ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<filter_positions_v4>:
  endbr64
  push   %r15
  mov    %rdi,%rax
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    0x40(%rsp),%edi
  mov    0x50(%rsp),%rbp
  mov    0x58(%rsp),%r12
  test   %edx,%edx
  jle    L93
  mov    %rsi,%r10
  mov    0x48(%rsp),%r15d
  mov    %edx,%esi
  mov    %ecx,%r11d
  sub    $0x1,%esi
  lea    0x1(%r10),%rcx
  lea    0x4(%rax),%rdx
  xor    %r10d,%r10d
  lea    0x8(%rax,%rsi,4),%rsi
  add    %edi,%r15d
  jmp    L99
  nopl   0x0(%rax)
L95:
  cmp    %r11d,%eax
  jg     L43
  add    $0x1,%r10d
  movslq %r10d,%rbx
  lea    0x0(%rbp,%rbx,4),%r13
  test   %r9d,%r9d
  jne    L23
  lea    (%rax,%rdi,1),%r14d
  test   %r8d,%r8d
  cmovne %r14d,%eax
L70:
  mov    %eax,0x0(%r13)
  movzbl (%rcx),%eax
  mov    %al,(%r12,%rbx,1)
L43:
  add    $0x4,%rdx
  add    $0x1,%rcx
  cmp    %rsi,%rdx
  je     L34
L99:
  mov    (%rdx),%eax
  test   %eax,%eax
  jg     L95
  mov    %eax,%ebx
  neg    %ebx
  cmp    %edi,%ebx
  jg     L118
  cmp    %r8d,%ebx
  jg     L43
  add    $0x1,%r10d
  add    %edi,%eax
  add    0x48(%rsp),%eax
  add    $0x4,%rdx
  movslq %r10d,%rbx
  add    $0x1,%rcx
  mov    %eax,0x0(%rbp,%rbx,4)
  movzbl -0x1(%rcx),%eax
  mov    %al,(%r12,%rbx,1)
  cmp    %rsi,%rdx
  jne    L99
L34:
  pop    %rbx
  mov    %r10d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L118:
  sub    %edi,%ebx
  cmp    %r9d,%ebx
  jg     L43
  mov    0x48(%rsp),%r13d
  add    $0x1,%r10d
  movslq %r10d,%rax
  sub    %ebx,%r13d
  mov    %r13d,0x0(%rbp,%rax,4)
  movzbl (%rcx),%ebx
  mov    %bl,(%r12,%rax,1)
  jmp    L43
  nopl   0x0(%rax)
L23:
  add    %r15d,%eax
  jmp    L70
  nopl   0x0(%rax,%rax,1)
L93:
  xor    %r10d,%r10d
  pop    %rbx
  pop    %rbp
  mov    %r10d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int filter_positions_v4(int *pos, char *strand, int n, int max_ul, int max_el, int max_il, int ul, int el, int il, int *filtered_pos, char *filtered_strand) {
  int i;
  int in;
  int nout = 0;
  int current_pos;
  for (i = 1; i <= n; i++) {
    current_pos = pos[i];
    if (current_pos > 0) {
      if (current_pos <= max_ul) {
        nout++;
        if (max_il == 0) {
          if (max_el == 0) {
            filtered_pos[nout] = current_pos;
          } else {
            filtered_pos[nout] = current_pos + el;
          }
        } else {
          filtered_pos[nout] = el + il + current_pos;
        }
        filtered_strand[nout] = strand[i];
      }
    } else {
      current_pos = -current_pos;
      if (current_pos <= el) {
        if (current_pos <= max_el) {
          current_pos = el - current_pos;
          nout++;
          filtered_pos[nout] = current_pos + il;
          filtered_strand[nout] = strand[i];
        }
      } else {
        current_pos = current_pos - el;
        if (current_pos <= max_il) {
          nout++;
          filtered_pos[nout] = il - current_pos;
          filtered_strand[nout] = strand[i];
        }
      }
    }
  }
  return nout;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<testB>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x0,%eax
  call   <getb@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L2
  mov    $0x1,%eax
  jmp    L4
L2:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <ungetb@plt>
  mov    $0x0,%eax
L4:
  leave
  ret
```

assistant: ```c
int testB(x) const int x;
{
  int i;
  if ((i = getb()) == x)
    return 1;
  ungetb(i);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<testB>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <getb@plt>
  mov    %eax,%edi
  mov    $0x1,%eax
  cmp    %ebx,%edi
  je     L0
  mov    $0x0,%eax
  call   <ungetb@plt>
  mov    $0x0,%eax
L0:
  pop    %rbx
  ret
```

