ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<array_push>:
  endbr64
  test   %edx,%edx
  jle    L22
  lea    -0x1(%rdx),%eax
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x8(%rsi,%rax,8),%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L13
  nopl   0x0(%rax)
L2:
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  je     L36
L13:
  mov    (%rbx),%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jne    L2
  mov    %r12,(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L36:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L22:
  xor    %eax,%eax
  ret
```

assistant: ```c
int array_push(char *v, char **array, int count) {
  for (int i = 0; i < count; ++i) {
    if (!strlen(array[i])) {
      array[i] = v;
      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
<array_push>:
  endbr64
  test   %edx,%edx
  jle    L25
  lea    -0x1(%rdx),%eax
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x8(%rsi,%rax,8),%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L4
  nopl   0x0(%rax)
L14:
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  je     L21
L4:
  mov    (%rbx),%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jne    L14
  mov    %r12,(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L21:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L25:
  xor    %eax,%eax
  ret
```

assistant: ```c
int array_push(char *v, char **array, int count) {
  for (int i = 0; i < count; ++i) {
    if (!strlen(array[i])) {
      array[i] = v;
      return 1;
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <test_arg_parse@plt>
  mov    $0x0,%eax
  call   <test_qarg_parse@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  test_arg_parse();
  test_qarg_parse();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test_arg_parse@plt>
  mov    $0x0,%eax
  call   <test_qarg_parse@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_arg_parse@plt>
  xor    %eax,%eax
  call   <test_qarg_parse@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test_arg_parse();
  test_qarg_parse();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_arg_parse@plt>
  xor    %eax,%eax
  call   <test_qarg_parse@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x9c(%rbp)
  jmp    L48
L53:
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  lea    -0xa0(%rbp),%rdx
  lea    -0x90(%rbp),%rax
  lea    D7(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  mov    %eax,-0x9c(%rbp)
L48:
  cmpl   $0x0,-0x9c(%rbp)
  jle    L53
  movl   $0x0,-0x9c(%rbp)
  jmp    L59
L36:
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x90(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  lea    -0xa4(%rbp),%rdx
  lea    -0x90(%rbp),%rax
  lea    D7(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  mov    %eax,-0x9c(%rbp)
  mov    -0xa4(%rbp),%eax
  test   %eax,%eax
  jne    L59
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x9c(%rbp)
L59:
  cmpl   $0x0,-0x9c(%rbp)
  jle    L36
  mov    -0xa0(%rbp),%eax
  mov    -0xa4(%rbp),%esi
  cltd
  idiv   %esi
  mov    %eax,-0x98(%rbp)
  mov    -0xa0(%rbp),%eax
  mov    -0xa4(%rbp),%ecx
  cltd
  idiv   %ecx
  mov    %edx,-0x94(%rbp)
  mov    -0xa4(%rbp),%edi
  mov    -0xa4(%rbp),%edx
  mov    -0xa0(%rbp),%eax
  mov    -0x94(%rbp),%esi
  mov    -0x98(%rbp),%ecx
  mov    %edi,%r9d
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D3
type: string
name: read_data
arguments: data_label: D7
type: string
name: read_data
arguments: data_label: D33
type: string
name: read_data
arguments: data_label: D23
type: string
name: read_data
arguments: data_label: D29
type: string

tool: D3: Enter the dividend: 

tool: D7: %d

tool: D33: Enter the divisor: 

tool: D23: Division by zero is not allowed

tool: D29: %d / %d = %d and %d/%d


assistant: ```c
main() {
  int divisor, dividend, quotient, remainder;
  int valid_input = 0;
  char inbuf[133];
  while (valid_input < 1) {
    printf("Enter the dividend: ");
    gets(inbuf);
    valid_input = sscanf(inbuf, "%d", &dividend);
  }
  valid_input = 0;
  while (valid_input < 1) {
    printf("Enter the divisor: ");
    gets(inbuf);
    valid_input = sscanf(inbuf, "%d", &divisor);
    if (divisor == 0) {
      printf("Division by zero is not allowed\n");
      valid_input = 0;
    }
  }
  quotient = dividend / divisor;
  remainder = dividend % divisor;
  printf("%d / %d = %d and %d/%d\n", dividend, divisor, quotient, remainder, divisor);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    D5(%rip),%r13
  lea    0x10(%rsp),%rbx
  lea    0xc(%rsp),%r12
  lea    D15(%rip),%rbp
L59:
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <gets@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  test   %eax,%eax
  jle    L59
  lea    D18(%rip),%r12
  lea    D15(%rip),%rbp
  lea    D45(%rip),%r13
  jmp    L47
L18:
  mov    %r13,%rdi
  call   <puts@plt>
L25:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <gets@plt>
  lea    0x8(%rsp),%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  mov    0x8(%rsp),%r9d
  test   %r9d,%r9d
  je     L18
  test   %eax,%eax
  jg     L16
L47:
  lea    0x10(%rsp),%rbx
  jmp    L25
L16:
  mov    0xc(%rsp),%esi
  mov    %esi,%eax
  cltd
  idiv   %r9d
  mov    %edx,%r8d
  mov    %eax,%ecx
  mov    %r9d,%edx
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0xa8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
data_type: string
name: extract_symbolic_data
arguments: label: D15
data_type: string
name: extract_symbolic_data
arguments: label: D18
data_type: string
name: extract_symbolic_data
arguments: label: D45
data_type: string
name: extract_symbolic_data
arguments: label: D19
data_type: string

