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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x810,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x610(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x40,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movw   $0x3,-0x60e(%rbp)
  movw   $0x6,-0x60c(%rbp)
  movw   $0x9,-0x60a(%rbp)
  movw   $0xc,-0x608(%rbp)
  movw   $0xf,-0x606(%rbp)
  movw   $0x12,-0x604(%rbp)
  movw   $0x15,-0x602(%rbp)
  movw   $0x18,-0x600(%rbp)
  movw   $0x1b,-0x5fe(%rbp)
  movw   $0x1e,-0x5fc(%rbp)
  movw   $0x21,-0x5fa(%rbp)
  movw   $0x24,-0x5f8(%rbp)
  movw   $0x27,-0x5f6(%rbp)
  movw   $0x2a,-0x5f4(%rbp)
  movw   $0x2d,-0x5f2(%rbp)
  movw   $0x30,-0x5f0(%rbp)
  movw   $0x33,-0x5ee(%rbp)
  movw   $0x36,-0x5ec(%rbp)
  movw   $0x39,-0x5ea(%rbp)
  lea    -0x410(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x40,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movw   $0x1,-0x40e(%rbp)
  movw   $0x2,-0x40c(%rbp)
  movw   $0x3,-0x40a(%rbp)
  movw   $0x4,-0x408(%rbp)
  movw   $0x5,-0x406(%rbp)
  movw   $0x6,-0x404(%rbp)
  movw   $0x7,-0x402(%rbp)
  movw   $0x8,-0x400(%rbp)
  movw   $0x9,-0x3fe(%rbp)
  movw   $0xa,-0x3fc(%rbp)
  movw   $0xb,-0x3fa(%rbp)
  movw   $0xc,-0x3f8(%rbp)
  movw   $0xd,-0x3f6(%rbp)
  movw   $0xe,-0x3f4(%rbp)
  movw   $0xf,-0x3f2(%rbp)
  movw   $0x10,-0x3f0(%rbp)
  movw   $0x11,-0x3ee(%rbp)
  movw   $0x12,-0x3ec(%rbp)
  movw   $0x13,-0x3ea(%rbp)
  lea    -0x410(%rbp),%rdi
  lea    -0x610(%rbp),%rdx
  lea    -0x810(%rbp),%rax
  lea    -0x210(%rbp),%rcx
  lea    0x2(%rcx),%rsi
  mov    %rdi,%r8
  mov    %rdx,%rcx
  mov    %rax,%rdx
  mov    $0x100,%edi
  call   <main1@plt>
  lea    -0x410(%rbp),%rdi
  lea    -0x610(%rbp),%rdx
  lea    -0x810(%rbp),%rax
  lea    -0x210(%rbp),%rcx
  lea    0x2(%rcx),%rsi
  mov    %rdi,%r8
  mov    %rdx,%rcx
  mov    %rax,%rdx
  mov    $0xfe,%edi
  call   <main1@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int main(void) {
  int i;
  short a[256] __attribute__((__aligned__(16)));
  short d[256 + 1] __attribute__((__aligned__(16)));
  short b[256] = {0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57};
  short c[256] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19};
  main1(256, &d[1], a, b, c);
  main1(256 - 2, &d[1], a, b, c);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  lea    0x200(%rsp),%rdi
  mov    $0x40,%ecx
  rep stos %rax,%es:(%rdi)
  movw   $0x3,0x202(%rsp)
  movw   $0x6,0x204(%rsp)
  movw   $0x9,0x206(%rsp)
  movw   $0xc,0x208(%rsp)
  movw   $0xf,0x20a(%rsp)
  movw   $0x12,0x20c(%rsp)
  movw   $0x15,0x20e(%rsp)
  movw   $0x18,0x210(%rsp)
  movw   $0x1b,0x212(%rsp)
  movw   $0x1e,0x214(%rsp)
  movw   $0x21,0x216(%rsp)
  movw   $0x24,0x218(%rsp)
  movw   $0x27,0x21a(%rsp)
  movw   $0x2a,0x21c(%rsp)
  movw   $0x2d,0x21e(%rsp)
  movw   $0x30,0x220(%rsp)
  movw   $0x33,0x222(%rsp)
  movw   $0x36,0x224(%rsp)
  movw   $0x39,0x226(%rsp)
  lea    0x400(%rsp),%rdi
  mov    $0x40,%ecx
  rep stos %rax,%es:(%rdi)
  movw   $0x1,0x402(%rsp)
  movw   $0x2,0x404(%rsp)
  movw   $0x3,0x406(%rsp)
  movw   $0x4,0x408(%rsp)
  movw   $0x5,0x40a(%rsp)
  movw   $0x6,0x40c(%rsp)
  movw   $0x7,0x40e(%rsp)
  movw   $0x8,0x410(%rsp)
  movw   $0x9,0x412(%rsp)
  movw   $0xa,0x414(%rsp)
  movw   $0xb,0x416(%rsp)
  movw   $0xc,0x418(%rsp)
  movw   $0xd,0x41a(%rsp)
  movw   $0xe,0x41c(%rsp)
  movw   $0xf,0x41e(%rsp)
  movw   $0x10,0x420(%rsp)
  movw   $0x11,0x422(%rsp)
  movw   $0x12,0x424(%rsp)
  movw   $0x13,0x426(%rsp)
  lea    0x400(%rsp),%r13
  lea    0x200(%rsp),%r12
