ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  lea    D0(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x300000001,%rax
  lea    0xc(%rsp),%rsi
  movq   $0x9,0x20(%rsp)
  mov    %rax,0x10(%rsp)
  movabs $0x700000005,%rax
  mov    %rax,0x18(%rsp)
  call   <scanf@plt>
  mov    0xc(%rsp),%r12d
  cmp    $0x1,%r12d
  jle    L68
  mov    $0x1,%eax
  mov    $0x3,%edx
  lea    0x10(%rsp),%rbx
  movslq %eax,%rbp
  cmp    %r12d,%edx
  jge    L54
L45:
  add    $0x1,%rax
  cmp    $0x6,%rax
  je     L16
  mov    (%rbx,%rax,4),%edx
  movslq %eax,%rbp
  cmp    %r12d,%edx
  jl     L45
L54:
  cmp    $0x5,%ebp
  je     L32
L44:
  mov    $0x4,%eax
  sub    %ebp,%eax
  imul   $0xfffffffffffffffc,%rax,%rdi
  lea    0x4(,%rax,4),%rdx
  cmp    $0x4,%ebp
  mov    $0x4,%eax
  cmovg  %rax,%rdx
  mov    $0x10,%eax
  lea    0x10(%rdi),%rsi
  cmovg  %rax,%rsi
  add    $0x14,%rdi
  mov    $0x14,%eax
  add    %rbx,%rsi
  cmp    $0x4,%ebp
  cmovg  %rax,%rdi
  add    %rbx,%rdi
  call   <memmove@plt>
  jmp    L32
L16:
  mov    $0x5,%ebp
L32:
  mov    %r12d,0x10(%rsp,%rbp,4)
  lea    0x28(%rsp),%r12
  lea    D6(%rip),%rbp
  nopl   0x0(%rax)
L15:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L15
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L76
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L68:
  xor    %ebp,%ebp
  lea    0x10(%rsp),%rbx
  jmp    L44
L76:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D6
type_guessed: string

tool: D0: %d

tool: D6: %3d

assistant: ```c
int main() {
  int a[6] = {1, 3, 5, 7, 9}, num, i, j;
  scanf("%d", &num);
  for (i = 0; a[i] < num && i < 5; i++)
    ;
  for (j = 5; j > i; j--)
    a[j] = a[j - 1];
  a[i] = num;
  for (i = 0; i < 6; i++)
    printf("%3d", a[i]);
  getchar();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  lea    D16(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D15(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  movq   $0x9,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <scanf@plt>
  mov    0xc(%rsp),%ebp
  cmp    $0x1,%ebp
  jle    L37
  cmp    $0x3,%ebp
  jle    L79
  cmp    $0x5,%ebp
  jle    L51
  cmp    $0x7,%ebp
  jle    L82
  mov    $0x5,%r12d
  lea    0x10(%rsp),%rbx
  cmp    $0x9,%ebp
  jle    L68
L26:
  mov    %ebp,0x10(%rsp,%r12,4)
  lea    0x28(%rsp),%r12
  lea    D25(%rip),%rbp
  nopl   0x0(%rax)
L57:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L57
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L37:
  xor    %r12d,%r12d
L93:
  mov    $0x4,%eax
  lea    0x10(%rsp),%rbx
  sub    %r12d,%eax
  imul   $0xfffffffffffffffc,%rax,%rdi
  lea    0x4(,%rax,4),%rdx
  cmp    $0x4,%r12d
  mov    $0x4,%eax
  cmovg  %rax,%rdx
  mov    $0x10,%eax
  lea    0x10(%rdi),%rsi
  cmovg  %rax,%rsi
  add    $0x14,%rdi
  mov    $0x14,%eax
  add    %rbx,%rsi
  cmp    $0x4,%r12d
  cmovg  %rax,%rdi
  add    %rbx,%rdi
  call   <memmove@plt>
  jmp    L26
L79:
  mov    $0x1,%r12d
  jmp    L93
L51:
  mov    $0x2,%r12d
  jmp    L93
L82:
  mov    $0x3,%r12d
  jmp    L93
L68:
  mov    $0x4,%r12d
  jmp    L93
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D16
datatype: string
name: parse_data
arguments: data_tag: D15
datatype: byte[16]
name: parse_data
arguments: data_tag: D25
datatype: string

