ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<remove_from_chrs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x34(%rbp),%edx
  mov    -0x30(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strchrs@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L14
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  jmp    L12
L14:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    -0x28(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x14(%rbp),%eax
L12:
  leave
  ret
```

assistant: ```c
int remove_from_chrs(char **string, char filters_array[], int negate) {
  char *start = *string;
  char *pos;
  int len;
  pos = strchrs(*string, filters_array, negate);
  if (!pos)
    return strlen(*string);
  *pos = 0;
  len = pos - start;
  *string = realloc(*string, len + 1);
  return len;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<remove_from_chrs>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    (%rdi),%rbp
  mov    %rbp,%rdi
  call   <strchrs@plt>
  test   %rax,%rax
  je     L7
  movb   $0x0,(%rax)
  sub    %rbp,%rax
  mov    %eax,%ebp
  lea    0x1(%rax),%esi
  mov    (%rbx),%rdi
  call   <realloc@plt>
  mov    %rax,(%rbx)
L1:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  mov    (%rbx),%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  jmp    L1
```

assistant: ```c
int remove_from_chrs(char **string, char filters_array[], int negate) {
  char *start = *string;
  char *pos;
  int len;
  pos = strchrs(*string, filters_array, negate);
  if (!pos)
    return strlen(*string);
  *pos = 0;
  len = pos - start;
  *string = realloc(*string, len + 1);
  return len;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<remove_from_chrs>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%rbp
  mov    %rdi,%rbx
  mov    %rbp,%rdi
  call   <strchrs@plt>
  test   %rax,%rax
  je     L1
  movb   $0x0,(%rax)
  sub    %rbp,%rax
  mov    (%rbx),%rdi
  lea    0x1(%rax),%esi
  mov    %eax,%r12d
  call   <realloc@plt>
  mov    %rax,(%rbx)
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    (%rbx),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <strlen@plt>
```

assistant: ```c
int remove_from_chrs(char **string, char filters_array[], int negate) {
  char *start = *string;
  char *pos;
  int len;
  pos = strchrs(*string, filters_array, negate);
  if (!pos)
    return strlen(*string);
  *pos = 0;
  len = pos - start;
  *string = realloc(*string, len + 1);
  return len;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<remove_from_chrs>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    (%rdi),%rbp
  mov    %rdi,%rbx
  mov    %rbp,%rdi
  call   <strchrs@plt>
  test   %rax,%rax
  je     L2
  movb   $0x0,(%rax)
  sub    %rbp,%rax
  mov    (%rbx),%rdi
  lea    0x1(%rax),%esi
  mov    %eax,%r12d
  call   <realloc@plt>
  mov    %rax,(%rbx)
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    (%rbx),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <strlen@plt>
```

assistant: ```c
int remove_from_chrs(char **string, char filters_array[], int negate) {
  char *start = *string;
  char *pos;
  int len;
  pos = strchrs(*string, filters_array, negate);
  if (!pos)
    return strlen(*string);
  *pos = 0;
  len = pos - start;
  *string = realloc(*string, len + 1);
  return len;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<IsMathConstant>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x70,%al
  jne    L38
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x69,%al
  jne    L38
  mov    $0x2,%eax
  jmp    L11
L38:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x65,%al
  jne    L24
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x78,%al
  je     L24
  mov    $0x1,%eax
  jmp    L11
L24:
  mov    $0x0,%eax
L11:
  pop    %rbp
  ret
```

assistant: ```c
int IsMathConstant(const char *cursor) {
  if (cursor[0] == 'p' && cursor[1] == 'i')
    return 2;
  if (cursor[0] == 'e' && cursor[1] != 'x')
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<IsMathConstant>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x70,%dl
  je     L23
  mov    $0x0,%eax
  cmp    $0x65,%dl
  je     L18
L11:
  ret
L23:
  cmpb   $0x69,0x1(%rdi)
  sete   %al
  movzbl %al,%eax
  add    %eax,%eax
  ret
L18:
  cmpb   $0x78,0x1(%rdi)
  setne  %al
  movzbl %al,%eax
  jmp    L11
```

