ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x3,%edi
  jne    L31
  mov    0x10(%rsi),%rdi
  mov    %rsi,%rbx
  lea    D45(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L37
  mov    0x8(%rbx),%rdi
  lea    D17(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L3
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <transformInfile@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L51:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  lea    D38(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L51
L3:
  mov    0x8(%rbx),%rsi
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L51
L37:
  mov    0x10(%rbx),%rsi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L51
```

tool_calls: name: parse_data
arguments: data_label: D45
type: string
name: parse_data
arguments: data_label: D17
type: string
name: parse_data
arguments: data_label: D38
type: string
name: parse_data
arguments: data_label: D42
type: string
name: parse_data
arguments: data_label: D8
type: string

tool: D45: r

tool: D17: w

tool: D38: Usage: a.out outputFileName inputFileName 

tool: D42: 
Unable to open output file: %s!


tool: D8: 
Unable to open file: %s!


assistant: ```c
int main(int argc, char *argv[]) {
  FILE *fin;
  FILE *fout;
  if (argc != 3) {
    printf("Usage: a.out outputFileName inputFileName \n");
    return -1;
  }
  fin = fopen(argv[2], "r");
  if (fin == NULL) {
    printf("\nUnable to open file: %s!\n", argv[2]);
    return -1;
  }
  fout = fopen(argv[1], "w");
  if (fout == NULL) {
    printf("\nUnable to open output file: %s!\n", argv[1]);
    return -1;
  }
  transformInfile(fin, fout);
  fclose(fin);
  fclose(fout);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  push   $0x0
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <foo@plt>
  add    $0x10,%rsp
  mov    $0x0,%edi
  call   <exit@plt>
```

assistant: ```c
main(argc, argv) int argc;
char **argv;
{
  foo(0, 0, 0, 0, 0, 0, 0, 0);
  exit(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  push   $0x0
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <foo@plt>
  mov    $0x0,%edi
  call   <exit@plt>
```

assistant: ```c
main(argc, argv) int argc;
char **argv;
{
  foo(0, 0, 0, 0, 0, 0, 0, 0);
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  push   $0x0
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  push   $0x0
  xor    %eax,%eax
  call   <foo@plt>
  xor    %edi,%edi
  call   <exit@plt>
```

assistant: ```c
main(argc, argv) int argc;
char **argv;
{
  foo(0, 0, 0, 0, 0, 0, 0, 0);
  exit(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  push   $0x0
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  push   $0x0
  xor    %eax,%eax
  call   <foo@plt>
  xor    %edi,%edi
  call   <exit@plt>
```

assistant: ```c
main(argc, argv) int argc;
char **argv;
{
  foo(0, 0, 0, 0, 0, 0, 0, 0);
  exit(0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<finddigits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movl   $0x1,(%rax)
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L40
L75:
  movl   $0x0,-0x18(%rbp)
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x14(%rbp)
  jmp    L1
L7:
  mov    -0x1c(%rbp),%eax
  mov    -0x14(%rbp),%edx
  lea    0x1(%rdx),%ecx
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %eax,%eax
  sete   %cl
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  movzbl %cl,%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x14(%rbp)
L1:
  cmpl   $0x9,-0x14(%rbp)
  jle    L7
  jmp    L47
L15:
  mov    -0x1c(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  movslq %edx,%rax
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L20
  addl   $0x1,-0x18(%rbp)
L20:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
L47:
  mov    -0x1c(%rbp),%eax
  test   %eax,%eax
  jg     L15
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L40:
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  test   %eax,%eax
