ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
L15:
  mov    %eax,(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x9,%rax
  jne    L15
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  lea    D9(%rip),%rbp
L14:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L14
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  call   <__stack_chk_fail@plt>
```

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

tool: D9: %d 

assistant: ```c
int main(int argc, char **argv) {
  int i;
  int a[10];
  for (i = 0; i < 9; i++)
    a[i] = i;
  for (i = 0; i < 10; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  nop
L11:
  mov    %eax,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    $0x9,%rax
  jne    L11
  lea    0x28(%rsp),%r12
  lea    D8(%rip),%rbp
  nopl   0x0(%rax)
L4:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L4
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
datatype: string

tool: D8: %d 

assistant: ```c
int main(int argc, char **argv) {
  int i;
  int a[10];
  for (i = 0; i < 9; i++)
    a[i] = i;
  for (i = 0; i < 10; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  lea    D11(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D4(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x8,0x20(%rsp)
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D20(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  nopl   0x0(%rax)
L8:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D11
type_guessed: string
name: unpack_data
arguments: data_tag: D4
type_guessed: byte[16]
name: unpack_data
arguments: data_tag: D20
type_guessed: byte[16]

tool: D11: %d 

tool: {"D4": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D20": [0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, char **argv) {
  int i;
  int a[10];
  for (i = 0; i < 9; i++)
    a[i] = i;
  for (i = 0; i < 10; i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fistp_preg32_qword>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0xdf,%edi
  call   <put8@plt>
  mov    -0x4(%rbp),%eax
  add    $0x38,%eax
  mov    %eax,%edi
  call   <put8@plt>
  nop
  leave
  ret
```

assistant: ```c
void fistp_preg32_qword(int reg32) {
  put8(0xDF);
  put8(0x38 + reg32);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fistp_preg32_qword>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0xdf,%edi
  call   <put8@plt>
  lea    0x38(%rbx),%edi
  call   <put8@plt>
  pop    %rbx
  ret
```

assistant: ```c
void fistp_preg32_qword(int reg32) {
  put8(0xDF);
  put8(0x38 + reg32);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<fistp_preg32_qword>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0xdf,%edi
  call   <put8@plt>
  lea    0x38(%rbx),%edi
  pop    %rbx
  jmp    1050 <put8@plt>
```

assistant: ```c
void fistp_preg32_qword(int reg32) {
  put8(0xDF);
  put8(0x38 + reg32);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<insert_space_tabs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L38
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0x20,%edi
  call   <insert_char@plt>
  jmp    L24
L38:
  mov    -0x14(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L42
  mov    $0x1,%esi
  mov    $0x9,%edi
  call   <insert_char@plt>
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %eax,%esi
  mov    $0x9,%edi
  call   <insert_char@plt>
  mov    -0x4(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %edx,%eax
  mov    %eax,%esi
  mov    $0x20,%edi
  call   <insert_char@plt>
  jmp    L24
L42:
  mov    -0x14(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  cmp    %eax,%ecx
  jge    L7
  mov    -0x10(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %edx,%eax
  sub    %eax,-0x10(%rbp)
L7:
  mov    -0xc(%rbp),%eax
  sub    -0x10(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %eax,%esi
  mov    $0x9,%edi
  call   <insert_char@plt>
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %edx,%eax
  mov    %eax,%esi
  mov    $0x20,%edi
  call   <insert_char@plt>
L24:
  nop
  leave
  ret
```

assistant: ```c
void insert_space_tabs(int from, int to, int m, int n) {
  if (to < m)
    insert_char(' ', from - to);
  else if (from < m) {
    insert_char('\t', 1);
    int k = to - m;
    insert_char('\t', k / n);
    insert_char(' ', k % n);
  } else {
    int a = from - m;
    int b = to - m;
    if (a / n < b / n)
      a = a - (a % n);
    int k = b - a;
    insert_char('\t', k / n);
    insert_char(' ', k % n);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<insert_space_tabs>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  cmp    %edx,%esi
  jl     L47
  mov    %edx,%r12d
  mov    %ecx,%ebp
  cmp    %edi,%edx
  jg     L13
  sub    %edx,%edi
  sub    %edx,%ebx
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ebp
  cmp    %eax,%ecx
  jge    L17
  mov    %edi,%eax
  cltd
  idiv   %ebp
  sub    %edx,%edi
L17:
  sub    %edi,%ebx
  mov    %ebx,%eax
  cltd
  idiv   %ebp
  mov    %edx,%r12d
  mov    %eax,%esi
  mov    $0x9,%edi
  call   <insert_char@plt>
  mov    %r12d,%esi
  mov    $0x20,%edi
  call   <insert_char@plt>
L48:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L47:
  mov    %edi,%esi
  sub    %ebx,%esi
  mov    $0x20,%edi
  call   <insert_char@plt>
  jmp    L48
L13:
  mov    $0x1,%esi
  mov    $0x9,%edi
  call   <insert_char@plt>
  mov    %ebx,%eax
  sub    %r12d,%eax
  cltd
  idiv   %ebp
  mov    %edx,%ebx
  mov    %eax,%esi
  mov    $0x9,%edi
  call   <insert_char@plt>
  mov    %ebx,%esi
  mov    $0x20,%edi
  call   <insert_char@plt>
  jmp    L48
```

