ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<do_north_west>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <get_north_penetration@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <get_west_penetration@plt>
  cmp    %rax,%rbx
  mov    $0x0,%eax
  jge    L4
  pop    %rbx
  jmp    10b0 <bounce_south@plt>
  cs nopw 0x0(%rax,%rax,1)
L4:
  pop    %rbx
  jmp    1090 <bounce_east@plt>
```

assistant: ```c
void do_north_west(void) {
  if (get_north_penetration() < get_west_penetration())
    bounce_south();
  else
    bounce_east();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<do_north_west>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <get_north_penetration@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <get_west_penetration@plt>
  cmp    %rax,%rbx
  mov    $0x0,%eax
  jge    L0
  pop    %rbx
  jmp    10b0 <bounce_south@plt>
  cs nopw 0x0(%rax,%rax,1)
L0:
  pop    %rbx
  jmp    1090 <bounce_east@plt>
```

assistant: ```c
void do_north_west(void) {
  if (get_north_penetration() < get_west_penetration())
    bounce_south();
  else
    bounce_east();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<byg_detect>:
  endbr64
  sub    $0x488,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x478(%rsp)
  xor    %eax,%eax
  lea    0x70(%rsp),%rax
  lea    0x470(%rsp),%rdx
L50:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L50
  movl   $0x0,(%rsp)
  movl   $0x1,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0x3,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x5,0x14(%rsp)
  movl   $0x6,0x18(%rsp)
  movl   $0x7,0x1c(%rsp)
  movl   $0x8,0x20(%rsp)
  movl   $0xffffffff,0x24(%rsp)
  movl   $0x9,0x28(%rsp)
  movl   $0xa,0x2c(%rsp)
  movl   $0xb,0x30(%rsp)
  movl   $0xc,0x34(%rsp)
  movl   $0x17,0x38(%rsp)
  movl   $0xd,0x3c(%rsp)
  movl   $0xe,0x40(%rsp)
  movl   $0xf,0x44(%rsp)
  movl   $0x10,0x48(%rsp)
  movl   $0x11,0x4c(%rsp)
  movl   $0x11,0x50(%rsp)
  movl   $0x12,0x54(%rsp)
  movl   $0x13,0x58(%rsp)
  movl   $0x14,0x5c(%rsp)
  movl   $0x15,0x60(%rsp)
  movl   $0x16,0x64(%rsp)
  lea    D5(%rip),%rdx
  lea    0x14(%rdx),%rcx
  jmp    L114
L85:
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  je     L109
L114:
  movsbl (%rdx),%eax
  sub    $0x41,%eax
  cltq
  mov    (%rsp,%rax,4),%eax
  test   %eax,%eax
  js     L85
  cltq
  orl    $0x1,0x70(%rsp,%rax,4)
  jmp    L85
L109:
  test   %esi,%esi
  jle    L51
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rsi
  mov    $0x0,%ecx
  jmp    L19
L33:
  add    $0x4,%rax
  cmp    %rsi,%rax
  je     L117
L19:
  mov    (%rax),%edx
  cmp    $0xffffffff,%edx
  je     L33
  lea    0x1(%rcx,%rcx,1),%ecx
  movslq %edx,%rdx
  and    0x70(%rsp,%rdx,4),%ecx
  test   $0x1,%cl
  je     L33
  mov    $0x0,%eax
  jmp    L12
L117:
  mov    $0x1,%eax
L12:
  mov    0x478(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L56
  add    $0x488,%rsp
  ret
L51:
  mov    $0x1,%eax
  jmp    L12
L56:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type: string