  jne    L75
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L50
  call   <__stack_chk_fail@plt>
L50:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D14
type: string
name: disassemble_data
arguments: data_tag: D6
type: string

tool: D14: %d

tool: D6: %d


assistant: ```c
int finddigits(void) {
  int numCases;
  int *divisors = malloc(sizeof(int) * 10);
  divisors[0] = 1;
  scanf("%d", &numCases);
  while (numCases--) {
    int number;
    int divisorDigits = 0;
    scanf("%d", &number);
    for (int i = 1; i < 10; i++) {
      divisors[i] = (number % (i + 1)) == 0;
    }
    while (number > 0) {
      if (divisors[(number % 10) - 1]) {
        divisorDigits++;
      }
      number /= 10;
    }
    printf("%d\n", divisorDigits);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<finddigits>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x28,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  movl   $0x1,(%rax)
  mov    %rsp,%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L31
  lea    0x4(%rsp),%r13
  lea    D5(%rip),%r12
  lea    D13(%rip),%rbp
  jmp    L81
L67:
  mov    %eax,%edx
L19:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cltq
  cmpl   $0x1,-0x4(%rbx,%rax,4)
  sbb    $0xffffffff,%esi
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,0x4(%rsp)
  cmp    $0x9,%edx
  jg     L67
L14:
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L31
L81:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    $0x2,%ecx
L0:
  mov    0x4(%rsp),%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  sete   %al
  movzbl %al,%eax
  mov    %eax,-0x4(%rbx,%rcx,4)
  add    $0x1,%rcx
  cmp    $0xb,%rcx
  jne    L0
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L88
  mov    $0x0,%esi
  jmp    L19
L88:
  mov    $0x0,%esi
  jmp    L14
L31:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
datatype: string
name: extract_symbolic_data
arguments: tag: D13
datatype: string

tool: D5: %d

tool: D13: %d


assistant: ```c
int finddigits(void) {
  int numCases;
  int *divisors = malloc(sizeof(int) * 10);
  divisors[0] = 1;
  scanf("%d", &numCases);
  while (numCases--) {
    int number;
    int divisorDigits = 0;
    scanf("%d", &number);
    for (int i = 1; i < 10; i++) {
      divisors[i] = (number % (i + 1)) == 0;
    }
    while (number > 0) {
      if (divisors[(number % 10) - 1]) {
        divisorDigits++;
      }
      number /= 10;
    }
    printf("%d\n", divisorDigits);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<finddigits>:
  endbr64
  push   %r14
  mov    $0x28,%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D2(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  movl   $0x1,(%rax)
  mov    %rax,%r13
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L51
  lea    0x4(%rsp),%r12
  lea    D8(%rip),%rbp
  mov    $0xcccccccd,%r14d
  nopl   0x0(%rax,%rax,1)
L23:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    $0x2,%ecx
L57:
  mov    0x4(%rsp),%eax
  cltd
  idiv   %ecx
  xor    %eax,%eax
  test   %edx,%edx
  sete   %al
  mov    %eax,-0x4(%r13,%rcx,4)
  add    $0x1,%rcx
  cmp    $0xb,%rcx
  jne    L57
  mov    0x4(%rsp),%edx
  xor    %r8d,%r8d
  test   %edx,%edx
  jg     L62
  jmp    L46
  nopw   0x0(%rax,%rax,1)
L61:
  mov    %eax,%edx
L62:
  mov    %edx,%eax
  mov    %edx,%ecx
  imul   %r14,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%ecx
  movslq %ecx,%rcx
  cmpl   $0x1,-0x4(%r13,%rcx,4)
  mov    %eax,0x4(%rsp)
  sbb    $0xffffffff,%r8d
  cmp    $0x9,%edx
  jg     L61
L46:
  mov    %r8d,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jne    L23
L51:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D2
type_infered: string
name: unpack_data
arguments: label: D8
type_infered: string

tool: D2: %d

tool: D8: %d


assistant: ```c
int finddigits(void) {
  int numCases;
  int *divisors = malloc(sizeof(int) * 10);
  divisors[0] = 1;
  scanf("%d", &numCases);
