ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %eax,%esi
  movslq %eax,%rcx
  lea    0xf(%rcx),%rax
  mov    %rax,%rdi
  and    $0xfffffffffffffff0,%rdi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L29:
  cmp    %rdx,%rsp
  je     L12
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L29
L12:
  mov    %rdi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L19
  orq    $0x0,-0x8(%rsp,%rax,1)
L19:
  mov    %rsp,%r8
  lea    -0x1(%rsi),%eax
  test   %esi,%esi
  jle    L13
  cltq
  add    %rbx,%rax
  mov    %r8,%rdx
  lea    -0x2(%rbx,%rcx,1),%rdi
  lea    -0x1(%rsi),%ecx
  sub    %rcx,%rdi
L39:
  movzbl (%rax),%ecx
  mov    %cl,(%rdx)
  sub    $0x1,%rax
  add    $0x1,%rdx
  cmp    %rdi,%rax
  jne    L39
  mov    %esi,%ecx
  mov    $0x0,%eax
L43:
  movzbl (%r8,%rax,1),%edx
  mov    %dl,(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L43
L13:
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  mov    -0x8(%rbp),%rbx
  leave
  ret
L58:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverse(char *A) {
  int sizeOfArray = strlen(A);
  char bufferHold[sizeOfArray];
  for (int i = 0, j = sizeOfArray - 1; i < sizeOfArray; i++, j--) {
    bufferHold[i] = A[j];
  }
  for (int i = 0; i < sizeOfArray; i++) {
    A[i] = bufferHold[i];
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  mov    %rdi,%r12
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rsi
  movslq %eax,%rdi
  lea    0xf(%rdi),%rax
  mov    %rdi,%rdx
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L46
L29:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L29
L46:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L61
L32:
  mov    %rsp,%r8
  lea    -0x1(%rdx),%esi
  test   %edx,%edx
  jle    L6
  movslq %esi,%rax
  lea    -0x2(%r12,%rdi,1),%rdi
  mov    %esi,%esi
  mov    %rsp,%rcx
  add    %r12,%rax
  sub    %rsi,%rdi
  nopw   0x0(%rax,%rax,1)
L57:
  movzbl (%rax),%esi
  sub    $0x1,%rax
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %rdi,%rax
  jne    L57
  mov    %edx,%edx
  mov    %r8,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
L6:
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  mov    -0x8(%rbp),%r12
  leave
  ret
  cs nopw 0x0(%rax,%rax,1)
L61:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L32
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverse(char *A) {
  int sizeOfArray = strlen(A);
  char bufferHold[sizeOfArray];
  for (int i = 0, j = sizeOfArray - 1; i < sizeOfArray; i++, j--) {
    bufferHold[i] = A[j];
  }
  for (int i = 0; i < sizeOfArray; i++) {
    A[i] = bufferHold[i];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  mov    %rdi,%r12
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rsi
  movslq %eax,%rdi
  lea    0xf(%rdi),%rax
  mov    %rdi,%rdx
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L9
L40:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L40
L9:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L6
L39:
  mov    %rsp,%r8
  lea    -0x1(%rdx),%esi
  test   %edx,%edx
  jle    L62
  movslq %esi,%rax
  lea    -0x2(%r12,%rdi,1),%rdi
  mov    %esi,%esi
  mov    %rsp,%rcx
  add    %r12,%rax
  sub    %rsi,%rdi
  nopw   0x0(%rax,%rax,1)
L31:
  movzbl (%rax),%esi
  sub    $0x1,%rax
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %rdi,%rax
  jne    L31
  mov    %edx,%edx
  mov    %r8,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
L62:
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  mov    -0x8(%rbp),%r12
  leave
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L39
