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

```asm
<check_lines>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x358,%rsp
  mov    %rdi,%r8
  mov    %rsi,%r9
  mov    %rdx,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x348(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  mov    $0x69,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  lea    0x348(%r8),%rcx
  mov    $0x1b,%ebx
  mov    $0x0,%r11d
  mov    $0x0,%edi
  mov    %rsp,%rbp
  jmp    L169
L133:
  add    $0x1,%r11d
L120:
  sub    $0x1e,%rcx
  cmp    %r8,%rcx
  je     L23
L169:
  lea    -0x1e(%rcx),%rax
  mov    %rax,%rsi
  mov    %edi,%edx
L81:
  cmpb   $0x1,(%rax)
  sbb    $0xff,%dl
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L81
  cmp    $0x1e,%dl
  je     L133
  test   %dl,%dl
  je     L120
  movslq %ebx,%rdx
  mov    %rdx,%rax
  shl    $0x4,%rax
  sub    %rdx,%rax
  lea    0x0(%rbp,%rax,2),%rax
  movdqu (%rsi),%xmm0
  movups %xmm0,(%rax)
  mov    0x10(%rsi),%rdx
  mov    %rdx,0x10(%rax)
  mov    0x18(%rsi),%edx
  mov    %edx,0x18(%rax)
  movzwl 0x1c(%rsi),%edx
  mov    %dx,0x1c(%rax)
  lea    -0x1(%rbx),%ebx
  jmp    L120
L23:
  test   %r11b,%r11b
  je     L127
  mov    %rsp,%rsi
  mov    (%rsp),%rax
  mov    %rax,(%r8)
  mov    0x340(%rsp),%rax
  mov    %rax,0x340(%r8)
  lea    0x8(%r8),%rdi
  and    $0xfffffffffffffff8,%rdi
  sub    %rdi,%r8
  sub    %r8,%rsi
  lea    0x348(%r8),%ecx
  shr    $0x3,%ecx
  mov    %ecx,%ecx
  rep movsq %ds:(%rsi),%es:(%rdi)
  movsbl %r11b,%eax
  add    %eax,(%r10)
  cmp    $0x3,%r11b
  je     L139
  jg     L150
  cmp    $0x1,%r11b
  je     L26
  cmp    $0x2,%r11b
  jne    L129
  addl   $0x14,(%r9)
  jmp    L127
L150:
  cmp    $0x4,%r11b
  jne    L129
  addl   $0x50,(%r9)
  jmp    L127
L26:
  addl   $0xa,(%r9)
L127:
  movsbl %r11b,%eax
  mov    0x348(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L77
  add    $0x358,%rsp
  pop    %rbx
  pop    %rbp
  ret
L139:
  addl   $0x32,(%r9)
  jmp    L127
L129:
  addl   $0x50,(%r9)
  jmp    L127
L77:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int check_lines(char board[(30 - 2)][(32 - 2)], int *score, int *lines) {
  char tmp_board[(30 - 2)][(32 - 2)];
  memset(tmp_board, 0, (30 - 2) * (32 - 2));
  char cont = 0;
  char lns = 0;
  int i, j;
  int ti = (30 - 2) - 1;
  for (i = (30 - 2) - 1; i >= 0; i--) {
    for (j = 0; j < (32 - 2); j++) {
      if (board[i][j]) {
        cont++;
      }
    }
    if (cont == (32 - 2)) {
      cont = 0;
      lns++;
    }
    if (cont != 0) {
      memcpy(tmp_board[ti--], board[i], (32 - 2));
      cont = 0;
    }
  }
  if (lns != 0) {
    memcpy(board, tmp_board, (30 - 2) * (32 - 2));
    *lines += lns;
    switch (lns) {
    case 1:
      *score += 10;
      break;
    case 2:
      *score += 10 * 2;
      break;
    case 3:
      *score += 10 * 5;
      break;
    case 4:
      *score += 10 * 8;
      break;
    default:
      *score += 10 * 8;
      break;
    }
  }
  return lns;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<check_lines>:
  endbr64
  push   %rbx
  mov    %rdi,%r8
  mov    $0x69,%ecx
  mov    %rsi,%r9
  mov    %rdx,%r10
  lea    0x348(%r8),%rsi
  xor    %r11d,%r11d
  sub    $0x350,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x348(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    $0x1b,%edi
  nopl   (%rax)
L47:
  mov    %rsi,%rcx
  sub    $0x1e,%rsi
  xor    %edx,%edx
  mov    %rsi,%rax
  nopl   0x0(%rax)
L125:
  cmpb   $0x1,(%rax)
  sbb    $0xff,%dl
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L125
  cmp    $0x1e,%dl
  je     L135
  test   %dl,%dl
  jne    L74
L20:
  cmp    %r8,%rsi
  jne    L47
  movsbl %r11b,%r8d
  test   %r11b,%r11b
  je     L65
  mov    (%rsp),%rax
  lea    0x8(%rsi),%rdi
  and    $0xfffffffffffffff8,%rdi
  mov    %rax,(%rsi)
  mov    0x340(%rsp),%rax
  mov    %rax,0x340(%rsi)
  mov    %rsi,%rax
  sub    %rdi,%rax
  sub    %rax,%rbx
  lea    0x348(%rax),%ecx
  mov    %rbx,%rsi
  shr    $0x3,%ecx
  rep movsq %ds:(%rsi),%es:(%rdi)
  add    %r8d,(%r10)
  cmp    $0x2,%r11b
  je     L55
  cmp    $0x3,%r11b
  je     L33
  cmp    $0x1,%r11b
  je     L116
  addl   $0x50,(%r9)
L65:
  mov    0x348(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  add    $0x350,%rsp
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L135:
  add    $0x1,%r11d
  jmp    L20
L74:
  movslq %edi,%rdx
  movdqu (%rsi),%xmm0
  sub    $0x1,%edi
  mov    %rdx,%rax
  shl    $0x4,%rax
  sub    %rdx,%rax
  mov    0x10(%rsi),%rdx
  lea    (%rbx,%rax,2),%rax
  mov    %rdx,0x10(%rax)
  mov    0x18(%rsi),%edx
  movups %xmm0,(%rax)
  mov    %edx,0x18(%rax)
  movzwl 0x1c(%rsi),%edx
  mov    %dx,0x1c(%rax)
  jmp    L20
L33:
  addl   $0x32,(%r9)
  jmp    L65
L116:
  addl   $0xa,(%r9)
  jmp    L65
L55:
  addl   $0x14,(%r9)
  jmp    L65
L82:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int check_lines(char board[(30 - 2)][(32 - 2)], int *score, int *lines) {
  char tmp_board[(30 - 2)][(32 - 2)];
  memset(tmp_board, 0, (30 - 2) * (32 - 2));
  char cont = 0;
  char lns = 0;
  int i, j;
  int ti = (30 - 2) - 1;
  for (i = (30 - 2) - 1; i >= 0; i--) {
    for (j = 0; j < (32 - 2); j++) {
      if (board[i][j]) {
        cont++;
      }
    }
    if (cont == (32 - 2)) {
      cont = 0;
      lns++;
    }
    if (cont != 0) {
      memcpy(tmp_board[ti--], board[i], (32 - 2));
      cont = 0;
    }
  }
  if (lns != 0) {
    memcpy(board, tmp_board, (30 - 2) * (32 - 2));
    *lines += lns;
    switch (lns) {
    case 1:
      *score += 10;
      break;
    case 2:
      *score += 10 * 2;
      break;
    case 3:
      *score += 10 * 5;
      break;
    case 4:
      *score += 10 * 8;
      break;
    default:
      *score += 10 * 8;
      break;
    }
  }
  return lns;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<check_lines>:
  endbr64
  push   %rbx
  mov    %rdi,%r8
  mov    $0x69,%ecx
  mov    %rsi,%r9
  mov    %rdx,%r10
  lea    0x348(%r8),%rsi
  xor    %r11d,%r11d
  sub    $0x350,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x348(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    $0x1b,%edi
  nopl   (%rax)
L137:
  mov    %rsi,%rcx
  sub    $0x1e,%rsi
  xor    %edx,%edx
  mov    %rsi,%rax
  nopl   0x0(%rax)
L136:
  cmpb   $0x1,(%rax)
  sbb    $0xff,%dl
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L136
  cmp    $0x1e,%dl
  je     L68
  test   %dl,%dl
  jne    L106
L20:
  cmp    %rsi,%r8
  jne    L137
  movsbl %r11b,%eax
  test   %r11b,%r11b
  je     L67
  mov    (%rsp),%rdx
  lea    0x8(%r8),%rdi
  and    $0xfffffffffffffff8,%rdi
  mov    %rdx,(%r8)
  mov    0x340(%rsp),%rdx
  mov    %rdx,0x340(%r8)
  sub    %rdi,%r8
  sub    %r8,%rbx
  lea    0x348(%r8),%ecx
  mov    %rbx,%rsi
  shr    $0x3,%ecx
  rep movsq %ds:(%rsi),%es:(%rdi)
  add    %eax,(%r10)
  cmp    $0x2,%r11b
  je     L40
  cmp    $0x3,%r11b
  je     L25
  cmp    $0x1,%r11b
  je     L105
  addl   $0x50,(%r9)