  while (numCases--) {
    int number;
    int divisorDigits = 0;
    scanf("%d", &number);
    for (int i = 1; i < 10; i++) {
      divisors[i] = (number % (i + 1)) == 0;
    }
    while (number > 0) {
      if (divisors[(number % 10) - 1]) {
        divisorDigits++;
      }
      number /= 10;
    }
    printf("%d\n", divisorDigits);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L79
  jmp    L54
L78:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L73
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
L63:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L78
  jmp    L20
L73:
  nop
L20:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L39
  mov    -0x8(%rbp),%rax
  jmp    L40
L39:
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L54:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L63
L79:
  mov    $0x0,%eax
L40:
  pop    %rbp
  ret
```

assistant: ```c
char *strstr(const char *haystack, const char *needle) {
  const char *needleStart = (const char *)needle;
  const char *haystackStart = haystack;
  if (haystack) {
    while (*haystack) {
      while (*needle) {
        if (*haystack == *needle) {
          haystack++;
          needle++;
        } else
          break;
      }
      if (!*needle)
        return (char *)haystackStart;
      else {
        haystack = haystackStart + 1;
        needle = needleStart;
      }
    }
  }
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strstr>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L87
  movzbl (%rsi),%r8d
  mov    %rdi,%rcx
  lea    0x1(%rdi),%r9
  jmp    L50
L71:
  test   %dl,%dl
  je     L58
  mov    %r9,%rcx
  cmpb   $0x0,0x1(%rdi)
  je     L42
L50:
  test   %r8b,%r8b
  je     L58
  mov    %r8d,%edx
  mov    $0x0,%eax
L84:
  cmp    %dl,(%rcx,%rax,1)
  jne    L71
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L84
L58:
  mov    %rdi,%rax
  ret
L87:
  mov    $0x0,%edi
  jmp    L58
L42:
  mov    $0x0,%edi
  jmp    L58
```

assistant: ```c
char *strstr(const char *haystack, const char *needle) {
  const char *needleStart = (const char *)needle;
  const char *haystackStart = haystack;
  if (haystack) {
    while (*haystack) {
      while (*needle) {
        if (*haystack == *needle) {
          haystack++;
          needle++;
        } else
          break;
      }
      if (!*needle)
        return (char *)haystackStart;
      else {
        haystack = haystackStart + 1;
        needle = needleStart;
      }
    }
  }
  return NULL;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strstr>:
  endbr64
  cmpb   $0x0,(%rdi)
  mov    %rdi,%r9
  je     L25
  movzbl (%rsi),%r8d
  mov    %rdi,%rcx
  lea    0x1(%rdi),%rdi
  nopw   0x0(%rax,%rax,1)
L33:
  test   %r8b,%r8b
  je     L58
  mov    %r8d,%edx
  xor    %eax,%eax
  jmp    L6
  nopl   0x0(%rax)
L27:
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L58
L6:
  cmp    %dl,(%rcx,%rax,1)
  je     L27
  cmpb   $0x0,0x1(%r9)
  mov    %rdi,%rcx
  jne    L33
L25:
  xor    %r9d,%r9d
L58:
  mov    %r9,%rax
  ret
```

assistant: ```c
char *strstr(const char *haystack, const char *needle) {
  const char *needleStart = (const char *)needle;
  const char *haystackStart = haystack;
  if (haystack) {
    while (*haystack) {
      while (*needle) {
        if (*haystack == *needle) {
          haystack++;
          needle++;
        } else
          break;
      }
      if (!*needle)
        return (char *)haystackStart;
      else {
        haystack = haystackStart + 1;
        needle = needleStart;
      }
    }
  }
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strstr>:
  endbr64
  cmpb   $0x0,(%rdi)
  mov    %rdi,%r9
  je     L7
  movzbl (%rsi),%r8d
  mov    %rdi,%rcx
  lea    0x1(%rdi),%rdi
  nopw   0x0(%rax,%rax,1)
L46:
  test   %r8b,%r8b
  je     L37
  mov    %r8d,%edx
  xor    %eax,%eax
  jmp    L10
  nopl   0x0(%rax)
L33:
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L37
L10:
  cmp    %dl,(%rcx,%rax,1)
  je     L33
  cmpb   $0x0,0x1(%r9)
  mov    %rdi,%rcx
  jne    L46
L7:
  xor    %r9d,%r9d
L37:
  mov    %r9,%rax
  ret
```

assistant: ```c
char *strstr(const char *haystack, const char *needle) {
  const char *needleStart = (const char *)needle;
  const char *haystackStart = haystack;
  if (haystack) {
    while (*haystack) {
      while (*needle) {
        if (*haystack == *needle) {
          haystack++;
          needle++;
        } else
          break;
      }
      if (!*needle)
        return (char *)haystackStart;
      else {
        haystack = haystackStart + 1;
        needle = needleStart;
      }
    }
  }
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<setAllowedFiles>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x430,%rsp
  mov    %rdi,-0x428(%rbp)
  mov    %rsi,-0x430(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x41c(%rbp)
  movl   $0x0,-0x418(%rbp)
  movl   $0x0,-0x414(%rbp)
  jmp    L2
L46:
  mov    -0x41c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x428(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x3b,%al
  jne    L40
  mov    -0x418(%rbp),%eax
  cltq
  movb   $0x0,-0x410(%rbp,%rax,1)
  mov    -0x414(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x414(%rbp)
  cltq
  shl    $0xa,%rax
  mov    %rax,%rdx
  mov    -0x430(%rbp),%rax
  add    %rax,%rdx
  lea    -0x410(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  movl   $0x0,-0x418(%rbp)
  jmp    L38
L40:
  mov    -0x41c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x428(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x418(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x418(%rbp)
  movzbl (%rcx),%edx
  cltq
  mov    %dl,-0x410(%rbp,%rax,1)
L38:
  addl   $0x1,-0x41c(%rbp)
L2:
  mov    -0x41c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x428(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L46
  mov    -0x418(%rbp),%eax
  cltq
  movb   $0x0,-0x410(%rbp,%rax,1)
  mov    -0x414(%rbp),%eax
  cltq
  shl    $0xa,%rax
  mov    %rax,%rdx
  mov    -0x430(%rbp),%rax
  add    %rax,%rdx
  lea    -0x410(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
void setAllowedFiles(char *str, char arr[10][1024]) {
  char buf_tmp[1024];
  int i = 0, j = 0, a = 0;
  while (str[i] != '\0') {
    if (str[i] == ';') {
      buf_tmp[j] = '\0';
      strcpy(arr[a++], buf_tmp);
      j = 0;
    } else {
      buf_tmp[j++] = str[i];
    }
    ++i;
  }
  buf_tmp[j] = '\0';
  strcpy(arr[a], buf_tmp);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<setAllowedFiles>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  movzbl (%rdi),%eax
  test   %al,%al
  je     L7
  mov    %rdi,%rbx
  add    $0x1,%rbx
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rsp,%r12
  jmp    L11
L38:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  lea    0x1(%rcx),%r13d
  movslq %ecx,%rdi
  mov    %rdi,%rcx
  shl    $0xa,%rcx
  lea    0x0(%rbp,%rcx,1),%rdi
  mov    %r12,%rsi
  call   <strcpy@plt>
  mov    %r13d,%ecx
  mov    $0x0,%edx
L10:
  add    $0x1,%rbx
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  je     L47
L11:
  cmp    $0x3b,%al
  je     L38
  movslq %edx,%rsi
  mov    %al,(%rsp,%rsi,1)
  lea    0x1(%rdx),%edx
  jmp    L10
L7:
  mov    $0x0,%ecx
  mov    $0x0,%edx
L47:
  movslq %edx,%rdx
  movb   $0x0,(%rsp,%rdx,1)
  mov    %rsp,%rsi
  movslq %ecx,%rcx
  shl    $0xa,%rcx
  lea    0x0(%rbp,%rcx,1),%rdi
  call   <strcpy@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void setAllowedFiles(char *str, char arr[10][1024]) {
  char buf_tmp[1024];
  int i = 0, j = 0, a = 0;
  while (str[i] != '\0') {
    if (str[i] == ';') {
      buf_tmp[j] = '\0';
      strcpy(arr[a++], buf_tmp);
      j = 0;
    } else {
      buf_tmp[j++] = str[i];
    }
    ++i;
  }
  buf_tmp[j] = '\0';
  strcpy(arr[a], buf_tmp);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<setAllowedFiles>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  movzbl (%rdi),%eax