assistant: ```c
int testB(x) const int x;
{
  int i;
  if ((i = getb()) == x)
    return 1;
  ungetb(i);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<testB>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  mov    %edi,%ebx
  call   <getb@plt>
  mov    %eax,%edi
  mov    $0x1,%eax
  cmp    %ebx,%edi
  je     L3
  xor    %eax,%eax
  call   <ungetb@plt>
  xor    %eax,%eax
L3:
  pop    %rbx
  ret
```

assistant: ```c
int testB(x) const int x;
{
  int i;
  if ((i = getb()) == x)
    return 1;
  ungetb(i);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<ybase64_encode_alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  movq   $0x0,-0x18(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L51
  cmpq   $0x0,-0x30(%rbp)
  je     L51
  cmpq   $0x0,-0x38(%rbp)
  je     L51
  mov    -0x30(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <ybase64_encode@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L51
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x30(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <ybase64_encode@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jne    L1
  mov    -0x38(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  jmp    L51
L1:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x18(%rbp)
L51:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *ybase64_encode_alloc(const void *from, const size_t from_len, size_t *to_len) {
  void *to = NULL;
  if (from && from_len > 0 && to_len) {
    size_t size = ybase64_encode(from, from_len, NULL, 0);
    to = malloc(size);
    if (to) {
      size_t sz = ybase64_encode(from, from_len, to, size);
      if (sz == size) {
        *to_len = size;
      } else {
        free(to);
        to = NULL;
      }
    }
  }
  return to;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ybase64_encode_alloc>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%r13
  test   %rsi,%rsi
  setne  %dl
  test   %r13,%r13
  setne  %al
  test   %al,%dl
  je     L48
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  test   %rdi,%rdi
  je     L48
  mov    $0x0,%ecx
  mov    $0x0,%edx
  call   <ybase64_encode@plt>
  mov    %rax,%r14
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L33
  mov    %r14,%rcx
  mov    %rax,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <ybase64_encode@plt>
  cmp    %rax,%r14
  jne    L11
  mov    %r14,0x0(%r13)
  jmp    L33
L11:
  mov    %r12,%rdi
  call   <free@plt>
  mov    $0x0,%r12d
  jmp    L33
L48:
  mov    $0x0,%r12d
L33:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void *ybase64_encode_alloc(const void *from, const size_t from_len, size_t *to_len) {
  void *to = NULL;
  if (from && from_len > 0 && to_len) {
    size_t size = ybase64_encode(from, from_len, NULL, 0);
    to = malloc(size);
    if (to) {
      size_t sz = ybase64_encode(from, from_len, to, size);
      if (sz == size) {
        *to_len = size;
      } else {
        free(to);
        to = NULL;
      }
    }
  }
  return to;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ybase64_encode_alloc>:
  endbr64
  push   %r14
  test   %rsi,%rsi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  setne  %dl
  test   %rbx,%rbx
  setne  %al
  test   %al,%dl
  je     L4
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L4
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rsi,%rbp
  call   <ybase64_encode@plt>
  mov    %rax,%rdi
  mov    %rax,%r14
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L4
  mov    %r14,%rcx
  mov    %rax,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <ybase64_encode@plt>
  cmp    %rax,%r14
  jne    L23
  mov    %r14,(%rbx)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  mov    %r13,%rdi
  call   <free@plt>
L4:
  xor    %r13d,%r13d
  pop    %rbx
  pop    %rbp
  mov    %r13,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void *ybase64_encode_alloc(const void *from, const size_t from_len, size_t *to_len) {
  void *to = NULL;
  if (from && from_len > 0 && to_len) {
    size_t size = ybase64_encode(from, from_len, NULL, 0);
    to = malloc(size);
    if (to) {
      size_t sz = ybase64_encode(from, from_len, to, size);
      if (sz == size) {
        *to_len = size;
      } else {
        free(to);
        to = NULL;
      }
    }
  }
  return to;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ybase64_encode_alloc>:
  endbr64
  push   %r14
  test   %rsi,%rsi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  setne  %dl
  test   %rbx,%rbx
  setne  %al
  test   %al,%dl
  je     L29
  mov    %rdi,%r12
  test   %rdi,%rdi
  je     L29
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rsi,%rbp
  call   <ybase64_encode@plt>
  mov    %rax,%rdi
  mov    %rax,%r14
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L29
  mov    %r14,%rcx
  mov    %rax,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <ybase64_encode@plt>
  cmp    %rax,%r14
  jne    L12
  mov    %r14,(%rbx)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %r13,%rdi
  call   <free@plt>
L29:
  xor    %r13d,%r13d
  pop    %rbx
  pop    %rbp
  mov    %r13,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void *ybase64_encode_alloc(const void *from, const size_t from_len, size_t *to_len) {
  void *to = NULL;
  if (from && from_len > 0 && to_len) {
    size_t size = ybase64_encode(from, from_len, NULL, 0);
    to = malloc(size);
    if (to) {
      size_t sz = ybase64_encode(from, from_len, to, size);
      if (sz == size) {
        *to_len = size;
      } else {
        free(to);
        to = NULL;
      }
    }
  }
  return to;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<EEPROMInit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x2,%edi
  call   <SysCtlDelay@plt>
  mov    $0x0,%eax
  call   <_EEPROMWaitForDone@plt>
  mov    $0x400af01c,%eax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0xc,%eax
  test   %eax,%eax
  je     L23
  mov    $0xf0005800,%edi
  call   <SysCtlPeripheralReset@plt>
  mov    $0x2,%edi
  call   <SysCtlDelay@plt>
  mov    $0x0,%eax
  call   <_EEPROMWaitForDone@plt>
  mov    $0x400af01c,%eax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0xc,%eax
  test   %eax,%eax
  je     L29
  mov    $0x2,%eax
  jmp    L6
L29:
  mov    $0x1,%eax
  jmp    L6
L23:
  mov    $0x0,%eax
L6:
  leave
  ret
```

