ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D27(%rip),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x38(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jne    L44
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0xffffffff,%eax
  jmp    L49
L44:
  lea    -0x40(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  lea    D7(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fscanf@plt>
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  lea    D31(%rip),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x30(%rbp)
  movl   $0x1,-0x3c(%rbp)
  jmp    L48
L56:
  mov    -0x40(%rbp),%edx
  mov    -0x3c(%rbp),%eax
  imul   %eax,%eax
  lea    -0x17(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <convert@plt>
  lea    -0x17(%rbp),%rax
  mov    %rax,%rdi
  call   <is_palindrome@plt>
  test   %rax,%rax
  je     L25
  mov    -0x40(%rbp),%edx
  lea    -0x26(%rbp),%rcx
  mov    -0x3c(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <convert@plt>
  lea    -0x17(%rbp),%rcx
  lea    -0x26(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  lea    D56(%rip),%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
L25:
  addl   $0x1,-0x3c(%rbp)
L48:
  cmpl   $0x12c,-0x3c(%rbp)
  jle    L56
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L49:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D27
datatype: string
name: parse_data
arguments: tag: D54
datatype: string
name: parse_data
arguments: tag: D18
datatype: string
name: parse_data
arguments: tag: D7
datatype: string
name: parse_data
arguments: tag: D31
datatype: string
name: parse_data
arguments: tag: D22
datatype: string
name: parse_data
arguments: tag: D56
datatype: string

tool: D27: r

tool: D54: palsquare.in

tool: D18: open file palsquare.in failed

tool: D7: %d

tool: D31: w

tool: D22: palsquare.out

tool: D56: %s %s


assistant: ```c
int main() {
  FILE *fin, *fout;
  int base, mod;
  fin = fopen("palsquare.in", "r");
  if (!fin) {
    perror("open file palsquare.in failed");
    return -1;
  }
  fscanf(fin, "%d", &base);
  fclose(fin);
  fout = fopen("palsquare.out", "w");
  char orig[15];
  char list[15];
  int i, j;
  for (i = 1; i <= 300; ++i) {
    convert(i * i, list, base);
    if (is_palindrome(list)) {
      convert(i, orig, base);
      fprintf(fin, "%s %s\n", orig, list);
    }
  }
  fclose(fin);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D12(%rip),%rsi
  lea    D68(%rip),%rdi
  call   <fopen@plt>
  test   %rax,%rax
  je     L18
  mov    %rax,%r12
  lea    0x4(%rsp),%rdx
  lea    D51(%rip),%rsi
  mov    %rax,%rdi
  call   <fscanf@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  lea    D6(%rip),%rsi
  lea    D45(%rip),%rdi
  call   <fopen@plt>
  mov    $0x1,%ebx
  lea    D48(%rip),%r13
  jmp    L59
L18:
  lea    D41(%rip),%rdi
  call   <perror@plt>
  mov    $0xffffffff,%eax
  jmp    L40
L16:
  add    $0x1,%ebx
  cmp    $0x12d,%ebx
  je     L46
L59:
  lea    0x19(%rsp),%rbp
  mov    %ebx,%edi
  imul   %ebx,%edi
  mov    0x4(%rsp),%edx
  mov    %rbp,%rsi
  call   <convert@plt>
  mov    %rbp,%rdi
  call   <is_palindrome@plt>
  test   %rax,%rax
  je     L16
  lea    0xa(%rsp),%rbp
  mov    0x4(%rsp),%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <convert@plt>
  lea    0x19(%rsp),%rcx
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <fprintf@plt>
  jmp    L16
L46:
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L40:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L21
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D12
datatype: string
name: extract_symbolic_data
arguments: data_label: D68
datatype: string
name: extract_symbolic_data
arguments: data_label: D51
datatype: string
name: extract_symbolic_data
arguments: data_label: D6
datatype: string
name: extract_symbolic_data
arguments: data_label: D45
datatype: string
name: extract_symbolic_data
arguments: data_label: D48
datatype: string
name: extract_symbolic_data
arguments: data_label: D41
datatype: string