tool: D5: BDEFHIJKLMNOPQRSVWYZ

assistant: ```c
int byg_detect(int *text, int n) {
  int Tc;
  int i = 0;
  int s = 0;
  int T[256];
  for (i = 0; i < 256; i++) {
    T[i] = 0;
  }
  int mb = 1;
  char *unique_aa = "BDEFHIJKLMNOPQRSVWYZ";
  int aacode[26] = {0, 1, 2, 3, 4, 5, 6, 7, 8, -1, 9, 10, 11, 12, 23, 13, 14, 15, 16, 17, 17, 18, 19, 20, 21, 22};
  for (i = 0; i < 20; i++) {
    if (aacode[unique_aa[i] - 65] >= 0) {
      T[(int)aacode[unique_aa[i] - 65]] |= 1;
    }
  }
  for (i = 0; i < n; i++) {
    if (text[i] != -1) {
      s <<= 1;
      s |= 1;
      Tc = T[text[i]];
      s &= Tc;
      if (s & mb) {
        return 0;
      }
    }
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<byg_detect>:
  endbr64
  sub    $0x488,%rsp
  mov    %rdi,%r8
  mov    $0x80,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x478(%rsp)
  xor    %eax,%eax
  lea    0x70(%rsp),%rdx
  mov    %rdx,%rdi
  lea    D5(%rip),%rdx
  rep stos %rax,%es:(%rdi)
  lea    0x13(%rdx),%rcx
  movabs $0x100000000,%rax
  mov    %rax,(%rsp)
  movabs $0x300000002,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x500000004,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x700000006,%rax
  mov    %rax,0x18(%rsp)
  movabs $0xffffffff00000008,%rax
  mov    %rax,0x20(%rsp)
  movabs $0xa00000009,%rax
  mov    %rax,0x28(%rsp)
  movabs $0xc0000000b,%rax
  mov    %rax,0x30(%rsp)
  movabs $0xd00000017,%rax
  mov    %rax,0x38(%rsp)
  movabs $0xf0000000e,%rax
  mov    %rax,0x40(%rsp)
  movabs $0x1100000010,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x1200000011,%rax
  mov    %rax,0x50(%rsp)
  movabs $0x1400000013,%rax
  mov    %rax,0x58(%rsp)
  movabs $0x1600000015,%rax
  mov    %rax,0x60(%rsp)
  mov    $0x1,%eax
  jmp    L9
  nopl   (%rax)
L62:
  movsbl (%rdx),%eax
  add    $0x1,%rdx
  sub    $0x41,%eax
  cltq
  movslq (%rsp,%rax,4),%rax
L9:
  test   %eax,%eax
  js     L44
  orl    $0x1,0x70(%rsp,%rax,4)
L44:
  cmp    %rdx,%rcx
  jne    L62
  test   %esi,%esi
  jle    L17
  lea    -0x1(%rsi),%edx
  mov    %r8,%rax
  xor    %ecx,%ecx
  lea    0x4(%r8,%rdx,4),%rsi
  nopw   0x0(%rax,%rax,1)
L35:
  movslq (%rax),%rdx
  cmp    $0xffffffff,%edx
  je     L84
  lea    0x1(%rcx,%rcx,1),%ecx
  and    0x70(%rsp,%rdx,4),%ecx
  test   $0x1,%cl
  jne    L22
L84:
  add    $0x4,%rax
  cmp    %rax,%rsi
  jne    L35
L17:
  mov    $0x1,%eax
L64:
  mov    0x478(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L77
  add    $0x488,%rsp
  ret
L22:
  xor    %eax,%eax
  jmp    L64
L77:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
type_guessed: string

tool: D5: DEFHIJKLMNOPQRSVWYZ

assistant: ```c
int byg_detect(int *text, int n) {
  int Tc;
  int i = 0;
  int s = 0;
  int T[256];
  for (i = 0; i < 256; i++) {
    T[i] = 0;
  }
  int mb = 1;
  char *unique_aa = "BDEFHIJKLMNOPQRSVWYZ";
  int aacode[26] = {0, 1, 2, 3, 4, 5, 6, 7, 8, -1, 9, 10, 11, 12, 23, 13, 14, 15, 16, 17, 17, 18, 19, 20, 21, 22};
  for (i = 0; i < 20; i++) {
    if (aacode[unique_aa[i] - 65] >= 0) {
      T[(int)aacode[unique_aa[i] - 65]] |= 1;
    }
  }
  for (i = 0; i < n; i++) {
    if (text[i] != -1) {
      s <<= 1;
      s |= 1;
      Tc = T[text[i]];
      s &= Tc;
      if (s & mb) {
        return 0;
      }
    }
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hanoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L19
  mov    -0x20(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <move@plt>
  jmp    L17
L19:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edi
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  call   <hanoi>
  mov    -0x20(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <move@plt>
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edi
  mov    -0x20(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  call   <hanoi>
L17:
  nop
  leave
  ret
```

assistant: ```c
void hanoi(int n, struct stack *from, struct stack *mid, struct stack *to) {
  if (n == 1) {
    move(from, to);
  } else {
    hanoi(n - 1, from, to, mid);
    move(from, to);
    hanoi(n - 1, mid, from, to);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hanoi>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  mov    %rcx,%r12
  cmp    $0x1,%edi
  je     L8
  mov    %rdx,%r13
  lea    -0x1(%rdi),%ebx
  mov    %rdx,%rcx
  mov    %r12,%rdx
  mov    %ebx,%edi
  call   <hanoi>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <move@plt>
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %ebx,%edi
  call   <hanoi>
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  mov    %rcx,%rsi
  mov    %rbp,%rdi
  call   <move@plt>
  jmp    L18
```

