ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<KmlStringCopy>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *KmlStringCopy(const char *str) {
  char *res;
  res = (char *)malloc((strlen(str) + 1) * sizeof(char));
  strcpy(res, str);
  return res;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<SetBit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x1f,-0x8(%rbp)
  jg     L6
  mov    -0x8(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x4(%rbp)
  jmp    L10
L6:
  nop
L10:
  pop    %rbp
  ret
```

assistant: ```c
void SetBit(int data, int bit) {
  if (bit > 31) {
    return;
  }
  data = data | (0x1 << bit);
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<SetBit>:
  endbr64
  ret
```

assistant: ```c
void SetBit(int data, int bit) {
  if (bit > 31) {
    return;
  }
  data = data | (0x1 << bit);
  return;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<SetBit>:
  endbr64
  ret
```

assistant: ```c
void SetBit(int data, int bit) {
  if (bit > 31) {
    return;
  }
  data = data | (0x1 << bit);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<builtin_div>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L27
  mov    $0x0,%edi
  call   <push_from_long@plt>
  jmp    L5
L27:
  cmpl   $0x1,-0x1c(%rbp)
  jne    L17
  mov    $0x0,%eax
  call   <pop_to_long@plt>
  mov    %rax,%rdi
  call   <push_from_long@plt>
  jmp    L5
L17:
  mov    $0x0,%eax
  call   <pop_to_long@plt>
  mov    %rax,-0x8(%rbp)
  subl   $0x1,-0x1c(%rbp)
  jmp    L52
L33:
  mov    $0x0,%eax
  call   <pop_to_long@plt>
  mov    %rax,%rcx
  mov    -0x8(%rbp),%rax
  cqto
  idiv   %rcx
  mov    %rax,-0x8(%rbp)
  subl   $0x1,-0x1c(%rbp)
L52:
  cmpl   $0x0,-0x1c(%rbp)
  jne    L33
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <push_from_long@plt>
L5:
  leave
  ret
```

assistant: ```c
void builtin_div(struct Context *ctx, unsigned int stack_track) {
  if (stack_track == 0) {
    push_from_long(0L);
    return;
  }
  if (stack_track == 1) {
    push_from_long(pop_to_long());
    return;
  }
  long result = pop_to_long();
  for (stack_track--; stack_track; stack_track--)
    result /= pop_to_long();
  push_from_long(result);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<builtin_div>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  je     L2
  mov    %esi,%ebx
  cmp    $0x1,%esi
  je     L40
  mov    $0x0,%eax
  call   <pop_to_long@plt>
  mov    %rax,%rbp
  sub    $0x1,%ebx
L17:
  mov    $0x0,%eax
  call   <pop_to_long@plt>
  mov    %rax,%rcx
  mov    %rbp,%rax
  cqto
  idiv   %rcx
  mov    %rax,%rbp
  sub    $0x1,%ebx
  jne    L17
  mov    %rax,%rdi
  call   <push_from_long@plt>
L30:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  mov    $0x0,%edi
  call   <push_from_long@plt>
  jmp    L30
L40:
  mov    $0x0,%eax
  call   <pop_to_long@plt>
  mov    %rax,%rdi
  call   <push_from_long@plt>
  jmp    L30
```

assistant: ```c
void builtin_div(struct Context *ctx, unsigned int stack_track) {
  if (stack_track == 0) {
    push_from_long(0L);
    return;
  }
  if (stack_track == 1) {
    push_from_long(pop_to_long());
    return;
  }
  long result = pop_to_long();
  for (stack_track--; stack_track; stack_track--)
    result /= pop_to_long();
  push_from_long(result);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<builtin_div>:
  endbr64
  test   %esi,%esi
  je     L13
  push   %r12
  xor    %eax,%eax
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  cmp    $0x1,%esi
  je     L21
  call   <pop_to_long@plt>
  sub    $0x1,%ebx
  mov    %rax,%r12
  nopl   0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  call   <pop_to_long@plt>
  mov    %rax,%r8