tool: D16: %d

tool: {"D15": [0x01,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: D25: %3d

assistant: ```c
int main() {
  int a[6] = {1, 3, 5, 7, 9}, num, i, j;
  scanf("%d", &num);
  for (i = 0; a[i] < num && i < 5; i++)
    ;
  for (j = 5; j > i; j--)
    a[j] = a[j - 1];
  a[i] = num;
  for (i = 0; i < 6; i++)
    printf("%3d", a[i]);
  getchar();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<read_escaped_data>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <tor_malloc_@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  jmp    L61
L72:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L35
  addq   $0x1,-0x28(%rbp)
L35:
  mov    -0x10(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <memchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L68
  mov    -0x8(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L45
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0xd,%al
  jne    L45
  subq   $0x1,-0x18(%rbp)
L45:
  mov    -0x18(%rbp),%rax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rax
  add    %rax,-0x20(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x28(%rbp)
  jmp    L71
L68:
  mov    -0x10(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rax
  sub    -0x28(%rbp),%rax
  add    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rax
  sub    %rdx,%rax
  jmp    L50
L71:
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movb   $0xa,(%rax)
L61:
  mov    -0x28(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L72
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rax
  sub    %rdx,%rax
L50:
  leave
  ret
```

assistant: ```c
size_t read_escaped_data(const char *data, size_t len, char **out) {
  char *outp;
  const char *next;
  const char *end;
  *out = outp = tor_malloc_(len + 1);
  end = data + len;
  while (data < end) {
    if (*data == '.')
      ++data;
    next = memchr(data, '\n', end - data);
    if (next) {
      size_t n_to_copy = next - data;
      if (n_to_copy && *(next - 1) == '\r')
        --n_to_copy;
      memcpy(outp, data, n_to_copy);
      outp += n_to_copy;
      data = next + 1;
    } else {
      memcpy(outp, data, end - data);
      outp += (end - data);
      *outp = '\0';
      return outp - *out;
    }
    *outp++ = '\n';
  }
  *outp = '\0';
  return outp - *out;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_escaped_data>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    %rdx,%r15
  mov    %rdx,0x8(%rsp)
  lea    0x1(%rsi),%rdi
  call   <tor_malloc_@plt>
  mov    %rax,%r12
  mov    %rax,(%r15)
  add    %rbx,%r13
  cmp    %r13,%rbx
  jb     L9
L19:
  movb   $0x0,(%r12)
  mov    0x8(%rsp),%rcx
  mov    %r12,%rax
  sub    (%rcx),%rax
L43:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L58:
  mov    %r14d,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  add    %r12,%r14
  lea    0x1(%rbp),%rbx
  lea    0x1(%r14),%r12
  movb   $0xa,(%r14)
  cmp    %rbx,%r13
  jbe    L19
L9:
  cmpb   $0x2e,(%rbx)
  sete   %al
  movzbl %al,%eax
  add    %rax,%rbx
  mov    %r13,%r14
  sub    %rbx,%r14
  mov    %r14d,%edx
  mov    $0xa,%esi
  mov    %rbx,%rdi
  call   <memchr@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L22
  mov    %rax,%r14
  sub    %rbx,%r14
  je     L58
  cmpb   $0xd,-0x1(%rax)
  sete   %al
  movzbl %al,%eax
  sub    %rax,%r14
  jmp    L58
L22:
  mov    %r14d,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  add    %r14,%r12
  movb   $0x0,(%r12)
  mov    0x8(%rsp),%rcx
  mov    %r12,%rax
  sub    (%rcx),%rax
  jmp    L43
```

