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   %rbp
  mov    $0x1c2,%ebp
  push   %rbx
  mov    $0x104,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L0:
  mov    %ebx,%esi
  mov    %ebp,%edi
  mov    %ebx,%ecx
  mov    %ebp,%edx
  shr    $0x1f,%esi
  shr    $0x1f,%edi
  add    %ebx,%esi
  add    %ebp,%edi
  sub    $0x28,%ebx
  sub    $0x32,%ebp
  sar    %esi
  sar    %edi
  neg    %esi
  neg    %edi
  add    $0x96,%esi
  add    $0xfa,%edi
  call   <drawOval@plt>
  cmp    $0xffffff9c,%ebx
  jne    L0
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int i = 1;
  for (i; i < 10; i++) {
    int width = 500 - i * 50;
    int height = 300 - i * 40;
    drawOval(250 - (int)width / 2, 150 - (int)height / 2, width, height);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<check_parity>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xbb1f3480,-0x20(%rbp)
  movl   $0x5d8f9a40,-0x1c(%rbp)
  movl   $0xaec7cd00,-0x18(%rbp)
  movl   $0x5763e680,-0x14(%rbp)
  movl   $0x6bb1f340,-0x10(%rbp)
  movl   $0x8b7a89c0,-0xc(%rbp)
  movl   $0x0,-0x2c(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L52
L63:
  shll   -0x2c(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  and    -0x34(%rbp),%eax
  shr    $0x6,%eax
  mov    %eax,-0x28(%rbp)
  jmp    L33
L20:
  mov    -0x28(%rbp),%eax
  and    $0x1,%eax
  xor    %eax,-0x2c(%rbp)
  shrl   -0x28(%rbp)
L33:
  cmpl   $0x0,-0x28(%rbp)
  jne    L20
  addl   $0x1,-0x24(%rbp)
L52:
  cmpl   $0x5,-0x24(%rbp)
  jle    L63
  mov    -0x34(%rbp),%eax
  and    $0x3f,%eax
  cmp    %eax,-0x2c(%rbp)
  je     L44
  mov    $0x0,%eax
  jmp    L18
L44:
  movl   $0x0,-0x24(%rbp)
  jmp    L62
L23:
  mov    -0x24(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    $0x16,%eax
  sub    %edx,%eax
  mov    -0x34(%rbp),%edx
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%ecx
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x24(%rbp)
L62:
  cmpl   $0x2,-0x24(%rbp)
  jle    L23
  mov    $0x1,%eax
L18:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L30
  call   <__stack_chk_fail@plt>
L30:
  leave
  ret
```

assistant: ```c
extern int check_parity(unsigned int word, unsigned char *data) {
  const unsigned int hamming[] = {0xBB1F3480, 0x5D8F9A40, 0xAEC7CD00, 0x5763E680, 0x6BB1F340, 0x8B7A89C0};
  unsigned int parity = 0, w;
  int i;
  for (i = 0; i < 6; i++) {
    parity <<= 1;
    for (w = (word & hamming[i]) >> 6; w; w >>= 1)
      parity ^= w & 1;
  }
  if (parity != (word & 0x3F))
    return 0;
  for (i = 0; i < 3; i++)
    data[i] = (unsigned char)(word >> (22 - i * 8));
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<check_parity>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0xbb1f3480,(%rsp)
  movl   $0x5d8f9a40,0x4(%rsp)
  movl   $0xaec7cd00,0x8(%rsp)
  movl   $0x5763e680,0xc(%rsp)
  movl   $0x6bb1f340,0x10(%rsp)
  movl   $0x8b7a89c0,0x14(%rsp)
  mov    %rsp,%r8
  lea    0x18(%rsp),%r9
  mov    $0x0,%edx
L34:
  add    %edx,%edx
  mov    %edi,%eax
  and    (%r8),%eax
  shr    $0x6,%eax
  je     L32
L45:
  mov    %eax,%ecx
  and    $0x1,%ecx
  xor    %ecx,%edx
