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
<_scan_pattern>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  mov    $0x0,%ebx
  lea    0x8(%rsp),%rbp
L26:
  mov    %rbp,%rsi
  call   <_copy_string@plt>
  mov    %rax,%rdi
  cmpb   $0x0,(%rax)
  je     L27
  add    $0x1,%ebx
  mov    %rbp,%rsi
  call   <_copy_string@plt>
  mov    %rax,%rdi
  cmpb   $0x0,(%rax)
  jne    L26
  jmp    L7
L27:
  mov    $0x0,%ebx
L7:
  mov    %ebx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int _scan_pattern(const char *format, char *temp) {
  int n = 0;
  for (;;) {
    format = _copy_string(format, &temp);
    if (format[0] == '\0')
      return 0;
    ++n;
    format = _copy_string(format, &temp);
    if (format[0] == '\0')
      return n;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_scan_pattern>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x8(%rsp),%rbx
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %rbx,%rsi
  add    $0x1,%r12d
  call   <_copy_string@plt>
  cmpb   $0x0,(%rax)
  mov    %rax,%rdi
  je     L9
L18:
  mov    %rbx,%rsi
  call   <_copy_string@plt>
  cmpb   $0x0,(%rax)
  mov    %rax,%rdi
  jne    L20
  xor    %r12d,%r12d
L9:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
int _scan_pattern(const char *format, char *temp) {
  int n = 0;
  for (;;) {
    format = _copy_string(format, &temp);
    if (format[0] == '\0')
      return 0;
    ++n;
    format = _copy_string(format, &temp);
    if (format[0] == '\0')
      return n;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_scan_pattern>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,0x8(%rsp)
  lea    0x8(%rsp),%rbx
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L19:
  mov    %rbx,%rsi
  add    $0x1,%r12d
  call   <_copy_string@plt>
  cmpb   $0x0,(%rax)
  mov    %rax,%rdi
  je     L28
L0:
  mov    %rbx,%rsi
  call   <_copy_string@plt>
  cmpb   $0x0,(%rax)
  mov    %rax,%rdi
  jne    L19
  xor    %r12d,%r12d
L28:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
int _scan_pattern(const char *format, char *temp) {
  int n = 0;
  for (;;) {
    format = _copy_string(format, &temp);
    if (format[0] == '\0')
      return 0;
    ++n;
    format = _copy_string(format, &temp);
    if (format[0] == '\0')
      return n;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<JFT_count_bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int JFT_count_bits(uint64_t v) {
  v = (v - ((v >> 1) & (uint64_t) ~(uint64_t)0 / 3));
  v = (v & (uint64_t) ~(uint64_t)0 / 15 * 3) + ((v >> 2) & (uint64_t) ~(uint64_t)0 / 15 * 3);
  v = (v + (v >> 4)) & (uint64_t) ~(uint64_t)0 / 255 * 15;
  v = (v * ((uint64_t) ~(uint64_t)0 / 255)) >> (sizeof(uint64_t) - 1) * 8;
  return v;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<JFT_count_bits>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int JFT_count_bits(uint64_t v) {
  v = (v - ((v >> 1) & (uint64_t) ~(uint64_t)0 / 3));
  v = (v & (uint64_t) ~(uint64_t)0 / 15 * 3) + ((v >> 2) & (uint64_t) ~(uint64_t)0 / 15 * 3);
  v = (v + (v >> 4)) & (uint64_t) ~(uint64_t)0 / 255 * 15;