assistant: ```c
void insert_space_tabs(int from, int to, int m, int n) {
  if (to < m)
    insert_char(' ', from - to);
  else if (from < m) {
    insert_char('\t', 1);
    int k = to - m;
    insert_char('\t', k / n);
    insert_char(' ', k % n);
  } else {
    int a = from - m;
    int b = to - m;
    if (a / n < b / n)
      a = a - (a % n);
    int k = b - a;
    insert_char('\t', k / n);
    insert_char(' ', k % n);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<insert_space_tabs>:
  endbr64
  cmp    %edx,%esi
  jl     L4
  push   %r12
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    %edx,%ebx
  cmp    %edi,%edx
  jg     L17
  sub    %edx,%edi
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %eax,%ecx
  mov    %ebx,%eax
  mov    %edx,%esi
  cltd
  idiv   %ebp
  mov    %edi,%edx
  sub    %esi,%edx
  cmp    %eax,%ecx
  cmovl  %edx,%edi
  sub    %edi,%ebx
  mov    %ebx,%eax
L21:
  cltd
  mov    $0x9,%edi
  idiv   %ebp
  mov    %edx,%r12d
  mov    %eax,%esi
  call   <insert_char@plt>
  pop    %rbx
  mov    %r12d,%esi
  pop    %rbp
  mov    $0x20,%edi
  pop    %r12
  jmp    1050 <insert_char@plt>
  cs nopw 0x0(%rax,%rax,1)
L17:
  mov    $0x1,%esi
  mov    $0x9,%edi
  call   <insert_char@plt>
  mov    %ebx,%eax
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L4:
  sub    %esi,%edi
  mov    %edi,%esi
  mov    $0x20,%edi
  jmp    1050 <insert_char@plt>
```

assistant: ```c
void insert_space_tabs(int from, int to, int m, int n) {
  if (to < m)
    insert_char(' ', from - to);
  else if (from < m) {
    insert_char('\t', 1);
    int k = to - m;
    insert_char('\t', k / n);
    insert_char(' ', k % n);
  } else {
    int a = from - m;
    int b = to - m;
    if (a / n < b / n)
      a = a - (a % n);
    int k = b - a;
    insert_char('\t', k / n);
    insert_char(' ', k % n);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<insert_space_tabs>:
  endbr64
  cmp    %edx,%esi
  jl     L2
  push   %r12
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  mov    %esi,%ebx
  sub    %edx,%ebx
  cmp    %edi,%edx
  jg     L15
  sub    %edx,%edi
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %eax,%ecx
  mov    %ebx,%eax
  mov    %edx,%esi
  cltd
  idiv   %ebp
  mov    %edi,%edx
  sub    %esi,%edx
  cmp    %eax,%ecx
  cmovl  %edx,%edi
  sub    %edi,%ebx
  mov    %ebx,%eax
L21:
  cltd
  mov    $0x9,%edi
  idiv   %ebp
  mov    %edx,%r12d
  mov    %eax,%esi
  call   <insert_char@plt>
  pop    %rbx
  mov    %r12d,%esi
  pop    %rbp
  mov    $0x20,%edi
  pop    %r12
  jmp    1050 <insert_char@plt>
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    $0x1,%esi
  mov    $0x9,%edi
  call   <insert_char@plt>
  mov    %ebx,%eax
  jmp    L21
  nopl   0x0(%rax,%rax,1)
L2:
  sub    %esi,%edi
  mov    %edi,%esi
  mov    $0x20,%edi
  jmp    1050 <insert_char@plt>
```

