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

```asm
<VHB_DrawTile8M>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %edx,%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  call   <VL_GetScrollX@plt>
  and    $0x8,%eax
  add    %eax,%ebp
  xor    %eax,%eax
  call   <VL_GetScrollY@plt>
  lea    0x7(%rbp),%edx
  mov    %ebp,%edi
  add    %eax,%r12d
  lea    0x7(%r12),%ecx
  mov    %r12d,%esi
  call   <VH_MarkUpdateBlock@plt>
  test   %rax,%rax
  jne    L7
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L7:
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    10a0 <VH_DrawTile8M@plt>
```

assistant: ```c
void VHB_DrawTile8M(int x, int y, int tile) {
  x += VL_GetScrollX() & 8;
  y += VL_GetScrollY();
  if (VH_MarkUpdateBlock(x, y, x + 7, y + 7))
    VH_DrawTile8M(x, y, tile);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<placement>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %esi,%r12d
  mov    %edx,%r13d
  mov    %ecx,%ebx
  mov    %r8d,%ebp
  cmp    $0x5,%ecx
  ja     L133
  mov    %ecx,%eax
  lea    D7(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x5,%r15d
L133:
  movsbl %bpl,%r8d
  mov    %r15d,%ecx
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  call   <sanityCheck@plt>
  cmp    $0x1,%eax
  je     L138
  cmp    $0x2,%eax
  je     L138
  cmp    $0x53,%bpl
  je     L5
  jg     L27
  cmp    $0x45,%bpl
  je     L195
  cmp    $0x4e,%bpl
  jne    L113
  movslq %r13d,%r13
  movslq %r12d,%r12
  lea    (%r12,%r12,4),%rax
  lea    0x0(%r13,%rax,2),%rax
  lea    (%r14,%rax,4),%rax
  lea    -0x1(%r15),%edx
  add    %r12,%rdx
  lea    (%rdx,%rdx,4),%rdx
  lea    0x0(%r13,%rdx,2),%rdx
  lea    0x28(%r14,%rdx,4),%rdx
L115:
  mov    %ebx,(%rax)
  add    $0x28,%rax
  cmp    %rdx,%rax
  jne    L115
  mov    $0x0,%eax
L138:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  mov    $0x4,%r15d
  jmp    L133
  mov    %ecx,%r15d
  jmp    L133
  mov    $0x3,%r15d
  jmp    L133
  mov    $0x2,%r15d
  jmp    L133
L113:
  mov    $0x0,%eax
  jmp    L138
L27:
  mov    $0x0,%eax
  cmp    $0x57,%bpl
  jne    L138
  movslq %r12d,%r12
  lea    (%r12,%r12,4),%rax
  movslq %r13d,%r13
  lea    0x0(%r13,%rax,2),%rax
  lea    (%r14,%rax,4),%rdx
  mov    $0x0,%eax
L162:
  mov    %ebx,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r15d
  jg     L162
  mov    $0x0,%eax
  jmp    L138
L5:
  movslq %r12d,%r12
  lea    (%r12,%r12,4),%rax
  movslq %r13d,%r13
  lea    0x0(%r13,%rax,2),%rax
  lea    (%r14,%rax,4),%rdx
  mov    $0x0,%eax
L99:
  mov    %ebx,(%rdx)
  add    $0x1,%eax
  sub    $0x28,%rdx
  cmp    %eax,%r15d
  jne    L99
  mov    $0x0,%eax
  jmp    L138
L195:
  movslq %r12d,%r12
  lea    (%r12,%r12,4),%rax
  movslq %r13d,%r13
  lea    0x0(%r13,%rax,2),%rax
  lea    (%r14,%rax,4),%rdx
  mov    $0x0,%eax
L39:
  mov    %ebx,(%rdx)
  add    $0x1,%eax
  sub    $0x4,%rdx
  cmp    %eax,%r15d
  jne    L39
  mov    $0x0,%eax
  jmp    L138
```

assistant: ```c
int placement(int (*board)[][10], int row, int col, int ship, char direction) {
  int length;
  switch (ship) {
  case 1:
    length = 5;
    break;
  case 2:
    length = 4;
    break;
  case 3:
    length = 3;
    break;
  case 4:
    length = 3;
    break;
  case 5:
    length = 2;
    break;
    break;
  }
  switch (sanityCheck(board, row, col, length, direction)) {
  case 0:
    break;
  case 1:
    return 1;
  case 2:
    return 2;
  }
  switch (direction) {
  case 'N':
    for (int i = 0; i < length; i++) {
      (*board)[row + i][col] = ship;
    }
    break;
  case 'S':
    for (int i = 0; i < length; i++) {
      (*board)[row - i][col] = ship;
    }
    break;
  case 'E':
    for (int i = 0; i < length; i++) {
      (*board)[row][col - i] = ship;
    }
    break;
  case 'W':
    for (int i = 0; i < length; i++) {
      (*board)[row][col + i] = ship;
    }
    break;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<placement>:
  endbr64
  push   %r15
  lea    -0x1(%rcx),%eax
  push   %r14
  mov    %rdi,%r14
  push   %r13
  movslq %edx,%r13
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %r8d,%ebp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x8,%rsp
  cmp    $0x4,%eax
  ja     L23
  lea    D9(%rip),%rdx
  mov    (%rdx,%rax,4),%r15d
L23:
  movsbl %bpl,%r8d
  mov    %r15d,%ecx
  mov    %r13d,%edx
  mov    %r12d,%esi
  mov    %r14,%rdi
  call   <sanityCheck@plt>
  cmp    $0x1,%eax
  je     L32
  cmp    $0x2,%eax
  je     L32
  cmp    $0x53,%bpl
  je     L64
  jg     L176
  cmp    $0x45,%bpl
  je     L85
  cmp    $0x4e,%bpl
  jne    L182
  test   %r15d,%r15d
  jle    L182
  lea    -0x1(%r15),%edx
  lea    (%r12,%r12,4),%rax
  add    %r12,%rdx
  lea    0x0(%r13,%rax,2),%rax
  lea    (%rdx,%rdx,4),%rdx
  lea    (%r14,%rax,4),%rax
  lea    0x0(%r13,%rdx,2),%rdx
  lea    0x28(%r14,%rdx,4),%rdx
  nopl   (%rax)
L139:
  mov    %ebx,(%rax)
  add    $0x28,%rax
  cmp    %rdx,%rax
  jne    L139
L182:
  xor    %eax,%eax
L32:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L176:
  xor    %eax,%eax
  cmp    $0x57,%bpl
  jne    L32
  test   %r15d,%r15d
  jle    L182
  lea    (%r12,%r12,4),%rax
  lea    0x0(%r13,%rax,2),%rax
  lea    (%r14,%rax,4),%rdx
  xor    %eax,%eax
  nopl   0x0(%rax)
L105:
  mov    %ebx,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r15d
  jg     L105
  xor    %eax,%eax
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L64:
  test   %r15d,%r15d
  jle    L182
  lea    (%r12,%r12,4),%rax
  lea    0x0(%r13,%rax,2),%rax
  lea    (%r14,%rax,4),%rdx
  xor    %eax,%eax
  nopl   0x0(%rax)
L58:
  add    $0x1,%eax
  mov    %ebx,(%rdx)
  sub    $0x28,%rdx
  cmp    %eax,%r15d
  jne    L58
  xor    %eax,%eax
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L85:
  test   %r15d,%r15d
  jle    L182
  lea    (%r12,%r12,4),%rax
  lea    0x0(%r13,%rax,2),%rax
  lea    (%r14,%rax,4),%rdx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L127:
  add    $0x1,%eax
  mov    %ebx,(%rdx)
  sub    $0x4,%rdx
  cmp    %eax,%r15d
  jne    L127
  xor    %eax,%eax
  jmp    L32
```

tool_calls: name: read_data
arguments: label: D9
data_type: string

tool: D9: 

assistant: ```c
int placement(int (*board)[][10], int row, int col, int ship, char direction) {
  int length;
  switch (ship) {
  case 1:
    length = 5;
    break;
  case 2:
    length = 4;
    break;
  case 3:
    length = 3;
    break;
  case 4:
    length = 3;
    break;
  case 5:
    length = 2;
    break;
    break;
  }
  switch (sanityCheck(board, row, col, length, direction)) {
  case 0:
    break;
  case 1:
    return 1;
  case 2:
    return 2;
  }
  switch (direction) {
  case 'N':
    for (int i = 0; i < length; i++) {
      (*board)[row + i][col] = ship;
    }
    break;
  case 'S':
    for (int i = 0; i < length; i++) {
      (*board)[row - i][col] = ship;