  v = (v * ((uint64_t) ~(uint64_t)0 / 255)) >> (sizeof(uint64_t) - 1) * 8;
  return v;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<JFT_count_bits>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int JFT_count_bits(uint64_t v) {
  v = (v - ((v >> 1) & (uint64_t) ~(uint64_t)0 / 3));
  v = (v & (uint64_t) ~(uint64_t)0 / 15 * 3) + ((v >> 2) & (uint64_t) ~(uint64_t)0 / 15 * 3);
  v = (v + (v >> 4)) & (uint64_t) ~(uint64_t)0 / 255 * 15;
  v = (v * ((uint64_t) ~(uint64_t)0 / 255)) >> (sizeof(uint64_t) - 1) * 8;
  return v;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<JFT_count_bits>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int JFT_count_bits(uint64_t v) {
  v = (v - ((v >> 1) & (uint64_t) ~(uint64_t)0 / 3));
  v = (v & (uint64_t) ~(uint64_t)0 / 15 * 3) + ((v >> 2) & (uint64_t) ~(uint64_t)0 / 15 * 3);
  v = (v + (v >> 4)) & (uint64_t) ~(uint64_t)0 / 255 * 15;
  v = (v * ((uint64_t) ~(uint64_t)0 / 255)) >> (sizeof(uint64_t) - 1) * 8;
  return v;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<postyn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x1c,-0x14(%rbp)
  movl   $0x0,-0x1c(%rbp)
  lea    D15(%rip),%rax
  mov    %rax,-0x10(%rbp)
  lea    D13(%rip),%rax
  mov    %rax,-0x8(%rbp)
L76:
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x28(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x28(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x24(%rbp),%ecx
  mov    %eax,%esi
  mov    %ecx,%edi
  call   <io_printborder@plt>
  mov    -0x28(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x28(%rbp)
  mov    -0x30(%rbp),%rcx
  mov    -0x14(%rbp),%edx
  mov    -0x24(%rbp),%edi
  mov    %eax,%esi
  call   <io_printtext@plt>
  mov    %eax,-0x28(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L60
  mov    -0x10(%rbp),%rcx
  mov    -0x14(%rbp),%edx
  mov    -0x28(%rbp),%esi
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <io_printtext@plt>
  mov    %eax,-0x28(%rbp)
L60:
  cmpl   $0x1,-0x1c(%rbp)
  jne    L2
  mov    -0x8(%rbp),%rcx
  mov    -0x14(%rbp),%edx
  mov    -0x28(%rbp),%esi
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <io_printtext@plt>
  mov    %eax,-0x28(%rbp)
L2:
  mov    -0x28(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x28(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x24(%rbp),%ecx
  mov    %eax,%esi
  mov    %ecx,%edi
  call   <io_printborder@plt>
  mov    $0x0,%eax
  call   <io_getkey@plt>
  cmp    $0x4,%eax
  je     L82
  cmp    $0x4,%eax
  jg     L77
  test   %eax,%eax
  je     L23
  cmp    $0x1,%eax
  je     L31
  jmp    L77
L23:
  subl   $0x1,-0x1c(%rbp)
  jmp    L68
L31:
  addl   $0x1,-0x1c(%rbp)
  jmp    L68
L82:
  mov    -0x1c(%rbp),%eax
  jmp    L97
L77:
  nop
L68:
  cmpl   $0x0,-0x1c(%rbp)
  jns    L6
  movl   $0x1,-0x1c(%rbp)
L6:
  mov    -0x1c(%rbp),%eax
  cltd
  shr    $0x1f,%edx
  add    %edx,%eax
  and    $0x1,%eax
  sub    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L76
L97:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
data_type: string
name: extract_symbolic_data
arguments: tag: D13
data_type: string