assistant: ```c
size_t read_escaped_data(const char *data, size_t len, char **out) {
  char *outp;
  const char *next;
  const char *end;
  *out = outp = tor_malloc_(len + 1);
  end = data + len;
  while (data < end) {
    if (*data == '.')
      ++data;
    next = memchr(data, '\n', end - data);
    if (next) {
      size_t n_to_copy = next - data;
      if (n_to_copy && *(next - 1) == '\r')
        --n_to_copy;
      memcpy(outp, data, n_to_copy);
      outp += n_to_copy;
      data = next + 1;
    } else {
      memcpy(outp, data, end - data);
      outp += (end - data);
      *outp = '\0';
      return outp - *out;
    }
    *outp++ = '\n';
  }
  *outp = '\0';
  return outp - *out;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<read_escaped_data>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rsi),%rdi
  push   %rbx
  add    %rbp,%r13
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  call   <tor_malloc_@plt>
  mov    %rax,(%rbx)
  mov    %rax,%r15
  cmp    %r13,%rbp
  jb     L54
  jmp    L18
  nopl   0x0(%rax)
L31:
  cmpb   $0xd,-0x1(%rax)
  jne    L56
  sub    $0x1,%r12
L56:
  mov    %r12d,%edx
  add    %r15,%r12
L36:
  mov    %rbp,%rsi
  mov    %r15,%rdi
  lea    0x1(%rbx),%rbp
  call   <memcpy@plt>
  lea    0x1(%r12),%r15
  movb   $0xa,(%r12)
  cmp    %rbp,%r13
  jbe    L18
L54:
  xor    %eax,%eax
  cmpb   $0x2e,0x0(%rbp)
  mov    %r13,%r12
  mov    $0xa,%esi
  sete   %al
  add    %rax,%rbp
  sub    %rbp,%r12
  mov    %rbp,%rdi
  mov    %r12d,%edx
  call   <memchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L61
  mov    %rax,%r12
  sub    %rbp,%r12
  jne    L31
  mov    %r15,%r12
  xor    %edx,%edx
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L18:
  movb   $0x0,(%r15)
  mov    0x8(%rsp),%rcx
  mov    %r15,%rax
  sub    (%rcx),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L61:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r15,%rdi
  add    %r12,%r15
  call   <memcpy@plt>
  movb   $0x0,(%r15)
  mov    0x8(%rsp),%rcx
  mov    %r15,%rax
  sub    (%rcx),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
size_t read_escaped_data(const char *data, size_t len, char **out) {
  char *outp;
  const char *next;
  const char *end;
  *out = outp = tor_malloc_(len + 1);
  end = data + len;
  while (data < end) {
    if (*data == '.')
      ++data;
    next = memchr(data, '\n', end - data);
    if (next) {
      size_t n_to_copy = next - data;
      if (n_to_copy && *(next - 1) == '\r')
        --n_to_copy;
      memcpy(outp, data, n_to_copy);
      outp += n_to_copy;
      data = next + 1;
    } else {
      memcpy(outp, data, end - data);
      outp += (end - data);
      *outp = '\0';
      return outp - *out;
    }
    *outp++ = '\n';
  }
  *outp = '\0';
  return outp - *out;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<read_escaped_data>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rsi),%rdi
  push   %rbx
  add    %rbp,%r13
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    %rdx,0x8(%rsp)
  call   <tor_malloc_@plt>
  mov    %rax,(%rbx)
  mov    %rax,%r15
  cmp    %r13,%rbp
  jb     L36
  jmp    L47
  nopl   0x0(%rax)
L12:
  cmpb   $0xd,-0x1(%rax)
  jne    L8
  sub    $0x1,%r12
L8:
  mov    %r12d,%edx
  add    %r15,%r12
L54:
  mov    %rbp,%rsi
  mov    %r15,%rdi
  lea    0x1(%rbx),%rbp
  call   <memcpy@plt>
  lea    0x1(%r12),%r15
  movb   $0xa,(%r12)
  cmp    %rbp,%r13
  jbe    L47
L36:
  xor    %eax,%eax
  cmpb   $0x2e,0x0(%rbp)
  mov    %r13,%r12
  mov    $0xa,%esi
  sete   %al
  add    %rax,%rbp
  sub    %rbp,%r12
  mov    %rbp,%rdi
  mov    %r12d,%edx
  call   <memchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L10
  mov    %rax,%r12
  sub    %rbp,%r12
  jne    L12
  mov    %r15,%r12
  xor    %edx,%edx
  jmp    L54
  nopw   0x0(%rax,%rax,1)
L47:
  movb   $0x0,(%r15)
  mov    0x8(%rsp),%rcx
  mov    %r15,%rax
  sub    (%rcx),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L10:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %r15,%rdi
  add    %r12,%r15
  call   <memcpy@plt>
  movb   $0x0,(%r15)
  mov    0x8(%rsp),%rcx
  mov    %r15,%rax
  sub    (%rcx),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
size_t read_escaped_data(const char *data, size_t len, char **out) {
  char *outp;
  const char *next;
  const char *end;
  *out = outp = tor_malloc_(len + 1);
  end = data + len;
  while (data < end) {
    if (*data == '.')
      ++data;
    next = memchr(data, '\n', end - data);
    if (next) {
      size_t n_to_copy = next - data;
      if (n_to_copy && *(next - 1) == '\r')
        --n_to_copy;
      memcpy(outp, data, n_to_copy);
      outp += n_to_copy;
      data = next + 1;
    } else {
      memcpy(outp, data, end - data);
      outp += (end - data);
      *outp = '\0';
      return outp - *out;
    }
    *outp++ = '\n';
  }
  *outp = '\0';
  return outp - *out;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  cmpl   $0x0,-0xc(%rbp)
  jne    L29
  cmpl   $0x0,-0x8(%rbp)
  je     L26
L29:
  mov    $0x1,%eax
  jmp    L14
L26:
  mov    $0x0,%eax
L14:
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int c, b;
  int d = c || b;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