  shr    %eax
  jne    L45
L32:
  add    $0x4,%r8
  cmp    %r9,%r8
  jne    L34
  mov    %edi,%ecx
  and    $0x3f,%ecx
  mov    $0x0,%eax
  cmp    %edx,%ecx
  jne    L38
  mov    %edi,%eax
  shr    $0x16,%eax
  mov    %al,(%rsi)
  mov    %edi,%eax
  shr    $0xe,%eax
  mov    %al,0x1(%rsi)
  shr    $0x6,%edi
  mov    %dil,0x2(%rsi)
  mov    $0x1,%eax
L38:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x28,%rsp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
extern int check_parity(unsigned int word, unsigned char *data) {
  const unsigned int hamming[] = {0xBB1F3480, 0x5D8F9A40, 0xAEC7CD00, 0x5763E680, 0x6BB1F340, 0x8B7A89C0};
  unsigned int parity = 0, w;
  int i;
  for (i = 0; i < 6; i++) {
    parity <<= 1;
    for (w = (word & hamming[i]) >> 6; w; w >>= 1)
      parity ^= w & 1;
  }
  if (parity != (word & 0x3F))
    return 0;
  for (i = 0; i < 3; i++)
    data[i] = (unsigned char)(word >> (22 - i * 8));
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<check_parity>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,%r8d
  xor    %edx,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x5d8f9a40,0x4(%rsp)
  mov    %rsp,%rdi
  lea    0x18(%rsp),%r9
  movabs $0x5763e680aec7cd00,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x8b7a89c06bb1f340,%rax
  mov    %rax,0x10(%rsp)
  mov    $0xbb1f3480,%eax
L36:
  and    %r8d,%eax
  add    %edx,%edx
  shr    $0x6,%eax
  je     L43
  nopw   0x0(%rax,%rax,1)
L33:
  mov    %eax,%ecx
  and    $0x1,%ecx
  xor    %ecx,%edx
  shr    %eax
  jne    L33
L43:
  add    $0x4,%rdi
  cmp    %r9,%rdi
  je     L26
  mov    (%rdi),%eax
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L26:
  mov    %r8d,%ecx
  xor    %eax,%eax
  and    $0x3f,%ecx
  cmp    %edx,%ecx
  jne    L14
  mov    %r8d,%eax
  shr    $0x16,%eax
  mov    %al,(%rsi)
  mov    %r8d,%eax
  shr    $0x6,%r8d
  mov    %r8b,0x2(%rsi)
  shr    $0xe,%eax
  mov    %al,0x1(%rsi)
  mov    $0x1,%eax
L14:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L45
  add    $0x28,%rsp
  ret
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
extern int check_parity(unsigned int word, unsigned char *data) {
  const unsigned int hamming[] = {0xBB1F3480, 0x5D8F9A40, 0xAEC7CD00, 0x5763E680, 0x6BB1F340, 0x8B7A89C0};
  unsigned int parity = 0, w;
  int i;
  for (i = 0; i < 6; i++) {
    parity <<= 1;
    for (w = (word & hamming[i]) >> 6; w; w >>= 1)
      parity ^= w & 1;
  }
  if (parity != (word & 0x3F))
    return 0;
  for (i = 0; i < 3; i++)
    data[i] = (unsigned char)(word >> (22 - i * 8));
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<check_parity>:
  endbr64
  sub    $0x28,%rsp
  movdqa D14(%rip),%xmm0
  mov    %edi,%r8d
  xor    %edx,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D2(%rip),%rax
  mov    %rsp,%rdi
  lea    0x18(%rsp),%r9
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  nopl   (%rax)
L45:
  mov    (%rdi),%eax
  add    %edx,%edx
  and    %r8d,%eax
  shr    $0x6,%eax
  je     L30
  nopl   0x0(%rax)
L44:
  mov    %eax,%ecx
  and    $0x1,%ecx
  xor    %ecx,%edx
  shr    %eax
  jne    L44
L30:
  add    $0x4,%rdi
  cmp    %rdi,%r9
  jne    L45
  mov    %r8d,%ecx
  xor    %eax,%eax
  and    $0x3f,%ecx
  cmp    %edx,%ecx
  jne    L0
  mov    %r8d,%eax
  shr    $0x16,%eax
  mov    %al,(%rsi)
  mov    %r8d,%eax
  shr    $0x6,%r8d
  mov    %r8b,0x2(%rsi)
  shr    $0xe,%eax
  mov    %al,0x1(%rsi)
  mov    $0x1,%eax
L0:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L29
  add    $0x28,%rsp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D14
type_infered: byte[16]
name: parse_data
arguments: data_tag: D2
type_infered: byte[8]

tool: {"D14": [0x80,0x34,0x1f,0xbb,0x40,0x9a,0x8f,0x5d,0x00,0xcd,0xc7,0xae,0x80,0xe6,0x63,0x57]}

tool: {"D2": [0x40,0xf3,0xb1,0x6b,0xc0,0x89,0x7a,0x8b]}

assistant: ```c
extern int check_parity(unsigned int word, unsigned char *data) {
  const unsigned int hamming[] = {0xBB1F3480, 0x5D8F9A40, 0xAEC7CD00, 0x5763E680, 0x6BB1F340, 0x8B7A89C0};
  unsigned int parity = 0, w;
  int i;
  for (i = 0; i < 6; i++) {
    parity <<= 1;
    for (w = (word & hamming[i]) >> 6; w; w >>= 1)
      parity ^= w & 1;
  }
  if (parity != (word & 0x3F))
    return 0;
  for (i = 0; i < 3; i++)
    data[i] = (unsigned char)(word >> (22 - i * 8));
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<itoa_xml>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x38(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  movl   $0xa,-0x20(%rbp)
  cmpl   $0x64,-0x38(%rbp)
  jne    L63
  cmpl   $0x0,-0x34(%rbp)
  jns    L63
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movb   $0x2d,(%rax)
  addq   $0x1,-0x40(%rbp)
  mov    -0x34(%rbp),%eax
  neg    %eax
  mov    %eax,-0x24(%rbp)
  jmp    L16
L63:
  cmpl   $0x78,-0x38(%rbp)
  jne    L16
  movl   $0x10,-0x20(%rbp)
L16:
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %edx,-0x1c(%rbp)
  cmpl   $0x9,-0x1c(%rbp)
  jg     L33
  mov    -0x1c(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%ecx
  jmp    L61
L33:
  mov    -0x1c(%rbp),%eax
  add    $0x57,%eax
  mov    %eax,%ecx
L61:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %cl,(%rax)
  mov    -0x24(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jne    L16
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L36
L22:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x25(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl -0x25(%rbp),%edx
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
  subq   $0x1,-0x8(%rbp)
L36:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L22
  mov    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int itoa_xml(int d, char *buf, int base) {
  char *p = buf;
  char *p1, *p2;
  uint64_t ud = d;
  int divisor = 10;
  if (base == 'd' && d < 0) {
    *p++ = '-';
    buf++;
    ud = -d;
  } else if (base == 'x')
    divisor = 16;
  do {
    int remainder = ud % divisor;
    *p++ = (remainder < 10) ? remainder + '0' : remainder + 'a' - 10;
  } while (ud /= divisor);
  *p = 0;
  p1 = buf;
  p2 = p - 1;
  while (p1 < p2) {
    char tmp = *p1;
    *p1 = *p2;
    *p2 = tmp;
    p1++;
    p2--;
  }
  return (1);
}
```
user: What is the c source code of the assembly code below?

