ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<owWriteBytePower>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <hasPowerDelivery@plt>
  test   %eax,%eax
  jne    L16
L0:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L16:
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <owTouchByte@plt>
  cmp    %ebx,%eax
  jne    L0
  mov    %ebp,%edi
  mov    $0x2,%esi
  call   <owLevel@plt>
  cmp    $0x2,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int owWriteBytePower(int portnum, int sendbyte) {
  if (!hasPowerDelivery(portnum))
    return 0;
  if (owTouchByte(portnum, sendbyte) != sendbyte)
    return 0;
  if (owLevel(portnum, 0x02) != 0x02)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<owWriteBytePower>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <hasPowerDelivery@plt>
  test   %eax,%eax
  jne    L14
L7:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L14:
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <owTouchByte@plt>
  cmp    %ebx,%eax
  jne    L7
  mov    %ebp,%edi
  mov    $0x2,%esi
  call   <owLevel@plt>
  cmp    $0x2,%eax
  sete   %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int owWriteBytePower(int portnum, int sendbyte) {
  if (!hasPowerDelivery(portnum))
    return 0;
  if (owTouchByte(portnum, sendbyte) != sendbyte)
    return 0;
  if (owLevel(portnum, 0x02) != 0x02)
    return 0;
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<check_grid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x51,-0xc(%rbp)
  jmp    L2
L5:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x38e38e39,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x38e38e39,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x3,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  cmp    $0x9,%eax
  jle    L103
  mov    $0x0,%eax
  jmp    L101
L1:
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <count_key_line@plt>
  cmp    $0x1,%eax
  jle    L8
  mov    $0x0,%eax
  jmp    L101
L8:
  mov    -0x10(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <count_key_column@plt>
  cmp    $0x1,%eax
  jle    L69
  mov    $0x0,%eax
  jmp    L101
L69:
  mov    -0x10(%rbp),%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <count_key_block@plt>
  cmp    $0x1,%eax
  jle    L103
  mov    $0x0,%eax
  jmp    L101
L103:
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cmp    $0x8,%eax
  jle    L1
L2:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  test   %eax,%eax
  jne    L5
  mov    $0x1,%eax
L101:
  leave
  ret
```

assistant: ```c
bool check_grid(int sudoku[9][9]) {
  int k;
  int position;
  int v;
  int h;
  position = 81;
  while (position--) {
    v = position / 9;
    h = position % 9;
    k = 0;
    if (sudoku[v][h] > 9)
      return false;
    while (k++ < 9) {
      if (count_key_line(sudoku, v, k) > 1)
        return false;
      if (count_key_column(sudoku, h, k) > 1)
        return false;
      if (count_key_block(sudoku, v, h, k) > 1)
        return false;
    }
  }
  return true;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<check_grid>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x50,%r14d
L50:
  movslq %r14d,%rbp
  imul   $0x38e38e39,%rbp,%rbp
  sar    $0x21,%rbp
  mov    %r14d,%eax
  sar    $0x1f,%eax
  sub    %eax,%ebp
  lea    0x0(%rbp,%rbp,8),%eax
  mov    %r14d,%r13d
  sub    %eax,%r13d
  movslq %ebp,%rax
  lea    (%rax,%rax,8),%rax
  lea    (%r12,%rax,4),%rax
  movslq %r13d,%rdx
  cmpl   $0x9,(%rax,%rdx,4)
  jg     L76
  mov    $0x1,%ebx
L38:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <count_key_line@plt>
  cmp    $0x1,%eax
  jg     L19
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <count_key_column@plt>
  cmp    $0x1,%eax
  jg     L49
  mov    %ebx,%ecx
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <count_key_block@plt>
  cmp    $0x1,%eax
  jg     L69
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L38
  sub    $0x1,%r14d
  cmp    $0xffffffff,%r14d
  jne    L50
  mov    $0x1,%eax
  jmp    L36
L76:
  mov    $0x0,%eax
  jmp    L36
L19:
  mov    $0x0,%eax
L36:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L49:
  mov    $0x0,%eax
  jmp    L36
L69:
  mov    $0x0,%eax
  jmp    L36
```

assistant: ```c
bool check_grid(int sudoku[9][9]) {
  int k;
  int position;
  int v;
  int h;
  position = 81;
  while (position--) {
    v = position / 9;
    h = position % 9;
    k = 0;
    if (sudoku[v][h] > 9)
      return false;
    while (k++ < 9) {
      if (count_key_line(sudoku, v, k) > 1)
        return false;
      if (count_key_column(sudoku, h, k) > 1)
        return false;
      if (count_key_block(sudoku, v, h, k) > 1)
        return false;
    }
  }
  return true;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<check_grid>:
  endbr64
  push   %r14
