ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%edx
  sub    $0x25,%edx
  mov    %edx,%eax
  shl    $0x4,%eax
  or     %edx,%eax
  xor    $0xffffffec,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)219;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[0UL] |= (state[0UL] & (unsigned char)63) << 4UL;
      local1++;
    }
    output[0UL] = state[0UL] ^ (unsigned char)236;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<url2int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L14:
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x5,%eax
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x4(%rbp)
  jl     L14
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int url2int(const char *url) {
  int h = 0, i = 0;
  for (i = 0; i < strlen(url); i++) {
    h = 31 * h + url[i];
  }
  return h;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<url2int>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x0,%ebp
  mov    $0x0,%ebx
  jmp    L2
L5:
  mov    %ebx,%eax
  shl    $0x5,%eax
  sub    %ebx,%eax
  movsbl (%r12,%rbp,1),%ebx
  add    %eax,%ebx
  add    $0x1,%rbp
L2:
  mov    %r12,%rdi
  call   <strlen@plt>
  cmp    %ebp,%eax
  jg     L5
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int url2int(const char *url) {
  int h = 0, i = 0;
  for (i = 0; i < strlen(url); i++) {
    h = 31 * h + url[i];
  }
  return h;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<url2int>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L3
  nopw   0x0(%rax,%rax,1)
L6:
  shl    $0x5,%eax
  sub    %r12d,%eax
  movsbl 0x0(%rbp,%rbx,1),%r12d
  add    $0x1,%rbx
  add    %eax,%r12d
L3:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  mov    %r12d,%eax
  jg     L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int url2int(const char *url) {
  int h = 0, i = 0;
  for (i = 0; i < strlen(url); i++) {
    h = 31 * h + url[i];
  }
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<url2int>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L0:
  shl    $0x5,%eax
  sub    %r12d,%eax
  movsbl 0x0(%rbp,%rbx,1),%r12d
  add    $0x1,%rbx
  add    %eax,%r12d
L10:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  mov    %r12d,%eax
  jg     L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int url2int(const char *url) {
  int h = 0, i = 0;
  for (i = 0; i < strlen(url); i++) {
    h = 31 * h + url[i];
  }
  return h;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bcopy0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x4030201,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0xc(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <bcopy@plt>
  movzbl -0xc(%rbp),%eax
  test   %al,%al
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <clang_analyzer_eval@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void bcopy0() {
  char src[] = {1, 2, 3, 4};
  char dst[4] = {0};
  bcopy(src, dst, 4);
  clang_analyzer_eval(dst[0] != 0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bcopy0>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x4030201,(%rsp)
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rsi
  mov    %rsp,%rdi
  mov    $0x4,%edx
  call   <bcopy@plt>
  cmpb   $0x0,0x4(%rsp)
  setne  %dil
  movzbl %dil,%edi
  call   <clang_analyzer_eval@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bcopy0() {
  char src[] = {1, 2, 3, 4};
  char dst[4] = {0};
  bcopy(src, dst, 4);
  clang_analyzer_eval(dst[0] != 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bcopy0>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x4(%rsp),%rsi
  movl   $0x4030201,(%rsp)
  movl   $0x0,0x4(%rsp)
  call   <bcopy@plt>
  xor    %edi,%edi
  cmpb   $0x0,0x4(%rsp)
  setne  %dil
  call   <clang_analyzer_eval@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bcopy0() {
  char src[] = {1, 2, 3, 4};
  char dst[4] = {0};
  bcopy(src, dst, 4);
  clang_analyzer_eval(dst[0] != 0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bcopy0>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    0x4(%rsp),%rsi
  movl   $0x4030201,(%rsp)
  movl   $0x0,0x4(%rsp)
  call   <bcopy@plt>
  xor    %edi,%edi
  cmpb   $0x0,0x4(%rsp)
  setne  %dil
  call   <clang_analyzer_eval@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bcopy0() {
  char src[] = {1, 2, 3, 4};
  char dst[4] = {0};
  bcopy(src, dst, 4);
  clang_analyzer_eval(dst[0] != 0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<my_strchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  nop
L24:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  je     L27
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0xc(%rbp)
  jne    L24
L27:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0xc(%rbp)
  jne    L3
  mov    -0x8(%rbp),%rax
  jmp    L26
L3:
  mov    $0x0,%eax
L26:
  pop    %rbp
  ret
```

assistant: ```c
char *my_strchr(const char *str, const int x) {
  while (*(str++) && *str != x)
    ;
  if (*str == x)
    return ((char *)str);
  return (NULL);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<my_strchr>:
  endbr64
L9:
  add    $0x1,%rdi
  cmpb   $0x0,-0x1(%rdi)
  je     L18
  movsbl (%rdi),%eax
  cmp    %esi,%eax
  jne    L9
  mov    %rdi,%rax
  ret
L18:
  movsbl (%rdi),%eax
  cmp    %esi,%eax
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
```

assistant: ```c
char *my_strchr(const char *str, const int x) {
  while (*(str++) && *str != x)
    ;
  if (*str == x)
    return ((char *)str);
  return (NULL);
}
```
user: What is the c source code of the assembly code below?