```asm
<itoa_xml>:
  endbr64
  cmp    $0x64,%edx
  jne    L38
  test   %edi,%edi
  jns    L38
  movb   $0x2d,(%rsi)
  neg    %edi
  lea    0x1(%rsi),%rsi
  mov    $0xa,%r8d
L30:
  mov    %rsi,%rcx
  jmp    L37
L38:
  cmp    $0x78,%edx
  mov    $0xa,%r8d
  mov    $0x10,%eax
  cmove  %eax,%r8d
  jmp    L30
L67:
  mov    %r9,%rcx
L37:
  mov    %edi,%eax
  cltd
  idiv   %r8d
  lea    0x30(%rdx),%r9d
  lea    0x57(%rdx),%eax
  cmp    $0x9,%edx
  cmovle %r9d,%eax
  lea    0x1(%rcx),%r9
  mov    %al,-0x1(%r9)
  mov    %edi,%eax
  cltd
  idiv   %r8d
  mov    %eax,%edi
  test   %eax,%eax
  jne    L67
  movb   $0x0,(%r9)
  cmp    %rsi,%rcx
  jbe    L20
L61:
  movzbl (%rsi),%eax
  movzbl (%rcx),%edx
  mov    %dl,(%rsi)
  mov    %al,(%rcx)
  add    $0x1,%rsi
  sub    $0x1,%rcx
  cmp    %rcx,%rsi
  jb     L61
L20:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int itoa_xml(int d, char *buf, int base) {
  char *p = buf;
  char *p1, *p2;
  uint64_t ud = d;
  int divisor = 10;
  if (base == 'd' && d < 0) {
    *p++ = '-';
    buf++;
    ud = -d;
  } else if (base == 'x')
    divisor = 16;
  do {
    int remainder = ud % divisor;
    *p++ = (remainder < 10) ? remainder + '0' : remainder + 'a' - 10;
  } while (ud /= divisor);
  *p = 0;
  p1 = buf;
  p2 = p - 1;
  while (p1 < p2) {
    char tmp = *p1;
    *p1 = *p2;
    *p2 = tmp;
    p1++;
    p2--;
  }
  return (1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<itoa_xml>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%rdi
  cmp    $0x64,%edx
  jne    L68
  test   %eax,%eax
  jns    L68
  movb   $0x2d,(%rsi)
  neg    %eax
  add    $0x1,%rdi
  mov    $0xa,%r8d
L2:
  mov    %rdi,%rcx
  jmp    L23
  cs nopw 0x0(%rax,%rax,1)
L51:
  mov    %rsi,%rcx
L23:
  cltd
  idiv   %r8d
  lea    0x30(%rdx),%r9d
  lea    0x57(%rdx),%esi
  cmp    $0x9,%edx
  mov    %r9d,%edx
  cmovg  %esi,%edx
  lea    0x1(%rcx),%rsi
  mov    %dl,-0x1(%rsi)
  test   %eax,%eax
  jne    L51
  movb   $0x0,(%rsi)
  cmp    %rdi,%rcx
  ja     L36
  jmp    L53
  nopl   0x0(%rax)
L9:
  movzbl (%rcx),%edx
L36:
  movzbl (%rdi),%eax
  sub    $0x1,%rcx
  mov    %dl,(%rdi)
  add    $0x1,%rdi
  mov    %al,0x1(%rcx)
  cmp    %rcx,%rdi
  jb     L9
L53:
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L68:
  cmp    $0x78,%edx
  mov    $0xa,%r8d
  mov    $0x10,%edx
  cmove  %edx,%r8d
  jmp    L2
```

assistant: ```c
int itoa_xml(int d, char *buf, int base) {
  char *p = buf;
  char *p1, *p2;
  uint64_t ud = d;
  int divisor = 10;
  if (base == 'd' && d < 0) {
    *p++ = '-';
    buf++;
    ud = -d;
  } else if (base == 'x')
    divisor = 16;
  do {
    int remainder = ud % divisor;
    *p++ = (remainder < 10) ? remainder + '0' : remainder + 'a' - 10;
  } while (ud /= divisor);
  *p = 0;
  p1 = buf;
  p2 = p - 1;
  while (p1 < p2) {
    char tmp = *p1;
    *p1 = *p2;
    *p2 = tmp;
    p1++;
    p2--;
  }
  return (1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<itoa_xml>:
  endbr64
  mov    %edi,%eax
  cmp    $0x64,%edx
  jne    L25
  test   %edi,%edi
  jns    L25
  movb   $0x2d,(%rsi)
  neg    %eax
  add    $0x1,%rsi
  mov    $0xa,%r8d
L73:
  mov    %rsi,%rcx
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L75:
  add    $0x30,%edx
  mov    %dl,-0x1(%rdi)
  test   %eax,%eax
  je     L80
L14:
  mov    %rdi,%rcx
L23:
  cltd
  lea    0x1(%rcx),%rdi
  idiv   %r8d
  cmp    $0x9,%edx
  jle    L75
  add    $0x57,%edx
  mov    %dl,-0x1(%rdi)
  test   %eax,%eax
  jne    L14
L80:
  movb   $0x0,(%rdi)
  cmp    %rcx,%rsi
  jae    L29
  nopl   0x0(%rax)
L53:
  movzbl (%rsi),%eax
  movzbl (%rcx),%edx
  add    $0x1,%rsi
  sub    $0x1,%rcx
  mov    %dl,-0x1(%rsi)
  mov    %al,0x1(%rcx)
  cmp    %rcx,%rsi
  jb     L53
L29:
  mov    $0x1,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L25:
  cmp    $0x78,%edx
  mov    $0xa,%r8d
