ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  lea    D31(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xd(%rsp),%rcx
  lea    0x10(%rsp),%rsi
  lea    0x14(%rsp),%rdx
  sub    $0x8,%rsp
  lea    0x17(%rsp),%rax
  push   %rax
  lea    0x1e(%rsp),%rax
  push   %rax
  lea    0x3c(%rsp),%rax
  push   %rax
  lea    0x40(%rsp),%rax
  push   %rax
  push   %rcx
  lea    0x4c(%rsp),%r9
  lea    0x48(%rsp),%r8
  call   <scanf@plt>
  add    $0x30,%rsp
  mov    0x10(%rsp),%esi
  cmpb   $0x41,0xe(%rsp)
  je     L75
  mov    0x14(%rsp),%eax
  mov    0x24(%rsp),%r8d
  mov    0x20(%rsp),%r9d
  mov    0x1c(%rsp),%edx
  mov    0x18(%rsp),%r11d
  cmp    $0x1,%esi
  jne    L2
  cmp    $0x2,%eax
  je     L48
L2:
  lea    0x2(%rax),%r10d
  add    $0x1,%esi
L9:
  sub    $0x8,%rsp
  mov    %r11d,%ecx
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  push   %r8
  mov    %edx,%r8d
  mov    %r10d,%edx
  call   <printf@plt>
  pop    %rax
  pop    %rdx
L44:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L75:
  mov    0x14(%rsp),%r10d
  mov    0x24(%rsp),%r8d
  mov    0x20(%rsp),%r9d
  mov    0x1c(%rsp),%edx
  mov    0x18(%rsp),%r11d
  cmp    $0x1,%esi
  jne    L9
  cmp    $0x2,%r10d
  jne    L9
  mov    %r9d,%ecx
  mov    %r11d,%esi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L44
L48:
  mov    %r9d,%ecx
  mov    %r11d,%esi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L44
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D31
datatype: string
name: read_data
arguments: label: D13
datatype: string
name: read_data
arguments: label: D8
datatype: string
name: read_data
arguments: label: D17
datatype: string

tool: D31: %1d%1d%c%1d%1d%c%1d%1d%c%c

tool: D13: %d%d:%d%d:%d%d

tool: D8: 00:%d%d:%d%d

tool: D17: 12:%d%d:%d%d

assistant: ```c
int main() {
  int h1, h2, m1, m2, s1, s2;
  char a, b, c;
  scanf("%1d%1d%c%1d%1d%c%1d%1d%c%c", &h1, &h2, &a, &m1, &m2, &a, &s1, &s2, &b, &c);
  if (b == 'A') {
    if (h1 == 1 && h2 == 2)
      printf("00:%d%d:%d%d", m1, m2, s1, s2);
    else
      printf("%d%d:%d%d:%d%d", h1, h2, m1, m2, s1, s2);
  } else {
    if (h1 == 1 && h2 == 2)
      printf("12:%d%d:%d%d", m1, m2, s1, s2);
    else
      printf("%d%d:%d%d:%d%d", h1 + 1, h2 + 2, m1, m2, s1, s2);
  }
  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
  sub    $0x38,%rsp
  lea    D19(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xd(%rsp),%rcx
  lea    0x10(%rsp),%rsi
  lea    0x14(%rsp),%rdx
  sub    $0x8,%rsp
  lea    0x17(%rsp),%rax
  push   %rax
  lea    0x1e(%rsp),%rax
  push   %rax
  lea    0x3c(%rsp),%rax
  push   %rax
  lea    0x40(%rsp),%rax
  push   %rax
  push   %rcx
  lea    0x4c(%rsp),%r9
  lea    0x48(%rsp),%r8
  call   <scanf@plt>
  add    $0x30,%rsp
  mov    0x10(%rsp),%esi
  cmpb   $0x41,0xe(%rsp)
  je     L49
  mov    0x14(%rsp),%eax
  mov    0x24(%rsp),%r8d
  mov    0x20(%rsp),%r9d
  mov    0x1c(%rsp),%edx
  mov    0x18(%rsp),%r11d
  cmp    $0x1,%esi
  jne    L72
  cmp    $0x2,%eax
  je     L32
L72:
  lea    0x2(%rax),%r10d
  add    $0x1,%esi
L33:
  sub    $0x8,%rsp
  mov    %r11d,%ecx
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  push   %r8
  mov    %edx,%r8d
  mov    %r10d,%edx
  call   <printf@plt>
  pop    %rax
  pop    %rdx
L63:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L61
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L49:
  mov    0x14(%rsp),%r10d
  mov    0x24(%rsp),%r8d
  mov    0x20(%rsp),%r9d
  mov    0x1c(%rsp),%edx
  mov    0x18(%rsp),%r11d
  cmp    $0x1,%esi
  jne    L33
  cmp    $0x2,%r10d
  jne    L33
  mov    %r9d,%ecx
  mov    %r11d,%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L63
L32:
  mov    %r9d,%ecx
  mov    %r11d,%esi
  lea    D31(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L63
L61:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D19
type_guessed: string
name: extract_symbolic_data
arguments: label: D3
type_guessed: string
name: extract_symbolic_data
arguments: label: D13
type_guessed: string
name: extract_symbolic_data
arguments: label: D31
type_guessed: string