  mov    $0x50,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
L2:
  mov    %r14d,%ebp
  mov    %r14d,%r13d
  imul   $0x38e38e39,%rbp,%rbp
  shr    $0x21,%rbp
  lea    0x0(%rbp,%rbp,8),%eax
  sub    %eax,%r13d
  movslq %ebp,%rax
  lea    (%rax,%rax,8),%rax
  movslq %r13d,%rdx
  lea    (%r12,%rax,4),%rax
  cmpl   $0x9,(%rax,%rdx,4)
  jg     L17
  mov    $0x1,%ebx
  cs nopw 0x0(%rax,%rax,1)
L67:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <count_key_line@plt>
  cmp    $0x1,%eax
  jg     L17
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <count_key_column@plt>
  cmp    $0x1,%eax
  jg     L17
  mov    %ebx,%ecx
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <count_key_block@plt>
  cmp    $0x1,%eax
  jg     L17
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L67
  sub    $0x1,%r14d
  jae    L2
  mov    $0x1,%eax
  jmp    L46
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
L46:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
bool check_grid(int sudoku[9][9]) {
  int k;
  int position;
  int v;
  int h;
  position = 81;
  while (position--) {
    v = position / 9;
    h = position % 9;
    k = 0;
    if (sudoku[v][h] > 9)
      return false;
    while (k++ < 9) {
      if (count_key_line(sudoku, v, k) > 1)
        return false;
      if (count_key_column(sudoku, h, k) > 1)
        return false;
      if (count_key_block(sudoku, v, h, k) > 1)
        return false;
    }
  }
  return true;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<check_grid>:
  endbr64
  push   %r14
  mov    $0x50,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
L44:
  mov    %r14d,%ebp
  mov    %r14d,%r13d
  imul   $0x38e38e39,%rbp,%rbp
  shr    $0x21,%rbp
  lea    0x0(%rbp,%rbp,8),%eax
  sub    %eax,%r13d
  movslq %ebp,%rax
  lea    (%rax,%rax,8),%rax
  movslq %r13d,%rdx
  lea    (%r12,%rax,4),%rax
  cmpl   $0x9,(%rax,%rdx,4)
  jg     L11
  mov    $0x1,%ebx
  cs nopw 0x0(%rax,%rax,1)
L31:
  mov    %ebx,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <count_key_line@plt>
  cmp    $0x1,%eax
  jg     L11
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <count_key_column@plt>
  cmp    $0x1,%eax
  jg     L11
  mov    %ebx,%ecx
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <count_key_block@plt>
  cmp    $0x1,%eax
  jg     L11
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L31
  sub    $0x1,%r14d
  jae    L44
  mov    $0x1,%eax
  jmp    L38
  nopl   0x0(%rax)
L11:
  xor    %eax,%eax
L38:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
bool check_grid(int sudoku[9][9]) {
  int k;
  int position;
  int v;
  int h;
  position = 81;
  while (position--) {
    v = position / 9;
    h = position % 9;
    k = 0;
    if (sudoku[v][h] > 9)
      return false;
    while (k++ < 9) {
      if (count_key_line(sudoku, v, k) > 1)
        return false;
      if (count_key_column(sudoku, h, k) > 1)
        return false;
      if (count_key_block(sudoku, v, h, k) > 1)
        return false;
    }
  }
  return true;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getline2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L24:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L8:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jge    L5
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x8(%rbp)
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  je     L5
  cmpl   $0xa,-0x8(%rbp)
  jne    L24
L5:
  cmpl   $0xa,-0x8(%rbp)
  jne    L27
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L27:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getline2(char *s, int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<getline2>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  cmp    $0x1,%esi
  jle    L12
  mov    %rdi,%rbp
  lea    -0x1(%rsi),%r12d
  mov    $0x0,%ebx
L38:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D3(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L60
  cmp    $0xa,%eax
  je     L60
  mov    %al,0x0(%rbp)
  add    $0x1,%ebx
  add    $0x1,%rbp
  cmp    %r12d,%ebx
  jne    L38
  mov    %r12d,%ebx
  jmp    L11
L12:
  mov    $0x0,%ebx
L60:
  cmp    $0xa,%eax
  je     L23
L11:
  movslq %ebx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  movslq %ebx,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  add    $0x1,%ebx
  jmp    L11
```

assistant: ```c
int getline2(char *s, int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<getline2>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L16
  mov    D4(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L21:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L23
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L20
  mov    %edx,%r14d
L25:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L21
  cmp    $0xa,%eax
  je     L5