assistant: ```c
uint32_t EEPROMInit(void) {
  uint32_t ui32Status;
  SysCtlDelay(2);
  _EEPROMWaitForDone();
  ui32Status = (*((volatile uint32_t *)(0x400AF01C)));
  if (ui32Status & (0x00000008 | 0x00000004)) {
    SysCtlPeripheralReset(0xf0005800);
    SysCtlDelay(2);
    _EEPROMWaitForDone();
    ui32Status = (*((volatile uint32_t *)(0x400AF01C)));
    if (ui32Status & (0x00000008 | 0x00000004)) {
      return (2);
    } else {
      return (1);
    }
  }
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<EEPROMInit>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <SysCtlDelay@plt>
  mov    $0x0,%eax
  call   <_EEPROMWaitForDone@plt>
  mov    0x400af01c,%eax
  and    $0xc,%eax
  jne    L17
L19:
  add    $0x8,%rsp
  ret
L17:
  mov    $0xf0005800,%edi
  call   <SysCtlPeripheralReset@plt>
  mov    $0x2,%edi
  call   <SysCtlDelay@plt>
  mov    $0x0,%eax
  call   <_EEPROMWaitForDone@plt>
  mov    0x400af01c,%eax
  and    $0xc,%eax
  cmp    $0x1,%eax
  mov    $0x1,%eax
  sbb    $0xffffffff,%eax
  jmp    L19
```

assistant: ```c
uint32_t EEPROMInit(void) {
  uint32_t ui32Status;
  SysCtlDelay(2);
  _EEPROMWaitForDone();
  ui32Status = (*((volatile uint32_t *)(0x400AF01C)));
  if (ui32Status & (0x00000008 | 0x00000004)) {
    SysCtlPeripheralReset(0xf0005800);
    SysCtlDelay(2);
    _EEPROMWaitForDone();
    ui32Status = (*((volatile uint32_t *)(0x400AF01C)));
    if (ui32Status & (0x00000008 | 0x00000004)) {
      return (2);
    } else {
      return (1);
    }
  }
  return (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<EEPROMInit>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <SysCtlDelay@plt>
  xor    %eax,%eax
  call   <_EEPROMWaitForDone@plt>
  mov    0x400af01c,%eax
  and    $0xc,%eax
  jne    L0
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    $0xf0005800,%edi
  call   <SysCtlPeripheralReset@plt>
  mov    $0x2,%edi
  call   <SysCtlDelay@plt>
  xor    %eax,%eax
  call   <_EEPROMWaitForDone@plt>
  mov    0x400af01c,%eax
  and    $0xc,%eax
  cmp    $0x1,%eax
  mov    $0x1,%eax
  sbb    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
uint32_t EEPROMInit(void) {
  uint32_t ui32Status;
  SysCtlDelay(2);
  _EEPROMWaitForDone();
  ui32Status = (*((volatile uint32_t *)(0x400AF01C)));
  if (ui32Status & (0x00000008 | 0x00000004)) {
    SysCtlPeripheralReset(0xf0005800);
    SysCtlDelay(2);
    _EEPROMWaitForDone();
    ui32Status = (*((volatile uint32_t *)(0x400AF01C)));
    if (ui32Status & (0x00000008 | 0x00000004)) {
      return (2);
    } else {
      return (1);
    }
  }
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<EEPROMInit>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <SysCtlDelay@plt>
  xor    %eax,%eax
  call   <_EEPROMWaitForDone@plt>
  mov    0x400af01c,%eax
  and    $0xc,%eax
  jne    L7
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  mov    $0xf0005800,%edi
  call   <SysCtlPeripheralReset@plt>
  mov    $0x2,%edi
  call   <SysCtlDelay@plt>
  xor    %eax,%eax
  call   <_EEPROMWaitForDone@plt>
  mov    0x400af01c,%eax
  and    $0xc,%eax
  cmp    $0x1,%eax
  mov    $0x1,%eax
  sbb    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
uint32_t EEPROMInit(void) {
  uint32_t ui32Status;
  SysCtlDelay(2);
  _EEPROMWaitForDone();
  ui32Status = (*((volatile uint32_t *)(0x400AF01C)));
  if (ui32Status & (0x00000008 | 0x00000004)) {
    SysCtlPeripheralReset(0xf0005800);
    SysCtlDelay(2);
    _EEPROMWaitForDone();
    ui32Status = (*((volatile uint32_t *)(0x400AF01C)));
    if (ui32Status & (0x00000008 | 0x00000004)) {
      return (2);
    } else {
      return (1);
    }
  }
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<grid_clear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x52,-0xc(%rbp)
  jmp    L18
L0:
  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)
  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
  movl   $0x0,(%rdx,%rax,4)
L18:
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  test   %eax,%eax
  jne    L0
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void grid_clear(int sudoku[9][9]) {
  int v;
  int h;
  int pos;
  pos = 82;
  while (pos--) {
    v = pos / 9;
    h = pos % 9;
    sudoku[v][h] = 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<grid_clear>:
  endbr64
  mov    $0x51,%edx
L0:
  movslq %edx,%rax
  imul   $0x38e38e39,%rax,%rax
  sar    $0x21,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  movslq %eax,%rcx
  lea    (%rcx,%rcx,8),%rcx
  lea    (%rdi,%rcx,4),%rcx
  lea    (%rax,%rax,8),%esi
  mov    %edx,%eax
  sub    %esi,%eax
  cltq
  movl   $0x0,(%rcx,%rax,4)
  sub    $0x1,%edx
  cmp    $0xffffffff,%edx
  jne    L0
  ret
```

