ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<local>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x616d6172676f7270,%rax
  mov    %rax,0xc(%rsp)
  movl   $0x726f64,0x14(%rsp)
  lea    0xc(%rsp),%rbp
  mov    %rbp,(%rsp)
  mov    %rbp,%rsi
  lea    D19(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    (%rsp),%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    (%rsp),%rsi
  lea    D4(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D19
type: string
name: extract_symbolic_data
arguments: data_label: D4
type: string

tool: D19: %s


tool: D4: %p


assistant: ```c
char *local(void) {
  static char nomex[12] = "aaaaaaaaaa";
  char nome[12] = "programador";
  char *ponteiro = nome;
  printf("%s\n", nome);
  printf("%s\n", ponteiro);
  printf("%p\n", (void *)ponteiro);
  printf("%p\n", (void *)nome);
  printf("%p\n", (void *)nomex);
  printf("%p\n", (void *)&ponteiro);
  return ponteiro;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<local>:
  endbr64
  push   %r12
  push   %rbp
  lea    D5(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x616d6172676f7270,%rax
  lea    0xc(%rsp),%r12
  movl   $0x726f64,0x14(%rsp)
  mov    %r12,%rsi
  mov    %r12,(%rsp)
  mov    %rax,0xc(%rsp)
  call   <printf@plt>
  mov    (%rsp),%rsi
  mov    %rbp,%rdi
  lea    D25(%rip),%rbp
  call   <printf@plt>
  mov    (%rsp),%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D22(%rip),%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D5
data_type: string
name: parse_data
arguments: data_label: D25
data_type: string

tool: D5: %s


tool: D25: %p


assistant: ```c
char *local(void) {
  static char nomex[12] = "aaaaaaaaaa";
  char nome[12] = "programador";
  char *ponteiro = nome;
  printf("%s\n", nome);
  printf("%s\n", ponteiro);
  printf("%p\n", (void *)ponteiro);
  printf("%p\n", (void *)nome);
  printf("%p\n", (void *)nomex);
  printf("%p\n", (void *)&ponteiro);
  return ponteiro;
}
```
user: Translate the below assembly code into C programming language:

```asm
<local>:
  endbr64
  push   %r12
  push   %rbp
  lea    D13(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x616d6172676f7270,%rax
  lea    0xc(%rsp),%r12
  movl   $0x726f64,0x14(%rsp)
  mov    %r12,%rsi
  mov    %r12,(%rsp)
  mov    %rax,0xc(%rsp)
  call   <printf@plt>
  mov    (%rsp),%rsi
  mov    %rbp,%rdi
  lea    D7(%rip),%rbp
  call   <printf@plt>
  mov    (%rsp),%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    D24(%rip),%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
type: string
name: retrieve_data
arguments: data_tag: D7
type: string

tool: D13: %s


tool: D7: %p


assistant: ```c
char *local(void) {
  static char nomex[12] = "aaaaaaaaaa";
  char nome[12] = "programador";
  char *ponteiro = nome;
  printf("%s\n", nome);
  printf("%s\n", ponteiro);
  printf("%p\n", (void *)ponteiro);
  printf("%p\n", (void *)nome);
  printf("%p\n", (void *)nomex);
  printf("%p\n", (void *)&ponteiro);
  return ponteiro;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ZnoAllocClear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  cmpq   $0x0,-0x18(%rbp)
  je     L9
  mov    -0x10(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rax,%rdi
  call   *%rdx
L9:
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *ZnoAllocClear(size_t len, void *p, void *(*init)(void *)) {
  memset(p, 0, len);
  if (init)
    init(p);
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ZnoAllocClear>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    %rdi,%rdx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  test   %rbp,%rbp
  je     L7
  mov    %rbx,%rdi
  call   *%rbp
L7:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *ZnoAllocClear(size_t len, void *p, void *(*init)(void *)) {
  memset(p, 0, len);
  if (init)
    init(p);
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ZnoAllocClear>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbx
  mov    %rdx,%rbx
  mov    %rdi,%rdx
  mov    %r12,%rdi
  sub    $0x8,%rsp
  call   <memset@plt>
  test   %rbx,%rbx
  je     L7
  mov    %r12,%rdi
  call   *%rbx
L7:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
void *ZnoAllocClear(size_t len, void *p, void *(*init)(void *)) {
  memset(p, 0, len);
  if (init)
    init(p);
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ZnoAllocClear>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbx
  mov    %rdx,%rbx
  mov    %rdi,%rdx
  mov    %r12,%rdi
  sub    $0x8,%rsp
  call   <memset@plt>
  test   %rbx,%rbx
  je     L3
  mov    %r12,%rdi
  call   *%rbx
L3:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
void *ZnoAllocClear(size_t len, void *p, void *(*init)(void *)) {
  memset(p, 0, len);
  if (init)
    init(p);
  return p;
}
```
user: What is the c source code of the assembly code below:

```asm
<etypesw2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x20,-0x4(%rbp)
  jne    L15
  mov    $0x1,%eax
  jmp    L8
L15:
  mov    $0x0,%eax
L8:
  pop    %rbp
  ret
```

assistant: ```c
int etypesw2(etype)
int etype;
{
  switch (etype) {
  case 32:
    return 1;
  default:
    return 0;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<etypesw2>:
  endbr64
  cmp    $0x20,%edi
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int etypesw2(etype)
int etype;
{
  switch (etype) {
  case 32:
    return 1;
  default:
    return 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<etypesw2>:
  endbr64
  xor    %eax,%eax
  cmp    $0x20,%edi
  sete   %al
  ret
```

assistant: ```c
int etypesw2(etype)
int etype;
{
  switch (etype) {
  case 32:
    return 1;
  default:
    return 0;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<etypesw2>:
  endbr64
  xor    %eax,%eax
  cmp    $0x20,%edi
  sete   %al
  ret
```

assistant: ```c
int etypesw2(etype)
int etype;
{
  switch (etype) {
  case 32:
    return 1;
  default:
    return 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x3,-0x8(%rbp)
  movl   $0x5,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int func(void) {
  int n = 3;
  int m = 5;
  n = (m--);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<func>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int func(void) {
  int n = 3;
  int m = 5;
  n = (m--);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<func>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int func(void) {
  int n = 3;
  int m = 5;
  n = (m--);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<func>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int func(void) {
  int n = 3;
  int m = 5;
  n = (m--);
  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    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x410(%rbp),%rax
  mov    $0x400,%esi
  mov    %rax,%rdi
  call   <GetLine@plt>
  lea    -0x410(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
type: string

tool: D6: %s

assistant: ```c
int main(void) {
  char s[1024];
  GetLine(s, 1024);
  printf("%s", s);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x400,%esi
  mov    %rbx,%rdi
  call   <GetLine@plt>
  mov    %rbx,%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x410,%rsp
  pop    %rbx
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D5
type: string

tool: D5: %s

assistant: ```c
int main(void) {
  char s[1024];
  GetLine(s, 1024);
  printf("%s", s);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x400,%esi
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <GetLine@plt>
  mov    %rbp,%rsi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type: string

tool: D8: %s

assistant: ```c
int main(void) {
  char s[1024];
  GetLine(s, 1024);
  printf("%s", 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
<main>:
  endbr64
  push   %rbp
  mov    $0x400,%esi
  sub    $0x410,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <GetLine@plt>
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x410,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: %s

assistant: ```c
int main(void) {
  char s[1024];
  GetLine(s, 1024);
  printf("%s", s);
  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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x10(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L38
L33:
  movl   $0x0,-0xc(%rbp)
  jmp    L30
L14:
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L30:
  cmpl   $0x2,-0xc(%rbp)
  jle    L14
  addl   $0x1,-0x10(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
L38:
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  jg     L33
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D11
type: string
name: extract_symbolic_data
arguments: label: D8
type: string
name: extract_symbolic_data
arguments: label: D2
type: string

tool: D11:  %d

tool: D8: %d 

tool: D2: PUM

assistant: ```c
int main() {
  int N, i = 1, k;
  scanf(" %d", &N);
  while (N > 0) {
    k = 0;
    while (k < 3) {
      printf("%d ", i);
      i++;
      k++;
    }
    i += 1;
    printf("PUM\n");
    N--;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```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    D16(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L7