assistant: ```c
int IsMathConstant(const char *cursor) {
  if (cursor[0] == 'p' && cursor[1] == 'i')
    return 2;
  if (cursor[0] == 'e' && cursor[1] != 'x')
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<IsMathConstant>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x70,%dl
  je     L12
  xor    %eax,%eax
  cmp    $0x65,%dl
  je     L11
  ret
  nopl   0x0(%rax)
L11:
  xor    %eax,%eax
  cmpb   $0x78,0x1(%rdi)
  setne  %al
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  xor    %eax,%eax
  cmpb   $0x69,0x1(%rdi)
  sete   %al
  add    %eax,%eax
  ret
```

assistant: ```c
int IsMathConstant(const char *cursor) {
  if (cursor[0] == 'p' && cursor[1] == 'i')
    return 2;
  if (cursor[0] == 'e' && cursor[1] != 'x')
    return 1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<IsMathConstant>:
  endbr64
  movzbl (%rdi),%edx
  cmp    $0x70,%dl
  je     L12
  xor    %eax,%eax
  cmp    $0x65,%dl
  je     L0
  ret
  nopl   0x0(%rax)
L0:
  xor    %eax,%eax
  cmpb   $0x78,0x1(%rdi)
  setne  %al
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  xor    %eax,%eax
  cmpb   $0x69,0x1(%rdi)
  sete   %al
  add    %eax,%eax
  ret
```

assistant: ```c
int IsMathConstant(const char *cursor) {
  if (cursor[0] == 'p' && cursor[1] == 'i')
    return 2;
  if (cursor[0] == 'e' && cursor[1] != 'x')
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<wack_float>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movss  %xmm1,-0x4(%rbp)
  call   <add_float@plt>
  addss  -0x4(%rbp),%xmm0
  leave
  ret
```

assistant: ```c
float wack_float(register float u, register float v) {
  register float l = u, r = v;
  l = add_float(l, r);
  return l + r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<wack_float>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm1,0xc(%rsp)
  call   <add_float@plt>
  addss  0xc(%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

assistant: ```c
float wack_float(register float u, register float v) {
  register float l = u, r = v;
  l = add_float(l, r);
  return l + r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<wack_float>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm1,0xc(%rsp)
  call   <add_float@plt>
  movss  0xc(%rsp),%xmm1
  add    $0x18,%rsp
  addss  %xmm1,%xmm0
  ret
```

assistant: ```c
float wack_float(register float u, register float v) {
  register float l = u, r = v;
  l = add_float(l, r);
  return l + r;
}
```
user: Translate the below assembly code into C programming language:

```asm
<wack_float>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm1,0xc(%rsp)
  call   <add_float@plt>
  movss  0xc(%rsp),%xmm1
  add    $0x18,%rsp
  addss  %xmm1,%xmm0
  ret
```

assistant: ```c
float wack_float(register float u, register float v) {
  register float l = u, r = v;
  l = add_float(l, r);
  return l + r;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<vp9_mb_lpf_horizontal_edge_w_avx2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    %r8,-0x28(%rbp)
  mov    %r9d,-0x10(%rbp)
  cmpl   $0x1,-0x10(%rbp)
  jne    L16
  mov    -0x28(%rbp),%rdi
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <mb_lpf_horizontal_edge_w_avx2_8@plt>
  jmp    L14
L16:
  mov    -0x28(%rbp),%rdi
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0xc(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <mb_lpf_horizontal_edge_w_avx2_16@plt>
L14:
  nop
  leave
  ret
```

assistant: ```c
void vp9_mb_lpf_horizontal_edge_w_avx2(unsigned char *s, int p, const unsigned char *_blimit, const unsigned char *_limit, const unsigned char *_thresh, int count) {
  if (count == 1)
    mb_lpf_horizontal_edge_w_avx2_8(s, p, _blimit, _limit, _thresh);
  else
    mb_lpf_horizontal_edge_w_avx2_16(s, p, _blimit, _limit, _thresh);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vp9_mb_lpf_horizontal_edge_w_avx2>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%r9d
  je     L14
  call   <mb_lpf_horizontal_edge_w_avx2_16@plt>
L13:
  add    $0x8,%rsp
  ret
L14:
  call   <mb_lpf_horizontal_edge_w_avx2_8@plt>
  jmp    L13
```

assistant: ```c
void vp9_mb_lpf_horizontal_edge_w_avx2(unsigned char *s, int p, const unsigned char *_blimit, const unsigned char *_limit, const unsigned char *_thresh, int count) {
  if (count == 1)
    mb_lpf_horizontal_edge_w_avx2_8(s, p, _blimit, _limit, _thresh);
  else
    mb_lpf_horizontal_edge_w_avx2_16(s, p, _blimit, _limit, _thresh);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vp9_mb_lpf_horizontal_edge_w_avx2>:
  endbr64
  cmp    $0x1,%r9d
  je     L5
  jmp    1070 <mb_lpf_horizontal_edge_w_avx2_16@plt>
  nop
L5:
  jmp    1060 <mb_lpf_horizontal_edge_w_avx2_8@plt>
```

