ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbx
  cmp    $0x3,%edi
  je     L3
  xor    %eax,%eax
  call   <usage@plt>
  mov    $0x1,%eax
  pop    %rbx
  ret
L3:
  lea    D86(%rip),%rdi
  mov    %rsi,%rbx
  call   <printf@plt>
  lea    D70(%rip),%rdi
  call   <printf@plt>
  lea    D77(%rip),%rdi
  call   <printf@plt>
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    D40(%rip),%rdi
  call   <printf@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D63(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rbx),%rdi
  lea    0x10(%rbx),%rsi
  call   <directory_buster@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: label: D86
type_guessed: string
name: disassemble_data
arguments: label: D70
type_guessed: string
name: disassemble_data
arguments: label: D77
type_guessed: string
name: disassemble_data
arguments: label: D36
type_guessed: string
name: disassemble_data
arguments: label: D40
type_guessed: string
name: disassemble_data
arguments: label: D35
type_guessed: string
name: disassemble_data
arguments: label: D27
type_guessed: string
name: disassemble_data
arguments: label: D21
type_guessed: string
name: disassemble_data
arguments: label: D63
type_guessed: string

tool: 

tool: D70:         __         .__          


tool: D77: _____  |  | ____ __|  | _____   


tool: D36: \__  \ |  |/ /  |  \  | \__  \  


tool: D40:  / __ \|    <|  |  /  |__/ __ \_


tool: D35: (____  /__|_ \____/|____(____  /


tool: D27:      \/     \/               \/ 


tool: D21:   MIT/X licensed HTTP Fuzzer


tool: D63:   copyright 2015 Stefan Edwards/lojikil


assistant: ```c
int main(int ac, char **al, char **el) {
  if (ac != 3) {
    usage();
    return 1;
  }
  printf("");
  printf("        __         .__          \n");
  printf("_____  |  | ____ __|  | _____   \n");
  printf("\\__  \\ |  |/ /  |  \\  | \\__  \\  \n");
  printf(" / __ \\|    <|  |  /  |__/ __ \\_\n");
  printf("(____  /__|_ \\____/|____(____  /\n");
  printf("     \\/     \\/               \\/ \n");
  printf("  MIT/X licensed HTTP Fuzzer\n");
  printf("  copyright 2015 Stefan Edwards/lojikil\n");
  directory_buster(al[1], &al[2]);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<replay_ansic_grammar_with_dparser>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <replay_malloc@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    D29(%rip),%r8
  lea    D13(%rip),%rdx
  mov    %rdx,%rcx
  lea    D23(%rip),%rdx
  mov    %rax,%rsi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <_assert_equal@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D24
type: string
name: retrieve_data
arguments: data_tag: D29
type: string
name: retrieve_data
arguments: data_tag: D13
type: string
name: retrieve_data
arguments: data_tag: D23
type: string

tool: D24: /tmp/trace.txt

tool: D29: replay_ansic_grammar_with_dparser

tool: D13: result

tool: D23: 0

assistant: ```c
void replay_ansic_grammar_with_dparser() {
  int result = replay_malloc("/tmp/trace.txt");
  _assert_equal((unsigned long)0, (unsigned long)result, "0", "result", __func__);
}
```
user: Translate the below assembly code into C programming language:

```asm
<replay_ansic_grammar_with_dparser>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <replay_malloc@plt>
  movslq %eax,%rsi
  lea    D17(%rip),%r8
  lea    D36(%rip),%rcx
  lea    D23(%rip),%rdx
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <_assert_equal@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D9
datatype: string
name: unpack_data
arguments: tag: D17
datatype: string
name: unpack_data
arguments: tag: D36
datatype: string
name: unpack_data
arguments: tag: D23
datatype: string

tool: D9: /tmp/trace.txt

tool: D17: replay_ansic_grammar_with_dparser

tool: D36: result

tool: D23: 0

assistant: ```c
void replay_ansic_grammar_with_dparser() {
  int result = replay_malloc("/tmp/trace.txt");
  _assert_equal((unsigned long)0, (unsigned long)result, "0", "result", __func__);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<replay_ansic_grammar_with_dparser>:
  endbr64
  sub    $0x8,%rsp
  lea    D36(%rip),%rdi
  xor    %eax,%eax
  call   <replay_malloc@plt>
  lea    D8(%rip),%r8
  lea    D22(%rip),%rcx
  xor    %edi,%edi
  movslq %eax,%rsi
  lea    D6(%rip),%rdx
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <_assert_equal@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D36
data_type: string
name: parse_data
arguments: data_tag: D8
data_type: string
name: parse_data
arguments: data_tag: D22
data_type: string
name: parse_data
arguments: data_tag: D6
data_type: string