assistant: ```c
void grid_clear(int sudoku[9][9]) {
  int v;
  int h;
  int pos;
  pos = 82;
  while (pos--) {
    v = pos / 9;
    h = pos % 9;
    sudoku[v][h] = 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<grid_clear>:
  endbr64
  mov    $0x51,%edx
  nopl   0x0(%rax)
L2:
  mov    %edx,%eax
  imul   $0x38e38e39,%rax,%rax
  shr    $0x21,%rax
  movslq %eax,%rcx
  lea    (%rax,%rax,8),%esi
  mov    %edx,%eax
  lea    (%rcx,%rcx,8),%rcx
  sub    %esi,%eax
  lea    (%rdi,%rcx,4),%rcx
  cltq
  movl   $0x0,(%rcx,%rax,4)
  sub    $0x1,%edx
  jae    L2
  ret
```

assistant: ```c
void grid_clear(int sudoku[9][9]) {
  int v;
  int h;
  int pos;
  pos = 82;
  while (pos--) {
    v = pos / 9;
    h = pos % 9;
    sudoku[v][h] = 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<grid_clear>:
  endbr64
  mov    $0x51,%edx
  nopl   0x0(%rax)
L1:
  mov    %edx,%eax
  imul   $0x38e38e39,%rax,%rax
  shr    $0x21,%rax
  movslq %eax,%rcx
  lea    (%rax,%rax,8),%esi
  mov    %edx,%eax
  lea    (%rcx,%rcx,8),%rcx
  sub    %esi,%eax
  lea    (%rdi,%rcx,4),%rcx
  cltq
  movl   $0x0,(%rcx,%rax,4)
  sub    $0x1,%edx
  jae    L1
  ret
```

