ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<strtoul>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  mov    %rsp,%rax
  test   %rsi,%rsi
  cmove  %rax,%rcx
  mov    %rdi,(%rcx)
  cmp    $0x24,%edx
  ja     L222
  mov    %edx,%r8d
  movsbw (%rdi),%dx
  cmp    $0x20,%dl
  ja     L72
  movabs $0xfffffffeffffd9ff,%rax
  nopl   (%rax)
L224:
  bt     %rdx,%rax
  jb     L108
  add    $0x1,%rdi
  mov    %rdi,(%rcx)
  movsbw (%rdi),%dx
  cmp    $0x20,%dl
  jbe    L224
L72:
  test   %r8d,%r8d
  je     L144
L204:
  cmp    $0x10,%r8d
  je     L115
L65:
  test   %dl,%dl
  je     L222
L75:
  add    $0x1,%rdi
  xor    %eax,%eax
  movslq %r8d,%r9
  jmp    L157
  nopl   0x0(%rax,%rax,1)
L82:
  sub    $0x30,%edx
L223:
  movswl %dx,%esi
  cmp    %r8d,%esi
  jge    L101
L171:
  imul   %r9,%rax
  mov    %rdi,(%rcx)
  movswq %dx,%rdx
  add    $0x1,%rdi
  add    %rdx,%rax
  movsbw -0x1(%rdi),%dx
  test   %dl,%dl
  je     L101
L157:
  lea    -0x30(%rdx),%esi
  cmp    $0x9,%sil
  jbe    L82
  lea    -0x61(%rdx),%esi
  cmp    $0x19,%sil
  ja     L77
  sub    $0x57,%edx
  movswl %dx,%esi
  cmp    %r8d,%esi
  jl     L171
  nopl   0x0(%rax,%rax,1)
L101:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L135
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L77:
  lea    -0x41(%rdx),%esi
  cmp    $0x19,%sil
  ja     L101
  sub    $0x37,%edx
  jmp    L223
  cs nopw 0x0(%rax,%rax,1)
L222:
  xor    %eax,%eax
  jmp    L101
  nopl   0x0(%rax)
L108:
  test   %r8d,%r8d
  jne    L204
  mov    $0xa,%r8d
  jmp    L65
L144:
  mov    $0xa,%r8d
  cmp    $0x30,%dl
  jne    L75
  cmpb   $0x78,0x1(%rdi)
  jne    L75
  lea    0x2(%rdi),%rax
  mov    %rax,(%rcx)
  movsbw 0x2(%rdi),%dx
  mov    %rax,%rdi
L115:
  mov    $0x10,%r8d
  cmp    $0x30,%dl
  jne    L65
  cmpb   $0x78,0x1(%rdi)
  jne    L75
  lea    0x2(%rdi),%rax
  mov    %rax,(%rcx)
  movsbw 0x2(%rdi),%dx
  mov    %rax,%rdi
  jmp    L65
L135:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long int strtoul(const char *S, char **PTR, int BASE) {
  unsigned long rval = 0;
  short int digit;
  char *ptr;
  if (PTR == 0) {
    PTR = &ptr;
  }
  *PTR = (char *)S;
  if ((BASE < 0) || BASE > 36) {
    return 0;
  }
  while ((**PTR == ' ') || (**PTR == '\t') || (**PTR == '\n') || (**PTR == '\r')) {
    (*PTR)++;
  }
  if (BASE == 0) {
    if ((**PTR == '0') && (*((*PTR) + 1) == 'x')) {
      BASE = 16;
      (*PTR)++;
      (*PTR)++;
    } else {
      BASE = 10;
    }
  }
  if (BASE == 16) {
    if ((**PTR == '0') && (*((*PTR) + 1) == 'x')) {
      (*PTR)++;
      (*PTR)++;
    }
  }
  while (**PTR) {
    if (((**PTR) >= '0') && ((**PTR) <= '9')) {
      digit = **PTR - '0';
    } else if (((**PTR) >= 'a') && ((**PTR) <= 'z')) {
      digit = (**PTR - 'a') + 10;
    } else if (((**PTR) >= 'A') && ((**PTR) <= 'Z')) {
      digit = (**PTR - 'A') + 10;
    } else {
      break;
    }
    if (digit < BASE) {
      rval = (rval * BASE) + digit;
    } else {
      break;
    }
    (*PTR)++;
  }
  return rval;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<getline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,%eax
  mov    %al,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L18:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x24(%rbp)
  je     L13
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  jmp    L35
L13:
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  jmp    L44
L35:
  addl   $0x1,-0x4(%rbp)
L3:
  cmpl   $0x11,-0x4(%rbp)
  jle    L18
L44:
  pop    %rbp
  ret
```

assistant: ```c
int getline(char *input, char *output, char delim) {
  for (int i = 0; i < 18; i++) {
    if (input[i] != delim) {
      output[i] = input[i];
      continue;
    }
    return i + 1;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getline>:
  endbr64
  mov    $0x0,%ecx
L18:
  mov    %ecx,%eax
  movzbl (%rdi,%rcx,1),%r8d
  cmp    %dl,%r8b
  jne    L0
  add    $0x1,%eax
  ret
L0:
  mov    %r8b,(%rsi,%rcx,1)
  add    $0x1,%rcx
  cmp    $0x12,%rcx
  jne    L18
  ret
```