assistant: ```c
void vp9_mb_lpf_horizontal_edge_w_avx2(unsigned char *s, int p, const unsigned char *_blimit, const unsigned char *_limit, const unsigned char *_thresh, int count) {
  if (count == 1)
    mb_lpf_horizontal_edge_w_avx2_8(s, p, _blimit, _limit, _thresh);
  else
    mb_lpf_horizontal_edge_w_avx2_16(s, p, _blimit, _limit, _thresh);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vp9_mb_lpf_horizontal_edge_w_avx2>:
  endbr64
  cmp    $0x1,%r9d
  je     L3
  jmp    1070 <mb_lpf_horizontal_edge_w_avx2_16@plt>
  nop
L3:
  jmp    1060 <mb_lpf_horizontal_edge_w_avx2_8@plt>
```

assistant: ```c
void vp9_mb_lpf_horizontal_edge_w_avx2(unsigned char *s, int p, const unsigned char *_blimit, const unsigned char *_limit, const unsigned char *_thresh, int count) {
  if (count == 1)
    mb_lpf_horizontal_edge_w_avx2_8(s, p, _blimit, _limit, _thresh);
  else
    mb_lpf_horizontal_edge_w_avx2_16(s, p, _blimit, _limit, _thresh);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  movsd  D3(%rip),%xmm0
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rax
  mov    %rax,%rdi
  call   <floor@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x2,-0x8(%rbp)
  jmp    L19
L8:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L4
  mov    $0x0,%eax
  jmp    L32
L4:
  addl   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L8
  mov    $0x1,%eax
L32:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
data_type: f64

tool: D3: 0.5

assistant: ```c
int is_prime(int n) {
  int m = floor(sqrt(n) + 0.5);
  for (int i = 2; i <= m; i++)
    if (n % i == 0)
      return 0;
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<is_prime>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  D9(%rip),%xmm0
  cvttsd2si %xmm0,%rdi
  call   <floor@plt>
  cmp    $0x1,%eax
  jle    L17
  mov    %eax,%esi
  test   $0x1,%bl
  je     L55
  mov    $0x2,%ecx
L58:
  add    $0x1,%ecx
  cmp    %ecx,%esi
  jl     L30
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %edx,%edx
  jne    L58
  jmp    L22
L30:
  mov    $0x1,%eax
L22:
  pop    %rbx
  ret
L17:
  mov    $0x1,%eax
  jmp    L22
L55:
  mov    $0x0,%eax
  jmp    L22
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type: f64

tool: D9: 0.5

assistant: ```c
int is_prime(int n) {
  int m = floor(sqrt(n) + 0.5);
  for (int i = 2; i <= m; i++)
    if (n % i == 0)
      return 0;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<is_prime>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  D0(%rip),%xmm0
  cvttsd2si %xmm0,%rdi
  call   <floor@plt>
  cmp    $0x1,%eax
  jle    L5
  test   $0x1,%r12b
  je     L37
  mov    %eax,%esi
  mov    $0x2,%ecx
  jmp    L9
  nopl   (%rax)
L47:
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L1
L9:
  add    $0x1,%ecx
  cmp    %ecx,%esi
  jge    L47
L5:
  mov    $0x1,%edx
L1:
  mov    %edx,%eax
  pop    %r12
  ret
L37:
  xor    %edx,%edx
  jmp    L1
```