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

assistant: ```c
int main() {
  int c, b;
  int d = c || b;
  return 0;
}
```
user: What is the c source code of the assembly code below?

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

assistant: ```c
int main() {
  int c, b;
  int d = c || b;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

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

assistant: ```c
int main() {
  int c, b;
  int d = c || b;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<WaitForDecodedFrame>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <pm_error@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D9
type_infered: string

tool: D9: This version of Ppmtompeg cannot run parallel mode because it does not have socket capability.

assistant: ```c
void WaitForDecodedFrame(int const id) {
  pm_error("This version of Ppmtompeg cannot run parallel mode because "
           "it does not have socket capability.");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<WaitForDecodedFrame>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <pm_error@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type_infered: string

tool: D8: This version of Ppmtompeg cannot run parallel mode because it does not have socket capability.

assistant: ```c
void WaitForDecodedFrame(int const id) {
  pm_error("This version of Ppmtompeg cannot run parallel mode because "
           "it does not have socket capability.");
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<WaitForDecodedFrame>:
  endbr64
  lea    D5(%rip),%rdi
  jmp    1050 <pm_error@plt>
```

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

tool: D5: This version of Ppmtompeg cannot run parallel mode because it does not have socket capability.

assistant: ```c
void WaitForDecodedFrame(int const id) {
  pm_error("This version of Ppmtompeg cannot run parallel mode because "
           "it does not have socket capability.");
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<WaitForDecodedFrame>:
  endbr64
  lea    D6(%rip),%rdi
  jmp    1050 <pm_error@plt>
```

tool_calls: name: read_data
arguments: tag: D6
type_infered: string

tool: D6: This version of Ppmtompeg cannot run parallel mode because it does not have socket capability.

assistant: ```c
void WaitForDecodedFrame(int const id) {
  pm_error("This version of Ppmtompeg cannot run parallel mode because "
           "it does not have socket capability.");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x30(%rsp)
  movl   $0x1,0x34(%rsp)
  movl   $0x2,0x38(%rsp)
  movl   $0x3,0x3c(%rsp)
  movl   $0x4,0x40(%rsp)
  movl   $0x5,0x44(%rsp)
  movl   $0x6,0x48(%rsp)
  movl   $0x7,0x4c(%rsp)
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    0x30(%rsp),%rcx
  lea    0xc(%rsp),%rsi
  mov    $0x8,%edx
  mov    $0x2,%edi
  call   <generate_combinations@plt>
  movl   $0x41,0x20(%rsp)
  movl   $0x42,0x24(%rsp)
  movl   $0x43,0x28(%rsp)
  movl   $0x44,0x2c(%rsp)
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    0x20(%rsp),%rbx
  lea    0x14(%rsp),%rsi
  mov    %rbx,%rcx
  mov    $0x4,%edx
  mov    $0x3,%edi
  call   <generate_combinations@plt>
  movl   $0x51,0x20(%rsp)
  movl   $0x52,0x24(%rsp)
  movl   $0x53,0x28(%rsp)
  movl   $0x54,0x2c(%rsp)
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x3,%edx
  mov    %rbx,%rsi
  mov    $0x4,%edi
  call   <iterative_combinations@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x60,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