assistant: ```c
int getline(char *input, char *output, char delim) {
  for (int i = 0; i < 18; i++) {
    if (input[i] != delim) {
      output[i] = input[i];
      continue;
    }
    return i + 1;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getline>:
  endbr64
  mov    $0x1,%eax
L10:
  movzbl -0x1(%rdi,%rax,1),%ecx
  mov    %eax,%r8d
  cmp    %dl,%cl
  je     L15
  mov    %cl,-0x1(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    $0x13,%rax
  jne    L10
L15:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int getline(char *input, char *output, char delim) {
  for (int i = 0; i < 18; i++) {
    if (input[i] != delim) {
      output[i] = input[i];
      continue;
    }
    return i + 1;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<getline>:
  endbr64
  mov    $0x1,%eax
L11:
  movzbl -0x1(%rdi,%rax,1),%ecx
  mov    %eax,%r8d
  cmp    %dl,%cl
  je     L15
  mov    %cl,-0x1(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    $0x13,%rax
  jne    L11
L15:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int getline(char *input, char *output, char delim) {
  for (int i = 0; i < 18; i++) {
    if (input[i] != delim) {
      output[i] = input[i];
      continue;
    }
    return i + 1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0xc(%rbp)
  movl   $0x3,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L6
L1:
  mov    -0x8(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x2,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L6:
  cmpl   $0xf4a10,-0x8(%rbp)
  jbe    L1
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
datatype: string

tool: D4: %u


assistant: ```c
int main(void) {
  unsigned int sum = 1;
  unsigned int i = 3;
  int skip = 1;
  while (i < 1001 * 1001) {
    sum += i;
    i += skip + 1;
    sum += i;
    i += skip + 1;
    sum += i;
    i += skip + 1;
    sum += i;
    skip += 2;
    i += skip + 1;
  }
  printf("%u\n", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edx
  mov    $0x3,%ecx
  mov    $0x1,%r8d
L7:
  lea    (%rdx,%rcx,1),%eax
  lea    (%rdx,%rax,1),%edi
  lea    (%rdx,%rdi,1),%esi
  add    %ecx,%eax
  add    %edi,%eax
  add    %esi,%eax
  add    %eax,%r8d
  add    $0x2,%edx
  lea    (%rsi,%rdx,1),%ecx
  cmp    $0xf4a10,%ecx
  jbe    L7
  mov    %r8d,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
datatype: string

