ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<getline2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    D49(%rip),%edx
  mov    D84(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L11
  mov    $0x0,%eax
  jmp    L47
L39:
  mov    D49(%rip),%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L11:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,D50(%rip)
  mov    D50(%rip),%eax
  cmp    $0xa,%eax
  je     L38
  mov    D50(%rip),%edx
  mov    D84(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L39
L38:
  mov    D50(%rip),%eax
  cmp    $0xa,%eax
  jne    L12
  mov    D50(%rip),%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L12:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    $0x1,%eax
L47:
  leave
  ret
```

assistant: ```c
int getline2(char s[]) {
  static int c;
  int i = 0;
  if (c == EOF) {
    return 0;
  }
  while ((c = getchar()) != '\n' && c != EOF) {
    s[i++] = c;
  }
  if (c == '\n') {
    s[i++] = c;
  }
  s[i] = '\0';
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getline2>:
  endbr64
  mov    $0x0,%eax
  mov    D18(%rip),%rdx
  mov    (%rdx),%ecx
  cmp    %ecx,D5(%rip)
  je     L34
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0x0,%ebx
  jmp    L41
L35:
  mov    %al,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
L41:
  mov    %ebx,%r12d
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,D19(%rip)
  cmp    $0xa,%eax
  je     L29
  mov    D18(%rip),%rdx
  cmp    (%rdx),%eax
  jne    L35
L4:
  movslq %r12d,%r12
  movb   $0x0,0x0(%rbp,%r12,1)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  movslq %ebx,%rax
  movb   $0xa,0x0(%rbp,%rax,1)
  lea    0x1(%rbx),%r12d
  jmp    L4
L34:
  ret
```

assistant: ```c
int getline2(char s[]) {
  static int c;
  int i = 0;
  if (c == EOF) {
    return 0;
  }
  while ((c = getchar()) != '\n' && c != EOF) {
    s[i++] = c;
  }
  if (c == '\n') {
    s[i++] = c;
  }
  s[i] = '\0';
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getline2>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D14(%rip),%r13
  mov    0x0(%r13),%edx
  cmp    %edx,D17(%rip)
  je     L29
  mov    %rdi,%r12
  mov    %rdi,%rbx
  xor    %ebp,%ebp
  jmp    L30
  nopl   (%rax)
L33:
  cmp    0x0(%r13),%eax
  je     L28
  mov    %al,(%rbx)
  add    $0x1,%ebp
  add    $0x1,%rbx
L30:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,D0(%rip)
  cmp    $0xa,%eax
  jne    L33
  movb   $0xa,(%rbx)
  add    $0x1,%ebp
L28:
  movslq %ebp,%rbp
  mov    $0x1,%eax
  movb   $0x0,(%r12,%rbp,1)
L29:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int getline2(char s[]) {
  static int c;
  int i = 0;
  if (c == EOF) {
    return 0;
  }
  while ((c = getchar()) != '\n' && c != EOF) {
    s[i++] = c;
  }
  if (c == '\n') {
    s[i++] = c;
  }
  s[i] = '\0';
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<getline2>:
  endbr64
  push   %r13
  xor    %eax,%eax
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D17(%rip),%r13
  mov    0x0(%r13),%edx
  cmp    %edx,D21(%rip)
  je     L13
  mov    %rdi,%r12
  mov    %rdi,%rbx
  xor    %ebp,%ebp
  jmp    L10
  nopl   (%rax)
L38:
  cmp    0x0(%r13),%eax
  je     L1
  mov    %al,(%rbx)
  add    $0x1,%ebp
  add    $0x1,%rbx
L10:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,D22(%rip)
  cmp    $0xa,%eax
  jne    L38
  movb   $0xa,(%rbx)
  add    $0x1,%ebp
L1:
  movslq %ebp,%rbp
  mov    $0x1,%eax
  movb   $0x0,(%r12,%rbp,1)
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int getline2(char s[]) {
  static int c;
  int i = 0;
  if (c == EOF) {
    return 0;
  }
  while ((c = getchar()) != '\n' && c != EOF) {
    s[i++] = c;
  }
  if (c == '\n') {
    s[i++] = c;
  }
  s[i] = '\0';
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x30(%rbp),%eax
  cltq
  mov    $0x8,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L98
L73:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x1,%esi
  mov    $0xb,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x2c(%rbp)
L98:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L73
L95:
  movl   $0x0,-0x28(%rbp)
  jmp    L37
L78:
  mov    -0x30(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x28(%rbp)
  jne    L50
  mov    $0xa,%edx
  jmp    L75
L50:
  mov    $0x20,%edx
L75:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x28(%rbp)
L37:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jl     L78
  mov    -0x30(%rbp),%eax
  mov    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <next_permutation@plt>
  test   %eax,%eax
  jne    L95
  movl   $0x0,-0x24(%rbp)
  jmp    L82
L13:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x24(%rbp)
L82:
  mov    -0x30(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  jl     L13
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L93
  call   <__stack_chk_fail@plt>
L93:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D15
type_infered: string
name: disassemble_data
arguments: label: D13
type_infered: string
name: disassemble_data
arguments: label: D2
type_infered: string