L67:
  mov    0x348(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x350,%rsp
  pop    %rbx
  ret
  nop
L68:
  add    $0x1,%r11d
  jmp    L20
L106:
  movslq %edi,%rdx
  movdqu (%rsi),%xmm0
  sub    $0x1,%edi
  mov    %rdx,%rax
  shl    $0x4,%rax
  sub    %rdx,%rax
  mov    0x10(%rsi),%rdx
  lea    (%rbx,%rax,2),%rax
  mov    %rdx,0x10(%rax)
  mov    0x18(%rsi),%edx
  movups %xmm0,(%rax)
  mov    %edx,0x18(%rax)
  movzwl 0x1c(%rsi),%edx
  mov    %dx,0x1c(%rax)
  jmp    L20
L25:
  addl   $0x32,(%r9)
  jmp    L67
L105:
  addl   $0xa,(%r9)
  jmp    L67
L40:
  addl   $0x14,(%r9)
  jmp    L67
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int check_lines(char board[(30 - 2)][(32 - 2)], int *score, int *lines) {
  char tmp_board[(30 - 2)][(32 - 2)];
  memset(tmp_board, 0, (30 - 2) * (32 - 2));
  char cont = 0;
  char lns = 0;
  int i, j;
  int ti = (30 - 2) - 1;
  for (i = (30 - 2) - 1; i >= 0; i--) {
    for (j = 0; j < (32 - 2); j++) {
      if (board[i][j]) {
        cont++;
      }
    }
    if (cont == (32 - 2)) {
      cont = 0;
      lns++;
    }
    if (cont != 0) {
      memcpy(tmp_board[ti--], board[i], (32 - 2));
      cont = 0;
    }
  }
  if (lns != 0) {
    memcpy(board, tmp_board, (30 - 2) * (32 - 2));
    *lines += lns;
    switch (lns) {
    case 1:
      *score += 10;
      break;
    case 2:
      *score += 10 * 2;
      break;
    case 3:
      *score += 10 * 5;
      break;
    case 4:
      *score += 10 * 8;
      break;
    default:
      *score += 10 * 8;
      break;
    }
  }
  return lns;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<softmax>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L10
L65:
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L31
L42:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <exp@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0x18(%rbp)
L31:
  mov    -0x18(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L42
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  D13(%rip),%xmm0
  divsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x10(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x1c(%rbp)
L10:
  mov    -0x1c(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L65
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L71
L40:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jne    L45
  movsd  D13(%rip),%xmm0
  jmp    L23
L45:
  pxor   %xmm0,%xmm0
L23:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  subsd  %xmm2,%xmm0
  subsd  -0x10(%rbp),%xmm0
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x14(%rbp)
L71:
  mov    -0x14(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L40
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D13
type_infered: f64

tool: D13: 1

assistant: ```c
void softmax(double input[10], double loss[10], int label, int count) {
  double inner = 0;
  for (int i = 0; i < count; ++i) {
    double res = 0;
    for (int j = 0; j < count; ++j) {
      res += exp(input[j] - input[i]);
    }
    loss[i] = 1. / res;
    inner -= loss[i] * loss[i];
  }
  inner += loss[label];
  for (int i = 0; i < count; ++i) {
    loss[i] *= (i == label) - loss[i] - inner;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<softmax>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edx,0x18(%rsp)
  mov    %ecx,0x1c(%rsp)
  test   %ecx,%ecx
  jle    L43
  mov    %rdi,%r15
  mov    %rsi,%r14
  mov    %rsi,%r13
  lea    -0x1(%rcx),%eax
  lea    0x8(%rdi,%rax,8),%r12