    }
    break;
  case 'E':
    for (int i = 0; i < length; i++) {
      (*board)[row][col - i] = ship;
    }
    break;
  case 'W':
    for (int i = 0; i < length; i++) {
      (*board)[row][col + i] = ship;
    }
    break;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<my_getnbr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L85
L80:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L6
  negl   -0xc(%rbp)
L6:
  addl   $0x1,-0x10(%rbp)
L85:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2b,%al
  je     L80
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  je     L80
  jmp    L20
L27:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L7
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x68,%al
  jg     L7
  mov    $0x0,%eax
  call   <print_only_number@plt>
  jmp    L78
L7:
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rcx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbq %al,%rax
  add    %rcx,%rax
  sub    $0x30,%rax
  mov    %rax,-0x8(%rbp)
L78:
  addl   $0x1,-0x10(%rbp)
L20:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L27
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <error_getnbr@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int my_getnbr(char *str) {
  int i;
  long int nb;
  int sign;
  i = 0;
  nb = 0;
  sign = 1;
  while (str[i] == '+' || str[i] == '-') {
    if (str[i] == '-')
      sign = sign * -1;
    i++;
  }
  while (str[i] != '\0') {
    if (str[i] - 48 > '0' && str[i] - 48 < '9')
      print_only_number();
    else
      nb = nb * 10 + str[i] - 48;
    i++;
  }
  nb = error_getnbr(nb, sign);
  return (nb);
}
```
user: What is the c source code of the assembly code below?

```asm
<my_getnbr>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  movzbl (%rdi),%edx
  lea    -0x2b(%rdx),%eax
  test   $0xfd,%al
  jne    L14
  mov    $0x1,%eax
  mov    $0x1,%ebx
L70:
  mov    %ebx,%ecx
  neg    %ecx
  cmp    $0x2d,%dl
  cmove  %ecx,%ebx
  mov    %eax,%esi
  movzbl (%rdi,%rax,1),%edx
  add    $0x1,%rax
  lea    -0x2b(%rdx),%ecx
  test   $0xfd,%cl
  je     L70
L74:
  movslq %esi,%rsi
  movzbl (%rdi,%rsi,1),%eax
  test   %al,%al
  je     L40
  lea    0x1(%rdi,%rsi,1),%rbp
  mov    $0x0,%r12d
  jmp    L16
L14:
  mov    $0x1,%ebx
  mov    $0x0,%esi
  jmp    L74
L69:
  lea    (%r12,%r12,4),%rdx
  movsbq %al,%rax
  lea    -0x30(%rax,%rdx,2),%r12
L10:
  add    $0x1,%rbp
  movzbl -0x1(%rbp),%eax
  test   %al,%al
  je     L55
L16:
  lea    -0x61(%rax),%edx
  cmp    $0x7,%dl
  ja     L69
  mov    $0x0,%eax
  call   <print_only_number@plt>
  jmp    L10
L40:
  mov    $0x0,%r12d
L55:
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <error_getnbr@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int my_getnbr(char *str) {
  int i;
  long int nb;
  int sign;
  i = 0;
  nb = 0;
  sign = 1;
  while (str[i] == '+' || str[i] == '-') {
    if (str[i] == '-')
      sign = sign * -1;
    i++;
  }
  while (str[i] != '\0') {
    if (str[i] - 48 > '0' && str[i] - 48 < '9')
      print_only_number();
    else
      nb = nb * 10 + str[i] - 48;
    i++;
  }
  nb = error_getnbr(nb, sign);
  return (nb);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_getnbr>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  movsbq (%rdi),%rax
  lea    -0x2b(%rax),%edx
  and    $0xfd,%edx
  jne    L42
  mov    $0x1,%edx
  mov    $0x1,%ebp
  nopl   (%rax)
L44:
  mov    %ebp,%ecx
  movslq %edx,%rsi
  neg    %ecx
  cmp    $0x2d,%al
  movsbq (%rdi,%rdx,1),%rax
  cmove  %ecx,%ebp
  add    $0x1,%rdx
  lea    -0x2b(%rax),%ecx
  and    $0xfd,%ecx
  je     L44
  test   %al,%al
  je     L14
L52:
  lea    0x1(%rdi,%rsi,1),%rbx
  xor    %r12d,%r12d
  jmp    L8
  cs nopw 0x0(%rax,%rax,1)
L62:
  lea    (%r12,%r12,4),%rdx
  add    $0x1,%rbx
  lea    -0x30(%rax,%rdx,2),%r12
  movsbq -0x1(%rbx),%rax
  test   %al,%al
  je     L51
L8:
  lea    -0x61(%rax),%edx
  cmp    $0x7,%dl
  ja     L62
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <print_only_number@plt>
  movsbq -0x1(%rbx),%rax
  test   %al,%al
  jne    L8
L51:
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <error_getnbr@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L42:
  mov    $0x1,%ebp
  xor    %esi,%esi
  test   %al,%al
  jne    L52
L14:
  xor    %r12d,%r12d
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <error_getnbr@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int my_getnbr(char *str) {
  int i;
  long int nb;
  int sign;
  i = 0;
  nb = 0;
  sign = 1;
  while (str[i] == '+' || str[i] == '-') {
    if (str[i] == '-')
      sign = sign * -1;
    i++;
  }
  while (str[i] != '\0') {
    if (str[i] - 48 > '0' && str[i] - 48 < '9')
      print_only_number();
    else
      nb = nb * 10 + str[i] - 48;
    i++;
  }
  nb = error_getnbr(nb, sign);
  return (nb);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_getnbr>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  movsbq (%rdi),%rax
  lea    -0x2b(%rax),%edx
  and    $0xfd,%edx
  jne    L33
  mov    $0x1,%edx
  mov    $0x1,%ebp
  nopl   (%rax)
L11:
  mov    %ebp,%ecx
  movslq %edx,%rsi
  neg    %ecx
  cmp    $0x2d,%al
  movsbq (%rdi,%rdx,1),%rax
  cmove  %ecx,%ebp
  add    $0x1,%rdx
  lea    -0x2b(%rax),%ecx
  and    $0xfd,%ecx
  je     L11
  test   %al,%al
  je     L54
L21:
  lea    0x1(%rdi,%rsi,1),%rbx
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L3:
  lea    -0x61(%rax),%edx
  cmp    $0x7,%dl
  jbe    L49
  lea    (%r12,%r12,4),%rdx
  add    $0x1,%rbx
  lea    -0x30(%rax,%rdx,2),%r12
  movsbq -0x1(%rbx),%rax
  test   %al,%al
  jne    L3
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <error_getnbr@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L49:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <print_only_number@plt>
  movsbq -0x1(%rbx),%rax
  test   %al,%al
  jne    L3
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <error_getnbr@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L33:
  mov    $0x1,%ebp
  xor    %esi,%esi
  test   %al,%al
  jne    L21
L54:
  xor    %r12d,%r12d
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <error_getnbr@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int my_getnbr(char *str) {
  int i;
  long int nb;
  int sign;
  i = 0;
  nb = 0;
  sign = 1;
  while (str[i] == '+' || str[i] == '-') {
    if (str[i] == '-')
      sign = sign * -1;
    i++;
  }
  while (str[i] != '\0') {
    if (str[i] - 48 > '0' && str[i] - 48 < '9')
      print_only_number();
    else
      nb = nb * 10 + str[i] - 48;
    i++;
  }
  nb = error_getnbr(nb, sign);
  return (nb);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L34
L49:
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L11
  addq   $0x1,-0x8(%rbp)
  jmp    L6
L19:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addq   $0x1,-0x8(%rbp)