tool: D15: %d

tool: D13: %s

tool: D2: %s%c

assistant: ```c
int main() {
  char **s;
  int n;
  scanf("%d", &n);
  s = calloc(n, sizeof(char *));
  for (int i = 0; i < n; i++) {
    s[i] = calloc(11, sizeof(char));
    scanf("%s", s[i]);
  }
  do {
    for (int i = 0; i < n; i++)
      printf("%s%c", s[i], i == n - 1 ? '\n' : ' ');
  } while (next_permutation(n, s));
  for (int i = 0; i < n; i++)
    free(s[i]);
  free(s);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ebx
  movslq %ebx,%rdi
  mov    $0x8,%esi
  call   <calloc@plt>
  mov    %rax,%rbp
  test   %ebx,%ebx
  jle    L71
  mov    $0x0,%ebx
  lea    D18(%rip),%r12
L0:
  mov    $0x1,%esi
  mov    $0xb,%edi
  call   <calloc@plt>
  mov    %rax,%rsi
  mov    %rax,0x0(%rbp,%rbx,8)
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x1,%rbx
  cmp    %ebx,0x4(%rsp)
  jg     L0
L71:
  lea    D16(%rip),%r12
L17:
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L76
  mov    $0x0,%ebx
L5:
  sub    $0x1,%edi
  cmp    %ebx,%edi
  mov    $0xa,%edx
  mov    $0x20,%eax
  cmovne %eax,%edx
  mov    0x0(%rbp,%rbx,8),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x4(%rsp),%edi
  add    $0x1,%rbx
  cmp    %ebx,%edi
  jg     L5
L76:
  mov    %rbp,%rsi
  mov    $0x0,%eax
  call   <next_permutation@plt>
  test   %eax,%eax
  jne    L17
  cmpl   $0x0,0x4(%rsp)
  jle    L11
  mov    $0x0,%ebx
L64:
  mov    0x0(%rbp,%rbx,8),%rdi
  call   <free@plt>
  add    $0x1,%rbx
  cmp    %ebx,0x4(%rsp)
  jg     L64
L11:
  mov    %rbp,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D8
datatype: string
name: read_data
arguments: data_label: D18
datatype: string
name: read_data
arguments: data_label: D16
datatype: string

tool: D8: %d

tool: D18: %s

tool: D16: %s%c

assistant: ```c
int main() {
  char **s;
  int n;
  scanf("%d", &n);
  s = calloc(n, sizeof(char *));
  for (int i = 0; i < n; i++) {
    s[i] = calloc(11, sizeof(char));
    scanf("%s", s[i]);
  }
  do {
    for (int i = 0; i < n; i++)
      printf("%s%c", s[i], i == n - 1 ? '\n' : ' ');
  } while (next_permutation(n, s));
  for (int i = 0; i < n; i++)
    free(s[i]);
  free(s);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  lea    D13(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rdi
  mov    $0x8,%esi
  mov    %rdi,%rbp
  call   <calloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L15
  xor    %ebx,%ebx
  lea    D8(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L85:
  mov    $0x1,%esi
  mov    $0xb,%edi
  call   <calloc@plt>
  mov    %r13,%rdi
  mov    %rax,(%r12,%rbx,8)
  mov    %rax,%rsi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <scanf@plt>
  mov    0x4(%rsp),%ebp
  cmp    %ebx,%ebp
  jg     L85
L15:
  lea    D9(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L73:
  test   %ebp,%ebp
  jle    L48
  xor    %ebx,%ebx
  mov    $0x20,%r13d
  nopl   0x0(%rax)
L57:
  sub    $0x1,%ebp
  mov    $0xa,%edx
  mov    (%r12,%rbx,8),%rsi
  mov    %r14,%rdi
  cmp    %ebx,%ebp
  cmovne %r13d,%edx
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    0x4(%rsp),%ebp
  cmp    %ebx,%ebp
  jg     L57
L48:
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <next_permutation@plt>
  test   %eax,%eax
  je     L86
  mov    0x4(%rsp),%ebp
  jmp    L73
  xchg   %ax,%ax
L86:
  mov    0x4(%rsp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L8
  nopw   0x0(%rax,%rax,1)
L87:
  mov    (%r12,%rbx,8),%rdi
  add    $0x1,%rbx
  call   <free@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L87
L8:
  mov    %r12,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D13
type_guessed: string
name: parse_data
arguments: tag: D8
type_guessed: string
name: parse_data
arguments: tag: D9
type_guessed: string