  test   %al,%al
  je     L19
  lea    0x1(%rdi),%rbx
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rsp,%r13
  jmp    L59
  nop
L58:
  movslq %edx,%rsi
  add    $0x1,%rbx
  add    $0x1,%edx
  mov    %al,(%rsp,%rsi,1)
  movzbl -0x1(%rbx),%eax
  test   %al,%al
  je     L63
L59:
  cmp    $0x3b,%al
  jne    L58
  movslq %ecx,%rdi
  lea    0x1(%rcx),%ebp
  movslq %edx,%rdx
  mov    %r13,%rsi
  mov    %rdi,%rcx
  add    $0x1,%rbx
  movb   $0x0,(%rsp,%rdx,1)
  shl    $0xa,%rcx
  lea    (%r12,%rcx,1),%rdi
  call   <strcpy@plt>
  movzbl -0x1(%rbx),%eax
  movslq %ebp,%rcx
  xor    %edx,%edx
  test   %al,%al
  jne    L59
L63:
  shl    $0xa,%rcx
  add    %rcx,%r12
L32:
  movslq %edx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  movb   $0x0,(%rsp,%rdx,1)
  call   <strcpy@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L19:
  xor    %edx,%edx
  mov    %rsp,%r13
  jmp    L32
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void setAllowedFiles(char *str, char arr[10][1024]) {
  char buf_tmp[1024];
  int i = 0, j = 0, a = 0;
  while (str[i] != '\0') {
    if (str[i] == ';') {
      buf_tmp[j] = '\0';
      strcpy(arr[a++], buf_tmp);
      j = 0;
    } else {
      buf_tmp[j++] = str[i];
    }
    ++i;
  }
  buf_tmp[j] = '\0';
  strcpy(arr[a], buf_tmp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<setAllowedFiles>:
  endbr64
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  movzbl (%rdi),%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  test   %dl,%dl
  je     L56
  lea    0x1(%rdi),%rbx
  xor    %ecx,%ecx
  mov    %rsp,%r13
  cmp    $0x3b,%dl
  je     L36
L42:
  lea    0x1(%rax),%esi
  add    $0x1,%rbx
  cltq
  mov    %dl,(%rsp,%rax,1)
  movzbl -0x1(%rbx),%edx
  test   %dl,%dl
  je     L57
L13:
  mov    %esi,%eax
  cmp    $0x3b,%dl
  jne    L42
L36:
  movslq %ecx,%rdi
  lea    0x1(%rcx),%ebp
  cltq
  mov    %r13,%rsi
  mov    %rdi,%rcx
  add    $0x1,%rbx
  movb   $0x0,(%rsp,%rax,1)
  shl    $0xa,%rcx
  lea    (%r12,%rcx,1),%rdi
  call   <strcpy@plt>
  movzbl -0x1(%rbx),%edx
  movslq %ebp,%rcx
  xor    %esi,%esi
  test   %dl,%dl
  jne    L13
L57:
  shl    $0xa,%rcx
  add    %rcx,%r12
L40:
  movslq %esi,%rsi
  mov    %r12,%rdi
  movb   $0x0,(%rsp,%rsi,1)
  mov    %r13,%rsi
  call   <strcpy@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L56:
  xor    %esi,%esi
  mov    %rsp,%r13
  jmp    L40
L65:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void setAllowedFiles(char *str, char arr[10][1024]) {
  char buf_tmp[1024];
  int i = 0, j = 0, a = 0;
  while (str[i] != '\0') {
    if (str[i] == ';') {
      buf_tmp[j] = '\0';
      strcpy(arr[a++], buf_tmp);
      j = 0;
    } else {
      buf_tmp[j++] = str[i];
    }
    ++i;
  }
  buf_tmp[j] = '\0';
  strcpy(arr[a], buf_tmp);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<beep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0x1234dc,%eax
  mov    $0x0,%edx
  divl   -0x14(%rbp)
  mov    %eax,-0x4(%rbp)
  mov    $0x61,%edi
  call   <inb@plt>
  or     $0x3,%eax
  mov    $0x61,%esi
  mov    %eax,%edi
  call   <outb@plt>
  mov    $0x43,%esi
  mov    $0xb6,%edi
  call   <outb@plt>
  mov    -0x4(%rbp),%eax
  movzbl %al,%eax
  mov    $0x42,%esi
  mov    %eax,%edi
  call   <outb@plt>
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  movzbl %al,%eax
  mov    $0x42,%esi
  mov    %eax,%edi
  call   <outb@plt>
  mov    $0x1,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x64,%edi
  call   <sleep@plt>
  mov    $0x61,%edi
  call   <inb@plt>
  and    $0xfc,%eax
  mov    $0x61,%esi
  mov    %eax,%edi
  call   <outb@plt>
  nop
  leave
  ret
```

assistant: ```c
void beep(unsigned int frequency) {
  unsigned int count = 1193180 / frequency;
  outb(inb(0x61) | 3, 0x61);
  outb(0xB6, 0x43);
  outb(count & 0xff, 0x42);
  outb((count >> 8) & 0xff, 0x42);
  sleep(100, 0, 0, 1);
  outb(inb(0x61) & 0xFC, 0x61);
}
```