assistant: ```c
void hanoi(int n, struct stack *from, struct stack *mid, struct stack *to) {
  if (n == 1) {
    move(from, to);
  } else {
    hanoi(n - 1, from, to, mid);
    move(from, to);
    hanoi(n - 1, mid, from, to);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<partition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L35
  cmpl   $0x0,-0x1c(%rbp)
  js     L35
  cmpl   $0x0,-0x20(%rbp)
  jns    L20
L35:
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  jmp    L19
L20:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L49
L57:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L23
  addl   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L23:
  addl   $0x1,-0xc(%rbp)
L49:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L57
  mov    -0x10(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  add    $0x1,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
L19:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type: string

tool: D5: 
ERROR: Incorrect inputs to partition, 			left = %d, right = %d, arr* = %p

assistant: ```c
int partition(int *a, int left, int right) {
  int pivot, i, j, temp;
  if (!a || left < 0 || right < 0) {
    printf("\nERROR: Incorrect inputs to partition, 			left = %d, right = %d, arr* = %p", left, right, a);
    return;
  }
  pivot = a[right];
  i = left - 1;
  for (j = left; j < right; j++) {
    if (a[j] <= pivot) {
      i++;
      temp = a[i];
      a[i] = a[j];
      a[j] = temp;
    }
  }
  temp = a[i + 1];
  a[i + 1] = a[right];
  a[right] = temp;
  return i + 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<partition>:
  endbr64
  mov    %esi,%eax
  or     %edx,%eax
  js     L43
  test   %rdi,%rdi
  je     L43
  movslq %edx,%rax
  lea    (%rdi,%rax,4),%r11
  mov    (%r11),%r8d
  lea    -0x1(%rsi),%eax
  cmp    %edx,%esi
  jge    L51
  movslq %esi,%r9
  lea    (%rdi,%r9,4),%rcx
  sub    $0x1,%edx
  sub    %esi,%edx
  add    %r9,%rdx
  lea    0x4(%rdi,%rdx,4),%r9
  jmp    L22
L43:
  sub    $0x8,%rsp
  mov    %rdi,%rcx
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
L13:
  add    $0x4,%rcx
  cmp    %r9,%rcx
  je     L51
L22:
  mov    (%rcx),%edx
  cmp    %r8d,%edx
  jg     L13
  add    $0x1,%eax
  movslq %eax,%rsi
  lea    (%rdi,%rsi,4),%rsi
  mov    (%rsi),%r10d
  mov    %edx,(%rsi)
  mov    %r10d,(%rcx)
  jmp    L13
L51:
  movslq %eax,%rdx
  lea    0x4(%rdi,%rdx,4),%rdx
  mov    (%rdx),%ecx
  mov    (%r11),%esi
  mov    %esi,(%rdx)
  mov    %ecx,(%r11)
  add    $0x1,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: label: D5
data_type: string

tool: D5: 
ERROR: Incorrect inputs to partition, 			left = %d, right = %d, arr* = %p

assistant: ```c
int partition(int *a, int left, int right) {
  int pivot, i, j, temp;
  if (!a || left < 0 || right < 0) {
    printf("\nERROR: Incorrect inputs to partition, 			left = %d, right = %d, arr* = %p", left, right, a);
    return;
  }
  pivot = a[right];
  i = left - 1;
  for (j = left; j < right; j++) {
    if (a[j] <= pivot) {
      i++;
      temp = a[i];
      a[i] = a[j];
      a[j] = temp;
    }
  }
  temp = a[i + 1];
  a[i + 1] = a[right];
  a[right] = temp;
  return i + 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<partition>:
  endbr64
  mov    %esi,%eax
  mov    %rdi,%r8
  or     %edx,%eax
  js     L50
  test   %rdi,%rdi
  je     L50
  movslq %edx,%rax
  lea    (%rdi,%rax,4),%r10
  lea    -0x1(%rsi),%edi
  mov    (%r10),%r9d
  cmp    %edx,%esi
  jge    L33
  sub    $0x1,%edx
  movslq %esi,%rax
  sub    %esi,%edx
  lea    (%r8,%rax,4),%rcx
  add    %rax,%rdx
  lea    0x4(%r8,%rdx,4),%rsi
  nopw   0x0(%rax,%rax,1)
L4:
  mov    (%rcx),%eax
  cmp    %r9d,%eax
  jg     L40
  add    $0x1,%edi
  movslq %edi,%rdx
  lea    (%r8,%rdx,4),%rdx
  mov    (%rdx),%r11d
  mov    %eax,(%rdx)
  mov    %r11d,(%rcx)
L40:
  add    $0x4,%rcx
  cmp    %rcx,%rsi
  jne    L4
  mov    (%r10),%r9d
  lea    0x1(%rdi),%eax
L52:
  movslq %edi,%rdi
  lea    0x4(%r8,%rdi,4),%rdx
  mov    (%rdx),%ecx
  mov    %r9d,(%rdx)
  mov    %ecx,(%r10)
  ret
  nopl   0x0(%rax)
L50:
  mov    %r8,%rcx
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
  nop
L33:
  mov    %esi,%eax
  jmp    L52
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type: string