L47:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void reverse(char *A) {
  int sizeOfArray = strlen(A);
  char bufferHold[sizeOfArray];
  for (int i = 0, j = sizeOfArray - 1; i < sizeOfArray; i++, j--) {
    bufferHold[i] = A[j];
  }
  for (int i = 0; i < sizeOfArray; i++) {
    A[i] = bufferHold[i];
  }
}
```
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    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L31
L91:
  mov    -0xc(%rbp),%eax
  cmp    $0x3,%eax
  je     L1
  cmp    $0x3,%eax
  jg     L18
  cmp    $0x1,%eax
  je     L17
  cmp    $0x2,%eax
  je     L8
  jmp    L18
L17:
  mov    $0x0,%eax
  call   <insert@plt>
  jmp    L18
L8:
  mov    $0x0,%eax
  call   <dequeue@plt>
  jmp    L18
L1:
  mov    $0x0,%eax
  call   <display@plt>
  nop
L18:
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L31:
  mov    -0xc(%rbp),%eax
  cmp    $0x3,%eax
  jle    L91
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L61
  call   <__stack_chk_fail@plt>
L61:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
data_type: string
name: parse_data
arguments: data_label: D74
data_type: string
name: parse_data
arguments: data_label: D24
data_type: string
name: parse_data
arguments: data_label: D42
data_type: string
name: parse_data
arguments: data_label: D54
data_type: string
name: parse_data
arguments: data_label: D13
data_type: string
name: parse_data
arguments: data_label: D53
data_type: string

tool: D2: MAIN MENU 


tool: D74: 1.INSERT 


tool: D24: 2.DELETE 


tool: D42: 3.DISPLAY 


tool: D54: ENTER YOUR CHOICE 


tool: D13: %d

tool: D53: 
 ENTER YOUR CHOICE 


assistant: ```c
int main() {
  char ch;
  int choice;
  printf("MAIN MENU \n");
  printf("1.INSERT \n");
  printf("2.DELETE \n");
  printf("3.DISPLAY \n");
  printf("ENTER YOUR CHOICE \n");
  scanf("%d", &choice);
  while (choice < 4) {
    switch (choice) {
    case 1:
      insert();
      break;
    case 2:
      dequeue();
      break;
    case 3:
      display();
      break;
    }
    printf("\n ENTER YOUR CHOICE \n");
    scanf("%d", &choice);
  }
  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    D59(%rip),%rdi
  call   <printf@plt>
  lea    D50(%rip),%rdi
  call   <printf@plt>
  lea    D19(%rip),%rdi
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <printf@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D49(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x3,%eax
  jg     L5
  lea    D22(%rip),%r12
  lea    0x4(%rsp),%rbp
  lea    D49(%rip),%rbx
  jmp    L86
L84:
  mov    $0x0,%eax
  call   <dequeue@plt>
L35:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x3,%eax
  jg     L5
L86:
  cmp    $0x2,%eax
  je     L84
  cmp    $0x3,%eax
  je     L12
  cmp    $0x1,%eax
  jne    L35
  mov    $0x0,%eax
  call   <insert@plt>
  jmp    L35
L12:
  mov    $0x0,%eax
  call   <display@plt>
  jmp    L35
L5:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D59
type_infered: string
name: disassemble_data
arguments: data_tag: D50
type_infered: string
name: disassemble_data
arguments: data_tag: D19
type_infered: string
name: disassemble_data
arguments: data_tag: D12
type_infered: string
name: disassemble_data
arguments: data_tag: D9
type_infered: string
name: disassemble_data
arguments: data_tag: D49
type_infered: string
name: disassemble_data
arguments: data_tag: D22
type_infered: string

tool: D59: MAIN MENU 


tool: D50: 1.INSERT 


tool: D19: 2.DELETE 


tool: D12: 3.DISPLAY 


tool: D9: ENTER YOUR CHOICE 


tool: D49: %d

tool: D22: 