  mov    %r12,%rax
  cqto
  idiv   %r8
  mov    %rax,%r12
  sub    $0x1,%ebx
  jne    L2
  add    $0x8,%rsp
  mov    %rax,%rdi
  pop    %rbx
  pop    %r12
  jmp    1060 <push_from_long@plt>
  nopl   0x0(%rax)
L13:
  xor    %edi,%edi
  jmp    1060 <push_from_long@plt>
  nop
L21:
  call   <pop_to_long@plt>
  add    $0x8,%rsp
  pop    %rbx
  mov    %rax,%rdi
  pop    %r12
  jmp    1060 <push_from_long@plt>
```

assistant: ```c
void builtin_div(struct Context *ctx, unsigned int stack_track) {
  if (stack_track == 0) {
    push_from_long(0L);
    return;
  }
  if (stack_track == 1) {
    push_from_long(pop_to_long());
    return;
  }
  long result = pop_to_long();
  for (stack_track--; stack_track; stack_track--)
    result /= pop_to_long();
  push_from_long(result);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<builtin_div>:
  endbr64
  test   %esi,%esi
  je     L27
  push   %r12
  xor    %eax,%eax
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  cmp    $0x1,%esi
  je     L13
  call   <pop_to_long@plt>
  sub    $0x1,%ebx
  mov    %rax,%r12
  nopl   0x0(%rax,%rax,1)
L10:
  xor    %eax,%eax
  call   <pop_to_long@plt>
  mov    %rax,%r8
  mov    %r12,%rax
  cqto
  idiv   %r8
  mov    %rax,%r12
  sub    $0x1,%ebx
  jne    L10
  add    $0x8,%rsp
  mov    %rax,%rdi
  pop    %rbx
  pop    %r12
  jmp    1060 <push_from_long@plt>
  nopl   0x0(%rax)
L27:
  xor    %edi,%edi
  jmp    1060 <push_from_long@plt>
  nop
L13:
  call   <pop_to_long@plt>
  add    $0x8,%rsp
  pop    %rbx
  mov    %rax,%rdi
  pop    %r12
  jmp    1060 <push_from_long@plt>
```

assistant: ```c
void builtin_div(struct Context *ctx, unsigned int stack_track) {
  if (stack_track == 0) {
    push_from_long(0L);
    return;
  }
  if (stack_track == 1) {
    push_from_long(pop_to_long());
    return;
  }
  long result = pop_to_long();
  for (stack_track--; stack_track; stack_track--)
    result /= pop_to_long();
  push_from_long(result);
}
```
user: What is the c source code of the assembly code below:

```asm
<get_codeword_type>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x230,%rsp
  mov    %rdi,-0x228(%rbp)
  mov    %rsi,-0x230(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x210(%rbp),%rdx
  mov    -0x230(%rbp),%rsi
  mov    -0x228(%rbp),%rax
  mov    $0x1,%ecx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <get_token@plt>
  mov    %eax,-0x214(%rbp)
  cmpl   $0x2,-0x214(%rbp)
  je     L68
  mov    $0x0,%eax
  jmp    L50
L68:
  lea    -0x210(%rbp),%rdx
  mov    -0x230(%rbp),%rsi
  mov    -0x228(%rbp),%rax
  mov    $0x2,%ecx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <get_token@plt>
  mov    %eax,-0x214(%rbp)
  cmpl   $0x4,-0x214(%rbp)
  je     L108
  mov    $0x0,%eax
  jmp    L50
L108:
  lea    -0x210(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L90
  mov    $0x1,%eax
  jmp    L50
L90:
  lea    -0x210(%rbp),%rax
  lea    D14(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L61
  mov    $0x2,%eax
  jmp    L50
L61:
  lea    -0x210(%rbp),%rax
  lea    D12(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L27
  mov    $0x3,%eax
  jmp    L50
L27:
  mov    $0x0,%eax
L50:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D0
type: string
name: read_data
arguments: label: D14
type: string
name: read_data
arguments: label: D12
type: string

tool: D0: H

tool: D14: TN

tool: D12: FN

assistant: ```c
int get_codeword_type(line, char_p)
char *line;
int *char_p;
{
  char token[512];
  int token_type;
  token_type = get_token(line, char_p, token, 1);
  if (token_type != 2)
    return (0);
  token_type = get_token(line, char_p, token, 2);
  if (token_type != 4)