  mov    %rsp,%rbp
  lea    0x602(%rsp),%rbx
  mov    %r13,%r8
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    $0x100,%edi
  call   <main1@plt>
  mov    %r13,%r8
  mov    %r12,%rcx
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    $0xfe,%edi
  call   <main1@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x818,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int i;
  short a[256] __attribute__((__aligned__(16)));
  short d[256 + 1] __attribute__((__aligned__(16)));
  short b[256] = {0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57};
  short c[256] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19};
  main1(256, &d[1], a, b, c);
  main1(256 - 2, &d[1], a, b, c);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x40,%ecx
  movabs $0x9000600030000,%rdx
  movabs $0x150012000f000c,%rsi
  push   %r12
  push   %rbp
  push   %rbx
  movabs $0x21001e001b0018,%rbx
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  lea    0x200(%rsp),%rdi
  mov    %rsp,%r12
  lea    0x200(%rsp),%r13
  lea    0x602(%rsp),%rbp
  rep stos %rax,%es:(%rdi)
  lea    0x400(%rsp),%rdi
  mov    $0x40,%ecx
  mov    %rdx,0x200(%rsp)
  rep stos %rax,%es:(%rdi)
  mov    %r13,%rcx
  movabs $0x3000200010000,%rax
  mov    %rsi,0x208(%rsp)
  mov    %rax,0x400(%rsp)
  mov    $0x100,%edi
  movabs $0x7000600050004,%rax
  movabs $0x2d002a00270024,%rdx
  mov    %rax,0x408(%rsp)
  movabs $0xb000a00090008,%rax
  movabs $0x39003600330030,%rsi
  mov    %rbx,0x210(%rsp)
  lea    0x400(%rsp),%rbx
  mov    %rax,0x410(%rsp)
  mov    %rbx,%r8
  movabs $0xf000e000d000c,%rax
  mov    %rdx,0x218(%rsp)
  mov    %r12,%rdx
  mov    %rsi,0x220(%rsp)
  mov    %rbp,%rsi
  mov    %rax,0x418(%rsp)
  movabs $0x13001200110010,%rax
  mov    %rax,0x420(%rsp)
  call   <main1@plt>
  mov    %rbx,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    $0xfe,%edi
  call   <main1@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x818,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int i;
  short a[256] __attribute__((__aligned__(16)));
  short d[256 + 1] __attribute__((__aligned__(16)));
  short b[256] = {0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57};
  short c[256] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19};
  main1(256, &d[1], a, b, c);
  main1(256 - 2, &d[1], a, b, c);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x40,%ecx
  movabs $0x3900360033,%rdx
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x818,%rsp
  movdqa D18(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  lea    0x200(%rsp),%rdi
  mov    %rsp,%r12
  lea    0x400(%rsp),%rbx
  lea    0x200(%rsp),%r13
  lea    0x602(%rsp),%rbp
  mov    %rbx,%r8
  mov    %rbp,%rsi
  rep stos %rax,%es:(%rdi)
  movups %xmm0,0x202(%rsp)
  mov    $0x40,%ecx
  movdqa D3(%rip),%xmm0
  lea    0x400(%rsp),%rdi
  mov    %rdx,0x222(%rsp)
  mov    %r12,%rdx
  rep stos %rax,%es:(%rdi)
  movups %xmm0,0x212(%rsp)
  mov    %r13,%rcx
  mov    $0x100,%edi
  movdqa D31(%rip),%xmm0
  movabs $0x1300120011,%rax
  mov    %rax,0x422(%rsp)
  movups %xmm0,0x402(%rsp)
  movdqa D14(%rip),%xmm0
  movups %xmm0,0x412(%rsp)
  call   <main1@plt>
  mov    %rbx,%r8
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    $0xfe,%edi
  call   <main1@plt>
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x818,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D18
type: byte[16]
name: unpack_data
arguments: label: D3
type: byte[16]
name: unpack_data
arguments: label: D31
type: byte[16]
name: unpack_data
arguments: label: D14
type: byte[16]