L6:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L19
L11:
  addq   $0x1,-0x8(%rbp)
L34:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L49
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strncpy(char *d, const char *s, size_t n) {
  size_t i;
  for (i = 0; i < n; i++) {
    if ((d[i] = s[i]) == '\0') {
      for (i++; i < n; i++) {
        d[i] = '\0';
      }
    }
  }
  return d;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L49
  mov    $0x0,%r8d
  lea    (%rdi,%rdx,1),%r9
  lea    -0x1(%rdx),%r10
  jmp    L37
L61:
  mov    %rdi,%r8
L58:
  add    $0x1,%r8
  cmp    %r8,%rdx
  jbe    L49
L37:
  movzbl (%rsi,%r8,1),%ecx
  mov    %cl,(%rax,%r8,1)
  test   %cl,%cl
  jne    L58
  lea    0x1(%r8),%rdi
  cmp    %rdi,%rdx
  jbe    L61
  lea    (%rax,%rdi,1),%rcx
L23:
  movb   $0x0,(%rcx)
  add    $0x1,%rcx
  cmp    %r9,%rcx
  jne    L23
  add    %r10,%rdi
  sub    %r8,%rdi
  mov    %rdi,%r8
  jmp    L58
L49:
  ret
```

assistant: ```c
char *strncpy(char *d, const char *s, size_t n) {
  size_t i;
  for (i = 0; i < n; i++) {
    if ((d[i] = s[i]) == '\0') {
      for (i++; i < n; i++) {
        d[i] = '\0';
      }
    }
  }
  return d;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strncpy>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  je     L20
  mov    %rsi,%r14
  mov    %rdx,%rbp
  lea    -0x1(%rdx),%r15
  xor    %ebx,%ebx
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L47:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jbe    L20
L33:
  movzbl (%r14,%rbx,1),%eax
  mov    %al,(%r12,%rbx,1)
  test   %al,%al
  jne    L47
  lea    0x1(%rbx),%r13
  cmp    %rbp,%r13
  jae    L4
  lea    (%r12,%r13,1),%rdi
  add    %r15,%r13
  mov    %r15,%rdx
  xor    %esi,%esi
  sub    %rbx,%r13
  sub    %rbx,%rdx
  call   <memset@plt>
  mov    %r13,%rbx
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  ja     L33
L20:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L4:
  mov    %r13,%rbx
  jmp    L47
```

assistant: ```c
char *strncpy(char *d, const char *s, size_t n) {
  size_t i;
  for (i = 0; i < n; i++) {
    if ((d[i] = s[i]) == '\0') {
      for (i++; i < n; i++) {
        d[i] = '\0';
      }
    }
  }
  return d;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strncpy>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  je     L37
  mov    %rsi,%r14
  mov    %rdx,%rbp
  lea    -0x1(%rdx),%r15
  xor    %ebx,%ebx
  jmp    L59
  nopl   0x0(%rax,%rax,1)
L23:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jbe    L37
L59:
  movzbl (%r14,%rbx,1),%eax
  mov    %al,(%r12,%rbx,1)
  test   %al,%al
  jne    L23
  lea    0x1(%rbx),%r13
  cmp    %r13,%rbp
  jbe    L67
  lea    (%r12,%r13,1),%rdi
  add    %r15,%r13
  mov    %r15,%rdx
  xor    %esi,%esi
  sub    %rbx,%r13
  sub    %rbx,%rdx
  call   <memset@plt>
  mov    %r13,%rbx
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  ja     L59
L37:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L67:
  mov    %r13,%rbx
  jmp    L23
```

assistant: ```c
char *strncpy(char *d, const char *s, size_t n) {
  size_t i;
  for (i = 0; i < n; i++) {
    if ((d[i] = s[i]) == '\0') {
      for (i++; i < n; i++) {
        d[i] = '\0';
      }
    }
  }
  return d;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x20(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    -0x28(%rbp),%ecx
  mov    -0x14(%rbp),%al