```asm
<my_strchr>:
  endbr64
  movzbl (%rdi),%eax
  jmp    L19
  nopl   0x0(%rax)
L4:
  cmp    %edx,%esi
  je     L24
L19:
  movsbl 0x1(%rdi),%edx
  mov    %eax,%ecx
  add    $0x1,%rdi
  mov    %edx,%eax
  test   %cl,%cl
  jne    L4
  xor    %eax,%eax
  cmp    %edx,%esi
  cmove  %rdi,%rax
  ret
  nopl   (%rax)
L24:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *my_strchr(const char *str, const int x) {
  while (*(str++) && *str != x)
    ;
  if (*str == x)
    return ((char *)str);
  return (NULL);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<my_strchr>:
  endbr64
  movzbl (%rdi),%eax
  jmp    L24
  nopl   0x0(%rax)
L8:
  cmp    %edx,%esi
  je     L27
L24:
  movsbl 0x1(%rdi),%edx
  mov    %eax,%ecx
  add    $0x1,%rdi
  mov    %edx,%eax
  test   %cl,%cl
  jne    L8
  xor    %eax,%eax
  cmp    %edx,%esi
  cmove  %rdi,%rax
  ret
  nopl   (%rax)
L27:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *my_strchr(const char *str, const int x) {
  while (*(str++) && *str != x)
    ;
  if (*str == x)
    return ((char *)str);
  return (NULL);
}
```
user: Translate the below assembly code into C programming language:

```asm
<greet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <clear@plt>
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x5,%edi
  call   <sleep@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
type: string

tool: D4: WELCOME TO GAME OF TIC_TAC_TOE


assistant: ```c
void greet(void) {
  clear();
  printf("WELCOME TO GAME OF TIC_TAC_TOE\n");
  sleep(5);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<greet>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <clear@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%edi
  call   <sleep@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D5: WELCOME TO GAME OF TIC_TAC_TOE


assistant: ```c
void greet(void) {
  clear();
  printf("WELCOME TO GAME OF TIC_TAC_TOE\n");
  sleep(5);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<greet>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clear@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%edi
  add    $0x8,%rsp
  jmp    1090 <sleep@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
data_type: string