tool: {"D18": [0x03,0x00,0x06,0x00,0x09,0x00,0x0c,0x00,0x0f,0x00,0x12,0x00,0x15,0x00,0x18,0x00]}

tool: {"D3": [0x1b,0x00,0x1e,0x00,0x21,0x00,0x24,0x00,0x27,0x00,0x2a,0x00,0x2d,0x00,0x30,0x00]}

tool: {"D31": [0x01,0x00,0x02,0x00,0x03,0x00,0x04,0x00,0x05,0x00,0x06,0x00,0x07,0x00,0x08,0x00]}

tool: {"D14": [0x09,0x00,0x0a,0x00,0x0b,0x00,0x0c,0x00,0x0d,0x00,0x0e,0x00,0x0f,0x00,0x10,0x00]}

assistant: ```c
int main(void) {
  int i;
  short a[256] __attribute__((__aligned__(16)));
  short d[256 + 1] __attribute__((__aligned__(16)));
  short b[256] = {0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57};
  short c[256] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19};
  main1(256, &d[1], a, b, c);
  main1(256 - 2, &d[1], a, b, c);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strerrno>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L3
  lea    D10(%rip),%rax
  jmp    L2
L3:
  lea    D1(%rip),%rax
L2:
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D10
data_type: string
name: disassemble_data
arguments: data_label: D1
data_type: string

tool: D10: OK

tool: D1: E??

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D18(%rip),%rax
  lea    D17(%rip),%rdx
  cmovne %rdx,%rax
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D18
datatype: string
name: disassemble_data
arguments: tag: D17
datatype: string

tool: D18: OK

tool: D17: E??