assistant: ```c
void grid_clear(int sudoku[9][9]) {
  int v;
  int h;
  int pos;
  pos = 82;
  while (pos--) {
    v = pos / 9;
    h = pos % 9;
    sudoku[v][h] = 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<solvemefirst>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
type: string
name: parse_data
arguments: data_label: D5
type: string

tool: D8: %d %d

tool: D5: %d

assistant: ```c
int solvemefirst(void) {
  int num1, num2;
  scanf("%d %d", &num1, &num2);
  printf("%d", num1 + num2);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<solvemefirst>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  add    (%rsp),%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D10
type_guessed: string
name: unpack_data
arguments: data_label: D14
type_guessed: string

tool: D10: %d %d

tool: D14: %d

assistant: ```c
int solvemefirst(void) {
  int num1, num2;
  scanf("%d %d", &num1, &num2);
  printf("%d", num1 + num2);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<solvemefirst>:
  endbr64
  sub    $0x18,%rsp
  lea    D18(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    D14(%rip),%rdi
  add    (%rsp),%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D18
type_guessed: string
name: parse_data
arguments: data_label: D14
type_guessed: string

tool: D18: %d %d

tool: D14: %d

assistant: ```c
int solvemefirst(void) {
  int num1, num2;
  scanf("%d %d", &num1, &num2);
  printf("%d", num1 + num2);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<solvemefirst>:
  endbr64
  sub    $0x18,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    0x4(%rsp),%rdx
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  lea    D6(%rip),%rdi
  add    (%rsp),%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D7
data_type: string
name: parse_data
arguments: data_label: D6
data_type: string