tool: D13: %d

tool: D8: %s

tool: D9: %s%c

assistant: ```c
int main() {
  char **s;
  int n;
  scanf("%d", &n);
  s = calloc(n, sizeof(char *));
  for (int i = 0; i < n; i++) {
    s[i] = calloc(11, sizeof(char));
    scanf("%s", s[i]);
  }
  do {
    for (int i = 0; i < n; i++)
      printf("%s%c", s[i], i == n - 1 ? '\n' : ' ');
  } while (next_permutation(n, s));
  for (int i = 0; i < n; i++)
    free(s[i]);
  free(s);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D26(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rdi
  mov    $0x8,%esi
  mov    %rdi,%rbp
  call   <calloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L11
  xor    %ebx,%ebx
  lea    D23(%rip),%r13
  nopl   0x0(%rax)
L44:
  mov    $0x1,%esi
  mov    $0xb,%edi
  call   <calloc@plt>
  mov    %r13,%rdi
  mov    %rax,(%r12,%rbx,8)
  mov    %rax,%rsi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <scanf@plt>
  mov    0x4(%rsp),%ebp
  cmp    %ebx,%ebp
  jg     L44
L11:
  lea    D21(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L12:
  test   %ebp,%ebp
  jle    L97
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L43:
  sub    $0x1,%ebp
  mov    (%r12,%rbx,8),%rsi
  mov    $0x20,%edx
  cmp    %ebx,%ebp
  jne    L92
  mov    $0xa,%edx
L92:
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    0x4(%rsp),%ebp
  cmp    %ebx,%ebp
  jg     L43
L97:
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <next_permutation@plt>
  test   %eax,%eax
  je     L28
  mov    0x4(%rsp),%ebp
  jmp    L12
  nopl   0x0(%rax)
L28:
  mov    0x4(%rsp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L65
  nopw   0x0(%rax,%rax,1)
L73:
  mov    (%r12,%rbx,8),%rdi
  add    $0x1,%rbx
  call   <free@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L73
L65:
  mov    %r12,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D26
type_guessed: string
name: disassemble_data
arguments: data_tag: D23
type_guessed: string
name: disassemble_data
arguments: data_tag: D21
type_guessed: string

tool: D26: %d

tool: D23: %s

tool: D21: %s%c

assistant: ```c
int main() {
  char **s;
  int n;
  scanf("%d", &n);
  s = calloc(n, sizeof(char *));
  for (int i = 0; i < n; i++) {
    s[i] = calloc(11, sizeof(char));
    scanf("%s", s[i]);
  }
  do {
    for (int i = 0; i < n; i++)
      printf("%s%c", s[i], i == n - 1 ? '\n' : ' ');
  } while (next_permutation(n, s));
  for (int i = 0; i < n; i++)
    free(s[i]);
  free(s);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sar_reg32_imm8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,%eax
  mov    %al,-0x8(%rbp)
  mov    $0xc1,%edi
  call   <put8@plt>
  mov    -0x4(%rbp),%eax
  sub    $0x8,%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <put8@plt>
  movzbl -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <put8@plt>
  nop
  leave
  ret
```

assistant: ```c
void sar_reg32_imm8(unsigned int reg32, unsigned char imm8) {
  put8(0xC1);
  put8(0xF8 + reg32);
  put8(imm8);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sar_reg32_imm8>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    $0xc1,%edi
  call   <put8@plt>
  lea    -0x8(%rbx),%edi
  movzbl %dil,%edi
  call   <put8@plt>
  movzbl %bpl,%edi
  call   <put8@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void sar_reg32_imm8(unsigned int reg32, unsigned char imm8) {
  put8(0xC1);
  put8(0xF8 + reg32);
  put8(imm8);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sar_reg32_imm8>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    $0xc1,%edi
  sub    $0x8,%rsp
  call   <put8@plt>
  lea    -0x8(%rbx),%edi
  movzbl %dil,%edi
  call   <put8@plt>
  add    $0x8,%rsp
  movzbl %bpl,%edi
  pop    %rbx
  pop    %rbp
  jmp    1050 <put8@plt>
```