tool: D5: Enter the dividend: 

tool: D15: %d

tool: D18: Enter the divisor: 

tool: D45: Division by zero is not allowed

tool: D19: %d / %d = %d and %d/%d


assistant: ```c
main() {
  int divisor, dividend, quotient, remainder;
  int valid_input = 0;
  char inbuf[133];
  while (valid_input < 1) {
    printf("Enter the dividend: ");
    gets(inbuf);
    valid_input = sscanf(inbuf, "%d", &dividend);
  }
  valid_input = 0;
  while (valid_input < 1) {
    printf("Enter the divisor: ");
    gets(inbuf);
    valid_input = sscanf(inbuf, "%d", &divisor);
    if (divisor == 0) {
      printf("Division by zero is not allowed\n");
      valid_input = 0;
    }
  }
  quotient = dividend / divisor;
  remainder = dividend % divisor;
  printf("%d / %d = %d and %d/%d\n", dividend, divisor, quotient, remainder, divisor);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  push   %rbp
  lea    D45(%rip),%rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0xc(%rsp),%r12
L33:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <gets@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  test   %eax,%eax
  jle    L33
  lea    D14(%rip),%r13
  lea    0x8(%rsp),%r12
  lea    D22(%rip),%r14
  jmp    L39
  nopw   0x0(%rax,%rax,1)
L41:
  mov    %r14,%rdi
  call   <puts@plt>
L39:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <gets@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  mov    0x8(%rsp),%r9d
  test   %r9d,%r9d
  je     L41
  test   %eax,%eax
  jle    L39
  mov    0xc(%rsp),%esi
  lea    D42(%rip),%rdi
  mov    %esi,%eax
  cltd
  idiv   %r9d
  mov    %edx,%r8d
  mov    %eax,%ecx
  mov    %r9d,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D16
type_guessed: string
name: parse_data
arguments: label: D45
type_guessed: string
name: parse_data
arguments: label: D14
type_guessed: string
name: parse_data
arguments: label: D22
type_guessed: string
name: parse_data
arguments: label: D42
type_guessed: string

tool: D16: Enter the dividend: 

tool: D45: %d

tool: D14: Enter the divisor: 

tool: D22: Division by zero is not allowed

tool: D42: %d / %d = %d and %d/%d


assistant: ```c
main() {
  int divisor, dividend, quotient, remainder;
  int valid_input = 0;
  char inbuf[133];
  while (valid_input < 1) {
    printf("Enter the dividend: ");
    gets(inbuf);
    valid_input = sscanf(inbuf, "%d", &dividend);
  }
  valid_input = 0;
  while (valid_input < 1) {
    printf("Enter the divisor: ");
    gets(inbuf);
    valid_input = sscanf(inbuf, "%d", &divisor);
    if (divisor == 0) {
      printf("Division by zero is not allowed\n");
      valid_input = 0;
    }
  }
  quotient = dividend / divisor;
  remainder = dividend % divisor;
  printf("%d / %d = %d and %d/%d\n", dividend, divisor, quotient, remainder, divisor);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  lea    D34(%rip),%r13
  push   %r12
  push   %rbp
  lea    D36(%rip),%rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  lea    0xc(%rsp),%r12
L7:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <gets@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  test   %eax,%eax
  jle    L7
  lea    D30(%rip),%r13
  lea    0x8(%rsp),%r12
  lea    D10(%rip),%r14
  jmp    L37
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r14,%rdi
  call   <puts@plt>
L37:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <gets@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  mov    0x8(%rsp),%r9d
  test   %r9d,%r9d
  je     L6
  test   %eax,%eax
  jle    L37
  mov    0xc(%rsp),%esi
  lea    D19(%rip),%rdi
  mov    %esi,%eax
  cltd
  idiv   %r9d
  mov    %edx,%r8d
  mov    %eax,%ecx
  mov    %r9d,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D34
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D36
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D30
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D10
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D19
type_infered: string