tool: D6: %u


assistant: ```c
int main(void) {
  unsigned int sum = 1;
  unsigned int i = 3;
  int skip = 1;
  while (i < 1001 * 1001) {
    sum += i;
    i += skip + 1;
    sum += i;
    i += skip + 1;
    sum += i;
    i += skip + 1;
    sum += i;
    skip += 2;
    i += skip + 1;
  }
  printf("%u\n", sum);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%ecx
  mov    $0x3,%esi
  mov    $0x1,%r8d
  nopl   0x0(%rax,%rax,1)
L6:
  lea    0x1(%rcx,%rsi,1),%eax
  lea    0x1(%rcx),%edx
  mov    %ecx,%r9d
  add    $0x2,%ecx
  lea    (%rax,%rdx,1),%edi
  add    %esi,%eax
  add    %edi,%edx
  add    %edi,%eax
  add    %edx,%eax
  lea    0x3(%rdx,%r9,1),%esi
  add    %eax,%r8d
  cmp    $0xf4a10,%esi
  jbe    L6
  mov    %r8d,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type_guessed: string

tool: D1: %u


assistant: ```c
int main(void) {
  unsigned int sum = 1;
  unsigned int i = 3;
  int skip = 1;
  while (i < 1001 * 1001) {
    sum += i;
    i += skip + 1;
    sum += i;
    i += skip + 1;
    sum += i;
    i += skip + 1;
    sum += i;
    skip += 2;
    i += skip + 1;
  }
  printf("%u\n", sum);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%ecx
  mov    $0x3,%esi
  mov    $0x1,%r8d
  nopl   0x0(%rax,%rax,1)
L6:
  lea    0x1(%rcx,%rsi,1),%eax
  lea    0x1(%rcx),%edx
  mov    %ecx,%r9d
  add    $0x2,%ecx
  lea    (%rax,%rdx,1),%edi
  add    %esi,%eax
  add    %edi,%edx
  add    %edi,%eax
  add    %edx,%eax
  lea    0x3(%rdx,%r9,1),%esi
  add    %eax,%r8d
  cmp    $0xf4a10,%esi
  jbe    L6
  mov    %r8d,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D6
type: string

tool: D6: %u


assistant: ```c
int main(void) {
  unsigned int sum = 1;
  unsigned int i = 3;
  int skip = 1;
  while (i < 1001 * 1001) {
    sum += i;
    i += skip + 1;
    sum += i;
    i += skip + 1;
    sum += i;
    i += skip + 1;
    sum += i;
    skip += 2;
    i += skip + 1;
  }
  printf("%u\n", sum);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

tool_calls: name: unpack_data
arguments: data_label: D2
type: string

tool: D2: Error loading saved game.

assistant: ```c
int lge() {
  draw_error("Error loading saved game.");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<lge>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <draw_error@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D5: Error loading saved game.

assistant: ```c
int lge() {
  draw_error("Error loading saved game.");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<lge>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <draw_error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D5: Error loading saved game.

assistant: ```c
int lge() {
  draw_error("Error loading saved game.");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<lge>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <draw_error@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
type_infered: string

tool: D3: Error loading saved game.