tool: D5: 
ERROR: Incorrect inputs to partition, 			left = %d, right = %d, arr* = %p

assistant: ```c
int partition(int *a, int left, int right) {
  int pivot, i, j, temp;
  if (!a || left < 0 || right < 0) {
    printf("\nERROR: Incorrect inputs to partition, 			left = %d, right = %d, arr* = %p", left, right, a);
    return;
  }
  pivot = a[right];
  i = left - 1;
  for (j = left; j < right; j++) {
    if (a[j] <= pivot) {
      i++;
      temp = a[i];
      a[i] = a[j];
      a[j] = temp;
    }
  }
  temp = a[i + 1];
  a[i + 1] = a[right];
  a[right] = temp;
  return i + 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<partition>:
  endbr64
  mov    %esi,%eax
  mov    %rdi,%r8
  or     %edx,%eax
  js     L7
  test   %rdi,%rdi
  je     L7
  movslq %edx,%rax
  lea    (%rdi,%rax,4),%r10
  lea    -0x1(%rsi),%edi
  mov    (%r10),%r9d
  cmp    %edx,%esi
  jge    L4
  sub    $0x1,%edx
  movslq %esi,%rax
  sub    %esi,%edx
  lea    (%r8,%rax,4),%rcx
  add    %rax,%rdx
  lea    0x4(%r8,%rdx,4),%rsi
  nopw   0x0(%rax,%rax,1)
L15:
  mov    (%rcx),%eax
  cmp    %r9d,%eax
  jg     L8
  add    $0x1,%edi
  movslq %edi,%rdx
  lea    (%r8,%rdx,4),%rdx
  mov    (%rdx),%r11d
  mov    %eax,(%rdx)
  mov    %r11d,(%rcx)
L8:
  add    $0x4,%rcx
  cmp    %rcx,%rsi
  jne    L15
  mov    (%r10),%r9d
  lea    0x1(%rdi),%eax
L14:
  movslq %edi,%rdi
  lea    0x4(%r8,%rdi,4),%rdx
  mov    (%rdx),%ecx
  mov    %r9d,(%rdx)
  mov    %ecx,(%r10)
  ret
  nopl   0x0(%rax)
L7:
  mov    %r8,%rcx
  lea    D2(%rip),%rdi
  jmp    1050 <printf@plt>
  nop
L4:
  mov    %esi,%eax
  jmp    L14
```

tool_calls: name: read_data
arguments: data_label: D2
type_infered: string

tool: D2: 
ERROR: Incorrect inputs to partition, 			left = %d, right = %d, arr* = %p

assistant: ```c
int partition(int *a, int left, int right) {
  int pivot, i, j, temp;
  if (!a || left < 0 || right < 0) {
    printf("\nERROR: Incorrect inputs to partition, 			left = %d, right = %d, arr* = %p", left, right, a);
    return;
  }
  pivot = a[right];
  i = left - 1;
  for (j = left; j < right; j++) {
    if (a[j] <= pivot) {
      i++;
      temp = a[i];
      a[i] = a[j];
      a[j] = temp;
    }
  }
  temp = a[i + 1];
  a[i + 1] = a[right];
  a[right] = temp;
  return i + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getInfo>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r12d
  lea    D17(%rip),%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%ebx
  test   %rbp,%rbp
  je     L26
  cmp    %r12d,%ebx
  jle    L11
L26:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  mov    %ebx,%edx
  lea    D17(%rip),%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  jmp    L26
```

tool_calls: name: unpack_data
arguments: data_tag: D17
data_type: string

tool: D17: info { name=IFS }

assistant: ```c
int getInfo(char *buf, int bufLen) {
  static const char *INFO = "info { name=IFS }";
  int reqLen = strlen(INFO) + 1;
  if (buf != 0 && reqLen <= bufLen) {
    memcpy(buf, INFO, reqLen);
  }
  return reqLen;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getInfo>:
  endbr64
  push   %r13
  lea    D7(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13,%rdi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  test   %rbp,%rbp
  je     L5
  cmp    %ebx,%r12d
  jle    L1
L5:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L1:
  mov    %r12d,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type: string