tool: D12: r

tool: D68: palsquare.in

tool: D51: %d

tool: D6: w

tool: D45: palsquare.out

tool: D48: %s %s


tool: D41: open file palsquare.in failed

assistant: ```c
int main() {
  FILE *fin, *fout;
  int base, mod;
  fin = fopen("palsquare.in", "r");
  if (!fin) {
    perror("open file palsquare.in failed");
    return -1;
  }
  fscanf(fin, "%d", &base);
  fclose(fin);
  fout = fopen("palsquare.out", "w");
  char orig[15];
  char list[15];
  int i, j;
  for (i = 1; i <= 300; ++i) {
    convert(i * i, list, base);
    if (is_palindrome(list)) {
      convert(i, orig, base);
      fprintf(fin, "%s %s\n", orig, list);
    }
  }
  fclose(fin);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  lea    D39(%rip),%rsi
  lea    D49(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  test   %rax,%rax
  je     L11
  lea    0x4(%rsp),%rdx
  mov    %rax,%r12
  mov    %rax,%rdi
  mov    $0x1,%ebx
  lea    D50(%rip),%rsi
  lea    0x19(%rsp),%rbp
  call   <fscanf@plt>
  mov    %r12,%rdi
  lea    D65(%rip),%r13
  call   <fclose@plt>
  lea    D14(%rip),%rsi
  lea    D22(%rip),%rdi
  call   <fopen@plt>
  jmp    L29
  nopl   0x0(%rax)
L54:
  add    $0x1,%ebx
  cmp    $0x12d,%ebx
  je     L49
L29:
  mov    %ebx,%edi
  mov    0x4(%rsp),%edx
  mov    %rbp,%rsi
  imul   %ebx,%edi
  call   <convert@plt>
  mov    %rbp,%rdi
  call   <is_palindrome@plt>
  test   %rax,%rax
  je     L54
  mov    0x4(%rsp),%edx
  lea    0xa(%rsp),%r14
  mov    %ebx,%edi
  add    $0x1,%ebx
  mov    %r14,%rsi
  call   <convert@plt>
  mov    %rbp,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <fprintf@plt>
  cmp    $0x12d,%ebx
  jne    L29
L49:
  mov    %r12,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L27:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  lea    D17(%rip),%rdi
  call   <perror@plt>
  or     $0xffffffff,%eax
  jmp    L27
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D39
type_guessed: string
name: retrieve_data
arguments: data_label: D49
type_guessed: string
name: retrieve_data
arguments: data_label: D50
type_guessed: string
name: retrieve_data
arguments: data_label: D65
type_guessed: string
name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D22
type_guessed: string
name: retrieve_data
arguments: data_label: D17
type_guessed: string

tool: D39: r

tool: D49: palsquare.in

tool: D50: %d

tool: D65: %s %s


tool: D14: w

tool: D22: palsquare.out

tool: D17: open file palsquare.in failed

assistant: ```c
int main() {
  FILE *fin, *fout;
  int base, mod;
  fin = fopen("palsquare.in", "r");
  if (!fin) {
    perror("open file palsquare.in failed");
    return -1;
  }
  fscanf(fin, "%d", &base);
  fclose(fin);
  fout = fopen("palsquare.out", "w");
  char orig[15];
  char list[15];
  int i, j;
  for (i = 1; i <= 300; ++i) {
    convert(i * i, list, base);
    if (is_palindrome(list)) {
      convert(i, orig, base);
      fprintf(fin, "%s %s\n", orig, list);
    }
  }
  fclose(fin);
  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   %r14
  lea    D26(%rip),%rsi
  lea    D67(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  test   %rax,%rax
  je     L0
  lea    0x4(%rsp),%rdx
  mov    %rax,%r12
  mov    %rax,%rdi
  mov    $0x1,%ebx
  lea    D9(%rip),%rsi
  lea    0x19(%rsp),%rbp
  call   <fscanf@plt>
  mov    %r12,%rdi
  lea    D57(%rip),%r13
  call   <fclose@plt>
  lea    D21(%rip),%rsi
  lea    D0(%rip),%rdi
  call   <fopen@plt>
  jmp    L51
  nopl   0x0(%rax)
L41:
  add    $0x1,%ebx
  cmp    $0x12d,%ebx
  je     L15
L51:
  mov    %ebx,%edi
  mov    0x4(%rsp),%edx
  mov    %rbp,%rsi
  imul   %ebx,%edi
  call   <convert@plt>
  mov    %rbp,%rdi
  call   <is_palindrome@plt>
  test   %rax,%rax
  je     L41
  mov    0x4(%rsp),%edx
  lea    0xa(%rsp),%r14
  mov    %ebx,%edi
  add    $0x1,%ebx
  mov    %r14,%rsi
  call   <convert@plt>
  mov    %rbp,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <fprintf@plt>
  cmp    $0x12d,%ebx
  jne    L51
L15:
  mov    %r12,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L33:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L63
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  lea    D32(%rip),%rdi
  call   <perror@plt>
  or     $0xffffffff,%eax
  jmp    L33
L63:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D26
type_infered: string
name: read_data
arguments: label: D67
type_infered: string
name: read_data
arguments: label: D9
type_infered: string
name: read_data
arguments: label: D57
type_infered: string
name: read_data
arguments: label: D21
type_infered: string
name: read_data
arguments: label: D0
type_infered: string
name: read_data
arguments: label: D32
type_infered: string