  mov    %rdi,%rbp
  pxor   %xmm3,%xmm3
  movsd  %xmm3,0x10(%rsp)
L49:
  mov    %r15,%rbx
  pxor   %xmm4,%xmm4
  movsd  %xmm4,0x8(%rsp)
L18:
  movsd  (%rbx),%xmm0
  subsd  0x0(%rbp),%xmm0
  call   <exp@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L18
  movsd  D11(%rip),%xmm0
  divsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x0(%r13)
  mulsd  %xmm0,%xmm0
  movsd  0x10(%rsp),%xmm6
  subsd  %xmm0,%xmm6
  movsd  %xmm6,0x10(%rsp)
  add    $0x8,%r13
  add    $0x8,%rbp
  cmp    %r12,%rbp
  jne    L49
  movslq 0x18(%rsp),%rax
  movsd  0x10(%rsp),%xmm1
  addsd  (%r14,%rax,8),%xmm1
  mov    0x1c(%rsp),%edx
  mov    $0x0,%eax
  movsd  D11(%rip),%xmm3
  pxor   %xmm4,%xmm4
  jmp    L20
L52:
  subsd  %xmm2,%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm2,%xmm0
  movsd  %xmm0,(%r14,%rax,8)
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L43
L20:
  movsd  (%r14,%rax,8),%xmm2
  movapd %xmm3,%xmm0
  cmp    %eax,0x18(%rsp)
  je     L52
  movapd %xmm4,%xmm0
  jmp    L52
L43:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D11
datatype: f64

tool: D11: 1

assistant: ```c
void softmax(double input[10], double loss[10], int label, int count) {
  double inner = 0;
  for (int i = 0; i < count; ++i) {
    double res = 0;
    for (int j = 0; j < count; ++j) {
      res += exp(input[j] - input[i]);
    }
    loss[i] = 1. / res;
    inner -= loss[i] * loss[i];
  }
  inner += loss[label];
  for (int i = 0; i < count; ++i) {
    loss[i] *= (i == label) - loss[i] - inner;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<softmax>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edx,0x18(%rsp)
  mov    %ecx,0x1c(%rsp)
  test   %ecx,%ecx
  jle    L54
  mov    %ecx,%eax
  mov    %rdi,%r14
  mov    %rsi,%r13
  mov    %rsi,%r12
  movq   $0x0,0x10(%rsp)
  sub    $0x1,%eax
  mov    %rdi,%rbx
  lea    0x8(%rdi,%rax,8),%rbp
  nopl   0x0(%rax)
L65:
  mov    %r14,%r15
  pxor   %xmm1,%xmm1
  nop
L18:
  movsd  (%r15),%xmm0
  subsd  (%rbx),%xmm0
  movsd  %xmm1,0x8(%rsp)
  add    $0x8,%r15
  call   <exp@plt>
  pxor   %xmm0,%xmm0
  movsd  0x8(%rsp),%xmm1
  cvtsi2sd %rax,%xmm0
  addsd  %xmm0,%xmm1
  cmp    %rbp,%r15
  jne    L18
  movsd  D6(%rip),%xmm0
  movsd  0x10(%rsp),%xmm3
  add    $0x8,%rbx
  add    $0x8,%r12
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%r12)
  mulsd  %xmm0,%xmm0
  subsd  %xmm0,%xmm3
  movsd  %xmm3,0x10(%rsp)
  cmp    %r15,%rbx
  jne    L65
  movslq 0x18(%rsp),%rax
  movsd  0x10(%rsp),%xmm1
  mov    0x1c(%rsp),%edx
  addsd  0x0(%r13,%rax,8),%xmm1
  xor    %eax,%eax
  jmp    L9
  xchg   %ax,%ax
L26:
  subsd  %xmm2,%xmm0
  subsd  %xmm1,%xmm0
  mulsd  %xmm2,%xmm0
  movsd  %xmm0,0x0(%r13,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L54
L9:
  movsd  0x0(%r13,%rax,8),%xmm2
  pxor   %xmm0,%xmm0
  cmp    %eax,0x18(%rsp)
  jne    L26
  mov    D6(%rip),%rsi
  movq   %rsi,%xmm0
  jmp    L26
  nopl   0x0(%rax,%rax,1)
L54:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D6
type_guessed: f64

tool: D6: 1

assistant: ```c
void softmax(double input[10], double loss[10], int label, int count) {
  double inner = 0;
  for (int i = 0; i < count; ++i) {
    double res = 0;
    for (int j = 0; j < count; ++j) {
      res += exp(input[j] - input[i]);
    }
    loss[i] = 1. / res;
    inner -= loss[i] * loss[i];
  }
  inner += loss[label];
  for (int i = 0; i < count; ++i) {