tool: D15:  * No         Yes  

tool: D13:    No       * Yes  

assistant: ```c
int postyn(int x, int y, char *message) {
  int yorig = y;
  int width = 28;
  int cursor = 0;
  char *a = " * No         Yes  ";
  char *b = "   No       * Yes  ";
  while (1) {
    y = yorig;
    io_printborder(x, y++, width);
    y = io_printtext(x, y++, width, message);
    if (cursor == 0)
      y = io_printtext(x, y, width, a);
    if (cursor == 1)
      y = io_printtext(x, y, width, b);
    io_printborder(x, y++, width);
    switch (io_getkey()) {
    case 0:
      cursor--;
      break;
    case 1:
      cursor++;
      break;
    case 4:
      return cursor;
      break;
    default:
      break;
    }
    if (cursor < 0)
      cursor = 1;
    cursor = cursor % 2;
  }
  return cursor;
}
```
user: What is the c source code of the assembly code below?

```asm
<postyn>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%r12d
  mov    %rdx,%r13
  mov    $0x0,%ebx
  lea    0x1(%rsi),%r14d
  mov    $0x1,%r15d
  jmp    L108
L87:
  lea    D9(%rip),%rcx
  mov    $0x1c,%edx
  mov    %ebp,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
L5:
  mov    $0x1c,%edx
  mov    %ebp,%edi
  call   <io_printborder@plt>
  mov    $0x0,%eax
  call   <io_getkey@plt>
  cmp    $0x1,%eax
  je     L49
  cmp    $0x4,%eax
  je     L26
  test   %eax,%eax
  je     L10
L106:
  test   %ebx,%ebx
  cmovs  %r15d,%ebx
L55:
  mov    %ebx,%eax
  shr    $0x1f,%eax
  add    %eax,%ebx
  and    $0x1,%ebx
  sub    %eax,%ebx
L108:
  mov    $0x1c,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <io_printborder@plt>
  mov    %r13,%rcx
  mov    $0x1c,%edx
  mov    %r14d,%esi
  mov    %ebp,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
  test   %ebx,%ebx
  je     L87
  cmp    $0x1,%ebx
  jne    L5
  lea    D5(%rip),%rcx
  mov    $0x1c,%edx
  mov    %ebp,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
  mov    $0x1c,%edx
  mov    %ebp,%edi
  call   <io_printborder@plt>
  mov    $0x0,%eax
  call   <io_getkey@plt>
  cmp    $0x1,%eax
  je     L49
  cmp    $0x4,%eax
  je     L26
  test   %eax,%eax
  jne    L55
L10:
  sub    $0x1,%ebx
  jmp    L106
L49:
  add    $0x1,%ebx
  jmp    L106
L26:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
datatype: string
name: disassemble_data
arguments: tag: D5
datatype: string