tool: D26: r

tool: D67: palsquare.in

tool: D9: %d

tool: D57: %s %s


tool: D21: w

tool: D0: palsquare.out

tool: D32: open file palsquare.in failed

assistant: ```c
int main() {
  FILE *fin, *fout;
  int base, mod;
  fin = fopen("palsquare.in", "r");
  if (!fin) {
    perror("open file palsquare.in failed");
    return -1;
  }
  fscanf(fin, "%d", &base);
  fclose(fin);
  fout = fopen("palsquare.out", "w");
  char orig[15];
  char list[15];
  int i, j;
  for (i = 1; i <= 300; ++i) {
    convert(i * i, list, base);
    if (is_palindrome(list)) {
      convert(i, orig, base);
      fprintf(fin, "%s %s\n", orig, list);
    }
  }
  fclose(fin);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_fff_pth_element>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r11
  mov    %rsi,%rbx
  lea    -0x1(%rcx),%r10
  lea    0x0(,%rdx,8),%r8
  mov    %r8,%rdi
  neg    %rdi
  mov    $0x0,%ebp
  mov    $0x0,%r12d
  jmp    L115
L58:
  ucomisd %xmm0,%xmm1
  setnp  %r13b
  movzbl %r13b,%r13d
  cmovne %r12d,%r13d
  jmp    L104
L23:
  mov    %rcx,%r10
L115:
  mov    %r8,%r9
  imul   %rbp,%r9
  add    %r11,%r9
  mov    %r8,%rax
  imul   %r10,%rax
  add    %r11,%rax
  movsd  (%r9),%xmm1
  movsd  (%rax),%xmm0
  comisd %xmm0,%xmm1
  jbe    L58
  movsd  %xmm0,(%r9)
  movsd  %xmm1,(%rax)
  mov    %r12d,%r13d
L104:
  movsd  (%r9),%xmm2
  cmp    %r10,%rbp
  je     L159
  lea    (%r9,%r8,1),%rdx
  lea    0x1(%rbp),%rsi
  mov    %r10,%rcx
L78:
  movsd  (%rdx),%xmm0
  comisd %xmm0,%xmm2
  jbe    L102
L91:
  add    $0x1,%rsi
  add    %r8,%rdx
  movsd  (%rdx),%xmm0
  comisd %xmm0,%xmm2
  ja     L91
L102:
  movsd  (%rax),%xmm1
  comisd %xmm2,%xmm1
  jbe    L113
L145:
  sub    $0x1,%rcx
  add    %rdi,%rax
  movsd  (%rax),%xmm1
  comisd %xmm2,%xmm1
  ja     L145
L113:
  cmp    %rcx,%rsi
  jae    L136
  movsd  %xmm1,(%rdx)
  movsd  %xmm0,(%rax)
  sub    $0x1,%rcx
  add    %rdi,%rax
  add    $0x1,%rsi
  add    %r8,%rdx
  cmp    %r10,%rcx
  jne    L78
  test   %r13d,%r13d
  je     L78
  jmp    L137
L136:
  cmp    %r10,%rcx
  jne    L95
  test   %r13d,%r13d
  je     L95
L137:
  sub    $0x1,%rcx
  add    %rdi,%rax
  movsd  (%r9),%xmm0
  movsd  (%rax),%xmm1
  movsd  %xmm1,(%r9)
  movsd  %xmm0,(%rax)
L95:
  cmp    %rbx,%rcx
  ja     L23
  jae    L159
  mov    %rsi,%rbp
  jmp    L115
L159:
  movapd %xmm2,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
double _fff_pth_element(double *x, size_t p, size_t stride, size_t n) {
  double a, tmp;
  double *bufl, *bufr;
  size_t i, j, il, jr, stop1, stop2;
  int same_extremities;
  stop1 = 0;
  il = 0;
  jr = n - 1;
  while (stop1 == 0) {
    same_extremities = 0;
    bufl = x + stride * il;
    bufr = x + stride * jr;
    if (*bufl > *bufr) {
      tmp = (*bufl);
      (*bufl) = (*bufr);
      (*bufr) = tmp;
    } else if (*bufl == *bufr)
      same_extremities = 1;
    a = *bufl;
    if (il == jr)
      return a;
    bufl += stride;
    i = il + 1;
    j = jr;
    stop2 = 0;
    while (stop2 == 0) {
      while (*bufl < a) {
        i++;
        bufl += stride;
      }
      while (*bufr > a) {
        j--;
        bufr -= stride;
      }
      if (j <= i)
        stop2 = 1;
      else {
        {
          tmp = (*bufl);
          (*bufl) = (*bufr);
          (*bufr) = tmp;
        }
        j--;
        bufr -= stride;
        i++;
        bufl += stride;
      }
      if ((same_extremities) && (j == jr)) {
        j--;
        bufr -= stride;
        {
          tmp = (x[il * stride]);
          (x[il * stride]) = (*bufr);
          (*bufr) = tmp;
        }
        stop2 = 1;
      }
    }
    if (j > p)
      jr = j;
    else if (j < p)
      il = i;
    else
      stop1 = 1;
  }
  return a;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<next_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L19
L18:
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <is_prime@plt>
  test   %eax,%eax
  je     L18
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned int next_prime(unsigned int start) {
  int i = start + 1;
  while (!is_prime(i))
    i++;
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<next_prime>:
  endbr64
  push   %rbx
  lea    0x1(%rdi),%ebx
  jmp    L4
L9:
  add    $0x1,%ebx
L4:
  mov    %ebx,%edi
  call   <is_prime@plt>
  test   %eax,%eax
  je     L9
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
unsigned int next_prime(unsigned int start) {
  int i = start + 1;
  while (!is_prime(i))
    i++;
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<next_prime>:
  endbr64
  push   %r12