tool: D19: %1d%1d%c%1d%1d%c%1d%1d%c%c

tool: D3: %d%d:%d%d:%d%d

tool: D13: 00:%d%d:%d%d

tool: D31: 12:%d%d:%d%d

assistant: ```c
int main() {
  int h1, h2, m1, m2, s1, s2;
  char a, b, c;
  scanf("%1d%1d%c%1d%1d%c%1d%1d%c%c", &h1, &h2, &a, &m1, &m2, &a, &s1, &s2, &b, &c);
  if (b == 'A') {
    if (h1 == 1 && h2 == 2)
      printf("00:%d%d:%d%d", m1, m2, s1, s2);
    else
      printf("%d%d:%d%d:%d%d", h1, h2, m1, m2, s1, s2);
  } else {
    if (h1 == 1 && h2 == 2)
      printf("12:%d%d:%d%d", m1, m2, s1, s2);
    else
      printf("%d%d:%d%d:%d%d", h1 + 1, h2 + 2, m1, m2, s1, s2);
  }
  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   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xa,-0x8(%rbp)
  movl   $0x2,-0x4(%rbp)
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%ecx
  mov    $0x2,%edx
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D24
type_infered: string
name: parse_data
arguments: tag: D10
type_infered: string
name: parse_data
arguments: tag: D11
type_infered: string

tool: D24: Doing it right: 

tool: D10: %d minus %d is %d


tool: D11: Doing it wrong: 

assistant: ```c
int main(int argc, const char *argv[]) {
  int ten = 10;
  int two = 2;
  printf("Doing it right: ");
  printf("%d minus %d is %d\n", ten, 2, ten - two);
  printf("Doing it wrong: ");
  printf("%d minus %d is %d\n", ten);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  lea    D23(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x8,%ecx
  mov    $0x2,%edx
  mov    $0xa,%esi
  lea    D27(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D23
datatype: string
name: disassemble_data
arguments: tag: D27
datatype: string
name: disassemble_data
arguments: tag: D24
datatype: string

tool: D23: Doing it right: 

tool: D27: %d minus %d is %d


tool: D24: Doing it wrong: 

assistant: ```c
int main(int argc, const char *argv[]) {
  int ten = 10;
  int two = 2;
  printf("Doing it right: ");
  printf("%d minus %d is %d\n", ten, 2, ten - two);
  printf("Doing it wrong: ");
  printf("%d minus %d is %d\n", ten);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rdi
  lea    D29(%rip),%rbp
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x8,%ecx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    $0x2,%edx
  mov    $0xa,%esi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xa,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D14
data_type: string
name: retrieve_data
arguments: data_label: D29
data_type: string
name: retrieve_data
arguments: data_label: D20
data_type: string

tool: D14: Doing it right: 

tool: D29: %d minus %d is %d


tool: D20: Doing it wrong: 