L16:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  lea    0x1(%r14),%edx
L23:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L20:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int getline2(char *s, int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getline2>:
  endbr64
  push   %r14
  mov    %rdi,%rcx
  xor    %r14d,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  lea    -0x1(%rsi),%ebp
  push   %rbx
  test   %ebp,%ebp
  jle    L28
  mov    D0(%rip),%r12
  mov    %rdi,%rbx
  xor    %r14d,%r14d
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L10:
  lea    0x1(%r14),%edx
  cmp    $0xa,%eax
  je     L6
  mov    %al,(%rcx)
  lea    0x1(%rcx),%rbx
  cmp    %edx,%ebp
  je     L8
  mov    %edx,%r14d
L9:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rcx
  cmp    %eax,(%r12)
  jne    L10
  cmp    $0xa,%eax
  je     L14
L28:
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L14:
  lea    0x1(%r14),%edx
L6:
  movslq %edx,%rcx
  movb   $0xa,(%rbx)
  mov    %rcx,%r14
  add    %r13,%rcx
  movb   $0x0,(%rcx)
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L8:
  movslq %ebp,%rcx
  mov    %ebp,%r14d
  add    %r13,%rcx
  mov    %r14d,%eax
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int getline2(char *s, int lim) {
  int c, i;
  for (i = 0; i < lim - 1 && (c = getchar()) != EOF && c != '\n'; ++i)
    s[i] = c;
  if (c == '\n') {
    s[i] = c;
    ++i;
  }
  s[i] = '\0';
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<myputchar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0xa,-0x4(%rbp)
  jne    L6
  mov    $0xd,%edi
  call   <putcharfunc@plt>
L6:
  movzbl -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putcharfunc@plt>
  nop
  leave
  ret
```

assistant: ```c
void myputchar(unsigned char c) {
  if (c == '\n')
    putcharfunc('\r');
  putcharfunc(c);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<myputchar>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0xa,%dil
  je     L12
L4:
  movzbl %bl,%edi
  call   <putcharfunc@plt>
  pop    %rbx
  ret
L12:
  mov    $0xd,%edi
  call   <putcharfunc@plt>
  jmp    L4
```

assistant: ```c
void myputchar(unsigned char c) {
  if (c == '\n')
    putcharfunc('\r');
  putcharfunc(c);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<myputchar>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0xa,%dil
  je     L2
  movzbl %bl,%edi
  pop    %rbx
  jmp    1050 <putcharfunc@plt>
  cs nopw 0x0(%rax,%rax,1)
L2:
  mov    $0xd,%edi
  call   <putcharfunc@plt>
  movzbl %bl,%edi
  pop    %rbx
  jmp    1050 <putcharfunc@plt>
```

assistant: ```c
void myputchar(unsigned char c) {
  if (c == '\n')
    putcharfunc('\r');
  putcharfunc(c);
}
```
user: What is the c source code of the assembly code below:

```asm
<myputchar>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0xa,%dil
  je     L0
  movzbl %bl,%edi
  pop    %rbx
  jmp    1050 <putcharfunc@plt>
  cs nopw 0x0(%rax,%rax,1)
L0:
  mov    $0xd,%edi
  call   <putcharfunc@plt>
  movzbl %bl,%edi
  pop    %rbx
  jmp    1050 <putcharfunc@plt>
```

assistant: ```c
void myputchar(unsigned char c) {
  if (c == '\n')
    putcharfunc('\r');
  putcharfunc(c);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <yyparse@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
data_type: string

tool: D2: Type expression like 5*3


assistant: ```c
int main() {
  printf("Type expression like 5*3\n");
  yyparse();
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <yyparse@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D1
data_type: string

tool: D1: Type expression like 5*3


assistant: ```c
int main() {
  printf("Type expression like 5*3\n");
  yyparse();
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <yyparse@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D6
type: string

tool: D6: Type expression like 5*3


assistant: ```c
int main() {
  printf("Type expression like 5*3\n");
  yyparse();
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <yyparse@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D6
type_guessed: string

tool: D6: Type expression like 5*3


assistant: ```c
int main() {
  printf("Type expression like 5*3\n");
  yyparse();
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x440,%rsp
  mov    %rdi,-0x438(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x438(%rbp),%rax
  mov    %rax,%rdi
  call   <input_line@plt>
  mov    %rax,-0x420(%rbp)
  mov    -0x420(%rbp),%rsi
  lea    -0x410(%rbp),%rax
  mov    $0x3,%ecx
  lea    D19(%rip),%rdx
  mov    %rax,%rdi
  call   <nthtoken@plt>
  lea    -0x410(%rbp),%rax
  lea    D3(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x418(%rbp)
  cmpq   $0x0,-0x418(%rbp)
  jne    L29
  mov    $0x1,%eax
  jmp    L28
L29:
  mov    -0x418(%rbp),%rdx