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D4(%rip),%rdx
  lea    D14(%rip),%rax
  cmovne %rdx,%rax
  ret
```

tool_calls: name: unpack_data
arguments: label: D4
type_infered: string
name: unpack_data
arguments: label: D14
type_infered: string

tool: D4: E??

tool: D14: OK

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strerrno>:
  endbr64
  test   %edi,%edi
  lea    D5(%rip),%rdx
  lea    D18(%rip),%rax
  cmovne %rdx,%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type: string
name: unpack_data
arguments: data_tag: D18
type: string

tool: D5: E??

tool: D18: OK

assistant: ```c
const char *strerrno(int err) {
  switch (err) {
  case 0:
    return "OK";
  default:
    return "E??";
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_min>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <ft_strlen@plt>
  mov    %eax,%edi
  call   <ft_strnew@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    D12(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ft_strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D12
type_infered: string

tool: D12: -2147483648

assistant: ```c
char *ft_min(void) {
  char *s;
  s = ft_strnew(ft_strlen("-2147483648"));
  ft_strcpy(s, "-2147483648");
  return (s);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_min>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D9(%rip),%rbp
  mov    %rbp,%rdi
  call   <ft_strlen@plt>
  mov    %eax,%edi
  call   <ft_strnew@plt>
  mov    %rax,%rbx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <ft_strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D9
data_type: string

tool: D9: -2147483648

assistant: ```c
char *ft_min(void) {
  char *s;
  s = ft_strnew(ft_strlen("-2147483648"));
  ft_strcpy(s, "-2147483648");
  return (s);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ft_min>:
  endbr64
  push   %r12
  push   %rbp
  lea    D0(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  mov    %eax,%edi
  call   <ft_strnew@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <ft_strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: label: D0
datatype: string

tool: D0: -2147483648

assistant: ```c
char *ft_min(void) {
  char *s;
  s = ft_strnew(ft_strlen("-2147483648"));
  ft_strcpy(s, "-2147483648");
  return (s);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_min>:
  endbr64
  push   %r12
  push   %rbp
  lea    D4(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <ft_strlen@plt>
  mov    %eax,%edi
  call   <ft_strnew@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <ft_strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D4: -2147483648

assistant: ```c
char *ft_min(void) {
  char *s;
  s = ft_strnew(ft_strlen("-2147483648"));
  ft_strcpy(s, "-2147483648");
  return (s);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rd_sorted>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  cmpq   $0x1,-0x30(%rbp)
  ja     L11
  mov    $0x1,%eax
  jmp    L23
L11:
  mov    -0x28(%rbp),%r12
  mov    -0x38(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x38(%rbp),%rax
  imul   -0x30(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%r13
  jmp    L0
L32:
  mov    -0x40(%rbp),%rax
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   *%rax
  test   %eax,%eax
  jle    L4
  mov    $0x0,%eax
  jmp    L23
L4:
  mov    %rbx,%r12
  mov    -0x38(%rbp),%rax
  shl    $0x2,%rax
  add    %rax,%rbx
L0:
  cmp    %r13,%rbx
  jne    L32
  mov    $0x1,%eax
L23:
  add    $0x28,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

assistant: ```c
bool rd_sorted(const void *base, size_t nmemb, size_t size, int (*cmp)(const void *, const void *)) {
  if (nmemb < 2) {
    return true;
  }
  register uint8_t *a = (uint8_t *)base, *b = (uint8_t *)base + size, *end = (uint8_t *)base + (size * nmemb);
  while (b != end) {
    if (cmp((void *)a, (void *)b) > 0) {
      return false;
    }
    a = b, b += size;
  }
  return true;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rd_sorted>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%rsi
  jbe    L9
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rcx,%r14
  lea    0x0(,%rdx,4),%r12
  lea    (%rdi,%r12,1),%rbx
  imul   %r12,%rsi
  lea    (%rdi,%rsi,1),%r13
  cmp    %r13,%rbx
  je     L4
L3:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   *%r14
  test   %eax,%eax
  jg     L33
  add    %r12,%rbx
  add    %r12,%rbp
  cmp    %rbx,%r13
  jne    L3
  mov    $0x1,%eax
  jmp    L42
L4:
  mov    $0x1,%eax
  jmp    L42
L33:
  mov    $0x0,%eax
L42:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  ret
```