  mov    $0x1,%ebx
  lea    D14(%rip),%rbp
  lea    D17(%rip),%r12
L29:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x2(%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%ebx
  mov    %r12,%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L29
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D16
type: string
name: disassemble_data
arguments: tag: D14
type: string
name: disassemble_data
arguments: tag: D17
type: string

tool: D16:  %d

tool: D14: %d 

tool: D17: PUM

assistant: ```c
int main() {
  int N, i = 1, k;
  scanf(" %d", &N);
  while (N > 0) {
    k = 0;
    while (k < 3) {
      printf("%d ", i);
      i++;
      k++;
    }
    i += 1;
    printf("PUM\n");
    N--;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  lea    D12(%rip),%rdi
  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>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L17
  mov    $0x1,%ebx
  lea    D5(%rip),%rbp
  lea    D24(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x2(%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  add    $0x4,%ebx
  call   <puts@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L2
L17:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D12
datatype: string
name: unpack_data
arguments: label: D5
datatype: string
name: unpack_data
arguments: label: D24
datatype: string

tool: D12:  %d

tool: D5: %d 

tool: D24: PUM

assistant: ```c
int main() {
  int N, i = 1, k;
  scanf(" %d", &N);
  while (N > 0) {
    k = 0;
    while (k < 3) {
      printf("%d ", i);
      i++;
      k++;
    }
    i += 1;
    printf("PUM\n");
    N--;
  }
  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    D3(%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>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L7
  mov    $0x4,%ebp
  lea    D28(%rip),%r12
  lea    D11(%rip),%r13
  nopw   0x0(%rax,%rax,1)
L12:
  lea    -0x3(%rbp),%ebx
L23:
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebp,%ebx
  jne    L23
  mov    %r13,%rdi
  add    $0x4,%ebp
  call   <puts@plt>
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L12
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: string
name: unpack_data
arguments: data_tag: D28
data_type: string
name: unpack_data
arguments: data_tag: D11
data_type: string

tool: D3:  %d

tool: D28: %d 

tool: D11: PUM

assistant: ```c
int main() {
  int N, i = 1, k;
  scanf(" %d", &N);
  while (N > 0) {
    k = 0;
    while (k < 3) {
      printf("%d ", i);
      i++;
      k++;
    }
    i += 1;
    printf("PUM\n");
    N--;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<strncmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xe(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  jmp    L13
L3:
  movsbl -0xe(%rbp),%eax
  movsbl -0xd(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L68
  mov    -0xc(%rbp),%eax
  jmp    L12
L68:
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jae    L56
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xe(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
L13:
  cmpb   $0x0,-0xe(%rbp)
  je     L5
  cmpb   $0x0,-0xd(%rbp)
  jne    L3
  jmp    L5
L56:
  nop
L5:
  movsbl -0xe(%rbp),%eax
  movsbl -0xd(%rbp),%edx
  sub    %edx,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int strncmp(const char *str1, const char *str2, size_t num) {
  size_t i = 0;
  char ch1 = str1[i];
  char ch2 = str2[i];
  while (ch1 != '\0' && ch2 != '\0') {
    int diff = ch1 - ch2;
    if (diff != 0) {
      return diff;
    }
    ++i;
    if (i >= num) {
      break;
    }
    ch1 = str1[i];
    ch2 = str2[i];
  }
  return ch1 - ch2;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<strncmp>:
  endbr64
  mov    %rdi,%r10
  movzbl (%rdi),%r8d
  movzbl (%rsi),%edi
  test   %r8b,%r8b
  je     L10
  test   %dil,%dil
  je     L10
  movsbl %r8b,%eax
  movsbl %dil,%ecx
  sub    %ecx,%eax
  jne    L53
  cmp    $0x1,%rdx
  jbe    L10
  mov    $0x1,%ecx
L60:
  movzbl (%r10,%rcx,1),%r8d
  movzbl (%rsi,%rcx,1),%edi
  test   %r8b,%r8b
  je     L10
  test   %dil,%dil
  je     L10
  movsbl %r8b,%eax
  movsbl %dil,%r9d
  sub    %r9d,%eax
  jne    L53
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L60
L10:
  movsbl %r8b,%eax
  movsbl %dil,%edi
  sub    %edi,%eax
L53:
  ret
```