assistant: ```c
void insert_space_tabs(int from, int to, int m, int n) {
  if (to < m)
    insert_char(' ', from - to);
  else if (from < m) {
    insert_char('\t', 1);
    int k = to - m;
    insert_char('\t', k / n);
    insert_char(' ', k % n);
  } else {
    int a = from - m;
    int b = to - m;
    if (a / n < b / n)
      a = a - (a % n);
    int k = b - a;
    insert_char('\t', k / n);
    insert_char(' ', k % n);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bf_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L9
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <srand@plt>
  jmp    L0
L9:
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
L0:
  nop
  leave
  ret
```

assistant: ```c
void bf_init(unsigned seed) {
  if (seed) {
    srand(seed);
  } else {
    srand(time(0));
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bf_init>:
  endbr64
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L5
  call   <srand@plt>
L14:
  add    $0x8,%rsp
  ret
L5:
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  jmp    L14
```

assistant: ```c
void bf_init(unsigned seed) {
  if (seed) {
    srand(seed);
  } else {
    srand(time(0));
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<bf_init>:
  endbr64
  test   %edi,%edi
  jne    L4
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <time@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
L4:
  jmp    1060 <srand@plt>
```

assistant: ```c
void bf_init(unsigned seed) {
  if (seed) {
    srand(seed);
  } else {
    srand(time(0));
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bf_init>:
  endbr64
  test   %edi,%edi
  jne    L7
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <time@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
L7:
  jmp    1060 <srand@plt>
```