tool: D34: Enter the dividend: 

tool: D36: %d

tool: D30: Enter the divisor: 

tool: D10: Division by zero is not allowed

tool: D19: %d / %d = %d and %d/%d


assistant: ```c
main() {
  int divisor, dividend, quotient, remainder;
  int valid_input = 0;
  char inbuf[133];
  while (valid_input < 1) {
    printf("Enter the dividend: ");
    gets(inbuf);
    valid_input = sscanf(inbuf, "%d", &dividend);
  }
  valid_input = 0;
  while (valid_input < 1) {
    printf("Enter the divisor: ");
    gets(inbuf);
    valid_input = sscanf(inbuf, "%d", &divisor);
    if (divisor == 0) {
      printf("Division by zero is not allowed\n");
      valid_input = 0;
    }
  }
  quotient = dividend / divisor;
  remainder = dividend % divisor;
  printf("%d / %d = %d and %d/%d\n", dividend, divisor, quotient, remainder, divisor);
}
```
user: Translate the below assembly code into C programming language:

```asm
<int_as_float>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  pop    %rbp
  ret
```

assistant: ```c
float int_as_float(uint32_t intval) {
  union intfloat {
    uint32_t i;
    float f;
  } uval;
  uval.i = intval;
  return uval.f;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<int_as_float>:
  endbr64
  movd   %edi,%xmm0
  ret
```

assistant: ```c
float int_as_float(uint32_t intval) {
  union intfloat {
    uint32_t i;
    float f;
  } uval;
  uval.i = intval;
  return uval.f;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<int_as_float>:
  endbr64
  movd   %edi,%xmm0
  ret
```

assistant: ```c
float int_as_float(uint32_t intval) {
  union intfloat {
    uint32_t i;
    float f;
  } uval;
  uval.i = intval;
  return uval.f;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<int_as_float>:
  endbr64
  movd   %edi,%xmm0
  ret
```