    return (0);
  if (strcmp(token, "H") == 0)
    return (1);
  else if (strcmp(token, "TN") == 0)
    return (2);
  else if (strcmp(token, "FN") == 0)
    return (3);
  else {
    return (0);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_codeword_type>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdx
  mov    $0x1,%ecx
  call   <get_token@plt>
  mov    $0x0,%ebx
  cmp    $0x2,%eax
  je     L42
L31:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  mov    %ebx,%eax
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L42:
  mov    %eax,%ebx
  mov    %rsp,%rdx
  mov    $0x2,%ecx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <get_token@plt>
  cmp    $0x4,%eax
  jne    L66
  mov    %rsp,%rdi
  lea    D16(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L28
  mov    %rsp,%rdi
  lea    D13(%rip),%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L31
  mov    %rsp,%rdi
  lea    D20(%rip),%rsi
  call   <strcmp@plt>
  cmp    $0x1,%eax
  sbb    %ebx,%ebx
  and    $0x3,%ebx
  jmp    L31
L66:
  mov    $0x0,%ebx
  jmp    L31
L28:
  mov    $0x1,%ebx
  jmp    L31
L65:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D16
type: string
name: parse_data
arguments: data_label: D13
type: string
name: parse_data
arguments: data_label: D20
type: string

tool: D16: H

tool: D13: TN

tool: D20: FN

assistant: ```c
int get_codeword_type(line, char_p)
char *line;
int *char_p;
{
  char token[512];
  int token_type;
  token_type = get_token(line, char_p, token, 1);
  if (token_type != 2)
    return (0);
  token_type = get_token(line, char_p, token, 2);
  if (token_type != 4)
    return (0);
  if (strcmp(token, "H") == 0)
    return (1);
  else if (strcmp(token, "TN") == 0)
    return (2);
  else if (strcmp(token, "FN") == 0)
    return (3);
  else {
    return (0);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<get_codeword_type>:
  endbr64
  push   %r14
  mov    $0x1,%ecx
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x218,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdx
  call   <get_token@plt>
  cmp    $0x2,%eax
  je     L98
L60:
  xor    %r12d,%r12d
L54:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  add    $0x218,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L98:
  mov    %eax,%r12d
  mov    $0x2,%ecx
  xor    %eax,%eax
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <get_token@plt>
  cmp    $0x4,%eax
  jne    L60
  cmpw   $0x48,(%rsp)
  je     L12
  cmpw   $0x4e54,(%rsp)
  je     L53
L78:
  cmpw   $0x4e46,(%rsp)
  jne    L60
  cmpb   $0x0,0x2(%rsp)
  mov    $0x3,%r12d
  je     L54
  jmp    L60
  nopl   0x0(%rax,%rax,1)
L12:
  mov    $0x1,%r12d
  jmp    L54
L53:
  cmpb   $0x0,0x2(%rsp)
  je     L54
  jmp    L78
L83:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_codeword_type(line, char_p)
char *line;
int *char_p;
{
  char token[512];
  int token_type;
  token_type = get_token(line, char_p, token, 1);
  if (token_type != 2)
    return (0);
  token_type = get_token(line, char_p, token, 2);
  if (token_type != 4)
    return (0);
  if (strcmp(token, "H") == 0)
    return (1);
  else if (strcmp(token, "TN") == 0)
    return (2);
  else if (strcmp(token, "FN") == 0)
    return (3);
  else {
    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
<get_codeword_type>:
  endbr64
  push   %r14
  mov    $0x1,%ecx
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x218,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdx
  call   <get_token@plt>
  cmp    $0x2,%eax
  je     L66
L70:
  xor    %r12d,%r12d
L104:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x218,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L66:
  mov    %eax,%r12d
  mov    $0x2,%ecx
  xor    %eax,%eax
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <get_token@plt>
  cmp    $0x4,%eax
  jne    L70
  cmpw   $0x48,(%rsp)
  je     L34
  cmpw   $0x4e54,(%rsp)
  je     L98
L80:
  cmpw   $0x4e46,(%rsp)
  jne    L70
  cmpb   $0x0,0x2(%rsp)
  mov    $0x3,%r12d
  je     L104
  jmp    L70
  nopl   0x0(%rax,%rax,1)
L34:
  mov    $0x1,%r12d
  jmp    L104
L98:
  cmpb   $0x0,0x2(%rsp)
  je     L104
  jmp    L80
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_codeword_type(line, char_p)
char *line;
int *char_p;
{
  char token[512];
  int token_type;
  token_type = get_token(line, char_p, token, 1);
  if (token_type != 2)
    return (0);
  token_type = get_token(line, char_p, token, 2);
  if (token_type != 4)
    return (0);
  if (strcmp(token, "H") == 0)
    return (1);
  else if (strcmp(token, "TN") == 0)
    return (2);
  else if (strcmp(token, "FN") == 0)
    return (3);
  else {
    return (0);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memcpy1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L87
  mov    -0x18(%rbp),%rax
  sub    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <abs@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  jle    L52
  jmp    L47
L32:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x18(%rbp)
  addq   $0x4,-0x20(%rbp)
  subq   $0x4,-0x28(%rbp)
L47:
  cmpq   $0x3,-0x28(%rbp)
  ja     L32
L52:
  cmpl   $0x1,-0x4(%rbp)
  jle    L46
  cmpq   $0x1,-0x28(%rbp)
  jbe    L46
  jmp    L66
L124:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  addq   $0x2,-0x18(%rbp)
  addq   $0x2,-0x20(%rbp)
  subq   $0x2,-0x28(%rbp)
L66:
  cmpq   $0x1,-0x28(%rbp)
  ja     L124
L46:
  cmpl   $0x0,-0x4(%rbp)
  jle    L115
  movq   $0x1,-0x28(%rbp)
  jmp    L69
L29:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
  subq   $0x1,-0x28(%rbp)
L69:
  cmpq   $0x0,-0x28(%rbp)
  jne    L29
  jmp    L115
L87:
  nop
  jmp    L23
L115:
  nop
L23:
  leave
  ret
```

assistant: ```c
void memcpy1(void *dest, void *src, size_t count) {
  if (!count) {
    return;
  }
  int diff = abs((long)(dest - src));
  if (diff >= 4)
    while (count >= 4) {
      *(unsigned int *)dest = *(unsigned int *)src;
      dest += 4;
      src += 4;
      count -= 4;
    }
  if ((diff >= 2) && (count >= 2))
    while (count >= 2) {
      *(unsigned long long *)dest = *(unsigned long long *)src;
      dest += 2;
      src += 2;
      count -= 2;
    }
  if ((diff >= 1) && (count = 1))
    while (count >= 1) {
      *(unsigned long long *)dest = *(unsigned long long *)src;
      dest += 1;
      src += 1;
      count -= 1;
    }
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<memcpy1>:
  endbr64
  test   %rdx,%rdx
  jne    L20
  ret
L20:
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdx,%r12
  sub    %rsi,%rdi
  call   <abs@plt>
  cmp    $0x3,%eax
  jle    L39
  cmp    $0x3,%r12
  jbe    L36
  lea    -0x4(%r12),%rdx
  shr    $0x2,%rdx
  lea    0x4(,%rdx,4),%rsi
  mov    $0x0,%edx
L81:
  mov    0x0(%rbp,%rdx,1),%ecx
  mov    %ecx,(%rbx,%rdx,1)
  add    $0x4,%rdx
  cmp    %rsi,%rdx
  jne    L81
  add    %rsi,%rbx
  add    %rsi,%rbp
  and    $0x3,%r12d
L36:
  cmp    $0x1,%eax
  jle    L42
  cmp    $0x1,%r12
  jbe    L42
L83:
  sub    $0x2,%r12
  shr    %r12
  lea    0x2(%r12,%r12,1),%rcx
  mov    $0x0,%eax
L31:
  mov    0x0(%rbp,%rax,1),%rdx
  mov    %rdx,(%rbx,%rax,1)
  add    $0x2,%rax
  cmp    %rax,%rcx
  jne    L31
  add    %rcx,%rbx
  add    %rcx,%rbp
L42:
  mov    0x0(%rbp),%rax
  mov    %rax,(%rbx)
L43:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L39:
  cmp    $0x1,%eax