tool: D2: WELCOME TO GAME OF TIC_TAC_TOE


assistant: ```c
void greet(void) {
  clear();
  printf("WELCOME TO GAME OF TIC_TAC_TOE\n");
  sleep(5);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<greet>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <clear@plt>
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x5,%edi
  add    $0x8,%rsp
  jmp    1090 <sleep@plt>
```

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

tool: D3: WELCOME TO GAME OF TIC_TAC_TOE


assistant: ```c
void greet(void) {
  clear();
  printf("WELCOME TO GAME OF TIC_TAC_TOE\n");
  sleep(5);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<msg_out_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x180,%rsp
  mov    %rdi,-0x168(%rbp)
  mov    %rsi,-0x170(%rbp)
  mov    %rdx,-0x178(%rbp)
  mov    %ecx,-0x17c(%rbp)
  mov    %r8d,-0x180(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x150(%rbp),%rdx
  mov    -0x17c(%rbp),%esi
  mov    -0x168(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0xf,%edx
  mov    %rax,%rdi
  call   <cnfImpn@plt>
  lea    -0x140(%rbp),%rdx
  mov    -0x180(%rbp),%esi
  mov    -0x170(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x12c,%edx
  mov    %rax,%rdi
  call   <cnfImpn@plt>
  mov    -0x178(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x154(%rbp)
  lea    -0x154(%rbp),%rdx
  lea    -0x140(%rbp),%rcx
  lea    -0x150(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <msgOut@plt>
  mov    -0x154(%rbp),%edx
  mov    -0x178(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
void msg_out_(char *const PARAM, char *const TEXT, int *const STATUS, int PARAM_length, int TEXT_length) {
  char param[15 + 1];
  char text[300 + 1];
  int status;
  cnfImpn(PARAM, PARAM_length, 15, param);
  cnfImpn(TEXT, TEXT_length, 300, text);
  status = *STATUS;
  msgOut(param, text, &status);
  *STATUS = status;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<msg_out_>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x160,%rsp
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %ecx,%esi
  mov    %r8d,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x158(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r13
  mov    %r13,%rcx
  mov    $0xf,%edx
  call   <cnfImpn@plt>
  lea    0x20(%rsp),%r14
  mov    %r14,%rcx
  mov    $0x12c,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <cnfImpn@plt>
  mov    (%rbx),%eax
  mov    %eax,0xc(%rsp)
  lea    0xc(%rsp),%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <msgOut@plt>
  mov    0xc(%rsp),%eax
  mov    %eax,(%rbx)
  mov    0x158(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x160,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void msg_out_(char *const PARAM, char *const TEXT, int *const STATUS, int PARAM_length, int TEXT_length) {
  char param[15 + 1];
  char text[300 + 1];
  int status;
  cnfImpn(PARAM, PARAM_length, 15, param);
  cnfImpn(TEXT, TEXT_length, 300, text);
  status = *STATUS;
  msgOut(param, text, &status);
  *STATUS = status;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<msg_out_>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %r8d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  mov    %ecx,%esi
  push   %rbx
  mov    %rdx,%rbx
  mov    $0xf,%edx
  sub    $0x160,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x158(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r13
  lea    0x20(%rsp),%r14
  mov    %r13,%rcx
  call   <cnfImpn@plt>
  mov    %r14,%rcx
  mov    $0x12c,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <cnfImpn@plt>
  mov    (%rbx),%eax
  lea    0xc(%rsp),%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    %eax,0xc(%rsp)
  call   <msgOut@plt>
  mov    0xc(%rsp),%eax
  mov    %eax,(%rbx)
  mov    0x158(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x160,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void msg_out_(char *const PARAM, char *const TEXT, int *const STATUS, int PARAM_length, int TEXT_length) {
  char param[15 + 1];
  char text[300 + 1];
  int status;
  cnfImpn(PARAM, PARAM_length, 15, param);
  cnfImpn(TEXT, TEXT_length, 300, text);
  status = *STATUS;
  msgOut(param, text, &status);
  *STATUS = status;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<msg_out_>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %r8d,%r12d
  push   %rbp
  mov    %rsi,%rbp
  mov    %ecx,%esi
  push   %rbx
  mov    %rdx,%rbx
  mov    $0xf,%edx
  sub    $0x160,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x158(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r13
  lea    0x20(%rsp),%r14
  mov    %r13,%rcx
  call   <cnfImpn@plt>
  mov    %r14,%rcx
  mov    $0x12c,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <cnfImpn@plt>
  mov    (%rbx),%eax
  lea    0xc(%rsp),%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    %eax,0xc(%rsp)
  call   <msgOut@plt>
  mov    0xc(%rsp),%eax
  mov    %eax,(%rbx)
  mov    0x158(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x160,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void msg_out_(char *const PARAM, char *const TEXT, int *const STATUS, int PARAM_length, int TEXT_length) {
  char param[15 + 1];
  char text[300 + 1];
  int status;
  cnfImpn(PARAM, PARAM_length, 15, param);
  cnfImpn(TEXT, TEXT_length, 300, text);
  status = *STATUS;
  msgOut(param, text, &status);
  *STATUS = status;
}
```
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   $0x6f6f66,-0x18(%rbp)
  movabs $0x2072616220726162,%rax
  mov    %rax,-0x14(%rbp)
  movl   $0x726162,-0xc(%rbp)
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D71(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap_str_arr@plt>
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D71(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D27
datatype: string
name: disassemble_data
arguments: label: D50
datatype: string
name: disassemble_data
arguments: label: D40
datatype: string
name: disassemble_data
arguments: label: D71
datatype: string
name: disassemble_data
arguments: label: D20
datatype: string

tool: D27: Before swap:

tool: D50: ============

tool: D40: s1: %s


tool: D71: s2: %s


tool: D20: 
After swap:

assistant: ```c
int main() {
  char s1[] = "foo";
  char s2[] = "bar bar bar";
  printf("Before swap:\n");
  printf("============\n");
  printf("s1: %s\n", s1);
  printf("s2: %s\n", s2);
  swap_str_arr(s1, s2);
  printf("\nAfter swap:\n");
  printf("============\n");
  printf("s1: %s\n", s1);
  printf("s2: %s\n", s2);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x6f6f66,0x8(%rsp)
  movabs $0x2072616220726162,%rax
  mov    %rax,0xc(%rsp)
  movl   $0x726162,0x14(%rsp)
  lea    D25(%rip),%rdi
  call   <puts@plt>
  lea    D31(%rip),%r14
  mov    %r14,%rdi
  call   <puts@plt>
  lea    0x8(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D40(%rip),%r13
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D42(%rip),%r12
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <swap_str_arr@plt>
  lea    D15(%rip),%rdi
  call   <puts@plt>
  mov    %r14,%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D25
type_guessed: string
name: read_data
arguments: tag: D31
type_guessed: string
name: read_data
arguments: tag: D40
type_guessed: string
name: read_data
arguments: tag: D42
type_guessed: string
name: read_data
arguments: tag: D15
type_guessed: string

tool: D25: Before swap:

tool: D31: ============

tool: D40: s1: %s


tool: D42: s2: %s


tool: D15: 
After swap:

assistant: ```c
int main() {
  char s1[] = "foo";
  char s2[] = "bar bar bar";
  printf("Before swap:\n");
  printf("============\n");
  printf("s1: %s\n", s1);
  printf("s2: %s\n", s2);
  swap_str_arr(s1, s2);
  printf("\nAfter swap:\n");
  printf("============\n");
  printf("s1: %s\n", s1);
  printf("s2: %s\n", s2);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  lea    D41(%rip),%rdi
  lea    D36(%rip),%r15
  push   %r14
  lea    D44(%rip),%r14
  push   %r13
  lea    D15(%rip),%r13
  push   %r12
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%r12
  movl   $0x6f6f66,0x8(%rsp)
  movabs $0x2072616220726162,%rax
  mov    %rax,0xc(%rsp)
  lea    0xc(%rsp),%rbp
  movl   $0x726162,0x14(%rsp)
  call   <puts@plt>
  mov    %r15,%rdi
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <swap_str_arr@plt>
  lea    D33(%rip),%rdi
  call   <puts@plt>
  mov    %r15,%rdi
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D41
type: string
name: parse_data
arguments: data_label: D36
type: string
name: parse_data
arguments: data_label: D44
type: string
name: parse_data
arguments: data_label: D15
type: string
name: parse_data
arguments: data_label: D33
type: string

tool: D41: Before swap:

tool: D36: ============

tool: D44: s1: %s


tool: D15: s2: %s


tool: D33: 
After swap:

assistant: ```c
int main() {
  char s1[] = "foo";
  char s2[] = "bar bar bar";
  printf("Before swap:\n");
  printf("============\n");
  printf("s1: %s\n", s1);
  printf("s2: %s\n", s2);
  swap_str_arr(s1, s2);
  printf("\nAfter swap:\n");
  printf("============\n");
  printf("s1: %s\n", s1);
  printf("s2: %s\n", s2);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D5(%rip),%rdi
  lea    D18(%rip),%r15
  push   %r14
  lea    D20(%rip),%r14
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%r12
  movl   $0x6f6f66,0x8(%rsp)
  movabs $0x2072616220726162,%rax
  mov    %rax,0xc(%rsp)
  lea    0xc(%rsp),%rbp
  movl   $0x726162,0x14(%rsp)
  call   <puts@plt>
  mov    %r15,%rdi
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <swap_str_arr@plt>
  lea    D48(%rip),%rdi
  call   <puts@plt>
  mov    %r15,%rdi
  call   <puts@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D5
type_guessed: string
name: disassemble_data
arguments: label: D18
type_guessed: string
name: disassemble_data
arguments: label: D20
type_guessed: string
name: disassemble_data
arguments: label: D29
type_guessed: string
name: disassemble_data
arguments: label: D48
type_guessed: string

tool: D5: Before swap:

tool: D18: ============

tool: D20: s1: %s


tool: D29: s2: %s


tool: D48: 
After swap:

assistant: ```c
int main() {
  char s1[] = "foo";
  char s2[] = "bar bar bar";
  printf("Before swap:\n");
  printf("============\n");
  printf("s1: %s\n", s1);
  printf("s2: %s\n", s2);
  swap_str_arr(s1, s2);
  printf("\nAfter swap:\n");
  printf("============\n");
  printf("s1: %s\n", s1);
  printf("s2: %s\n", s2);
  return 0;
}
```