assistant: ```c
void bf_init(unsigned seed) {
  if (seed) {
    srand(seed);
  } else {
    srand(time(0));
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<cram_md5_digest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %rdx,-0x58(%rbp)
  mov    %rcx,-0x60(%rbp)
  mov    %r8,-0x68(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x50(%rbp)
  jne    L34
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  jmp    L15
L34:
  mov    -0x50(%rbp),%rbx
L15:
  cmpq   $0x0,-0x60(%rbp)
  jne    L72
  mov    -0x58(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  jmp    L12
L72:
  mov    -0x60(%rbp),%rax
L12:
  lea    -0x30(%rbp),%rcx
  mov    -0x48(%rbp),%rdx
  mov    -0x58(%rbp),%rdi
  mov    %rcx,%r8
  mov    %rbx,%rcx
  mov    %rax,%rsi
  call   <gc_hmac_md5@plt>
  movq   $0x0,-0x38(%rbp)
  jmp    L67
L86:
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  sar    $0x4,%al
  movsbl %al,%eax
  and    $0xf,%eax
  cmp    $0x9,%eax
  jle    L26
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  shr    $0x4,%al
  add    $0x57,%eax
  mov    %eax,%ecx
  jmp    L59
L26:
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  shr    $0x4,%al
  add    $0x30,%eax
  mov    %eax,%ecx
L59:
  mov    -0x68(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x68(%rbp)
  mov    %cl,(%rax)
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  and    $0xf,%eax
  cmp    $0x9,%eax
  jle    L81
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  and    $0xf,%eax
  add    $0x57,%eax
  mov    %eax,%ecx
  jmp    L17
L81:
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  and    $0xf,%eax
  add    $0x30,%eax
  mov    %eax,%ecx
L17:
  mov    -0x68(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x68(%rbp)
  mov    %cl,(%rax)
  addq   $0x1,-0x38(%rbp)
L67:
  cmpq   $0xf,-0x38(%rbp)
  jbe    L86
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void cram_md5_digest(const char *challenge, size_t challengelen, const char *secret, size_t secretlen, char response[32]) {
  char hash[16];
  size_t i;
  gc_hmac_md5(secret, secretlen ? secretlen : strlen(secret), challenge, challengelen ? challengelen : strlen(challenge), hash);
  for (i = 0; i < 16; i++) {
    *response++ = ((hash[i] >> 4 & 0x0F) > 9 ? 'a' + (hash[i] >> 4 & 0x0F) - 10 : '0' + (hash[i] >> 4 & 0x0F));
    *response++ = ((hash[i] & 0x0F) > 9 ? 'a' + (hash[i] & 0x0F) - 10 : '0' + (hash[i] & 0x0F));
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cram_md5_digest>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    %rcx,%rbp
  mov    %r8,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  jne    L48
  call   <strlen@plt>
  mov    %rax,%r12
L48:
  test   %rbp,%rbp
  jne    L3
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%rbp
L3:
  mov    %rsp,%r8
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <gc_hmac_md5@plt>
  mov    $0x0,%edx
  jmp    L63
L17:
  and    $0xf,%eax
  add    $0x30,%eax
L5:
  mov    %al,0x1(%rbx,%rdx,2)
  add    $0x1,%rdx
  cmp    $0x10,%rdx
  je     L1
L63:
  movzbl (%rsp,%rdx,1),%eax
  mov    %eax,%esi
  shr    $0x4,%sil
  lea    0x57(%rsi),%edi
  lea    0x30(%rsi),%ecx
  cmp    $0xa,%sil
  cmovge %edi,%ecx
  mov    %cl,(%rbx,%rdx,2)
  mov    %eax,%ecx
  and    $0xf,%ecx
  cmp    $0x9,%cl
  jle    L17
  mov    %ecx,%eax
  add    $0x57,%eax
  jmp    L5
L1:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void cram_md5_digest(const char *challenge, size_t challengelen, const char *secret, size_t secretlen, char response[32]) {
  char hash[16];
  size_t i;
  gc_hmac_md5(secret, secretlen ? secretlen : strlen(secret), challenge, challengelen ? challengelen : strlen(challenge), hash);
  for (i = 0; i < 16; i++) {
    *response++ = ((hash[i] >> 4 & 0x0F) > 9 ? 'a' + (hash[i] >> 4 & 0x0F) - 10 : '0' + (hash[i] >> 4 & 0x0F));
    *response++ = ((hash[i] & 0x0F) > 9 ? 'a' + (hash[i] & 0x0F) - 10 : '0' + (hash[i] & 0x0F));
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cram_md5_digest>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  jne    L25
  call   <strlen@plt>
  mov    %rax,%r13
L25:
  test   %r12,%r12
  jne    L20
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %rax,%r12
L20:
  mov    %rsp,%rbp
  mov    %r14,%rdx
  mov    %r13,%rcx
  mov    %r12,%rsi
  mov    %rbp,%r8
  mov    %r15,%rdi
  call   <gc_hmac_md5@plt>
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L11:
  movzbl 0x0(%rbp,%rdx,1),%eax
  mov    %eax,%esi
  shr    $0x4,%sil
  cmp    $0xa,%sil
  lea    0x57(%rsi),%edi
  lea    0x30(%rsi),%ecx
  cmovae %edi,%ecx
  and    $0xf,%eax
  lea    0x57(%rax),%esi
  cmp    $0xa,%al
  mov    %cl,(%rbx,%rdx,2)
  lea    0x30(%rax),%ecx
  mov    %esi,%eax
  cmovl  %ecx,%eax
  mov    %al,0x1(%rbx,%rdx,2)
  add    $0x1,%rdx
  cmp    $0x10,%rdx
  jne    L11
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L38
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L38:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void cram_md5_digest(const char *challenge, size_t challengelen, const char *secret, size_t secretlen, char response[32]) {
  char hash[16];
  size_t i;
  gc_hmac_md5(secret, secretlen ? secretlen : strlen(secret), challenge, challengelen ? challengelen : strlen(challenge), hash);
  for (i = 0; i < 16; i++) {
    *response++ = ((hash[i] >> 4 & 0x0F) > 9 ? 'a' + (hash[i] >> 4 & 0x0F) - 10 : '0' + (hash[i] >> 4 & 0x0F));
    *response++ = ((hash[i] & 0x0F) > 9 ? 'a' + (hash[i] & 0x0F) - 10 : '0' + (hash[i] & 0x0F));
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cram_md5_digest>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rcx,%rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  jne    L1
  call   <strlen@plt>
  mov    %rax,%r12
L1:
  test   %rbp,%rbp
  jne    L12
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%rbp
L12:
  mov    %rsp,%r8
  mov    %r12,%rcx
  mov    %r14,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <gc_hmac_md5@plt>
  movdqa (%rsp),%xmm0
  movdqa D17(%rip),%xmm6
  movdqa D38(%rip),%xmm5
  movdqa D0(%rip),%xmm3
  psrlw  $0x4,%xmm0
  movdqa D18(%rip),%xmm2
  pand   %xmm6,%xmm0
  pand   (%rsp),%xmm6
  movdqa %xmm0,%xmm1
  movdqa %xmm0,%xmm4
  movdqa %xmm0,%xmm7
  pminub %xmm5,%xmm1
  paddb  %xmm3,%xmm4
  pcmpeqb %xmm0,%xmm1
  movdqa %xmm6,%xmm0
  paddb  %xmm2,%xmm7
  pcmpgtb %xmm5,%xmm0
  paddb  %xmm6,%xmm3
  paddb  %xmm6,%xmm2
  pand   %xmm1,%xmm4
  pandn  %xmm7,%xmm1
  pand   %xmm0,%xmm2
  por    %xmm4,%xmm1
  pandn  %xmm3,%xmm0
  por    %xmm2,%xmm0
  movdqa %xmm1,%xmm2
  punpcklbw %xmm0,%xmm2
  punpckhbw %xmm0,%xmm1
  movups %xmm2,(%rbx)
  movups %xmm1,0x10(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D17
type_guessed: byte[16]
name: unpack_data
arguments: label: D38
type_guessed: byte[16]
name: unpack_data
arguments: label: D0
type_guessed: byte[16]
name: unpack_data
arguments: label: D18
type_guessed: byte[16]