tool: D7: info { name=IFS }

assistant: ```c
int getInfo(char *buf, int bufLen) {
  static const char *INFO = "info { name=IFS }";
  int reqLen = strlen(INFO) + 1;
  if (buf != 0 && reqLen <= bufLen) {
    memcpy(buf, INFO, reqLen);
  }
  return reqLen;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getInfo>:
  endbr64
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13,%rdi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  test   %rbp,%rbp
  je     L13
  cmp    %ebx,%r12d
  jle    L4
L13:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L4:
  mov    %r12d,%edx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
data_type: string

tool: D0: info { name=IFS }

assistant: ```c
int getInfo(char *buf, int bufLen) {
  static const char *INFO = "info { name=IFS }";
  int reqLen = strlen(INFO) + 1;
  if (buf != 0 && reqLen <= bufLen) {
    memcpy(buf, INFO, reqLen);
  }
  return reqLen;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
L47:
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jle    L47
  mov    -0x14(%rbp),%eax
  cmp    $0x270f,%eax
  jg     L47
  jmp    L64
L41:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  add    %edx,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0xc(%rbp)
L64:
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jne    L41
  mov    -0x10(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  cmp    $0x7,%eax
  jle    L20
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L40
L20:
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L40:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type_infered: string
name: unpack_data
arguments: data_tag: D19
type_infered: string
name: unpack_data
arguments: data_tag: D25
type_infered: string

tool: D6: %d

tool: D19: heavy

tool: D25: light

assistant: ```c
int main() {
  int num, sum = 0, j = 0;
  do {
    scanf("%d", &num);
  } while (num <= 0 || num >= 10000);
  while (num != 0) {
    sum = sum + num % 10;
    num = num / 10;
    j++;
  }
  if (sum / j > 7) {
    printf("heavy");
  } else {
    printf("light");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D5(%rip),%rbx
L42:
  lea    0x4(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  lea    -0x1(%rcx),%eax
  cmp    $0x270e,%eax
  ja     L42
  mov    $0x0,%edi
  mov    $0x0,%eax
L43:
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  add    %ecx,%eax
  mov    %edx,%ecx
  add    $0x1,%edi
  test   %edx,%edx
  jne    L43
  movl   $0x0,0x4(%rsp)
  cltd
  idiv   %edi
  cmp    $0x7,%eax
  jle    L46
  lea    D8(%rip),%rdi
  call   <printf@plt>
L30:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L46:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L30
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D5
data_type: string
name: retrieve_data
arguments: data_tag: D8
data_type: string
name: retrieve_data
arguments: data_tag: D1
data_type: string

tool: D5: %d

tool: D8: heavy

tool: D1: light

assistant: ```c
int main() {
  int num, sum = 0, j = 0;
  do {
    scanf("%d", &num);
  } while (num <= 0 || num >= 10000);
  while (num != 0) {
    sum = sum + num % 10;
    num = num / 10;
    j++;
  }
  if (sum / j > 7) {
    printf("heavy");
  } else {
    printf("light");
  }
  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
  push   %rbp
  push   %rbx
  lea    D3(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  cs nopw 0x0(%rax,%rax,1)
L42:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  lea    -0x1(%rdx),%eax
  cmp    $0x270e,%eax
  ja     L42
  xor    %edi,%edi
  xor    %eax,%eax
  mov    $0xcccccccd,%r8d
  nopl   0x0(%rax,%rax,1)
L31:
  mov    %edx,%ecx
  add    $0x1,%edi
  imul   %r8,%rcx
  shr    $0x23,%rcx
  lea    (%rcx,%rcx,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    %edx,%eax
  mov    %ecx,%edx
  test   %ecx,%ecx
  jne    L31
  cltd
  movl   $0x0,0x4(%rsp)
  idiv   %edi
  cmp    $0x7,%eax
  jle    L8
  lea    D18(%rip),%rdi
  call   <printf@plt>
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L8:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L21
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
type: string
name: unpack_data
arguments: data_tag: D18
type: string
name: unpack_data
arguments: data_tag: D10
type: string

tool: D3: %d

tool: D18: heavy

tool: D10: light

assistant: ```c
int main() {
  int num, sum = 0, j = 0;
  do {
    scanf("%d", &num);
  } while (num <= 0 || num >= 10000);
  while (num != 0) {
    sum = sum + num % 10;
    num = num / 10;
    j++;
  }
  if (sum / j > 7) {
    printf("heavy");
  } else {
    printf("light");
  }
  return 0;
}
```