    loss[i] *= (i == label) - loss[i] - inner;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<grabMonth>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x4000(%rsp),%r11
L13:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L13
  sub    $0x30,%rsp
  mov    %rdi,-0x4028(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x4028(%rbp),%rdx
  lea    -0x4010(%rbp),%rax
  lea    D6(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x4010(%rbp),%rax
  add    $0x4,%rax
  lea    -0x2010(%rbp),%rcx
  mov    $0x2,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strncpy@plt>
  movb   $0x0,-0x200e(%rbp)
  lea    -0x4018(%rbp),%rdx
  lea    -0x2010(%rbp),%rax
  lea    D6(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  mov    -0x4018(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

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

tool: D6: %lu

assistant: ```c
unsigned long grabMonth(unsigned long encoding) {
  char buf[8192];
  char buf2[8192];
  unsigned long month;
  sprintf(buf, "%lu", encoding);
  strncpy(buf2, buf + 4, 2);
  buf2[2] = '\0';
  sscanf(buf2, "%lu", &month);
  return month;
}
```
user: Translate the below assembly code into C programming language:

```asm
<grabMonth>:
  endbr64
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L2:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L2
  sub    $0x28,%rsp
  mov    %rdi,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x4018(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  lea    D7(%rip),%rbp
  mov    %rbp,%rsi
  call   <sprintf@plt>
  lea    0x14(%rsp),%rsi
  lea    0x2010(%rsp),%rbx
  mov    $0x2,%edx
  mov    %rbx,%rdi
  call   <strncpy@plt>
  movb   $0x0,0x2012(%rsp)
  lea    0x8(%rsp),%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  mov    0x8(%rsp),%rax
  mov    0x4018(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x4028,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D7
type_infered: string

tool: D7: %lu

assistant: ```c
unsigned long grabMonth(unsigned long encoding) {
  char buf[8192];
  char buf2[8192];
  unsigned long month;
  sprintf(buf, "%lu", encoding);
  strncpy(buf2, buf + 4, 2);
  buf2[2] = '\0';
  sscanf(buf2, "%lu", &month);
  return month;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<grabMonth>:
  endbr64
  push   %r12
  push   %rbp
  lea    -0x4000(%rsp),%r11
L14:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L14
  sub    $0x28,%rsp
  lea    D7(%rip),%r12
  mov    %r12,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x4018(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rdx
  lea    0x2010(%rsp),%rbp
  lea    0x10(%rsp),%rdi
  call   <sprintf@plt>
  lea    0x14(%rsp),%rsi
  mov    $0x2,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  lea    0x8(%rsp),%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movb   $0x0,0x2012(%rsp)
  call   <sscanf@plt>
  mov    0x8(%rsp),%rax
  mov    0x4018(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x4028,%rsp
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
data_type: string

tool: D7: %lu

assistant: ```c
unsigned long grabMonth(unsigned long encoding) {
  char buf[8192];
  char buf2[8192];
  unsigned long month;
  sprintf(buf, "%lu", encoding);
  strncpy(buf2, buf + 4, 2);
  buf2[2] = '\0';
  sscanf(buf2, "%lu", &month);
  return month;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<grabMonth>:
  endbr64
  push   %r12
  push   %rbp
  lea    -0x4000(%rsp),%r11
L3:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L3