tool: {"D17": [0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f,0x0f]}

tool: {"D38": [0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09,0x09]}

tool: {"D0": [0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30,0x30]}

tool: {"D18": [0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57,0x57]}

assistant: ```c
void cram_md5_digest(const char *challenge, size_t challengelen, const char *secret, size_t secretlen, char response[32]) {
  char hash[16];
  size_t i;
  gc_hmac_md5(secret, secretlen ? secretlen : strlen(secret), challenge, challengelen ? challengelen : strlen(challenge), hash);
  for (i = 0; i < 16; i++) {
    *response++ = ((hash[i] >> 4 & 0x0F) > 9 ? 'a' + (hash[i] >> 4 & 0x0F) - 10 : '0' + (hash[i] >> 4 & 0x0F));
    *response++ = ((hash[i] & 0x0F) > 9 ? 'a' + (hash[i] & 0x0F) - 10 : '0' + (hash[i] & 0x0F));
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x2,-0x24(%rbp)
  jle    L4
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %eax,%ebx
  jmp    L37
L4:
  mov    $0x0,%ebx
L37:
  cmpl   $0x1,-0x24(%rbp)
  jle    L58
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  jmp    L59
L58:
  mov    $0x578,%eax
L59:
  mov    %ebx,%esi
  mov    %eax,%edi
  call   <nfq_operate@plt>
  test   %eax,%eax
  je     L3
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L28
L3:
  nop
L28:
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int rv = EXIT_FAILURE;
  if (!nfq_operate(argc > 1 ? strtoul(argv[1], NULL, 10) : 1400, argc > 2 ? strtoul(argv[2], NULL, 10) : 0))
    goto fail;
  rv = EXIT_SUCCESS;
fail:
  return rv;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    D13(%rip),%rax
  mov    (%rax),%ebp
  cmp    $0x2,%edi
  jg     L2
  mov    $0x0,%r12d
  mov    $0x578,%eax
  je     L8
L0:
  mov    %r12d,%esi
  mov    %eax,%edi
  call   <nfq_operate@plt>
  test   %eax,%eax
  je     L21
  mov    D6(%rip),%rax
  mov    (%rax),%ebp
L21:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  mov    0x10(%rsi),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  mov    %eax,%r12d
L8:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  jmp    L0
```

assistant: ```c
int main(int argc, char *argv[]) {
  int rv = EXIT_FAILURE;
  if (!nfq_operate(argc > 1 ? strtoul(argv[1], NULL, 10) : 1400, argc > 2 ? strtoul(argv[2], NULL, 10) : 0))
    goto fail;
  rv = EXIT_SUCCESS;
fail:
  return rv;
}
```