assistant: ```c
int lge() {
  draw_error("Error loading saved game.");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_mcalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    -0x18(%rbp),%rax
  imul   -0x20(%rbp),%rax
  mov    -0x38(%rbp),%rcx
  mov    -0x2c(%rbp),%edx
  mov    -0x28(%rbp),%rsi
  mov    %rax,%rdi
  call   <_mmalloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  imul   -0x20(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *_mcalloc(size_t num, size_t size, const char *file, int line, const char *func) {
  void *p = _mmalloc(num * size, file, line, func);
  memset(p, 0, num * size);
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<_mcalloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  mov    %rdx,%rsi
  mov    %ecx,%edx
  mov    %r8,%rcx
  imul   %rdi,%rbp
  mov    %rbp,%rdi
  call   <_mmalloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *_mcalloc(size_t num, size_t size, const char *file, int line, const char *func) {
  void *p = _mmalloc(num * size, file, line, func);
  memset(p, 0, num * size);
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<_mcalloc>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  mov    %rdx,%rsi
  mov    %ecx,%edx
  push   %r12
  imul   %rdi,%r13
  mov    %r8,%rcx
  sub    $0x8,%rsp
  mov    %r13,%rdi
  call   <_mmalloc@plt>
  mov    %r13,%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memset@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *_mcalloc(size_t num, size_t size, const char *file, int line, const char *func) {
  void *p = _mmalloc(num * size, file, line, func);
  memset(p, 0, num * size);
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_mcalloc>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  mov    %rdx,%rsi
  mov    %ecx,%edx
  push   %r12
  imul   %rdi,%r13
  mov    %r8,%rcx
  sub    $0x8,%rsp
  mov    %r13,%rdi
  call   <_mmalloc@plt>
  mov    %r13,%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memset@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *_mcalloc(size_t num, size_t size, const char *file, int line, const char *func) {
  void *p = _mmalloc(num * size, file, line, func);
  memset(p, 0, num * size);
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   %eax,%edx
  mov    -0x14(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D12
type_infered: string
name: unpack_data
arguments: label: D7
type_infered: string

tool: D12: %d

tool: D7: %d
%d


assistant: ```c
int main() {
  int n, x, y;
  scanf("%d", &n);
  x = n * n;
  y = n * n * n;
  printf("%d\n%d\n", x, y);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  mov    %edx,%esi
  imul   %edx,%esi
  imul   %esi,%edx
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D2
data_type: string
name: disassemble_data
arguments: label: D18
data_type: string

tool: D2: %d

tool: D18: %d
%d


assistant: ```c
int main() {
  int n, x, y;
  scanf("%d", &n);
  x = n * n;
  y = n * n * n;
  printf("%d\n%d\n", x, y);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  lea    D8(%rip),%rdi
  mov    %edx,%esi
  imul   %edx,%esi
  imul   %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D16
type_guessed: string
name: disassemble_data
arguments: data_tag: D8
type_guessed: string

tool: D16: %d

tool: D8: %d
%d


assistant: ```c
int main() {
  int n, x, y;
  scanf("%d", &n);
  x = n * n;
  y = n * n * n;
  printf("%d\n%d\n", x, y);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D8(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  lea    D9(%rip),%rdi
  mov    %edx,%esi
  imul   %edx,%esi
  imul   %esi,%edx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
data_type: string
name: extract_symbolic_data
arguments: label: D9
data_type: string

tool: D8: %d

tool: D9: %d
%d


assistant: ```c
int main() {
  int n, x, y;
  scanf("%d", &n);
  x = n * n;
  y = n * n * n;
  printf("%d\n%d\n", x, y);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cpu_write16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  movzbl %al,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <cpu_write8@plt>
  mov    -0x8(%rbp),%eax
  sar    $0x8,%eax
  movzbl %al,%eax
  mov    -0x4(%rbp),%edx
  add    $0x1,%edx
  mov    %eax,%esi
  mov    %edx,%edi
  call   <cpu_write8@plt>
  nop
  leave
  ret
```

assistant: ```c
void cpu_write16(int adr, int dat) {
  cpu_write8(adr, (unsigned char)dat);
  cpu_write8(adr + 1, (unsigned char)(dat >> 8));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cpu_write16>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  movzbl %sil,%esi
  call   <cpu_write8@plt>
  mov    %ebp,%eax
  movzbl %ah,%esi
  lea    0x1(%rbx),%edi
  call   <cpu_write8@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void cpu_write16(int adr, int dat) {
  cpu_write8(adr, (unsigned char)dat);
  cpu_write8(adr + 1, (unsigned char)(dat >> 8));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cpu_write16>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  movzbl %sil,%esi
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <cpu_write8@plt>
  mov    %ebp,%eax
  add    $0x8,%rsp
  lea    0x1(%rbx),%edi
  movzbl %ah,%esi
  pop    %rbx
  pop    %rbp
  jmp    1050 <cpu_write8@plt>
```