  jle    L102
  cmp    $0x1,%r12
  ja     L83
L102:
  test   %eax,%eax
  jle    L43
  jmp    L42
```

assistant: ```c
void memcpy1(void *dest, void *src, size_t count) {
  if (!count) {
    return;
  }
  int diff = abs((long)(dest - src));
  if (diff >= 4)
    while (count >= 4) {
      *(unsigned int *)dest = *(unsigned int *)src;
      dest += 4;
      src += 4;
      count -= 4;
    }
  if ((diff >= 2) && (count >= 2))
    while (count >= 2) {
      *(unsigned long long *)dest = *(unsigned long long *)src;
      dest += 2;
      src += 2;
      count -= 2;
    }
  if ((diff >= 1) && (count = 1))
    while (count >= 1) {
      *(unsigned long long *)dest = *(unsigned long long *)src;
      dest += 1;
      src += 1;
      count -= 1;
    }
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<memcpy1>:
  endbr64
  test   %rdx,%rdx
  jne    L94
  ret
  nopw   0x0(%rax,%rax,1)
L94:
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    %rsi,%rdi
  call   <abs@plt>
  cmp    $0x3,%eax
  jle    L109
  cmp    $0x3,%r12
  jbe    L19
  lea    -0x4(%r12),%rax
  shr    $0x2,%rax
  lea    0x4(,%rax,4),%rcx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L68:
  mov    0x0(%rbp,%rax,1),%edx
  mov    %edx,(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L68
  add    %rax,%rbx
  add    %rax,%rbp
  and    $0x3,%r12d
L19:
  cmp    $0x1,%r12
  jbe    L69
L36:
  sub    $0x2,%r12
  xor    %eax,%eax
  shr    %r12
  lea    0x2(%r12,%r12,1),%rcx
  cs nopw 0x0(%rax,%rax,1)
L41:
  mov    0x0(%rbp,%rax,1),%rdx
  mov    %rdx,(%rbx,%rax,1)
  add    $0x2,%rax
  cmp    %rcx,%rax
  jne    L41
  add    %rax,%rbx
  add    %rax,%rbp
L69:
  mov    0x0(%rbp),%rax
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L109:
  cmp    $0x1,%eax
  jg     L46
  test   %eax,%eax
  jg     L69
L18:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L46:
  cmp    $0x1,%r12
  ja     L36
  test   %eax,%eax
  jle    L18
  jmp    L69
```

assistant: ```c
void memcpy1(void *dest, void *src, size_t count) {
  if (!count) {
    return;
  }
  int diff = abs((long)(dest - src));
  if (diff >= 4)
    while (count >= 4) {
      *(unsigned int *)dest = *(unsigned int *)src;
      dest += 4;
      src += 4;
      count -= 4;
    }
  if ((diff >= 2) && (count >= 2))
    while (count >= 2) {
      *(unsigned long long *)dest = *(unsigned long long *)src;
      dest += 2;
      src += 2;
      count -= 2;
    }
  if ((diff >= 1) && (count = 1))
    while (count >= 1) {
      *(unsigned long long *)dest = *(unsigned long long *)src;
      dest += 1;
      src += 1;
      count -= 1;
    }
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<memcpy1>:
  endbr64
  test   %rdx,%rdx
  jne    L150
  ret
  nopw   0x0(%rax,%rax,1)
L150:
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    %rsi,%rdi
  call   <abs@plt>
  cmp    $0x3,%eax
  jle    L31
  cmp    $0x3,%r12
  jbe    L17
  lea    0x4(%rbp),%rdx
  mov    %rbx,%rax
  lea    -0x4(%r12),%rcx
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L86
  cmp    $0xb,%rcx
  jbe    L86
  mov    %rcx,%rsi
  xor    %eax,%eax
  shr    $0x2,%rsi
  add    $0x1,%rsi
  mov    %rsi,%rdx
  shr    $0x2,%rdx
  shl    $0x4,%rdx
  nopl   0x0(%rax,%rax,1)
L2:
  movdqu 0x0(%rbp,%rax,1),%xmm0
  movups %xmm0,(%rbx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L2
  mov    %rsi,%rdx
  and    $0xfffffffffffffffc,%rdx
  lea    0x0(,%rdx,4),%rax
  lea    (%rbx,%rax,1),%rdi
  lea    0x0(%rbp,%rax,1),%r8
  cmp    %rdx,%rsi
  je     L107
  mov    (%r8),%edx
  mov    %edx,(%rdi)
  mov    %rcx,%rdx
  sub    %rax,%rdx
  cmp    $0x3,%rdx
  jbe    L107
  mov    0x4(%r8),%eax
  sub    $0x4,%rdx