  mov    $0x10,%edx
  cmove  %edx,%r8d
  jmp    L73
```

assistant: ```c
int itoa_xml(int d, char *buf, int base) {
  char *p = buf;
  char *p1, *p2;
  uint64_t ud = d;
  int divisor = 10;
  if (base == 'd' && d < 0) {
    *p++ = '-';
    buf++;
    ud = -d;
  } else if (base == 'x')
    divisor = 16;
  do {
    int remainder = ud % divisor;
    *p++ = (remainder < 10) ? remainder + '0' : remainder + 'a' - 10;
  } while (ud /= divisor);
  *p = 0;
  p1 = buf;
  p2 = p - 1;
  while (p1 < p2) {
    char tmp = *p1;
    *p1 = *p2;
    *p2 = tmp;
    p1++;
    p2--;
  }
  return (1);
}
```
user: What is the c source code of the assembly code below:

```asm
<ArrayFunction>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D8
data_type: string

tool: D8: %d


assistant: ```c
void ArrayFunction(int *buffer) {
  int temp = buffer[0];
  printf("%d\n", temp);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ArrayFunction>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rdi),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
type_infered: string

tool: D7: %d


assistant: ```c
void ArrayFunction(int *buffer) {
  int temp = buffer[0];
  printf("%d\n", temp);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ArrayFunction>:
  endbr64
  mov    (%rdi),%esi
  lea    D3(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: label: D3
data_type: string

tool: D3: %d


assistant: ```c
void ArrayFunction(int *buffer) {
  int temp = buffer[0];
  printf("%d\n", temp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ArrayFunction>:
  endbr64
  mov    (%rdi),%esi
  lea    D5(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_infered: string

tool: D5: %d


assistant: ```c
void ArrayFunction(int *buffer) {
  int temp = buffer[0];
  printf("%d\n", temp);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x30(%rbp)
  movl   $0x5555,-0x20(%rbp)
  movl   $0x3333,-0x1c(%rbp)
  movl   $0xf0f,-0x18(%rbp)
  movl   $0xff,-0x14(%rbp)
  movl   $0x1111,-0x2c(%rbp)
  movl   $0x0,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L8
L12:
  mov    -0x30(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  and    -0x2c(%rbp),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%edx
  mov    -0x2c(%rbp),%eax
  mov    %edx,%ecx
  shr    %cl,%eax
  mov    %eax,%edx
  mov    -0x30(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  and    %edx,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x28(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x2c(%rbp)
  addl   $0x1,-0x30(%rbp)
L8:
  cmpl   $0x3,-0x30(%rbp)
  jle    L12
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edx
  mov    $0x1111,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
datatype: string

tool: D0: %04X has %d bits set.


assistant: ```c
int main(void) {
  int i = 0;
  unsigned int masks[4];
  masks[0] = 0x5555;
  masks[1] = 0x3333;
  masks[2] = 0x0F0F;
  masks[3] = 0x00FF;
  unsigned int running_sum = 0x1111;
  unsigned int x1 = 0;
  unsigned int x2 = 0;
  for (i = 0; i < 4; i++) {
    x1 = running_sum & masks[i];
    x2 = (running_sum >> (unsigned int)pow(2, i)) & masks[i];
    running_sum = x1 + x2;
  }
  printf("%04X has %d bits set.\n", 0x1111, running_sum);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x5555,(%rsp)
  movl   $0x3333,0x4(%rsp)
  movl   $0xf0f,0x8(%rsp)
  movl   $0xff,0xc(%rsp)
  mov    $0x0,%ebp
  mov    $0x1111,%ebx
L13:
  mov    (%rsp,%rbp,4),%r12d
  mov    %r12d,%r13d
  and    %ebx,%r13d
  mov    %ebp,%esi
  mov    $0x2,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  shr    %cl,%ebx
  and    %r12d,%ebx
  add    %r13d,%ebx
  add    $0x1,%rbp
  cmp    $0x4,%rbp
  jne    L13
  mov    %ebx,%edx
  mov    $0x1111,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D9
type_guessed: string

tool: D9: %04X has %d bits set.