tool: D36: /tmp/trace.txt

tool: D8: replay_ansic_grammar_with_dparser

tool: D22: result

tool: D6: 0

assistant: ```c
void replay_ansic_grammar_with_dparser() {
  int result = replay_malloc("/tmp/trace.txt");
  _assert_equal((unsigned long)0, (unsigned long)result, "0", "result", __func__);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<replay_ansic_grammar_with_dparser>:
  endbr64
  sub    $0x8,%rsp
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <replay_malloc@plt>
  lea    D27(%rip),%r8
  lea    D35(%rip),%rcx
  xor    %edi,%edi
  movslq %eax,%rsi
  lea    D3(%rip),%rdx
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <_assert_equal@plt>
```

tool_calls: name: disassemble_data
arguments: label: D20
data_type: string
name: disassemble_data
arguments: label: D27
data_type: string
name: disassemble_data
arguments: label: D35
data_type: string
name: disassemble_data
arguments: label: D3
data_type: string

tool: D20: /tmp/trace.txt

tool: D27: replay_ansic_grammar_with_dparser

tool: D35: result

tool: D3: 0

assistant: ```c
void replay_ansic_grammar_with_dparser() {
  int result = replay_malloc("/tmp/trace.txt");
  _assert_equal((unsigned long)0, (unsigned long)result, "0", "result", __func__);
}
```
user: What is the c source code of the assembly code below:

```asm
<my_itoa_char_char_size>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x18(%rbp)
  mov    %al,-0x1c(%rbp)
  movzbl -0x1c(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <my_itoa_buffer_size@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  movzbl -0x18(%rbp),%eax
  cmp    %eax,-0x4(%rbp)
  jge    L16
  mov    -0x4(%rbp),%eax
  sub    %al,-0x18(%rbp)
  jmp    L3
L16:
  movb   $0x0,-0x18(%rbp)
L3:
  movzbl -0x18(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned int my_itoa_char_char_size(unsigned int number, unsigned char nbc, unsigned char base) {
  int len;
  if ((len = (my_itoa_buffer_size(number, base) - 1)) < nbc)
    nbc = nbc - len;
  else
    nbc = 0;
  return nbc;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<my_itoa_char_char_size>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  movzbl %dl,%esi
  call   <my_itoa_buffer_size@plt>
  sub    $0x1,%eax
  movzbl %bl,%edx
  sub    %eax,%ebx
  cmp    %eax,%edx
  mov    $0x0,%eax
  cmovg  %ebx,%eax
  movzbl %al,%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned int my_itoa_char_char_size(unsigned int number, unsigned char nbc, unsigned char base) {
  int len;
  if ((len = (my_itoa_buffer_size(number, base) - 1)) < nbc)
    nbc = nbc - len;
  else
    nbc = 0;
  return nbc;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_itoa_char_char_size>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  movzbl %dl,%esi
  call   <my_itoa_buffer_size@plt>
  movzbl %bl,%ecx
  lea    -0x1(%rax),%edx
  sub    %edx,%ebx
  cmp    %edx,%ecx
  mov    $0x0,%edx
  movzbl %bl,%eax
  pop    %rbx
  cmovle %edx,%eax
  ret
```

assistant: ```c
unsigned int my_itoa_char_char_size(unsigned int number, unsigned char nbc, unsigned char base) {
  int len;
  if ((len = (my_itoa_buffer_size(number, base) - 1)) < nbc)
    nbc = nbc - len;
  else
    nbc = 0;
  return nbc;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<my_itoa_char_char_size>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  movzbl %dl,%esi
  call   <my_itoa_buffer_size@plt>
  movzbl %bl,%ecx
  lea    -0x1(%rax),%edx
  sub    %edx,%ebx
  cmp    %edx,%ecx
  mov    $0x0,%edx
  movzbl %bl,%eax
  pop    %rbx
  cmovle %edx,%eax
  ret
```

assistant: ```c
unsigned int my_itoa_char_char_size(unsigned int number, unsigned char nbc, unsigned char base) {
  int len;
  if ((len = (my_itoa_buffer_size(number, base) - 1)) < nbc)
    nbc = nbc - len;
  else
    nbc = 0;
  return nbc;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<toupper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x60,-0x4(%rbp)
  jle    L1
  cmpl   $0x7a,-0x4(%rbp)
  jg     L1
  mov    -0x4(%rbp),%eax
  sub    $0x20,%eax
  jmp    L11
L1:
  mov    -0x4(%rbp),%eax
L11:
  pop    %rbp
  ret
```