 ENTER YOUR CHOICE 


assistant: ```c
int main() {
  char ch;
  int choice;
  printf("MAIN MENU \n");
  printf("1.INSERT \n");
  printf("2.DELETE \n");
  printf("3.DISPLAY \n");
  printf("ENTER YOUR CHOICE \n");
  scanf("%d", &choice);
  while (choice < 4) {
    switch (choice) {
    case 1:
      insert();
      break;
    case 2:
      dequeue();
      break;
    case 3:
      display();
      break;
    }
    printf("\n ENTER YOUR CHOICE \n");
    scanf("%d", &choice);
  }
  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   %r12
  lea    D6(%rip),%rdi
  push   %rbp
  lea    D43(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    D36(%rip),%rdi
  call   <printf@plt>
  lea    D63(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x3,%eax
  jg     L50
  lea    D21(%rip),%r12
  jmp    L90
  nopl   0x0(%rax)
L86:
  cmp    $0x1,%eax
  je     L71
L48:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x3,%eax
  jg     L50
L90:
  cmp    $0x2,%eax
  je     L93
  cmp    $0x3,%eax
  jne    L86
  xor    %eax,%eax
  call   <display@plt>
  jmp    L48
  nopl   0x0(%rax)
L93:
  xor    %eax,%eax
  call   <dequeue@plt>
  jmp    L48
  nopl   0x0(%rax)
L71:
  xor    %eax,%eax
  call   <insert@plt>
  jmp    L48
  nopl   0x0(%rax)
L50:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D6
type_infered: string
name: unpack_data
arguments: tag: D43
type_infered: string
name: unpack_data
arguments: tag: D27
type_infered: string
name: unpack_data
arguments: tag: D42
type_infered: string
name: unpack_data
arguments: tag: D36
type_infered: string
name: unpack_data
arguments: tag: D63
type_infered: string
name: unpack_data
arguments: tag: D21
type_infered: string

tool: D6: MAIN MENU 


tool: D43: %d

tool: D27: 1.INSERT 


tool: D42: 2.DELETE 


tool: D36: 3.DISPLAY 


tool: D63: ENTER YOUR CHOICE 


tool: D21: 
 ENTER YOUR CHOICE 


assistant: ```c
int main() {
  char ch;
  int choice;
  printf("MAIN MENU \n");
  printf("1.INSERT \n");
  printf("2.DELETE \n");
  printf("3.DISPLAY \n");
  printf("ENTER YOUR CHOICE \n");
  scanf("%d", &choice);
  while (choice < 4) {
    switch (choice) {
    case 1:
      insert();
      break;
    case 2:
      dequeue();
      break;
    case 3:
      display();
      break;
    }
    printf("\n ENTER YOUR CHOICE \n");
    scanf("%d", &choice);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  lea    D14(%rip),%rdi
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    D48(%rip),%rdi
  call   <printf@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x3,%eax
  jg     L45
  lea    D46(%rip),%r12
  jmp    L95
  nopl   0x0(%rax)
L3:
  cmp    $0x1,%eax
  je     L72
L39:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    $0x3,%eax
  jg     L45
L95:
  cmp    $0x2,%eax
  je     L32
  cmp    $0x3,%eax
  jne    L3
  xor    %eax,%eax
  call   <display@plt>
  jmp    L39
  nopl   0x0(%rax)
L32:
  xor    %eax,%eax
  call   <dequeue@plt>
  jmp    L39
  nopl   0x0(%rax)
L72:
  xor    %eax,%eax
  call   <insert@plt>
  jmp    L39
  nopl   0x0(%rax)
L45:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D14
type_guessed: string
name: unpack_data
arguments: data_label: D8
type_guessed: string
name: unpack_data
arguments: data_label: D4
type_guessed: string
name: unpack_data
arguments: data_label: D42
type_guessed: string
name: unpack_data
arguments: data_label: D48
type_guessed: string
name: unpack_data
arguments: data_label: D25
type_guessed: string
name: unpack_data
arguments: data_label: D46
type_guessed: string