assistant: ```c
int main(void) {
  int i = 0;
  unsigned int masks[4];
  masks[0] = 0x5555;
  masks[1] = 0x3333;
  masks[2] = 0x0F0F;
  masks[3] = 0x00FF;
  unsigned int running_sum = 0x1111;
  unsigned int x1 = 0;
  unsigned int x2 = 0;
  for (i = 0; i < 4; i++) {
    x1 = running_sum & masks[i];
    x2 = (running_sum >> (unsigned int)pow(2, i)) & masks[i];
    running_sum = x1 + x2;
  }
  printf("%04X has %d bits set.\n", 0x1111, running_sum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  mov    $0x1111,%r12d
  push   %rbp
  mov    $0x5555,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x3333,0x4(%rsp)
  movabs $0xff00000f0f,%rax
  mov    %rax,0x8(%rsp)
L22:
  mov    %ebx,%esi
  mov    $0x2,%edi
  mov    %ebp,%r13d
  add    $0x1,%rbx
  call   <pow@plt>
  and    %r12d,%r13d
  mov    %eax,%ecx
  shr    %cl,%r12d
  and    %ebp,%r12d
  add    %r13d,%r12d
  cmp    $0x4,%rbx
  je     L29
  mov    (%rsp,%rbx,4),%ebp
  jmp    L22
  xchg   %ax,%ax
L29:
  mov    %r12d,%edx
  mov    $0x1111,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
type: string

tool: D3: %04X has %d bits set.


assistant: ```c
int main(void) {
  int i = 0;
  unsigned int masks[4];
  masks[0] = 0x5555;
  masks[1] = 0x3333;
  masks[2] = 0x0F0F;
  masks[3] = 0x00FF;
  unsigned int running_sum = 0x1111;
  unsigned int x1 = 0;
  unsigned int x2 = 0;
  for (i = 0; i < 4; i++) {
    x1 = running_sum & masks[i];
    x2 = (running_sum >> (unsigned int)pow(2, i)) & masks[i];
    running_sum = x1 + x2;
  }
  printf("%04X has %d bits set.\n", 0x1111, running_sum);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    $0x1111,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x20,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  movaps %xmm0,(%rsp)
L19:
  mov    0x0(%r13,%rbx,4),%r14d
  mov    %ebx,%esi
  mov    $0x2,%edi
  add    $0x1,%rbx
  call   <pow@plt>
  mov    %r14d,%ebp
  mov    %eax,%ecx
  and    %r12d,%ebp
  shr    %cl,%r12d
  and    %r14d,%r12d
  add    %ebp,%r12d
  cmp    $0x4,%rbx
  jne    L19
  mov    %r12d,%edx
  mov    $0x1111,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D1
type_infered: byte[16]
name: read_data
arguments: data_tag: D2
type_infered: string

tool: {"D1": [0x55,0x55,0x00,0x00,0x33,0x33,0x00,0x00,0x0f,0x0f,0x00,0x00,0xff,0x00,0x00,0x00]}

tool: D2: %04X has %d bits set.


assistant: ```c
int main(void) {
  int i = 0;
  unsigned int masks[4];
  masks[0] = 0x5555;
  masks[1] = 0x3333;
  masks[2] = 0x0F0F;
  masks[3] = 0x00FF;
  unsigned int running_sum = 0x1111;
  unsigned int x1 = 0;
  unsigned int x2 = 0;
  for (i = 0; i < 4; i++) {
    x1 = running_sum & masks[i];
    x2 = (running_sum >> (unsigned int)pow(2, i)) & masks[i];
    running_sum = x1 + x2;
  }
  printf("%04X has %d bits set.\n", 0x1111, running_sum);
  return 0;
}
```