assistant: ```c
bool rd_sorted(const void *base, size_t nmemb, size_t size, int (*cmp)(const void *, const void *)) {
  if (nmemb < 2) {
    return true;
  }
  register uint8_t *a = (uint8_t *)base, *b = (uint8_t *)base + size, *end = (uint8_t *)base + (size * nmemb);
  while (b != end) {
    if (cmp((void *)a, (void *)b) > 0) {
      return false;
    }
    a = b, b += size;
  }
  return true;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rd_sorted>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L13
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  lea    0x0(,%rdx,4),%r12
  imul   %r12,%rsi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    (%rdi,%r12,1),%rbx
  lea    (%rdi,%rsi,1),%r14
  cmp    %r14,%rbx
  jne    L34
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L22:
  add    %r12,%rbx
  add    %r12,%rbp
  cmp    %rbx,%r14
  je     L25
L34:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   *%r13
  test   %eax,%eax
  jle    L22
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L13:
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool rd_sorted(const void *base, size_t nmemb, size_t size, int (*cmp)(const void *, const void *)) {
  if (nmemb < 2) {
    return true;
  }
  register uint8_t *a = (uint8_t *)base, *b = (uint8_t *)base + size, *end = (uint8_t *)base + (size * nmemb);
  while (b != end) {
    if (cmp((void *)a, (void *)b) > 0) {
      return false;
    }
    a = b, b += size;
  }
  return true;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rd_sorted>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L21
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  lea    0x0(,%rdx,4),%r12
  imul   %r12,%rsi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    (%rdi,%r12,1),%rbx
  lea    (%rdi,%rsi,1),%r14
  cmp    %r14,%rbx
  jne    L36
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L9:
  add    %r12,%rbx
  add    %r12,%rbp
  cmp    %rbx,%r14
  je     L15
L36:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   *%r13
  test   %eax,%eax
  jle    L9
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L21:
  mov    $0x1,%eax
  ret
```

assistant: ```c
bool rd_sorted(const void *base, size_t nmemb, size_t size, int (*cmp)(const void *, const void *)) {
  if (nmemb < 2) {
    return true;
  }
  register uint8_t *a = (uint8_t *)base, *b = (uint8_t *)base + size, *end = (uint8_t *)base + (size * nmemb);
  while (b != end) {
    if (cmp((void *)a, (void *)b) > 0) {
      return false;
    }
    a = b, b += size;
  }
  return true;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  jmp    L15
L35:
  addq   $0x1,-0x8(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L8
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    -0xc(%rbp),%edx
  cmp    %dl,%al
  jne    L35
L8:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0xc(%rbp)
  jne    L17
  mov    -0x8(%rbp),%rax
  jmp    L23
L17:
  mov    $0x0,%eax
L23:
  pop    %rbp
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  while (*s != '\0' && *s != (char)c)
    s++;
  return (*s == c) ? (char *)s : ((void *)0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L9
L5:
  cmp    %sil,%al
  je     L9
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L5
L9:
  movsbl %al,%eax
  cmp    %esi,%eax
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  while (*s != '\0' && *s != (char)c)
    s++;
  return (*s == c) ? (char *)s : ((void *)0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L20
  mov    %esi,%edx
  jmp    L6
  nop
L10:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L20
L6:
  cmp    %al,%dl
  jne    L10
  movsbl %sil,%eax
  cmp    %eax,%esi
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
L20:
  xor    %eax,%eax
  cmp    %eax,%esi
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  while (*s != '\0' && *s != (char)c)
    s++;
  return (*s == c) ? (char *)s : ((void *)0);
}
```
user: What is the c source code of the assembly code below?

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L9
  mov    %esi,%edx
  jmp    L0
  nop
L6:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L9
L0:
  cmp    %al,%dl
  jne    L6
  movsbl %sil,%eax
  cmp    %eax,%esi
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
L9:
  xor    %eax,%eax
  cmp    %eax,%esi
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  while (*s != '\0' && *s != (char)c)
    s++;
  return (*s == c) ? (char *)s : ((void *)0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  cmp    %eax,%edx
  jne    L19
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L3
L19:
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <hanoi@plt>
  mov    $0x0,%eax
L3:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D53
type_infered: string
name: retrieve_data
arguments: tag: D57
type_infered: string
name: retrieve_data
arguments: tag: D58
type_infered: string
name: retrieve_data
arguments: tag: D10
type_infered: string
name: retrieve_data
arguments: tag: D36
type_infered: string

tool: D53: How Many Disk? : 

tool: D57: %d

tool: D58: From Where? [1-3] : 

tool: D10: To where? [1-3] : 

tool: D36: Have the same start and finish


assistant: ```c
int main() {
  int n, from, to;
  printf("How Many Disk? : ");
  scanf("%d", &n);
  printf("From Where? [1-3] : ");
  scanf("%d", &from);
  printf("To where? [1-3] : ");
  scanf("%d", &to);
  if (from == to)
    return printf("Have the same start and finish\n");
  else
    hanoi(n, from, to);
  return 0;
}
```