assistant: ```c
void sar_reg32_imm8(unsigned int reg32, unsigned char imm8) {
  put8(0xC1);
  put8(0xF8 + reg32);
  put8(imm8);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sar_reg32_imm8>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    $0xc1,%edi
  sub    $0x8,%rsp
  call   <put8@plt>
  lea    -0x8(%rbx),%edi
  movzbl %dil,%edi
  call   <put8@plt>
  add    $0x8,%rsp
  movzbl %bpl,%edi
  pop    %rbx
  pop    %rbp
  jmp    1050 <put8@plt>
```

assistant: ```c
void sar_reg32_imm8(unsigned int reg32, unsigned char imm8) {
  put8(0xC1);
  put8(0xF8 + reg32);
  put8(imm8);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rop3_68_0_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  not    %eax
  mov    %al,-0x2(%rbp)
  movzbl -0x1c(%rbp),%eax
  and    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_68_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = ~*D;
  stk1 = S & stk2;
  *D = stk1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rop3_68_0_0>:
  endbr64
  movzbl (%rdi),%eax
  not    %eax
  and    %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_68_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = ~*D;
  stk1 = S & stk2;
  *D = stk1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rop3_68_0_0>:
  endbr64
  movzbl (%rdi),%eax
  not    %eax
  and    %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_68_0_0(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  stk2 = ~*D;
  stk1 = S & stk2;
  *D = stk1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_signed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
  pop    %rbp
  ret
```

assistant: ```c
void test_signed() {
  signed char a;
  signed short b;
  signed int c;
  signed long d;
  signed long long e;
  signed short int f;
  signed long int g;
  signed long long int h;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_signed>:
  endbr64
  ret
```

assistant: ```c
void test_signed() {
  signed char a;
  signed short b;
  signed int c;
  signed long d;
  signed long long e;
  signed short int f;
  signed long int g;
  signed long long int h;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_signed>:
  endbr64
  ret
```

assistant: ```c
void test_signed() {
  signed char a;
  signed short b;
  signed int c;
  signed long d;
  signed long long e;
  signed short int f;
  signed long int g;
  signed long long int h;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_signed>:
  endbr64
  ret
```

assistant: ```c
void test_signed() {
  signed char a;
  signed short b;
  signed int c;
  signed long d;
  signed long long e;
  signed short int f;
  signed long int g;
  signed long long int h;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<pdu_relative_validity>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x2d0,-0x14(%rbp)
  ja     L69
  mov    -0x14(%rbp),%eax
  add    $0x4,%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x2,%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L3
L69:
  cmpl   $0x5a0,-0x14(%rbp)
  ja     L39
  mov    -0x14(%rbp),%eax
  add    $0x1d,%eax
  mov    %eax,%edx
  mov    $0x88888889,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x4,%eax
  add    $0x77,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L3
L39:
  cmpl   $0xa8c0,-0x14(%rbp)
  ja     L43
  mov    -0x14(%rbp),%eax
  add    $0x59f,%eax
  shr    $0x5,%eax
  mov    %eax,%eax
  imul   $0x5b05b06,%rax,%rax
  shr    $0x20,%rax
  add    $0xa6,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L3
L43:
  cmpl   $0x9b0a0,-0x14(%rbp)
  ja     L45
  mov    -0x14(%rbp),%eax
  add    $0x275f,%eax
  shr    $0x5,%eax
  mov    %eax,%eax
  imul   $0x6806807,%rax,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  add    $0xc0,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L3
L45:
  movl   $0xff,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int pdu_relative_validity(unsigned minutes) {
  int validity;
  if (minutes <= 720)
    validity = (((minutes) + (5) - 1) / (5)) - 1;
  else if (minutes <= 1440)
    validity = (((minutes) + (30) - 1) / (30)) + 119;
  else if (minutes <= 43200)
    validity = (((minutes) + (1440) - 1) / (1440)) + 166;
  else if (minutes <= 635040)
    validity = (((minutes) + (10080) - 1) / (10080)) + 192;
  else
    validity = 0xFF;
  return validity;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<pdu_relative_validity>:
  endbr64
  cmp    $0x2d0,%edi
  ja     L23
  lea    0x4(%rdi),%eax
  mov    $0xcccccccd,%edx
  imul   %rdx,%rax
  shr    $0x22,%rax
  sub    $0x1,%eax
  ret
L23:
  cmp    $0x5a0,%edi
  ja     L24
  lea    0x1d(%rdi),%eax
  mov    $0x88888889,%edx
  imul   %rdx,%rax
  shr    $0x24,%rax
  add    $0x77,%eax
  ret
L24:
  cmp    $0xa8c0,%edi
  ja     L29
  lea    0x59f(%rdi),%eax
  shr    $0x5,%eax
  mov    %eax,%eax
  imul   $0x5b05b06,%rax,%rax
  shr    $0x20,%rax
  add    $0xa6,%eax
  ret
L29:
  mov    $0xff,%eax
  cmp    $0x9b0a0,%edi
  ja     L20
  lea    0x275f(%rdi),%eax
  shr    $0x5,%eax
  mov    %eax,%eax
  imul   $0x6806807,%rax,%rax
  shr    $0x23,%rax
  add    $0xc0,%eax
L20:
  ret
```