  mov    %eax,0x4(%rdi)
  cmp    $0x3,%rdx
  jbe    L107
  mov    0x8(%r8),%eax
  mov    %eax,0x8(%rdi)
L107:
  shr    $0x2,%rcx
  and    $0x3,%r12d
  lea    0x4(,%rcx,4),%rax
  add    %rax,%rbx
  add    %rax,%rbp
L17:
  cmp    $0x1,%r12
  jbe    L56
L85:
  sub    $0x2,%r12
  xor    %eax,%eax
  shr    %r12
  lea    0x2(%r12,%r12,1),%rcx
  nopl   0x0(%rax,%rax,1)
L43:
  mov    0x0(%rbp,%rax,1),%rdx
  mov    %rdx,(%rbx,%rax,1)
  add    $0x2,%rax
  cmp    %rax,%rcx
  jne    L43
  add    %rcx,%rbx
  add    %rcx,%rbp
L56:
  mov    0x0(%rbp),%rax
  mov    %rax,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L31:
  cmp    $0x1,%eax
  jg     L138
L19:
  test   %eax,%eax
  jg     L56
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L138:
  cmp    $0x1,%r12
  jbe    L19
  jmp    L85
  nopl   0x0(%rax,%rax,1)
L86:
  mov    %rcx,%rax
  shr    $0x2,%rax
  lea    0x4(,%rax,4),%rsi
  xor    %eax,%eax
  nopl   0x0(%rax)
L61:
  mov    0x0(%rbp,%rax,1),%edx
  mov    %edx,(%rbx,%rax,1)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L61
  jmp    L107
```

assistant: ```c
void memcpy1(void *dest, void *src, size_t count) {
  if (!count) {
    return;
  }
  int diff = abs((long)(dest - src));
  if (diff >= 4)
    while (count >= 4) {
      *(unsigned int *)dest = *(unsigned int *)src;
      dest += 4;
      src += 4;
      count -= 4;
    }
  if ((diff >= 2) && (count >= 2))
    while (count >= 2) {
      *(unsigned long long *)dest = *(unsigned long long *)src;
      dest += 2;
      src += 2;
      count -= 2;
    }
  if ((diff >= 1) && (count = 1))
    while (count >= 1) {
      *(unsigned long long *)dest = *(unsigned long long *)src;
      dest += 1;
      src += 1;
      count -= 1;
    }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L16
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L4
L16:
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <keygen@plt>
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L4:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D12
type: string
name: extract_symbolic_data
arguments: data_tag: D16
type: string

tool: D12: Usage: %s <challenge>


tool: D16: B%lu


assistant: ```c
void main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("Usage: %s <challenge>\n", argv[0]);
    return;
  }
  printf("B%lu\n", keygen(argv[1]));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L0
  mov    0x8(%rsi),%rdi
  call   <keygen@plt>
  mov    %rax,%rsi
  lea    D12(%rip),%rdi
  call   <printf@plt>
L2:
  add    $0x8,%rsp
  ret
L0:
  mov    (%rsi),%rsi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  jmp    L2
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
datatype: string
name: disassemble_data
arguments: data_tag: D16
datatype: string

tool: D12: B%lu


tool: D16: Usage: %s <challenge>


assistant: ```c
void main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("Usage: %s <challenge>\n", argv[0]);
    return;
  }
  printf("B%lu\n", keygen(argv[1]));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  cmp    $0x1,%edi
  jle    L4
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rdi
  call   <keygen@plt>
  lea    D6(%rip),%rdi
  add    $0x8,%rsp
  mov    %rax,%rsi
  jmp    1070 <printf@plt>
L4:
  mov    (%rsi),%rsi
  lea    D1(%rip),%rdi
  jmp    1070 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type: string
name: disassemble_data
arguments: data_label: D1
type: string

tool: D6: B%lu


tool: D1: Usage: %s <challenge>


assistant: ```c
void main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("Usage: %s <challenge>\n", argv[0]);
    return;
  }
  printf("B%lu\n", keygen(argv[1]));
}
```