tool_calls: name: parse_data
arguments: data_label: D0
type_guessed: f64

tool: D0: 0.5

assistant: ```c
int is_prime(int n) {
  int m = floor(sqrt(n) + 0.5);
  for (int i = 2; i <= m; i++)
    if (n % i == 0)
      return 0;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<is_prime>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  addsd  D5(%rip),%xmm0
  cvttsd2si %xmm0,%rdi
  call   <floor@plt>
  cmp    $0x1,%eax
  jle    L37
  test   $0x1,%r12b
  je     L7
  mov    %eax,%esi
  mov    $0x2,%ecx
  jmp    L46
  nopl   (%rax)
L49:
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L11
L46:
  add    $0x1,%ecx
  cmp    %ecx,%esi
  jge    L49
L37:
  mov    $0x1,%edx
L11:
  mov    %edx,%eax
  pop    %r12
  ret
L7:
  xor    %edx,%edx
  jmp    L11
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type_guessed: f64

tool: D5: 0.5

assistant: ```c
int is_prime(int n) {
  int m = floor(sqrt(n) + 0.5);
  for (int i = 2; i <= m; i++)
    if (n % i == 0)
      return 0;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L22:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L22
  sub    $0xc60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x9c60(%rbp)
  lea    -0x9c50(%rbp),%rax
  mov    $0x2710,%esi
  mov    %rax,%rdi
  call   <sieve@plt>
  movl   $0x0,-0x9c5c(%rbp)
  jmp    L18
L17:
  mov    -0x9c5c(%rbp),%eax
  cltq
  mov    -0x9c50(%rbp,%rax,4),%eax
  cmp    $0x1,%eax
  jne    L13
  addl   $0x1,-0x9c60(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x9c5c(%rbp),%xmm1
  movq   %xmm1,%rdx
  mov    -0x9c60(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L13:
  addl   $0x1,-0x9c5c(%rbp)
L18:
  cmpl   $0x2710,-0x9c5c(%rbp)
  jle    L17
  mov    D24(%rip),%rdx
  mov    -0x9c60(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x9c50(%rbp),%rax
  mov    $0x2710,%esi
  mov    %rax,%rdi
  call   <twin_primes@plt>
  mov    %eax,-0x9c60(%rbp)
  mov    D24(%rip),%rdx
  mov    -0x9c60(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x9c50(%rbp),%rax
  mov    $0x2710,%esi
  mov    %rax,%rdi
  call   <brun_constant@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x9c58(%rbp)
  mov    -0x9c58(%rbp),%rdx
  mov    -0x9c60(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D13
type_infered: string
name: unpack_data
arguments: label: D24
type_infered: byte[8]
name: unpack_data
arguments: label: D11
type_infered: string
name: unpack_data
arguments: label: D39
type_infered: string
name: unpack_data
arguments: label: D51
type_infered: string

tool: D13: Prime #%04i:  %4i


tool: {"D24": [0x00,0x00,0x00,0x00,0x00,0x88,0xc3,0x40]}

tool: D11: %i primes found between 0 and %i.


tool: D39: %i twin primes found between 0 and %i.


tool: D51: Brun's constant with %i twin primes is %1.12f.


assistant: ```c
int main(void) {
  int array[10001], count = 0, i;
  double bruns;
  sieve(array, 10001 - 1);
  for (i = 0; i < 10001; i++) {
    if (array[i] == 1) {
      count++;
      printf("Prime #%04i:  %4i\n", count, i);
    }
  }
  printf("%i primes found between 0 and %i.\n", count, 10001 - 1);
  count = twin_primes(array, 10001 - 1);
  printf("%i twin primes found between 0 and %i.\n", count, 10001 - 1);
  bruns = brun_constant(array, 10001 - 1);
  printf("Brun's constant with %i twin primes is %1.12f.\n", count, bruns);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L27:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L27
  sub    $0xc50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x2710,%esi
  call   <sieve@plt>
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  lea    D9(%rip),%r12
  jmp    L30
L16:
  add    $0x1,%rbx
  cmp    $0x2711,%rbx
  je     L58
L30:
  cmpl   $0x1,(%rsp,%rbx,4)
  jne    L16
  add    $0x1,%ebp
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L16
L58:
  mov    D51(%rip),%rax
  movq   %rax,%xmm0
  mov    %ebp,%esi
  lea    D46(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbp
  mov    $0x2710,%esi
  mov    %rbp,%rdi
  call   <twin_primes@plt>
  mov    %eax,%ebx
  mov    D51(%rip),%rax
  movq   %rax,%xmm0
  mov    %ebx,%esi
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    $0x2710,%esi
  mov    %rbp,%rdi
  call   <brun_constant@plt>
  mov    %ebx,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x9c50,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D9
data_type: string
name: read_data
arguments: data_tag: D51
data_type: byte[8]
name: read_data
arguments: data_tag: D46
data_type: string
name: read_data
arguments: data_tag: D25
data_type: string
name: read_data
arguments: data_tag: D7
data_type: string