assistant: ```c
int pdu_relative_validity(unsigned minutes) {
  int validity;
  if (minutes <= 720)
    validity = (((minutes) + (5) - 1) / (5)) - 1;
  else if (minutes <= 1440)
    validity = (((minutes) + (30) - 1) / (30)) + 119;
  else if (minutes <= 43200)
    validity = (((minutes) + (1440) - 1) / (1440)) + 166;
  else if (minutes <= 635040)
    validity = (((minutes) + (10080) - 1) / (10080)) + 192;
  else
    validity = 0xFF;
  return validity;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pdu_relative_validity>:
  endbr64
  cmp    $0x2d0,%edi
  ja     L15
  lea    0x4(%rdi),%eax
  mov    $0xcccccccd,%edx
  imul   %rdx,%rax
  shr    $0x22,%rax
  sub    $0x1,%eax
  ret
L15:
  cmp    $0x5a0,%edi
  ja     L4
  lea    0x1d(%rdi),%eax
  mov    $0x88888889,%edx
  imul   %rdx,%rax
  shr    $0x24,%rax
  add    $0x77,%eax
  ret
  nopl   0x0(%rax)
L4:
  cmp    $0xa8c0,%edi
  jbe    L24
  mov    $0xff,%eax
  cmp    $0x9b0a0,%edi
  ja     L11
  lea    0x275f(%rdi),%eax
  shr    $0x5,%eax
  imul   $0x6806807,%rax,%rax
  shr    $0x23,%rax
  add    $0xc0,%eax
L11:
  ret
  nop
L24:
  lea    0x59f(%rdi),%eax
  shr    $0x5,%eax
  imul   $0x5b05b06,%rax,%rax
  shr    $0x20,%rax
  add    $0xa6,%eax
  ret
```

assistant: ```c
int pdu_relative_validity(unsigned minutes) {
  int validity;
  if (minutes <= 720)
    validity = (((minutes) + (5) - 1) / (5)) - 1;
  else if (minutes <= 1440)
    validity = (((minutes) + (30) - 1) / (30)) + 119;
  else if (minutes <= 43200)
    validity = (((minutes) + (1440) - 1) / (1440)) + 166;
  else if (minutes <= 635040)
    validity = (((minutes) + (10080) - 1) / (10080)) + 192;
  else
    validity = 0xFF;
  return validity;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<pdu_relative_validity>:
  endbr64
  cmp    $0x2d0,%edi
  ja     L37
  lea    0x4(%rdi),%eax
  mov    $0xcccccccd,%edx
  imul   %rdx,%rax
  shr    $0x22,%rax
  sub    $0x1,%eax
  ret
L37:
  cmp    $0x5a0,%edi
  ja     L28
  lea    0x1d(%rdi),%eax
  mov    $0x88888889,%edx
  imul   %rdx,%rax
  shr    $0x24,%rax
  add    $0x77,%eax
  ret
  nopl   0x0(%rax)
L28:
  cmp    $0xa8c0,%edi
  jbe    L9
  mov    $0xff,%eax
  cmp    $0x9b0a0,%edi
  ja     L35
  lea    0x275f(%rdi),%eax
  shr    $0x5,%eax
  imul   $0x6806807,%rax,%rax
  shr    $0x23,%rax
  add    $0xc0,%eax
L35:
  ret
  nop
L9:
  lea    0x59f(%rdi),%eax
  shr    $0x5,%eax
  imul   $0x5b05b06,%rax,%rax
  shr    $0x20,%rax
  add    $0xa6,%eax
  ret
```

assistant: ```c
int pdu_relative_validity(unsigned minutes) {
  int validity;
  if (minutes <= 720)
    validity = (((minutes) + (5) - 1) / (5)) - 1;
  else if (minutes <= 1440)
    validity = (((minutes) + (30) - 1) / (30)) + 119;
  else if (minutes <= 43200)
    validity = (((minutes) + (1440) - 1) / (1440)) + 166;
  else if (minutes <= 635040)
    validity = (((minutes) + (10080) - 1) / (10080)) + 192;
  else
    validity = 0xFF;
  return validity;
}
```