  lea    0x1(%rdi),%r12d
  jmp    L8
  nopl   0x0(%rax)
L10:
  add    $0x1,%r12d
L8:
  mov    %r12d,%edi
  call   <is_prime@plt>
  test   %eax,%eax
  je     L10
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
unsigned int next_prime(unsigned int start) {
  int i = start + 1;
  while (!is_prime(i))
    i++;
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<next_prime>:
  endbr64
  push   %r12
  lea    0x1(%rdi),%r12d
  jmp    L3
  nopl   0x0(%rax)
L19:
  add    $0x1,%r12d
L3:
  mov    %r12d,%edi
  call   <is_prime@plt>
  test   %eax,%eax
  je     L19
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
unsigned int next_prime(unsigned int start) {
  int i = start + 1;
  while (!is_prime(i))
    i++;
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x5,-0x10(%rbp)
  movl   $0xb,-0xc(%rbp)
  movl   $0xc,-0x8(%rbp)
  jmp    L13
L62:
  movl   $0x1,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L25
  movl   $0x0,-0x4(%rbp)
  jmp    L90
L25:
  movl   $0x3,-0x14(%rbp)
  jmp    L77
L99:
  mov    -0x8(%rbp),%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L76
  movl   $0x0,-0x4(%rbp)
  jmp    L90
L76:
  addl   $0x2,-0x14(%rbp)
L77:
  mov    -0x14(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L99
L90:
  cmpl   $0x0,-0x4(%rbp)
  je     L3
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L3:
  addl   $0x1,-0x8(%rbp)
L13:
  cmpl   $0x64,-0x10(%rbp)
  jne    L62
  cmpl   $0x21d,-0xc(%rbp)
  je     L34
  mov    $0x1,%eax
  jmp    L18
L34:
  mov    $0x0,%eax
L18:
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int i, n, p, next, isprime;
  n = 5;
  p = 11;
  next = 12;
  while (n != 100) {
    isprime = 1;
    if (next % 2 == 0) {
      isprime = 0;
    } else {
      for (i = 3; i < next; i = i + 2) {
        if (next % i == 0) {
          isprime = 0;
          break;
        }
      }
    }
    if (isprime) {
      p = next;
      n++;
    }
    next = next + 1;
  }
  if (p != 541)
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  mov    $0xc,%esi
  mov    $0xb,%r8d
  mov    $0x5,%edi
  jmp    L56
L72:
  add    $0x1,%edi
  mov    %esi,%r8d
L56:
  add    $0x1,%esi
  cmp    $0x64,%edi
  je     L20
  test   $0x1,%sil
  je     L56
  cmp    $0x3,%esi
  jle    L72
  movslq %esi,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%eax
  cmp    %eax,%esi
  je     L56
  mov    $0x3,%ecx
L45:
  add    $0x2,%ecx
  cmp    %esi,%ecx
  jge    L72
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L45
  jmp    L56
L20:
  cmp    $0x21d,%r8d
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int main() {
  int i, n, p, next, isprime;
  n = 5;
  p = 11;
  next = 12;
  while (n != 100) {
    isprime = 1;
    if (next % 2 == 0) {
      isprime = 0;
    } else {
      for (i = 3; i < next; i = i + 2) {
        if (next % i == 0) {
          isprime = 0;
          break;
        }
      }
    }
    if (isprime) {
      p = next;
      n++;
    }
    next = next + 1;
  }
  if (p != 541)
    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
<main>:
  endbr64
  mov    $0xc,%esi
  mov    $0xb,%r8d
  mov    $0x5,%edi
  nopl   0x0(%rax)
L16:
  add    $0x1,%esi
  cmp    $0x64,%edi
  je     L36
L33:
  test   $0x1,%sil
  je     L16
  mov    $0x3,%ecx
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L16
  add    $0x2,%ecx
  cmp    %esi,%ecx
  jl     L13
  add    $0x1,%edi
  mov    %esi,%r8d
  add    $0x1,%esi
  cmp    $0x64,%edi
  jne    L33
L36:
  xor    %eax,%eax
  cmp    $0x21d,%r8d
  setne  %al
  ret
```

assistant: ```c
int main() {
  int i, n, p, next, isprime;
  n = 5;
  p = 11;
  next = 12;
  while (n != 100) {
    isprime = 1;
    if (next % 2 == 0) {
      isprime = 0;
    } else {
      for (i = 3; i < next; i = i + 2) {
        if (next % i == 0) {
          isprime = 0;
          break;
        }
      }
    }
    if (isprime) {
      p = next;
      n++;
    }
    next = next + 1;
  }
  if (p != 541)
    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
<main>:
  endbr64
  mov    $0xd,%esi
  mov    $0xb,%r8d
  mov    $0x5,%edi
  jmp    L32
  cs nopw 0x0(%rax,%rax,1)
L14:
  add    $0x1,%esi
  cmp    $0x64,%edi
  je     L16
L32:
  test   $0x1,%sil
  je     L14
  mov    $0x3,%ecx
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %esi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L14
  add    $0x2,%ecx
  cmp    %ecx,%esi
  jg     L10
  add    $0x1,%edi
  mov    %esi,%r8d
  add    $0x1,%esi
  cmp    $0x64,%edi
  jne    L32
L16:
  xor    %eax,%eax
  cmp    $0x21d,%r8d
  setne  %al
  ret
```