tool: D9:  * No         Yes  

tool: D5:    No       * Yes  

assistant: ```c
int postyn(int x, int y, char *message) {
  int yorig = y;
  int width = 28;
  int cursor = 0;
  char *a = " * No         Yes  ";
  char *b = "   No       * Yes  ";
  while (1) {
    y = yorig;
    io_printborder(x, y++, width);
    y = io_printtext(x, y++, width, message);
    if (cursor == 0)
      y = io_printtext(x, y, width, a);
    if (cursor == 1)
      y = io_printtext(x, y, width, b);
    io_printborder(x, y++, width);
    switch (io_getkey()) {
    case 0:
      cursor--;
      break;
    case 1:
      cursor++;
      break;
    case 4:
      return cursor;
      break;
    default:
      break;
    }
    if (cursor < 0)
      cursor = 1;
    cursor = cursor % 2;
  }
  return cursor;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<postyn>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  lea    0x1(%rsi),%r14d
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L50:
  mov    %r13d,%edx
  sub    %r15d,%edx
  test   %eax,%eax
  cmove  %edx,%r15d
L44:
  mov    $0x1c,%edx
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <io_printborder@plt>
  mov    %r14d,%esi
  mov    %r12,%rcx
  mov    $0x1c,%edx
  mov    %ebx,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
  test   %r15d,%r15d
  je     L34
  lea    D7(%rip),%rcx
  mov    $0x1c,%edx
  mov    %ebx,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
L14:
  mov    $0x1c,%edx
  mov    %ebx,%edi
  call   <io_printborder@plt>
  xor    %eax,%eax
  call   <io_getkey@plt>
  cmp    $0x1,%eax
  je     L29
  cmp    $0x4,%eax
  jne    L50
  add    $0x8,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L34:
  lea    D17(%rip),%rcx
  mov    $0x1c,%edx
  mov    %ebx,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
  jmp    L14
  nopw   0x0(%rax,%rax,1)
L29:
  lea    0x1(%r15),%r8d
  and    $0x1,%r8d
  mov    %r8d,%r15d
  jmp    L44
```

tool_calls: name: parse_data
arguments: tag: D7
type_infered: string
name: parse_data
arguments: tag: D17
type_infered: string

tool: D7:    No       * Yes  

tool: D17:  * No         Yes  

assistant: ```c
int postyn(int x, int y, char *message) {
  int yorig = y;
  int width = 28;
  int cursor = 0;
  char *a = " * No         Yes  ";
  char *b = "   No       * Yes  ";
  while (1) {
    y = yorig;
    io_printborder(x, y++, width);
    y = io_printtext(x, y++, width, message);
    if (cursor == 0)
      y = io_printtext(x, y, width, a);
    if (cursor == 1)
      y = io_printtext(x, y, width, b);
    io_printborder(x, y++, width);
    switch (io_getkey()) {
    case 0:
      cursor--;
      break;
    case 1:
      cursor++;
      break;
    case 4:
      return cursor;
      break;
    default:
      break;
    }
    if (cursor < 0)
      cursor = 1;
    cursor = cursor % 2;
  }
  return cursor;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<postyn>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  lea    0x1(%rsi),%r14d
  push   %r13
  mov    $0x1,%r13d
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  jmp    L57
  nopl   0x0(%rax,%rax,1)
L36:
  mov    %r13d,%edx
  sub    %r15d,%edx
  test   %eax,%eax
  cmove  %edx,%r15d
L57:
  mov    $0x1c,%edx
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <io_printborder@plt>
  mov    %r14d,%esi
  mov    %r12,%rcx
  mov    $0x1c,%edx
  mov    %ebx,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
  test   %r15d,%r15d
  je     L35
  lea    D12(%rip),%rcx
  mov    $0x1c,%edx
  mov    %ebx,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
L41:
  mov    $0x1c,%edx
  mov    %ebx,%edi
  call   <io_printborder@plt>
  xor    %eax,%eax
  call   <io_getkey@plt>
  cmp    $0x1,%eax
  je     L52
  cmp    $0x4,%eax
  jne    L36
  add    $0x8,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L35:
  lea    D4(%rip),%rcx
  mov    $0x1c,%edx
  mov    %ebx,%edi
  call   <io_printtext@plt>
  mov    %eax,%esi
  jmp    L41
  nopw   0x0(%rax,%rax,1)
L52:
  lea    0x1(%r15),%r8d
  and    $0x1,%r8d
  mov    %r8d,%r15d
  jmp    L57
```

tool_calls: name: disassemble_data
arguments: tag: D12
data_type: string
name: disassemble_data
arguments: tag: D4
data_type: string