assistant: ```c
int toupper(int c) {
  if (c > 96 && c < 123) {
    return c - 32;
  }
  return c;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<toupper>:
  endbr64
  mov    %edi,%eax
  lea    -0x61(%rdi),%ecx
  lea    -0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int toupper(int c) {
  if (c > 96 && c < 123) {
    return c - 32;
  }
  return c;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<toupper>:
  endbr64
  lea    -0x61(%rdi),%ecx
  mov    %edi,%eax
  lea    -0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int toupper(int c) {
  if (c > 96 && c < 123) {
    return c - 32;
  }
  return c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<toupper>:
  endbr64
  lea    -0x61(%rdi),%ecx
  mov    %edi,%eax
  lea    -0x20(%rdi),%edx
  cmp    $0x1a,%ecx
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int toupper(int c) {
  if (c > 96 && c < 123) {
    return c - 32;
  }
  return c;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_quick_first>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x140,%rsp
  mov    %rdi,-0x138(%rbp)
  mov    %esi,-0x13c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x124(%rbp)
  movl   $0x0,-0x120(%rbp)
  jmp    L81
L96:
  cmpl   $0x40,-0x120(%rbp)
  jne    L34
  movl   $0x0,-0x120(%rbp)
  mov    -0x120(%rbp),%eax
  cltq
  mov    -0x110(%rbp,%rax,4),%eax
  mov    %eax,-0x13c(%rbp)
L34:
  mov    -0x124(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x138(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x118(%rbp)
  mov    -0x120(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x120(%rbp)
  cltq
  mov    -0x13c(%rbp),%edx
  mov    %edx,-0x110(%rbp,%rax,4)
  mov    -0x124(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x11c(%rbp)
L23:
  nop
L84:
  addl   $0x1,-0x11c(%rbp)
  mov    -0x11c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x138(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x118(%rbp)
  jg     L84
  nop
L5:
  subl   $0x1,-0x13c(%rbp)
  mov    -0x13c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x138(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x118(%rbp)
  jl     L5
  mov    -0x11c(%rbp),%eax
  cmp    -0x13c(%rbp),%eax
  jge    L79
  mov    -0x11c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x138(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x114(%rbp)
  mov    -0x13c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x138(%rbp),%rax
  add    %rdx,%rax
  mov    -0x11c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x138(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x13c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x138(%rbp),%rax
  add    %rax,%rdx
  mov    -0x114(%rbp),%eax
  mov    %eax,(%rdx)
  jmp    L23
L79:
  nop
  addl   $0x1,-0x13c(%rbp)
L81:
  mov    -0x124(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x13c(%rbp)
  jg     L96
  cmpl   $0x0,-0x120(%rbp)
  je     L74
  mov    -0x13c(%rbp),%eax
  mov    %eax,-0x124(%rbp)
  subl   $0x1,-0x120(%rbp)
  mov    -0x120(%rbp),%eax
  cltq
  mov    -0x110(%rbp,%rax,4),%eax
  mov    %eax,-0x13c(%rbp)
  jmp    L81
L74:
  nop
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void _quick_first(int *array, int len) {
  int left = 0;
  int stack[64];
  int pos = 0;
  while (1) {
    for (; left + 1 < len; len++) {
      if (pos == 64)
        len = stack[pos = 0];
      int pivot = array[left];
      stack[pos++] = len;
      int right;
      for (right = left - 1;;) {
        while (array[++right] < pivot)
          ;
        while (pivot < array[--len])
          ;
        if (right >= len)
          break;
        int temp = array[right];
        array[right] = array[len];
        array[len] = temp;
      }
    }
    if (pos == 0)
      break;
    left = len;
    len = stack[--pos];
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_quick_first>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %rdi,%r11
  mov    %esi,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  mov    $0x0,%r12d
L81:
  lea    0x1(%rbp),%eax
  cmp    %edi,%eax
  jl     L45
  test   %ebx,%ebx
  je     L11
  sub    $0x1,%ebx
  movslq %ebx,%rax
  mov    %edi,%ebp
  mov    (%rsp,%rax,4),%edi
  jmp    L81
L45:
  mov    %ebx,%eax
  cmp    $0x40,%ebx
  je     L14
L36:
  movslq %ebp,%r9
  mov    (%r11,%r9,4),%esi
  lea    0x1(%rax),%ebx
  cltq
  mov    %edi,(%rsp,%rax,4)
  mov    %edi,%edx
  jmp    L87
L14:
  mov    (%rsp),%edi
  mov    %r12d,%eax
  jmp    L36
L51:
  add    $0x1,%r9
L87:
  mov    (%r11,%r9,4),%r10d
  cmp    %esi,%r10d
  jl     L51
  movslq %edx,%rax
  lea    -0x4(%r11,%rax,4),%rax
L105:
  mov    %edx,%edi
  sub    $0x1,%edx
  mov    %rax,%r8
  sub    $0x4,%rax
  mov    0x4(%rax),%ecx
  cmp    %esi,%ecx
  jg     L105
  cmp    %r9d,%edx
  jle    L81
  mov    %ecx,(%r11,%r9,4)
  mov    %r10d,(%r8)
  jmp    L51
L11:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L35:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void _quick_first(int *array, int len) {
  int left = 0;
  int stack[64];
  int pos = 0;
  while (1) {
    for (; left + 1 < len; len++) {
      if (pos == 64)
        len = stack[pos = 0];
      int pivot = array[left];
      stack[pos++] = len;
      int right;
      for (right = left - 1;;) {
        while (array[++right] < pivot)
          ;
        while (pivot < array[--len])
          ;
        if (right >= len)
          break;
        int temp = array[right];