assistant: ```c
int main(int argc, const char *argv[]) {
  int ten = 10;
  int two = 2;
  printf("Doing it right: ");
  printf("%d minus %d is %d\n", ten, 2, ten - two);
  printf("Doing it wrong: ");
  printf("%d minus %d is %d\n", ten);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rdi
  lea    D20(%rip),%rbp
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x8,%ecx
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    $0x2,%edx
  mov    $0xa,%esi
  call   <printf@plt>
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0xa,%esi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D7
datatype: string
name: parse_data
arguments: tag: D20
datatype: string
name: parse_data
arguments: tag: D26
datatype: string

tool: D7: Doing it right: 

tool: D20: %d minus %d is %d


tool: D26: Doing it wrong: 

assistant: ```c
int main(int argc, const char *argv[]) {
  int ten = 10;
  int two = 2;
  printf("Doing it right: ");
  printf("%d minus %d is %d\n", ten, 2, ten - two);
  printf("Doing it wrong: ");
  printf("%d minus %d is %d\n", ten);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<is_smark>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,%eax
  mov    %al,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x21,-0x40(%rbp)
  movl   $0x22,-0x3c(%rbp)
  movl   $0x2c,-0x38(%rbp)
  movl   $0x3a,-0x34(%rbp)
  movl   $0x3b,-0x30(%rbp)
  movl   $0x3f,-0x2c(%rbp)
  movl   $0x60,-0x28(%rbp)
  mov    $0x0,%ebx
  jmp    L1
L40:
  movsbl -0x44(%rbp),%edx
  movslq %ebx,%rax
  mov    -0x40(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jne    L26
  mov    $0x1,%eax
  jmp    L49
L26:
  add    $0x1,%ebx
L1:
  mov    %ebx,%eax
  cmp    $0x6,%eax
  jbe    L40
  mov    $0x0,%eax
L49:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int is_smark(char c) {
  int mark[] = {33, 34, 44, 58, 59, 63, 96};
  register int i;
  for (i = 0; i < sizeof(mark) / sizeof(int); i++)
    if (c == mark[i])
      return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<is_smark>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x21,(%rsp)
  movl   $0x22,0x4(%rsp)
  movl   $0x2c,0x8(%rsp)
  movl   $0x3a,0xc(%rsp)
  movl   $0x3b,0x10(%rsp)
  movl   $0x3f,0x14(%rsp)
  movl   $0x60,0x18(%rsp)
  mov    %rsp,%rax
  lea    0x1c(%rsp),%rdx
  movsbl %dil,%edi
L24:
  cmp    (%rax),%edi
  je     L21
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L24
  mov    $0x0,%eax
L15:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x38,%rsp
  ret
L21:
  mov    $0x1,%eax
  jmp    L15
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_smark(char c) {
  int mark[] = {33, 34, 44, 58, 59, 63, 96};
  register int i;
  for (i = 0; i < sizeof(mark) / sizeof(int); i++)
    if (c == mark[i])
      return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<is_smark>:
  endbr64
  sub    $0x38,%rsp
  movsbl %dil,%edi
  mov    $0x21,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x2c00000022,%rax
  lea    0x1c(%rsp),%rcx
  mov    %rax,0x4(%rsp)
  movabs $0x3b0000003a,%rax
  mov    %rax,0xc(%rsp)
  movabs $0x600000003f,%rax
  mov    %rax,0x14(%rsp)
  lea    0x4(%rsp),%rax
  jmp    L22
  nopl   0x0(%rax,%rax,1)
L38:
  mov    (%rax),%edx
  add    $0x4,%rax
L22:
  cmp    %edx,%edi
  je     L23
  cmp    %rcx,%rax
  jne    L38
  xor    %eax,%eax
L34:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x38,%rsp
  ret
  cs nopw 0x0(%rax,%rax,1)
L23:
  mov    $0x1,%eax
  jmp    L34
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_smark(char c) {
  int mark[] = {33, 34, 44, 58, 59, 63, 96};
  register int i;
  for (i = 0; i < sizeof(mark) / sizeof(int); i++)
    if (c == mark[i])
      return 1;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_smark>:
  endbr64
  lea    -0x21(%rdi),%ecx
  xor    %eax,%eax
  cmp    $0x3f,%cl
  ja     L7
  movabs $0x8000000046000803,%rax
  shr    %cl,%rax
  and    $0x1,%eax
L7:
  ret
```

assistant: ```c
int is_smark(char c) {
  int mark[] = {33, 34, 44, 58, 59, 63, 96};
  register int i;
  for (i = 0; i < sizeof(mark) / sizeof(int); i++)
    if (c == mark[i])
      return 1;
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<fcitx_utils_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L7
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L7:
  nop
  leave
  ret
```

