  movzbl (%rcx),%edi
  lea    (%rax,%rax,4),%rax
  add    $0x1,%rcx
  add    %rax,%rax
  lea    -0x30(%rdi),%edx
  cmp    $0x9,%dl
  jbe    L81
  mov    %rax,0x10(%rsp)
L119:
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rdi
  call   <is_negative@plt>
  movslq 0xc(%rsp),%rdx
  movzbl (%rbx,%rdx,1),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L50
  mov    0x10(%rsp),%rcx
  lea    0x1(%rbx,%rdx,1),%rsi
  xor    %r8d,%r8d
  movabs $0x6666666666666667,%rdi
  nopw   0x0(%rax,%rax,1)
L29:
  sub    $0x30,%eax
  add    $0x1,%rsi
  movsbq %al,%rax
  imul   %rcx,%rax
  add    %eax,%r8d
  mov    %rcx,%rax
  imul   %rdi
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  mov    %rdx,%rcx
  sub    %rax,%rcx
  movzbl -0x1(%rsi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L29
L79:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L107
  add    $0x20,%rsp
  mov    %r8d,%eax
  pop    %rbx
  ret
  xchg   %ax,%ax
L50:
  xor    %r8d,%r8d
  jmp    L79
  nopl   (%rax)
L51:
  mov    $0x1,%ecx
  jmp    L113
L107:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_atoi(char *str) {
  int num_start;
  int mult_it;
  long int mult;
  int resault;
  num_start = 0;
  if (str[num_start] == 0)
    return (0);
  while (((str[num_start] > '\10' && str[num_start] <= '\16') || str[num_start] == '\40') && str[num_start] != '-')
    num_start++;
  mult = 1;
  mult_it = num_start;
  while (str[++mult_it] >= '0' && str[mult_it] <= '9')
    mult *= 10;
  is_negative(&mult, str[num_start], &num_start);
  resault = 0;
  while (str[num_start] >= '0' && str[num_start] <= '9') {
    resault += (str[num_start++] - 48) * mult;
    mult /= 10;
  }
  return (resault);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_atoi>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  movsbl (%rdi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  test   %sil,%sil
  je     L10
  lea    -0x9(%rsi),%eax
  mov    %rdi,%rbx
  cmp    $0x5,%al
  mov    $0x1,%eax
  jbe    L111
  cmp    $0x20,%sil
  jne    L34
  cs nopw 0x0(%rax,%rax,1)
L111:
  movsbl (%rbx,%rax,1),%esi
  mov    %eax,%ecx
  add    $0x1,%rax
  lea    -0x9(%rsi),%edx
  cmp    $0x5,%dl
  jbe    L111
  cmp    $0x20,%sil
  je     L111
  mov    %ecx,0xc(%rsp)
  add    $0x1,%ecx
  movslq %ecx,%rax
L40:
  movq   $0x1,0x10(%rsp)
  movzbl (%rbx,%rax,1),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L98
  movslq %ecx,%rcx
  mov    $0x1,%eax
  lea    0x1(%rbx,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L21:
  movzbl (%rcx),%edi
  lea    (%rax,%rax,4),%rax
  add    $0x1,%rcx
  add    %rax,%rax
  lea    -0x30(%rdi),%edx
  cmp    $0x9,%dl
  jbe    L21
  mov    %rax,0x10(%rsp)
L98:
  lea    0xc(%rsp),%rdx
  lea    0x10(%rsp),%rdi
  call   <is_negative@plt>
  movslq 0xc(%rsp),%rdx
  movzbl (%rbx,%rdx,1),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L10
  mov    0x10(%rsp),%rcx
  lea    0x1(%rbx,%rdx,1),%rsi
  xor    %r8d,%r8d
  movabs $0x6666666666666667,%rdi
  nopw   0x0(%rax,%rax,1)
L112:
  sub    $0x30,%eax
  add    $0x1,%rsi
  movsbq %al,%rax
  imul   %rcx,%rax
  add    %eax,%r8d
  mov    %rcx,%rax
  imul   %rdi
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  mov    %rdx,%rcx
  sub    %rax,%rcx
  movzbl -0x1(%rsi),%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  jbe    L112
L88:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L92
  add    $0x20,%rsp
  mov    %r8d,%eax
  pop    %rbx
  ret
  xchg   %ax,%ax
L10:
  xor    %r8d,%r8d
  jmp    L88
  nopl   (%rax)
L34:
  mov    $0x1,%ecx
  jmp    L40
L92:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ft_atoi(char *str) {
  int num_start;
  int mult_it;
  long int mult;
  int resault;
  num_start = 0;
  if (str[num_start] == 0)
    return (0);
  while (((str[num_start] > '\10' && str[num_start] <= '\16') || str[num_start] == '\40') && str[num_start] != '-')
    num_start++;
  mult = 1;
  mult_it = num_start;
  while (str[++mult_it] >= '0' && str[mult_it] <= '9')
    mult *= 10;
  is_negative(&mult, str[num_start], &num_start);
  resault = 0;
  while (str[num_start] >= '0' && str[num_start] <= '9') {
    resault += (str[num_start++] - 48) * mult;
    mult /= 10;
  }
  return (resault);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<parse_arg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    -0x20(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x3a,%al
  jne    L1
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    -0x28(%rbp),%rdx
  mov    %eax,(%rdx)
L1:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
void parse_arg(char *arg, int *paddr, int *psize) {
  char *end;
  *paddr = strtoul(arg, &end, 0);
  if (*end == ':')
    *psize = strtoul(++end, 0, 0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<parse_arg>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0x0,%edx
  call   <strtoul@plt>
  mov    %eax,(%rbx)
  mov    (%rsp),%rdi
  cmpb   $0x3a,(%rdi)
  je     L9
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  add    $0x1,%rdi
  mov    %rdi,(%rsp)
  mov    $0x0,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  mov    %eax,0x0(%rbp)
  jmp    L22
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void parse_arg(char *arg, int *paddr, int *psize) {
  char *end;
  *paddr = strtoul(arg, &end, 0);
  if (*end == ':')
    *psize = strtoul(++end, 0, 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<parse_arg>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  xor    %edx,%edx
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtoul@plt>
  mov    (%rsp),%rdi
  mov    %eax,(%rbx)
  cmpb   $0x3a,(%rdi)
  je     L3
L16:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L3:
  add    $0x1,%rdi
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %rdi,(%rsp)
  call   <strtoul@plt>
  mov    %eax,0x0(%rbp)
  jmp    L16
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void parse_arg(char *arg, int *paddr, int *psize) {
  char *end;
  *paddr = strtoul(arg, &end, 0);
  if (*end == ':')
    *psize = strtoul(++end, 0, 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_arg>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  xor    %edx,%edx
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <strtoul@plt>
  mov    (%rsp),%rdi
  mov    %eax,(%rbx)
  cmpb   $0x3a,(%rdi)
  je     L28
L12:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L28:
  add    $0x1,%rdi
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %rdi,(%rsp)
  call   <strtoul@plt>
  mov    %eax,0x0(%rbp)
  jmp    L12
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void parse_arg(char *arg, int *paddr, int *psize) {
  char *end;
  *paddr = strtoul(arg, &end, 0);
  if (*end == ':')
    *psize = strtoul(++end, 0, 0);
}
```