assistant: ```c
void cpu_write16(int adr, int dat) {
  cpu_write8(adr, (unsigned char)dat);
  cpu_write8(adr + 1, (unsigned char)(dat >> 8));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cpu_write16>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  movzbl %sil,%esi
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <cpu_write8@plt>
  mov    %ebp,%eax
  add    $0x8,%rsp
  lea    0x1(%rbx),%edi
  movzbl %ah,%esi
  pop    %rbx
  pop    %rbp
  jmp    1050 <cpu_write8@plt>
```

assistant: ```c
void cpu_write16(int adr, int dat) {
  cpu_write8(adr, (unsigned char)dat);
  cpu_write8(adr + 1, (unsigned char)(dat >> 8));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  call   <test_absent_codes@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <test_1xx@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <test_2xx@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <test_3xx@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <test_4xx@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <test_5xx@plt>
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  setne  %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int errcount = 0;
  (void)argc;
  (void)argv;
  errcount += test_absent_codes();
  errcount += test_1xx();
  errcount += test_2xx();
  errcount += test_3xx();
  errcount += test_4xx();
  errcount += test_5xx();
  return errcount == 0 ? 0 : 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <test_absent_codes@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <test_1xx@plt>
  add    %eax,%ebx
  mov    $0x0,%eax
  call   <test_2xx@plt>
  add    %eax,%ebx
  mov    $0x0,%eax
  call   <test_3xx@plt>
  add    %eax,%ebx
  mov    $0x0,%eax
  call   <test_4xx@plt>
  add    %eax,%ebx
  mov    $0x0,%eax
  call   <test_5xx@plt>
  add    %eax,%ebx
  setne  %al
  movzbl %al,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int errcount = 0;
  (void)argc;
  (void)argv;
  errcount += test_absent_codes();
  errcount += test_1xx();
  errcount += test_2xx();
  errcount += test_3xx();
  errcount += test_4xx();
  errcount += test_5xx();
  return errcount == 0 ? 0 : 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <test_absent_codes@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <test_1xx@plt>
  add    %eax,%ebx
  xor    %eax,%eax
  call   <test_2xx@plt>
  add    %eax,%ebx
  xor    %eax,%eax
  call   <test_3xx@plt>
  add    %eax,%ebx
  xor    %eax,%eax
  call   <test_4xx@plt>
  add    %eax,%ebx
  xor    %eax,%eax
  call   <test_5xx@plt>
  add    %eax,%ebx
  pop    %rbx
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int errcount = 0;
  (void)argc;
  (void)argv;
  errcount += test_absent_codes();
  errcount += test_1xx();
  errcount += test_2xx();
  errcount += test_3xx();
  errcount += test_4xx();
  errcount += test_5xx();
  return errcount == 0 ? 0 : 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <test_absent_codes@plt>
  mov    %rax,%rbx
  xor    %eax,%eax
  call   <test_1xx@plt>
  add    %eax,%ebx
  xor    %eax,%eax
  call   <test_2xx@plt>
  add    %eax,%ebx
  xor    %eax,%eax
  call   <test_3xx@plt>
  add    %eax,%ebx
  xor    %eax,%eax
  call   <test_4xx@plt>
  add    %eax,%ebx
  xor    %eax,%eax
  call   <test_5xx@plt>
  add    %eax,%ebx
  pop    %rbx
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int errcount = 0;
  (void)argc;
  (void)argv;
  errcount += test_absent_codes();
  errcount += test_1xx();
  errcount += test_2xx();
  errcount += test_3xx();
  errcount += test_4xx();
  errcount += test_5xx();
  return errcount == 0 ? 0 : 1;
}
```