assistant: ```c
int main() {
  int i, n, p, next, isprime;
  n = 5;
  p = 11;
  next = 12;
  while (n != 100) {
    isprime = 1;
    if (next % 2 == 0) {
      isprime = 0;
    } else {
      for (i = 3; i < next; i = i + 2) {
        if (next % i == 0) {
          isprime = 0;
          break;
        }
      }
    }
    if (isprime) {
      p = next;
      n++;
    }
    next = next + 1;
  }
  if (p != 541)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<poclcc_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <print_help@plt>
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type_guessed: string

tool: D2: ERROR: %s

assistant: ```c
int poclcc_error(char *msg) {
  printf("ERROR: %s", msg);
  return print_help();
}
```
user: Translate the below assembly code into C programming language:

```asm
<poclcc_error>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <print_help@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type: string

tool: D7: ERROR: %s

assistant: ```c
int poclcc_error(char *msg) {
  printf("ERROR: %s", msg);
  return print_help();
}
```
user: Translate the below assembly code into C programming language:

```asm
<poclcc_error>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <print_help@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
datatype: string

tool: D2: ERROR: %s

assistant: ```c
int poclcc_error(char *msg) {
  printf("ERROR: %s", msg);
  return print_help();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<poclcc_error>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <print_help@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D6
data_type: string

tool: D6: ERROR: %s

assistant: ```c
int poclcc_error(char *msg) {
  printf("ERROR: %s", msg);
  return print_help();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<com_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <com_malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *com_strdup(const char *s) {
  char *dup;
  int len = strlen(s) + 1;
  dup = com_malloc(len);
  memcpy(dup, s, len);
  return dup;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<com_strdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r12d
  mov    %r12d,%edi
  call   <com_malloc@plt>
  mov    %rax,%rbx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *com_strdup(const char *s) {
  char *dup;
  int len = strlen(s) + 1;
  dup = com_malloc(len);
  memcpy(dup, s, len);
  return dup;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<com_strdup>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%r13d
  mov    %r13d,%edi
  call   <com_malloc@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *com_strdup(const char *s) {
  char *dup;
  int len = strlen(s) + 1;
  dup = com_malloc(len);
  memcpy(dup, s, len);
  return dup;
}
```