        array[right] = array[len];
        array[len] = temp;
      }
    }
    if (pos == 0)
      break;
    left = len;
    len = stack[--pos];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_quick_first>:
  endbr64
  push   %r12
  mov    %rdi,%r11
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
L25:
  cmp    %esi,%ebp
  jl     L7
L94:
  test   %ebx,%ebx
  je     L67
  sub    $0x1,%ebx
  mov    %esi,%r12d
  movslq %ebx,%rax
  lea    0x1(%r12),%ebp
  mov    (%rsp,%rax,4),%esi
  cmp    %esi,%ebp
  jge    L94
L7:
  cmp    $0x40,%ebx
  je     L78
  movslq %ebx,%rax
  mov    %esi,%edx
  add    $0x1,%ebx
L71:
  movslq %r12d,%r9
  mov    %edx,(%rsp,%rax,4)
  mov    (%r11,%r9,4),%edi
  mov    %edi,%r10d
  cs nopw 0x0(%rax,%rax,1)
L84:
  cmp    %r10d,%edi
  jg     L64
  movslq %edx,%rax
  lea    -0x4(%r11,%rax,4),%rax
  nopl   (%rax)
L69:
  mov    %rax,%r8
  mov    (%rax),%ecx
  mov    %edx,%esi
  sub    $0x4,%rax
  sub    $0x1,%edx
  cmp    %edi,%ecx
  jg     L69
  cmp    %r9d,%edx
  jle    L25
  mov    %ecx,(%r11,%r9,4)
  mov    %r10d,(%r8)
L64:
  mov    0x4(%r11,%r9,4),%r10d
  add    $0x1,%r9
  jmp    L84
L78:
  mov    (%rsp),%edx
  mov    $0x1,%ebx
  xor    %eax,%eax
  jmp    L71
L67:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void _quick_first(int *array, int len) {
  int left = 0;
  int stack[64];
  int pos = 0;
  while (1) {
    for (; left + 1 < len; len++) {
      if (pos == 64)
        len = stack[pos = 0];
      int pivot = array[left];
      stack[pos++] = len;
      int right;
      for (right = left - 1;;) {
        while (array[++right] < pivot)
          ;
        while (pivot < array[--len])
          ;
        if (right >= len)
          break;
        int temp = array[right];
        array[right] = array[len];
        array[len] = temp;
      }
    }
    if (pos == 0)
      break;
    left = len;
    len = stack[--pos];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_quick_first>:
  endbr64
  push   %r12
  mov    %rdi,%r11
  xor    %r12d,%r12d
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
L45:
  cmp    %esi,%ebp
  jl     L82
L8:
  test   %ebx,%ebx
  je     L17
  sub    $0x1,%ebx
  mov    %esi,%r12d
  movslq %ebx,%rax
  lea    0x1(%r12),%ebp
  mov    (%rsp,%rax,4),%esi
  cmp    %esi,%ebp
  jge    L8
L82:
  cmp    $0x40,%ebx
  je     L30
  movslq %ebx,%rax
  mov    %esi,%edx
  add    $0x1,%ebx
L71:
  movslq %r12d,%r9
  mov    %edx,(%rsp,%rax,4)
  mov    (%r11,%r9,4),%edi
  nopl   0x0(%rax,%rax,1)
L85:
  mov    (%r11,%r9,4),%r10d
  cmp    %edi,%r10d
  jl     L39
  movslq %edx,%rax
  lea    -0x4(%r11,%rax,4),%rax
  nopl   0x0(%rax)
L27:
  mov    %rax,%r8
  mov    (%rax),%ecx
  mov    %edx,%esi
  sub    $0x4,%rax
  sub    $0x1,%edx
  cmp    %edi,%ecx
  jg     L27
  cmp    %r9d,%edx
  jle    L45
  mov    %ecx,(%r11,%r9,4)
  mov    %r10d,(%r8)
L39:
  add    $0x1,%r9
  jmp    L85
L30:
  mov    (%rsp),%edx
  mov    $0x1,%ebx
  xor    %eax,%eax
  jmp    L71
L17:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L52:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void _quick_first(int *array, int len) {
  int left = 0;
  int stack[64];
  int pos = 0;
  while (1) {
    for (; left + 1 < len; len++) {
      if (pos == 64)
        len = stack[pos = 0];
      int pivot = array[left];
      stack[pos++] = len;
      int right;
      for (right = left - 1;;) {
        while (array[++right] < pivot)
          ;
        while (pivot < array[--len])
          ;
        if (right >= len)
          break;
        int temp = array[right];
        array[right] = array[len];
        array[len] = temp;
      }
    }
    if (pos == 0)
      break;
    left = len;
    len = stack[--pos];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<msd>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x218,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  lea    D35(%rip),%r14
  lea    D27(%rip),%r13
  lea    D8(%rip),%r12
  lea    D37(%rip),%r15
L7:
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %rsp,%rbx
  mov    $0x200,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x4,%edx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  movl   $0x12330984,0x4(%rsp)
  movb   $0x24,0x8(%rsp)
  movb   $0x80,0xc(%rsp)
  movb   $0x6,0xe(%rsp)
  movb   $0x12,0xf(%rsp)
  movb   $0x24,0x13(%rsp)
  mov    %r12,%rdx
  mov    $0x1f,%esi
  mov    %rbx,%rdi
  call   <hex@plt>
  mov    $0x1f,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <send_bulk@plt>
  test   %rax,%rax
  js     L21
  mov    $0x200,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x24,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <recv_bulk@plt>
  mov    %eax,%esi
  mov    %r15,%rdx
  mov    %rbx,%rdi
  call   <hex@plt>
  mov    $0x200,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x24,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <recv_bulk@plt>
  mov    %eax,%esi
  lea    D31(%rip),%rdx
  mov    %rbx,%rdi
  call   <hex@plt>
  mov    $0x249f0,%edi
  call   <usleep@plt>
  add    $0x1,%ebp
  cmp    $0x3,%ebp
  jne    L7
L21:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x218,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D35
type_guessed: string
name: retrieve_data
arguments: tag: D27
type_guessed: string
name: retrieve_data
arguments: tag: D8
type_guessed: string
name: retrieve_data
arguments: tag: D37
type_guessed: string
name: retrieve_data
arguments: tag: D31
type_guessed: string