assistant: ```c
void fcitx_utils_free(void *ptr) {
  if (ptr)
    free(ptr);
}
```
user: What is the c source code of the assembly code below:

```asm
<fcitx_utils_free>:
  endbr64
  test   %rdi,%rdi
  je     L2
  sub    $0x8,%rsp
  call   <free@plt>
  add    $0x8,%rsp
  ret
L2:
  ret
```

assistant: ```c
void fcitx_utils_free(void *ptr) {
  if (ptr)
    free(ptr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fcitx_utils_free>:
  endbr64
  test   %rdi,%rdi
  je     L9
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L9:
  ret
```

assistant: ```c
void fcitx_utils_free(void *ptr) {
  if (ptr)
    free(ptr);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fcitx_utils_free>:
  endbr64
  test   %rdi,%rdi
  je     L9
  jmp    1050 <free@plt>
  xchg   %ax,%ax
L9:
  ret
```

assistant: ```c
void fcitx_utils_free(void *ptr) {
  if (ptr)
    free(ptr);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ulimit_expand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <getrlimit@plt>
  test   %rax,%rax
  je     L59
  mov    $0xffffffff,%eax
  jmp    L34
L59:
  mov    -0x20(%rbp),%rdx
  mov    D31(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L24
  mov    -0x18(%rbp),%rdx
  mov    D31(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L24
  mov    $0x0,%eax
  jmp    L34
L24:
  mov    -0x20(%rbp),%rdx
  mov    D31(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L23
  mov    -0x20(%rbp),%rax
  add    $0x5000,%rax
  mov    %rax,-0x20(%rbp)
L23:
  mov    -0x18(%rbp),%rdx
  mov    D31(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L71
  mov    -0x18(%rbp),%rax
  add    $0x5000,%rax
  mov    %rax,-0x18(%rbp)
L71:
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <setrlimit@plt>
  test   %rax,%rax
  je     L56
  mov    $0xffffffff,%eax
  jmp    L34
L56:
  mov    $0x0,%eax
L34:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

assistant: ```c
int ulimit_expand(int newfd) {
  struct rlimit rlim;
  if (getrlimit(RLIMIT_NOFILE, &rlim) != 0) {
    return (-1);
  }
  if ((rlim.rlim_cur == RLIM_INFINITY) && (rlim.rlim_max == RLIM_INFINITY)) {
    return (0);
  }
  if (rlim.rlim_cur != RLIM_INFINITY) {
    rlim.rlim_cur += (64 * 64) * 5;
  }
  if (rlim.rlim_max != RLIM_INFINITY) {
    rlim.rlim_max += (64 * 64) * 5;
  }
  if (setrlimit(RLIMIT_NOFILE, &rlim) != 0) {
    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
<ulimit_expand>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  call   <getrlimit@plt>
  test   %rax,%rax
  jne    L6
  mov    (%rsp),%rax
  mov    D6(%rip),%rdx
  mov    (%rdx),%rdx
  cmp    %rdx,%rax
  je     L24
  add    $0x5000,%rax
  mov    %rax,(%rsp)
  mov    0x8(%rsp),%rax
  cmp    %rax,%rdx
  je     L68
L45:
  add    $0x5000,%rax
  mov    %rax,0x8(%rsp)
L68:
  mov    %rsp,%rsi
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  call   <setrlimit@plt>
  neg    %rax
  sbb    %eax,%eax
L8:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L32
  add    $0x28,%rsp
  ret
L24:
  mov    $0x0,%eax
  cmp    0x8(%rsp),%rdx
  je     L8
  mov    0x8(%rsp),%rax
  jmp    L45
L6:
  mov    $0xffffffff,%eax
  jmp    L8
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ulimit_expand(int newfd) {
  struct rlimit rlim;
  if (getrlimit(RLIMIT_NOFILE, &rlim) != 0) {
    return (-1);
  }
  if ((rlim.rlim_cur == RLIM_INFINITY) && (rlim.rlim_max == RLIM_INFINITY)) {
    return (0);
  }
  if (rlim.rlim_cur != RLIM_INFINITY) {
    rlim.rlim_cur += (64 * 64) * 5;
  }
  if (rlim.rlim_max != RLIM_INFINITY) {
    rlim.rlim_max += (64 * 64) * 5;
  }
  if (setrlimit(RLIMIT_NOFILE, &rlim) != 0) {
    return (-1);
  }
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ulimit_expand>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    D3(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    (%rbx),%edi
  mov    %rbp,%rsi
  call   <getrlimit@plt>
  test   %rax,%rax
  jne    L59
  mov    D19(%rip),%rdx
  mov    (%rsp),%rax
  mov    (%rdx),%rcx
  mov    0x8(%rsp),%rdx
  cmp    %rcx,%rax
  je     L44
  add    $0x5000,%rax
  mov    %rax,(%rsp)
  cmp    %rdx,%rcx
  je     L22
L12:
  add    $0x5000,%rdx
  mov    %rdx,0x8(%rsp)
L22:
  mov    (%rbx),%edi
  mov    %rbp,%rsi
  call   <setrlimit@plt>
  neg    %rax
  sbb    %r8d,%r8d
L4:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  add    $0x28,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
L44:
  xor    %r8d,%r8d
  cmp    %rdx,%rax
  jne    L12
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L59:
  mov    $0xffffffff,%r8d
  jmp    L4
L54:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ulimit_expand(int newfd) {
  struct rlimit rlim;
  if (getrlimit(RLIMIT_NOFILE, &rlim) != 0) {
    return (-1);
  }
  if ((rlim.rlim_cur == RLIM_INFINITY) && (rlim.rlim_max == RLIM_INFINITY)) {
    return (0);
  }
  if (rlim.rlim_cur != RLIM_INFINITY) {
    rlim.rlim_cur += (64 * 64) * 5;
  }
  if (rlim.rlim_max != RLIM_INFINITY) {
    rlim.rlim_max += (64 * 64) * 5;
  }
  if (setrlimit(RLIMIT_NOFILE, &rlim) != 0) {
    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
<ulimit_expand>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    D6(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    (%rbx),%edi
  mov    %rbp,%rsi
  call   <getrlimit@plt>
  test   %rax,%rax
  jne    L13
  mov    D8(%rip),%rdx
  mov    (%rsp),%rax
  mov    (%rdx),%rcx
  mov    0x8(%rsp),%rdx
  cmp    %rcx,%rax
  je     L32
  add    $0x5000,%rax
  mov    %rax,(%rsp)
  cmp    %rdx,%rcx
  je     L0
L39:
  add    $0x5000,%rdx
  mov    %rdx,0x8(%rsp)
L0:
  mov    (%rbx),%edi
  mov    %rbp,%rsi
  call   <setrlimit@plt>
  neg    %rax
  sbb    %r8d,%r8d
L49:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x28,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
L32:
  xor    %r8d,%r8d
  cmp    %rdx,%rax
  jne    L39
  jmp    L49
  nopw   0x0(%rax,%rax,1)
L13:
  mov    $0xffffffff,%r8d
  jmp    L49
L43:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ulimit_expand(int newfd) {
  struct rlimit rlim;
  if (getrlimit(RLIMIT_NOFILE, &rlim) != 0) {
    return (-1);
  }
  if ((rlim.rlim_cur == RLIM_INFINITY) && (rlim.rlim_max == RLIM_INFINITY)) {
    return (0);
  }