  sub    $0x28,%rsp
  lea    D3(%rip),%r12
  mov    %r12,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x4018(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rdx
  lea    0x2010(%rsp),%rbp
  lea    0x10(%rsp),%rdi
  call   <sprintf@plt>
  lea    0x14(%rsp),%rsi
  mov    $0x2,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  lea    0x8(%rsp),%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movb   $0x0,0x2012(%rsp)
  call   <sscanf@plt>
  mov    0x8(%rsp),%rax
  mov    0x4018(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x4028,%rsp
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D3
type_infered: string

tool: D3: %lu

assistant: ```c
unsigned long grabMonth(unsigned long encoding) {
  char buf[8192];
  char buf2[8192];
  unsigned long month;
  sprintf(buf, "%lu", encoding);
  strncpy(buf2, buf + 4, 2);
  buf2[2] = '\0';
  sscanf(buf2, "%lu", &month);
  return month;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<legal_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L5:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x23,%al
  je     L9
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L9
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L9
  mov    $0x1,%eax
  jmp    L31
L9:
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L5
  mov    $0x0,%eax
L31:
  pop    %rbp
  ret
```

assistant: ```c
int legal_char(char *str) {
  int i;
  i = 0;
  while (str[i]) {
    if (str[i] != '#' && str[i] != '.' && str[i] != '\n')
      return (1);
    ++i;
  }
  return (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<legal_char>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L19
  add    $0x1,%rdi
  movabs $0x400800000400,%rdx
  jmp    L15
L45:
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%eax
  test   %al,%al
  je     L30
L15:
  cmp    $0x2e,%al
  ja     L49
  bt     %rax,%rdx
  jb     L45
  mov    $0x1,%eax
  ret
L30:
  mov    $0x0,%eax
  ret
L19:
  mov    $0x0,%eax
  ret
L49:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int legal_char(char *str) {
  int i;
  i = 0;
  while (str[i]) {
    if (str[i] != '#' && str[i] != '.' && str[i] != '\n')
      return (1);
    ++i;
  }
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<legal_char>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L5
  movabs $0xffffbff7fffffbff,%rdx
  add    $0x1,%rdi
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L33:
  bt     %rax,%rdx
  jb     L41
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L5
L27:
  cmp    $0x2e,%al
  jbe    L33
L41:
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L5:
  xor    %eax,%eax
  ret
```

assistant: ```c
int legal_char(char *str) {
  int i;
  i = 0;
  while (str[i]) {
    if (str[i] != '#' && str[i] != '.' && str[i] != '\n')
      return (1);
    ++i;
  }
  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
<legal_char>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L31
  movabs $0xffffbff7fffffbff,%rdx
  add    $0x1,%rdi
  jmp    L14
  nopl   0x0(%rax,%rax,1)
L24:
  bt     %rax,%rdx
  jb     L27
  movzbl (%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L31
L14:
  cmp    $0x2e,%al
  jbe    L24
L27:
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L31:
  xor    %eax,%eax
  ret
```

assistant: ```c
int legal_char(char *str) {
  int i;
  i = 0;
  while (str[i]) {
    if (str[i] != '#' && str[i] != '.' && str[i] != '\n')
      return (1);
    ++i;
  }
  return (0);
}
```