  mov    -0x18(%rbp),%cl
  sub    %cl,%al
  jno    L50
  movb   $0x1,(%edx)
L50:
  jns    L46
  movb   $0x1,0x1(%edx)
L46:
  jne    L44
  movb   $0x1,0x2(%edx)
L44:
  jnp    L20
  movb   $0x1,0x3(%edx)
L20:
  jae    L7
  movb   $0x1,0x4(%edx)
L7:
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int test(int a, int b, char flags[], int g1, int g2) {
  int ret;
  __asm__ __volatile__("mov %3, %%edx\n\t"
                       "mov %4, %%eax\n\t"
                       "mov %5, %%ecx\n\t"
                       "movb %b1, %%al\n\t"
                       "movb %b2, %%cl\n\t"
                       "subb %%cl, %%al\n\t"
                       "jno 1f\n\t"
                       "movb $1, (%%edx)\n\t"
                       "1:\n\t"
                       "jns 1f\n\t"
                       "movb $1, 1(%%edx)\n\t"
                       "1:\n\t"
                       "jnz 1f\n\t"
                       "movb $1, 2(%%edx)\n\t"
                       "1:\n\t"
                       "jnp 1f\n\t"
                       "movb $1, 3(%%edx)\n\t"
                       "1:\n\t"
                       "jnc 1f\n\t"
                       "movb $1, 4(%%edx)\n\t"
                       "1:\n\t"
                       "movzbl %%al, %%eax\n\t"
                       "mov %%eax, %0\n\t"
                       : "=m"(ret)
                       : "m"(a), "m"(b), "m"(flags), "m"(g1), "m"(g2)
                       : "eax", "ecx", "edx", "memory");
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test>:
  endbr64
  sub    $0x38,%rsp
  mov    %edi,0x1c(%rsp)
  mov    %esi,0x18(%rsp)
  mov    %rdx,0x10(%rsp)
  mov    %ecx,0xc(%rsp)
  mov    %r8d,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%eax
  mov    0x8(%rsp),%ecx
  mov    0x1c(%rsp),%al
  mov    0x18(%rsp),%cl
  sub    %cl,%al
  jno    L59
  movb   $0x1,(%edx)
L59:
  jns    L26
  movb   $0x1,0x1(%edx)
L26:
  jne    L40
  movb   $0x1,0x2(%edx)
L40:
  jnp    L47
  movb   $0x1,0x3(%edx)
L47:
  jae    L24
  movb   $0x1,0x4(%edx)
L24:
  movzbl %al,%eax
  mov    %eax,0x24(%rsp)
  mov    0x24(%rsp),%eax
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x38,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test(int a, int b, char flags[], int g1, int g2) {
  int ret;
  __asm__ __volatile__("mov %3, %%edx\n\t"
                       "mov %4, %%eax\n\t"
                       "mov %5, %%ecx\n\t"
                       "movb %b1, %%al\n\t"
                       "movb %b2, %%cl\n\t"
                       "subb %%cl, %%al\n\t"
                       "jno 1f\n\t"
                       "movb $1, (%%edx)\n\t"
                       "1:\n\t"
                       "jns 1f\n\t"
                       "movb $1, 1(%%edx)\n\t"
                       "1:\n\t"
                       "jnz 1f\n\t"
                       "movb $1, 2(%%edx)\n\t"
                       "1:\n\t"
                       "jnp 1f\n\t"
                       "movb $1, 3(%%edx)\n\t"
                       "1:\n\t"
                       "jnc 1f\n\t"
                       "movb $1, 4(%%edx)\n\t"
                       "1:\n\t"
                       "movzbl %%al, %%eax\n\t"
                       "mov %%eax, %0\n\t"
                       : "=m"(ret)
                       : "m"(a), "m"(b), "m"(flags), "m"(g1), "m"(g2)
                       : "eax", "ecx", "edx", "memory");
  return ret;
}
```