  if (rlim.rlim_cur != RLIM_INFINITY) {
    rlim.rlim_cur += (64 * 64) * 5;
  }
  if (rlim.rlim_max != RLIM_INFINITY) {
    rlim.rlim_max += (64 * 64) * 5;
  }
  if (setrlimit(RLIMIT_NOFILE, &rlim) != 0) {
    return (-1);
  }
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<Decrypt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    $0x0,%ebp
L10:
  mov    0x50(%r13,%rbp,1),%edi
  call   <READ_ROUND_KEY_BYTE@plt>
  xor    %eax,(%rbx,%rbp,1)
  add    $0x4,%rbp
  cmp    $0x10,%rbp
  jne    L10
  mov    %r13,%r12
  mov    $0x1,%ebp
L26:
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <rrr_enc_dec_round@plt>
  add    $0x1,%ebp
  add    $0x30,%r12
  cmp    $0xd,%ebp
  jne    L26
  mov    (%rbx),%r15d
  mov    0x4(%rbx),%r14d
  mov    0x8(%rbx),%eax
  mov    %eax,0xc(%rsp)
  mov    0xc(%rbx),%eax
  mov    %eax,0x8(%rsp)
  mov    $0x0,%ebp
L1:
  mov    0x10(%rbx,%rbp,1),%r12d
  mov    0x20(%r13,%rbp,1),%edi
  call   <READ_ROUND_KEY_BYTE@plt>
  xor    %eax,%r12d
  mov    %r12d,(%rbx,%rbp,1)
  add    $0x4,%rbp
  cmp    $0x10,%rbp
  jne    L1
  mov    %r15d,0x10(%rbx)
  mov    %r14d,0x14(%rbx)
  mov    0xc(%rsp),%eax
  mov    %eax,0x18(%rbx)
  mov    0x8(%rsp),%eax
  mov    %eax,0x1c(%rbx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void Decrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i, j = 0, temp[4] = {0};
  for (i = 0; i < 4; i++)
    block[i] ^= READ_ROUND_KEY_BYTE(roundKeys[i + 20]);
  for (i = 1; i <= 12; i++) {
    rrr_enc_dec_round(block, roundKeys + j, i);
    j += 12;
  }
  for (i = 0; i < 4; i++)
    temp[i] = block[i];
  for (i = 0; i < 4; i++)
    block[i] = block[i + 4] ^ READ_ROUND_KEY_BYTE(roundKeys[i + 8]);
  for (i = 0; i < 4; i++)
    block[i + 4] = temp[i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Decrypt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
L12:
  mov    0x50(%r13,%rbx,1),%edi
  call   <READ_ROUND_KEY_BYTE@plt>
  xor    %eax,(%r12,%rbx,1)
  add    $0x4,%rbx
  cmp    $0x10,%rbx
  jne    L12
  mov    %r13,%rbp
  mov    $0x1,%ebx
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <rrr_enc_dec_round@plt>
  add    $0x30,%rbp
  cmp    $0xd,%ebx
  jne    L5
  mov    (%r12),%r15
  mov    0x8(%r12),%r14
  xor    %ebx,%ebx
L27:
  mov    0x20(%r13,%rbx,1),%edi
  mov    0x10(%r12,%rbx,1),%ebp
  call   <READ_ROUND_KEY_BYTE@plt>
  xor    %eax,%ebp
  mov    %ebp,(%r12,%rbx,1)
  add    $0x4,%rbx
  cmp    $0x10,%rbx
  jne    L27
  mov    %r15d,0x10(%r12)
  sar    $0x20,%r15
  mov    %r14d,0x18(%r12)
  sar    $0x20,%r14
  mov    %r15d,0x14(%r12)
  mov    %r14d,0x1c(%r12)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void Decrypt(uint8_t *block, uint8_t *roundKeys) {
  uint8_t i, j = 0, temp[4] = {0};
  for (i = 0; i < 4; i++)
    block[i] ^= READ_ROUND_KEY_BYTE(roundKeys[i + 20]);
  for (i = 1; i <= 12; i++) {
    rrr_enc_dec_round(block, roundKeys + j, i);
    j += 12;
  }
  for (i = 0; i < 4; i++)
    temp[i] = block[i];
  for (i = 0; i < 4; i++)
    block[i] = block[i + 4] ^ READ_ROUND_KEY_BYTE(roundKeys[i + 8]);
  for (i = 0; i < 4; i++)
    block[i + 4] = temp[i];
}
```