type_guessed: string
name: disassemble_data
arguments: data_tag: D13
type_guessed: string
name: disassemble_data
arguments: data_tag: D8
type_guessed: string

tool: D0: generate combination of octal numbers


tool: D13: 
generate padlock combinations when it is 4 wheels of A to C


tool: D8: 
generate padlock combinations iteratively


assistant: ```c
int main() {
  int positions = 2;
  int combo[positions];
  int set_size = 8;
  int num_set[set_size];
  int i;
  for (i = 0; i < set_size; i++) {
    num_set[i] = i;
  }
  printf("generate combination of octal numbers\n");
  generate_combinations(positions, combo, set_size, num_set);
  positions = 3;
  set_size = 4;
  int second_combo[positions];
  int letter_set[set_size];
  for (i = 0; i < set_size; i++) {
    letter_set[i] = 'A' + i;
  }
  printf("\ngenerate padlock combinations when it is 4 wheels of A to C\n");
  generate_combinations(positions, second_combo, set_size, letter_set);
  for (i = 0; i < set_size; i++) {
    letter_set[i] = 'Q' + i;
  }
  printf("\ngenerate padlock combinations iteratively\n");
  iterative_combinations(set_size, letter_set, positions);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rdi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  movabs $0x100000000,%rax
  lea    0x20(%rsp),%rbp
  mov    %rax,0x30(%rsp)
  movabs $0x300000002,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x500000004,%rax
  mov    %rax,0x40(%rsp)
  movabs $0x700000006,%rax
  mov    %rax,0x48(%rsp)
  call   <printf@plt>
  lea    0x30(%rsp),%rcx
  lea    0xc(%rsp),%rsi
  mov    $0x8,%edx
  mov    $0x2,%edi
  call   <generate_combinations@plt>
  lea    D14(%rip),%rdi
  movabs $0x4200000041,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x4400000043,%rax
  mov    %rax,0x28(%rsp)
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  mov    $0x4,%edx
  mov    %rbp,%rcx
  mov    $0x3,%edi
  call   <generate_combinations@plt>
  lea    D0(%rip),%rdi
  movabs $0x5200000051,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x5400000053,%rax
  mov    %rax,0x28(%rsp)
  call   <printf@plt>
  mov    $0x3,%edx
  mov    %rbp,%rsi
  mov    $0x4,%edi
  call   <iterative_combinations@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D8
datatype: string
name: disassemble_data
arguments: data_label: D14
datatype: string
name: disassemble_data
arguments: data_label: D0
datatype: string

tool: D8: generate combination of octal numbers


tool: D14: 
generate padlock combinations when it is 4 wheels of A to C


tool: D0: 
generate padlock combinations iteratively


assistant: ```c
int main() {
  int positions = 2;
  int combo[positions];
  int set_size = 8;
  int num_set[set_size];
  int i;
  for (i = 0; i < set_size; i++) {
    num_set[i] = i;
  }
  printf("generate combination of octal numbers\n");
  generate_combinations(positions, combo, set_size, num_set);
  positions = 3;
  set_size = 4;
  int second_combo[positions];
  int letter_set[set_size];
  for (i = 0; i < set_size; i++) {
    letter_set[i] = 'A' + i;
  }
  printf("\ngenerate padlock combinations when it is 4 wheels of A to C\n");
  generate_combinations(positions, second_combo, set_size, letter_set);
  for (i = 0; i < set_size; i++) {
    letter_set[i] = 'Q' + i;
  }
  printf("\ngenerate padlock combinations iteratively\n");
  iterative_combinations(set_size, letter_set, positions);
  return 0;
}
```