tool: D12:    No       * Yes  

tool: D4:  * No         Yes  

assistant: ```c
int postyn(int x, int y, char *message) {
  int yorig = y;
  int width = 28;
  int cursor = 0;
  char *a = " * No         Yes  ";
  char *b = "   No       * Yes  ";
  while (1) {
    y = yorig;
    io_printborder(x, y++, width);
    y = io_printtext(x, y++, width, message);
    if (cursor == 0)
      y = io_printtext(x, y, width, a);
    if (cursor == 1)
      y = io_printtext(x, y, width, b);
    io_printborder(x, y++, width);
    switch (io_getkey()) {
    case 0:
      cursor--;
      break;
    case 1:
      cursor++;
      break;
    case 4:
      return cursor;
      break;
    default:
      break;
    }
    if (cursor < 0)
      cursor = 1;
    cursor = cursor % 2;
  }
  return cursor;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<handle_msg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x610,%rsp
  mov    %rdi,-0x608(%rbp)
  mov    %esi,-0x60c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffffff,-0x5f4(%rbp)
  mov    -0x608(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x608(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x71,%eax
  je     L9
  cmp    $0x71,%eax
  jg     L60
  cmp    $0x61,%eax
  je     L32
  cmp    $0x6e,%eax
  je     L39
  jmp    L60
L9:
  mov    $0x1,%eax
  jmp    L36
L39:
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L60
L32:
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x608(%rbp),%rax
  mov    %rax,%rsi
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
L60:
  lea    -0x5f0(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  lea    -0x5f0(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x5f0(%rbp),%rsi
  mov    -0x60c(%rbp),%eax
  mov    $0x0,%ecx
  mov    %eax,%edi
  call   <send@plt>
  mov    %eax,-0x5f4(%rbp)
  mov    -0x5f4(%rbp),%eax
  mov    %eax,%esi
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
L36:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L38
  call   <__stack_chk_fail@plt>
L38:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D44
type_guessed: string
name: disassemble_data
arguments: data_tag: D34
type_guessed: string
name: disassemble_data
arguments: data_tag: D2
type_guessed: string
name: disassemble_data
arguments: data_tag: D53
type_guessed: string
name: disassemble_data
arguments: data_tag: D0
type_guessed: string
name: disassemble_data
arguments: data_tag: D48
type_guessed: string

tool: D44: first char is %c


tool: D34: Server ask for my name...

tool: D2: Server ask for action...

tool: D53: Received from server: %s


tool: D0: 


tool: D48: Sent %d bytes


assistant: ```c
int handle_msg(char request[1500], int sockfd) {
  char response[1500];
  int sent_bytes = -1;
  printf("first char is %c\n", request[0]);
  switch (request[0]) {
  case 'q':
    return 1;
  case 'n':
    printf("Server ask for my name...\n");
    break;
  case 'a':
    printf("Server ask for action...\n");
    printf("Received from server: %s\n", request);
    break;
  }
  strcat(response, "\r\n");
  sent_bytes = send(sockfd, response, strlen(response), 0);
  printf("Sent %d bytes\n", sent_bytes);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<handle_msg>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x5f8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x5e8(%rsp)
  xor    %eax,%eax
  movsbl (%rdi),%esi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  movzbl (%rbx),%eax
  cmp    $0x6e,%al
  je     L1
  mov    $0x1,%edx
  cmp    $0x71,%al
  je     L4
  cmp    $0x61,%al
  je     L6
L30:
  mov    %rsp,%rbx
  lea    D40(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    $0x0,%ecx
  mov    %rbx,%rsi
  mov    %ebp,%edi
  call   <send@plt>
  mov    %eax,%esi
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%edx
L4:
  mov    0x5e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  mov    %edx,%eax
  add    $0x5f8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  lea    D2(%rip),%rdi
  call   <puts@plt>
  jmp    L30
L6:
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    %rbx,%rsi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L30
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D19
data_type: string
name: read_data
arguments: tag: D40
data_type: string
name: read_data
arguments: tag: D27
data_type: string
name: read_data
arguments: tag: D2
data_type: string
name: read_data
arguments: tag: D0
data_type: string
name: read_data
arguments: tag: D11
data_type: string

tool: D19: first char is %c


tool: D40: 


tool: D27: Sent %d bytes


tool: D2: Server ask for my name...

tool: D0: Server ask for action...