assistant: ```c
float int_as_float(uint32_t intval) {
  union intfloat {
    uint32_t i;
    float f;
  } uval;
  uval.i = intval;
  return uval.f;
}
```
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
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x20(%rbp),%eax
  mov    %eax,%edi
  call   <fact@plt>
  mov    %eax,%ebx
  mov    -0x20(%rbp),%eax
  mov    -0x24(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,%edi
  call   <fact@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0x1c(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    -0x1c(%rbp),%ecx
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D13
data_type: string
name: disassemble_data
arguments: tag: D19
data_type: string
name: disassemble_data
arguments: tag: D5
data_type: string
name: disassemble_data
arguments: tag: D38
data_type: string
name: disassemble_data
arguments: tag: D39
data_type: string

tool: D13:  Enter permutation number for the patern P(n,k)

tool: D19: Enter n value : 

tool: D5: %d

tool: D38: 
Enter k value : 

tool: D39: 
Factorial of P(%d,%d) = %d

assistant: ```c
int main() {
  int p, k, n;
  printf(" Enter permutation number for the patern P(n,k)\n");
  printf("Enter n value : ");
  scanf("%d", &n);
  printf("\nEnter k value : ");
  scanf("%d", &k);
  p = (fact(n) / fact(n - k));
  printf("\nFactorial of P(%d,%d) = %d", n, k, p);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D30(%rip),%rdi
  call   <puts@plt>
  lea    D42(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D22(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D44(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <fact@plt>
  mov    %eax,%ebx
  mov    0x4(%rsp),%edi
  sub    (%rsp),%edi
  call   <fact@plt>
  mov    %eax,%ecx
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %eax,%ecx
  mov    (%rsp),%edx
  mov    0x4(%rsp),%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D30
data_type: string
name: parse_data
arguments: data_tag: D42
data_type: string
name: parse_data
arguments: data_tag: D22
data_type: string
name: parse_data
arguments: data_tag: D44
data_type: string
name: parse_data
arguments: data_tag: D9
data_type: string

tool: D30:  Enter permutation number for the patern P(n,k)

tool: D42: Enter n value : 

tool: D22: %d

tool: D44: 
Enter k value : 

tool: D9: 
Factorial of P(%d,%d) = %d

assistant: ```c
int main() {
  int p, k, n;
  printf(" Enter permutation number for the patern P(n,k)\n");
  printf("Enter n value : ");
  scanf("%d", &n);
  printf("\nEnter k value : ");
  scanf("%d", &k);
  p = (fact(n) / fact(n - k));
  printf("\nFactorial of P(%d,%d) = %d", n, k, p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D15(%rip),%rdi
  lea    D37(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <fact@plt>
  mov    0x4(%rsp),%edi
  sub    (%rsp),%edi
  mov    %eax,%ebx
  call   <fact@plt>
  mov    0x4(%rsp),%esi
  lea    D34(%rip),%rdi
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  mov    (%rsp),%edx
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D15
type_infered: string
name: extract_symbolic_data
arguments: data_label: D37
type_infered: string
name: extract_symbolic_data
arguments: data_label: D11
type_infered: string
name: extract_symbolic_data
arguments: data_label: D24
type_infered: string
name: extract_symbolic_data
arguments: data_label: D34
type_infered: string

tool: D15:  Enter permutation number for the patern P(n,k)

tool: D37: %d

tool: D11: Enter n value : 

tool: D24: 
Enter k value : 

tool: D34: 
Factorial of P(%d,%d) = %d

assistant: ```c
int main() {
  int p, k, n;
  printf(" Enter permutation number for the patern P(n,k)\n");
  printf("Enter n value : ");
  scanf("%d", &n);
  printf("\nEnter k value : ");
  scanf("%d", &k);
  p = (fact(n) / fact(n - k));
  printf("\nFactorial of P(%d,%d) = %d", n, k, p);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D13(%rip),%rdi
  lea    D11(%rip),%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <fact@plt>
  mov    0x4(%rsp),%edi
  sub    (%rsp),%edi
  mov    %eax,%ebx
  call   <fact@plt>
  mov    0x4(%rsp),%esi
  lea    D31(%rip),%rdi
  mov    %eax,%r8d
  mov    %ebx,%eax
  cltd
  idiv   %r8d
  mov    (%rsp),%edx
  mov    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D13
data_type: string
name: retrieve_data
arguments: tag: D11
data_type: string
name: retrieve_data
arguments: tag: D17
data_type: string
name: retrieve_data
arguments: tag: D6
data_type: string
name: retrieve_data
arguments: tag: D31
data_type: string

tool: D13:  Enter permutation number for the patern P(n,k)

tool: D11: %d

tool: D17: Enter n value : 

tool: D6: 
Enter k value : 

tool: D31: 
Factorial of P(%d,%d) = %d

assistant: ```c
int main() {
  int p, k, n;
  printf(" Enter permutation number for the patern P(n,k)\n");
  printf("Enter n value : ");
  scanf("%d", &n);
  printf("\nEnter k value : ");
  scanf("%d", &k);
  p = (fact(n) / fact(n - k));
  printf("\nFactorial of P(%d,%d) = %d", n, k, p);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<arr_insert_ul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L18
L15:
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x3,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%rax
  mov    %rax,(%rdx)
  subl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jg     L15
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rax,(%rdx)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int arr_insert_ul(unsigned long arr[], const int len, const int i, const unsigned long v) {
  int x;
  for (x = len; x > i; x--) {
    arr[x] = arr[x - 1];
  }
  arr[x] = v;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<arr_insert_ul>:
  endbr64
  mov    %rdi,%r8
  cmp    %edx,%esi
  jle    L21
  movslq %esi,%rdi
  shl    $0x3,%rdi
  lea    (%r8,%rdi,1),%rax
  lea    -0x8(%r8,%rdi,1),%rdi
  sub    %edx,%esi
  lea    -0x1(%rsi),%esi
  shl    $0x3,%rsi
  sub    %rsi,%rdi
L10:
  mov    -0x8(%rax),%rsi
  mov    %rsi,(%rax)
  sub    $0x8,%rax
  cmp    %rdi,%rax
  jne    L10
L27:
  movslq %edx,%rdx
  mov    %rcx,(%r8,%rdx,8)
  mov    $0x0,%eax
  ret
L21:
  mov    %esi,%edx
  jmp    L27
```

assistant: ```c
int arr_insert_ul(unsigned long arr[], const int len, const int i, const unsigned long v) {
  int x;
  for (x = len; x > i; x--) {
    arr[x] = arr[x - 1];
  }
  arr[x] = v;
  return 0;
}
```