assistant: ```c
int strncmp(const char *str1, const char *str2, size_t num) {
  size_t i = 0;
  char ch1 = str1[i];
  char ch2 = str2[i];
  while (ch1 != '\0' && ch2 != '\0') {
    int diff = ch1 - ch2;
    if (diff != 0) {
      return diff;
    }
    ++i;
    if (i >= num) {
      break;
    }
    ch1 = str1[i];
    ch2 = str2[i];
  }
  return ch1 - ch2;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strncmp>:
  endbr64
  movsbl (%rdi),%eax
  movsbl (%rsi),%r8d
  mov    %rdi,%r10
  mov    %eax,%ecx
  sub    %r8d,%eax
  test   %cl,%cl
  je     L89
  test   %r8b,%r8b
  je     L89
  test   %eax,%eax
  jne    L89
  cmp    $0x1,%rdx
  jbe    L89
  mov    $0x1,%ecx
  jmp    L5
  nopl   (%rax)
L43:
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  je     L89
L5:
  movsbl (%r10,%rcx,1),%eax
  movsbl (%rsi,%rcx,1),%r9d
  mov    %eax,%edi
  sub    %r9d,%eax
  test   %dil,%dil
  je     L89
  test   %r9b,%r9b
  je     L89
  test   %eax,%eax
  je     L43
L89:
  ret
```

assistant: ```c
int strncmp(const char *str1, const char *str2, size_t num) {
  size_t i = 0;
  char ch1 = str1[i];
  char ch2 = str2[i];
  while (ch1 != '\0' && ch2 != '\0') {
    int diff = ch1 - ch2;
    if (diff != 0) {
      return diff;
    }
    ++i;
    if (i >= num) {
      break;
    }
    ch1 = str1[i];
    ch2 = str2[i];
  }
  return ch1 - ch2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strncmp>:
  endbr64
  movsbl (%rdi),%eax
  movsbl (%rsi),%r8d
  mov    %rdi,%r10
  mov    %eax,%ecx
  sub    %r8d,%eax
  test   %cl,%cl
  je     L54
  test   %r8b,%r8b
  je     L54
  test   %eax,%eax
  jne    L54
  cmp    $0x1,%rdx
  jbe    L54
  mov    $0x1,%ecx
  jmp    L57
  nopl   (%rax)
L20:
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  je     L54
L57:
  movsbl (%r10,%rcx,1),%eax
  movsbl (%rsi,%rcx,1),%r9d
  mov    %eax,%edi
  sub    %r9d,%eax
  test   %dil,%dil
  je     L54
  test   %r9b,%r9b
  je     L54
  test   %eax,%eax
  je     L20
L54:
  ret
```

assistant: ```c
int strncmp(const char *str1, const char *str2, size_t num) {
  size_t i = 0;
  char ch1 = str1[i];
  char ch2 = str2[i];
  while (ch1 != '\0' && ch2 != '\0') {
    int diff = ch1 - ch2;
    if (diff != 0) {
      return diff;
    }
    ++i;
    if (i >= num) {
      break;
    }
    ch1 = str1[i];
    ch2 = str2[i];
  }
  return ch1 - ch2;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<pkcs7pad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  lea    0x1(%rax),%edx
  mov    -0x20(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L14
L19:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L14:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L19
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int pkcs7pad(const uint8_t *const message, const int len, const int BK_SIZE, uint8_t *const out) {
  const int nsize = (len / BK_SIZE + 1) * BK_SIZE;
  memset(out, 0, nsize);
  memcpy(out, message, len);
  const int pad = nsize - len;
  for (int i = len; i < nsize; i++) {
    out[i] = pad;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pkcs7pad>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%ebp
  mov    %edx,%esi
  mov    %rcx,%r12
  mov    %ebp,%eax
  cltd
  idiv   %esi
  lea    0x1(%rax),%ebx
  imul   %esi,%ebx
  mov    %ebx,%edx
  mov    $0x0,%esi
  mov    %rcx,%rdi
  call   <memset@plt>
  mov    %ebp,%edx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %ebx,%edx
  sub    %ebp,%edx
  cmp    %ebx,%ebp
  jge    L0
  movslq %ebp,%rsi
  lea    (%r12,%rsi,4),%rcx
  lea    -0x1(%rdx),%eax
  add    %rsi,%rax
  lea    0x4(%r12,%rax,4),%rax
L17:
  mov    %edx,(%rcx)
  add    $0x4,%rcx
  cmp    %rax,%rcx
  jne    L17
L0:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int pkcs7pad(const uint8_t *const message, const int len, const int BK_SIZE, uint8_t *const out) {
  const int nsize = (len / BK_SIZE + 1) * BK_SIZE;
  memset(out, 0, nsize);
  memcpy(out, message, len);
  const int pad = nsize - len;
  for (int i = len; i < nsize; i++) {
    out[i] = pad;
  }
  return 0;
}
```