tool: D9: Prime #%04i:  %4i


tool: {"D51": [0x00,0x00,0x00,0x00,0x00,0x88,0xc3,0x40]}

tool: D46: %i primes found between 0 and %i.


tool: D25: %i twin primes found between 0 and %i.


tool: D7: Brun's constant with %i twin primes is %1.12f.


assistant: ```c
int main(void) {
  int array[10001], count = 0, i;
  double bruns;
  sieve(array, 10001 - 1);
  for (i = 0; i < 10001; i++) {
    if (array[i] == 1) {
      count++;
      printf("Prime #%04i:  %4i\n", count, i);
    }
  }
  printf("%i primes found between 0 and %i.\n", count, 10001 - 1);
  count = twin_primes(array, 10001 - 1);
  printf("%i twin primes found between 0 and %i.\n", count, 10001 - 1);
  bruns = brun_constant(array, 10001 - 1);
  printf("Brun's constant with %i twin primes is %1.12f.\n", count, bruns);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L17:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L17
  sub    $0xc58,%rsp
  mov    $0x2710,%esi
  xor    %ebx,%ebx
  xor    %r12d,%r12d
  lea    D2(%rip),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x9c48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sieve@plt>
  jmp    L42
  nopl   0x0(%rax,%rax,1)
L31:
  add    $0x1,%rbx
  cmp    $0x2711,%rbx
  je     L19
L42:
  cmpl   $0x1,0x0(%rbp,%rbx,4)
  jne    L31
  pxor   %xmm0,%xmm0
  add    $0x1,%r12d
  mov    %r13,%rdi
  cvtsi2sd %ebx,%xmm0
  mov    %r12d,%esi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x2711,%rbx
  jne    L42
L19:
  movsd  D19(%rip),%xmm0
  mov    %r12d,%esi
  lea    D56(%rip),%rdi
  call   <printf@plt>
  mov    $0x2710,%esi
  mov    %rbp,%rdi
  call   <twin_primes@plt>
  lea    D1(%rip),%rdi
  mov    %eax,%r12d
  mov    D19(%rip),%rax
  mov    %r12d,%esi
  movq   %rax,%xmm0
  call   <printf@plt>
  mov    $0x2710,%esi
  mov    %rbp,%rdi
  call   <brun_constant@plt>
  mov    %r12d,%esi
  lea    D31(%rip),%rdi
  call   <printf@plt>
  mov    0x9c48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x9c58,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D2
type_infered: string
name: unpack_data
arguments: label: D19
type_infered: byte[8]
name: unpack_data
arguments: label: D56
type_infered: string
name: unpack_data
arguments: label: D1
type_infered: string
name: unpack_data
arguments: label: D31
type_infered: string

tool: D2: Prime #%04i:  %4i


tool: {"D19": [0x00,0x00,0x00,0x00,0x00,0x88,0xc3,0x40]}

tool: D56: %i primes found between 0 and %i.


tool: D1: %i twin primes found between 0 and %i.


tool: D31: Brun's constant with %i twin primes is %1.12f.


assistant: ```c
int main(void) {
  int array[10001], count = 0, i;
  double bruns;
  sieve(array, 10001 - 1);
  for (i = 0; i < 10001; i++) {
    if (array[i] == 1) {
      count++;
      printf("Prime #%04i:  %4i\n", count, i);
    }
  }
  printf("%i primes found between 0 and %i.\n", count, 10001 - 1);
  count = twin_primes(array, 10001 - 1);
  printf("%i twin primes found between 0 and %i.\n", count, 10001 - 1);
  bruns = brun_constant(array, 10001 - 1);
  printf("Brun's constant with %i twin primes is %1.12f.\n", count, bruns);
  return 0;
}
```