tool: D7: %d %d

tool: D6: %d

assistant: ```c
int solvemefirst(void) {
  int num1, num2;
  scanf("%d %d", &num1, &num2);
  printf("%d", num1 + num2);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<inet_aton_>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  movl   $0x0,(%rsp)
  test   %eax,%eax
  jle    L9
  mov    %rbp,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbp,%rax,1),%rdi
  mov    $0x1,%r9d
  mov    $0x0,%ecx
  mov    $0x0,%r10d
  jmp    L160
L86:
  add    $0x1,%ecx
  cmp    $0x3,%ecx
  jg     L177
  movslq %ecx,%rax
  movl   $0x0,(%rsp,%rax,4)
  jmp    L150
L55:
  mov    %r10d,%r9d
L150:
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  je     L3
L160:
  movzbl (%rdx),%eax
  cmp    $0x2e,%al
  je     L86
  lea    -0x30(%rax),%esi
  cmp    $0x9,%sil
  ja     L55
  movslq %ecx,%rsi
  mov    (%rsp,%rsi,4),%r8d
  lea    (%r8,%r8,4),%r8d
  movsbl %al,%eax
  lea    -0x30(%rax,%r8,2),%eax
  mov    %eax,(%rsp,%rsi,4)
  jmp    L150
L177:
  mov    $0x0,%r9d
L3:
  cmp    $0x2,%ecx
  je     L124
  jg     L126
  test   %ecx,%ecx
  je     L1
  cmp    $0x1,%ecx
  jne    L53
  mov    (%rsp),%eax
  shl    $0x18,%eax
  mov    0x4(%rsp),%edx
  and    $0xffffff,%edx
  or     %edx,%eax
  jmp    L89
L53:
  mov    $0x0,%eax
  jmp    L89
L126:
  mov    $0x0,%eax
  cmp    $0x3,%ecx
  jne    L89
  mov    (%rsp),%eax
  shl    $0x18,%eax
  movzbl 0xc(%rsp),%edx
  or     %edx,%eax
  mov    0x4(%rsp),%edx
  shl    $0x10,%edx
  and    $0xff0000,%edx
  or     %edx,%eax
  mov    0x8(%rsp),%edx
  shl    $0x8,%edx
  movzwl %dx,%edx
  or     %edx,%eax
L89:
  mov    %eax,(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L172
  mov    %r9d,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L124:
  mov    (%rsp),%eax
  shl    $0x18,%eax
  movzwl 0x8(%rsp),%edx
  or     %edx,%eax
  mov    0x4(%rsp),%edx
  shl    $0x10,%edx
  and    $0xff0000,%edx
  or     %edx,%eax
  jmp    L89
L9:
  mov    $0x1,%r9d
L1:
  mov    (%rsp),%eax
  jmp    L89
L172:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int inet_aton_(char *s, int *retval) {
  int i, l, x, el[4], ret, good = 1;
  x = 0, ret = 0;
  l = strlen(s);
  el[0] = 0;
  for (i = 0; i < l; i++) {
    if (s[i] == '.') {
      x++;
      if (x > 3) {
        good = 0;
        break;
      }
      el[x] = 0;
    } else if (s[i] >= '0' && s[i] <= '9') {
      el[x] *= 10;
      el[x] += s[i] - '0';
    } else
      good = 0;
  }
  switch (x) {
  case 3:
    ret = (((el[0] << 24) & 0xff000000) | ((el[1] << 16) & 0x00ff0000) | ((el[2] << 8) & 0x0000ff00) | (el[3] & 0x000000ff));
    break;
  case 2:
    ret = (((el[0] << 24) & 0xff000000) | ((el[1] << 16) & 0x00ff0000) | (el[2] & 0x0000ffff));
    break;
  case 1:
    ret = (((el[0] << 24) & 0xff000000) | (el[1] & 0x00ffffff));
    break;
  case 0:
    ret = el[0];
    break;
  }
  *retval = ret;
  return good;
}
```