tool: D11: Received from server: %s


assistant: ```c
int handle_msg(char request[1500], int sockfd) {
  char response[1500];
  int sent_bytes = -1;
  printf("first char is %c\n", request[0]);
  switch (request[0]) {
  case 'q':
    return 1;
  case 'n':
    printf("Server ask for my name...\n");
    break;
  case 'a':
    printf("Server ask for action...\n");
    printf("Received from server: %s\n", request);
    break;
  }
  strcat(response, "\r\n");
  sent_bytes = send(sockfd, response, strlen(response), 0);
  printf("Sent %d bytes\n", sent_bytes);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<handle_msg>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x5f8,%rsp
  movsbl (%rdi),%esi
  lea    D23(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x5e8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  movzbl 0x0(%rbp),%eax
  cmp    $0x6e,%al
  je     L35
  mov    $0x1,%r8d
  cmp    $0x71,%al
  je     L40
  cmp    $0x61,%al
  je     L32
L3:
  mov    %rsp,%rbp
  lea    D41(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %ecx,%ecx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <send@plt>
  lea    D57(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %r8d,%r8d
L40:
  mov    0x5e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x5f8,%rsp
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L35:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  jmp    L3
  xchg   %ax,%ax
L32:
  lea    D59(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  lea    D49(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L3
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D23
data_type: string
name: extract_symbolic_data
arguments: data_label: D41
data_type: string
name: extract_symbolic_data
arguments: data_label: D57
data_type: string
name: extract_symbolic_data
arguments: data_label: D19
data_type: string
name: extract_symbolic_data
arguments: data_label: D59
data_type: string
name: extract_symbolic_data
arguments: data_label: D49
data_type: string

tool: D23: first char is %c


tool: D41: 


tool: D57: Sent %d bytes


tool: D19: Server ask for my name...

tool: D59: Server ask for action...

tool: D49: Received from server: %s


assistant: ```c
int handle_msg(char request[1500], int sockfd) {
  char response[1500];
  int sent_bytes = -1;
  printf("first char is %c\n", request[0]);
  switch (request[0]) {
  case 'q':
    return 1;
  case 'n':
    printf("Server ask for my name...\n");
    break;
  case 'a':
    printf("Server ask for action...\n");
    printf("Received from server: %s\n", request);
    break;
  }
  strcat(response, "\r\n");
  sent_bytes = send(sockfd, response, strlen(response), 0);
  printf("Sent %d bytes\n", sent_bytes);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<handle_msg>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x5f8,%rsp
  movsbl (%rdi),%esi
  lea    D19(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x5e8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  movzbl 0x0(%rbp),%eax
  cmp    $0x6e,%al
  je     L42
  mov    $0x1,%r8d
  cmp    $0x71,%al
  je     L51
  cmp    $0x61,%al
  je     L12
L17:
  mov    %rsp,%rbp
  lea    D25(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcat@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %ecx,%ecx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  mov    %eax,%edx
  call   <send@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %r8d,%r8d
L51:
  mov    0x5e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x5f8,%rsp
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L42:
  lea    D23(%rip),%rdi
  call   <puts@plt>
  jmp    L17
  xchg   %ax,%ax
L12:
  lea    D42(%rip),%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L17
L32:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D19
type: string
name: disassemble_data
arguments: data_tag: D25
type: string
name: disassemble_data
arguments: data_tag: D0
type: string
name: disassemble_data
arguments: data_tag: D23
type: string
name: disassemble_data
arguments: data_tag: D42
type: string
name: disassemble_data
arguments: data_tag: D20
type: string

tool: D19: first char is %c


tool: D25: 


tool: D0: Sent %d bytes


tool: D23: Server ask for my name...

tool: D42: Server ask for action...

tool: D20: Received from server: %s


assistant: ```c
int handle_msg(char request[1500], int sockfd) {
  char response[1500];
  int sent_bytes = -1;
  printf("first char is %c\n", request[0]);
  switch (request[0]) {
  case 'q':
    return 1;
  case 'n':
    printf("Server ask for my name...\n");
    break;
  case 'a':
    printf("Server ask for action...\n");
    printf("Received from server: %s\n", request);
    break;
  }
  strcat(response, "\r\n");
  sent_bytes = send(sockfd, response, strlen(response), 0);
  printf("Sent %d bytes\n", sent_bytes);
  return 0;
}
```