  mov    -0x438(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <match_files@plt>
  mov    %eax,-0x424(%rbp)
  mov    -0x418(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x424(%rbp),%eax
L28:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D19
data_type: string
name: retrieve_data
arguments: data_label: D3
data_type: string

tool: D19:  


tool: D3: r

assistant: ```c
int read_file(struct input *input) {
  char filename[1024];
  FILE *realinput;
  char *line = input_line(input);
  int result;
  nthtoken(filename, line, " \n", 3);
  if (!(realinput = fopen(filename, "r")))
    return 1;
  result = match_files(input, realinput);
  fclose(realinput);
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<read_file>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  call   <input_line@plt>
  mov    %rax,%rsi
  mov    %rsp,%rbx
  mov    $0x3,%ecx
  lea    D16(%rip),%rdx
  mov    %rbx,%rdi
  call   <nthtoken@plt>
  lea    D4(%rip),%rsi
  mov    %rbx,%rdi
  call   <fopen@plt>
  mov    $0x1,%r12d
  test   %rax,%rax
  je     L19
  mov    %rax,%rbx
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <match_files@plt>
  mov    %eax,%r12d
  mov    %rbx,%rdi
  call   <fclose@plt>
L19:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    %r12d,%eax
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D16
data_type: string
name: parse_data
arguments: data_tag: D4
data_type: string

tool: D16:  


tool: D4: r

assistant: ```c
int read_file(struct input *input) {
  char filename[1024];
  FILE *realinput;
  char *line = input_line(input);
  int result;
  nthtoken(filename, line, " \n", 3);
  if (!(realinput = fopen(filename, "r")))
    return 1;
  result = match_files(input, realinput);
  fclose(realinput);
  return result;
}
```
user: What is the c source code of the assembly code below:

```asm
<read_file>:
  endbr64
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <input_line@plt>
  mov    $0x3,%ecx
  lea    D5(%rip),%rdx
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <nthtoken@plt>
  lea    D19(%rip),%rsi
  mov    %rbp,%rdi
  call   <fopen@plt>
  test   %rax,%rax
  je     L18
  mov    %rax,%rbp
  mov    %r12,%rdi
  mov    %rax,%rsi
  call   <match_files@plt>
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <fclose@plt>
L18:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x410,%rsp
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D5
type_guessed: string
name: read_data
arguments: label: D19
type_guessed: string

tool: D5:  


tool: D19: r

assistant: ```c
int read_file(struct input *input) {
  char filename[1024];
  FILE *realinput;
  char *line = input_line(input);
  int result;
  nthtoken(filename, line, " \n", 3);
  if (!(realinput = fopen(filename, "r")))
    return 1;
  result = match_files(input, realinput);
  fclose(realinput);
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read_file>:
  endbr64
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <input_line@plt>
  mov    $0x3,%ecx
  lea    D13(%rip),%rdx
  mov    %rbp,%rdi
  mov    %rax,%rsi
  call   <nthtoken@plt>
  lea    D18(%rip),%rsi
  mov    %rbp,%rdi
  call   <fopen@plt>
  test   %rax,%rax
  je     L3
  mov    %rax,%rbp
  mov    %r12,%rdi
  mov    %rax,%rsi
  call   <match_files@plt>
  mov    %rbp,%rdi
  mov    %eax,%r13d
  call   <fclose@plt>
L3:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x410,%rsp
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D13
type_infered: string
name: parse_data
arguments: data_tag: D18
type_infered: string

tool: D13:  


tool: D18: r

assistant: ```c
int read_file(struct input *input) {
  char filename[1024];
  FILE *realinput;
  char *line = input_line(input);
  int result;
  nthtoken(filename, line, " \n", 3);
  if (!(realinput = fopen(filename, "r")))
    return 1;
  result = match_files(input, realinput);
  fclose(realinput);
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<peripheralClockFrequency>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    $0xe01fc100,%eax
  movzbl (%rax),%eax
  movzbl %al,%eax
  and    $0x3,%eax
  cmp    $0x2,%eax
  je     L48
  cmp    $0x2,%eax
  jg     L17
  test   %eax,%eax
  je     L2
  cmp    $0x1,%eax
  je     L5
  jmp    L17
L2:
  movl   $0x4,-0x4(%rbp)
  jmp    L17
L5:
  movl   $0x1,-0x4(%rbp)
  jmp    L17
L48:
  movl   $0x2,-0x4(%rbp)
  nop
L17:
  mov    $0x0,%eax
  call   <processorClockFrequency@plt>
  mov    $0x0,%edx
  divl   -0x4(%rbp)
  leave
  ret
```

assistant: ```c
unsigned int peripheralClockFrequency(void) {
  unsigned int divider = 0;
  switch ((*((volatile unsigned char *)0xE01FC100)) & 3) {
  case 0:
    divider = 4;
    break;
  case 1:
    divider = 1;
    break;
  case 2:
    divider = 2;
    break;
  }
  return processorClockFrequency() / divider;
}
```