tool: D35: i=%d


tool: D27: USBC

tool: D8: intquery req

tool: D37: inquiry result

tool: D31: inquiry sense?

assistant: ```c
void msd() {
  unsigned InEndPt = 1;
  unsigned OutEndPt = 1;
  unsigned char abBuf[512];
  int i;
  for (i = 0; i < 3; i++) {
    printf("i=%d\n", i);
    memset(abBuf, 0, sizeof(abBuf));
    memcpy(abBuf, "USBC", 4);
    *(uint32_t *)(&abBuf[4]) = 0x12330984;
    abBuf[8] = 0x24;
    abBuf[0xc] = 0x80;
    abBuf[0xe] = 0x06;
    abBuf[0x0f] = 0x12;
    abBuf[0x13] = 0x24;
    hex(abBuf, 31, "intquery req");
    if (send_bulk(OutEndPt, abBuf, 31) < 0)
      return;
    memset(abBuf, 0, sizeof(abBuf));
    int cb = recv_bulk(InEndPt, abBuf, 36);
    hex(abBuf, cb, "inquiry result");
    memset(abBuf, 0, sizeof(abBuf));
    cb = recv_bulk(InEndPt, abBuf, 36);
    hex(abBuf, cb, "inquiry sense?");
    usleep(150000);
  }
}
```