tool: D14: MAIN MENU 


tool: D8: %d

tool: D4: 1.INSERT 


tool: D42: 2.DELETE 


tool: D48: 3.DISPLAY 


tool: D25: ENTER YOUR CHOICE 


tool: D46: 
 ENTER YOUR CHOICE 


assistant: ```c
int main() {
  char ch;
  int choice;
  printf("MAIN MENU \n");
  printf("1.INSERT \n");
  printf("2.DELETE \n");
  printf("3.DISPLAY \n");
  printf("ENTER YOUR CHOICE \n");
  scanf("%d", &choice);
  while (choice < 4) {
    switch (choice) {
    case 1:
      insert();
      break;
    case 2:
      dequeue();
      break;
    case 3:
      display();
      break;
    }
    printf("\n ENTER YOUR CHOICE \n");
    scanf("%d", &choice);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<utf8ncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L27
L19:
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L22
  addq   $0x1,-0x18(%rbp)
L27:
  mov    -0x18(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jb     L19
  jmp    L10
L22:
  nop
  jmp    L10
L20:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addq   $0x1,-0x18(%rbp)
L10:
  mov    -0x18(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jb     L20
  mov    -0x28(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *utf8ncpy(void *__restrict__ dst, const void *__restrict__ src, size_t n) {
  char *d = (char *)dst;
  const char *s = (const char *)src;
  size_t index;
  for (index = 0; index < n; index++) {
    d[index] = s[index];
    if ('\0' == s[index]) {
      break;
    }
  }
  for (; index < n; index++) {
    d[index] = 0;
  }
  return dst;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<utf8ncpy>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L31
  mov    $0x0,%ecx
L9:
  movzbl (%rsi,%rcx,1),%r8d
  mov    %r8b,(%rax,%rcx,1)
  test   %r8b,%r8b
  je     L12
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  jne    L9
  ret
L12:
  cmp    %rcx,%rdx
  jbe    L5
  add    %rax,%rcx
  add    %rax,%rdx
L4:
  movb   $0x0,(%rcx)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L4
L31:
  ret
L5:
  ret
```

assistant: ```c
void *utf8ncpy(void *__restrict__ dst, const void *__restrict__ src, size_t n) {
  char *d = (char *)dst;
  const char *s = (const char *)src;
  size_t index;
  for (index = 0; index < n; index++) {
    d[index] = s[index];
    if ('\0' == s[index]) {
      break;
    }
  }
  for (; index < n; index++) {
    d[index] = 0;
  }
  return dst;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<utf8ncpy>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  test   %rdx,%rdx
  je     L19
  xor    %edi,%edi
  jmp    L42
  nopw   0x0(%rax,%rax,1)
L43:
  add    $0x1,%rdi
  cmp    %rdi,%rdx
  je     L19
L42:
  movzbl (%rsi,%rdi,1),%eax
  mov    %al,(%r12,%rdi,1)
  test   %al,%al
  jne    L43
  cmp    %rdi,%rdx
  jbe    L19
  sub    %rdi,%rdx
  xor    %esi,%esi
  add    %r12,%rdi
  call   <memset@plt>
L19:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *utf8ncpy(void *__restrict__ dst, const void *__restrict__ src, size_t n) {
  char *d = (char *)dst;
  const char *s = (const char *)src;
  size_t index;
  for (index = 0; index < n; index++) {
    d[index] = s[index];
    if ('\0' == s[index]) {
      break;
    }
  }
  for (; index < n; index++) {
    d[index] = 0;
  }
  return dst;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<utf8ncpy>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  test   %rdx,%rdx
  je     L36
  xor    %edi,%edi
  jmp    L49
  nopw   0x0(%rax,%rax,1)
L21:
  add    $0x1,%rdi
  cmp    %rdi,%rdx
  je     L36
L49:
  movzbl (%rsi,%rdi,1),%eax
  mov    %al,(%r12,%rdi,1)
  test   %al,%al
  jne    L21
  cmp    %rdi,%rdx
  jbe    L36
  sub    %rdi,%rdx
  xor    %esi,%esi
  add    %r12,%rdi
  call   <memset@plt>
L36:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
void *utf8ncpy(void *__restrict__ dst, const void *__restrict__ src, size_t n) {
  char *d = (char *)dst;
  const char *s = (const char *)src;
  size_t index;
  for (index = 0; index < n; index++) {
    d[index] = s[index];
    if ('\0' == s[index]) {
      break;
    }
  }
  for (; index < n; index++) {
    d[index] = 0;
  }
  return dst;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<splitLen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%eax
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L20
L24:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0xc(%rbp)
L20:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L24
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L18
L29:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L18:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L29
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void splitLen(int *lens, int threads, int n) {
  int i, l, m;
  l = n / threads;
  m = n % threads;
  for (i = 0; i < m; ++i)
    lens[i] = l + 1;
  for (i = m; i < threads; ++i)
    lens[i] = l;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<splitLen>:
  endbr64
  mov    %rdi,%r9
  mov    %edx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  jle    L21
  mov    %rdi,%rcx
  lea    -0x1(%rdx),%edi
  lea    0x4(%r9,%rdi,4),%r8
  lea    0x1(%rax),%edi
L31:
  mov    %edi,(%rcx)
  add    $0x4,%rcx
  cmp    %r8,%rcx
  jne    L31
L21:
  cmp    %edx,%esi
  jle    L18
  movslq %edx,%rdi
  lea    (%r9,%rdi,4),%rcx
  sub    $0x1,%esi
  sub    %edx,%esi
  add    %rdi,%rsi
  lea    0x4(%r9,%rsi,4),%rdx
L4:
  mov    %eax,(%rcx)
  add    $0x4,%rcx
  cmp    %rdx,%rcx
  jne    L4
L18:
  ret
```

assistant: ```c
void splitLen(int *lens, int threads, int n) {
  int i, l, m;
  l = n / threads;
  m = n % threads;
  for (i = 0; i < m; ++i)
    lens[i] = l + 1;
  for (i = m; i < threads; ++i)
    lens[i] = l;
}
```
user: Translate the below assembly code into C programming language:

```asm
<splitLen>:
  endbr64
  mov    %edx,%eax
  mov    %rdi,%r9
  cltd
  idiv   %esi
  test   %edx,%edx
  jle    L2
  mov    %rdi,%rcx
  lea    -0x1(%rdx),%edi
  lea    0x1(%rax),%r8d
  lea    0x4(%r9,%rdi,4),%rdi
  nop
L28:
  mov    %r8d,(%rcx)
  add    $0x4,%rcx
  cmp    %rdi,%rcx
  jne    L28
L2:
  cmp    %edx,%esi
  jle    L9
  sub    $0x1,%esi
  movslq %edx,%rdi
  sub    %edx,%esi
  lea    (%r9,%rdi,4),%rcx
  add    %rdi,%rsi
  lea    0x4(%r9,%rsi,4),%rdx
  nopl   0x0(%rax)
L14:
  mov    %eax,(%rcx)
  add    $0x4,%rcx
  cmp    %rdx,%rcx
  jne    L14
  ret
  nopl   0x0(%rax)
L9:
  ret
```

assistant: ```c
void splitLen(int *lens, int threads, int n) {
  int i, l, m;
  l = n / threads;
  m = n % threads;
  for (i = 0; i < m; ++i)
    lens[i] = l + 1;
  for (i = m; i < threads; ++i)
    lens[i] = l;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<splitLen>:
  endbr64
  mov    %edx,%eax
  mov    %esi,%r9d
  cltd
  idiv   %esi
  mov    %edx,%r8d
  test   %edx,%edx
  jle    L16
  lea    -0x1(%rdx),%ecx
  lea    0x1(%rax),%r10d
  cmp    $0x2,%ecx
  jbe    L123
  mov    %edx,%esi
  movd   %r10d,%xmm1
  mov    %rdi,%rcx
  shr    $0x2,%esi
  pshufd $0x0,%xmm1,%xmm0
  sub    $0x1,%esi
  shl    $0x4,%rsi
  lea    0x10(%rdi,%rsi,1),%rsi
  nopl   0x0(%rax)
L18:
  movups %xmm0,(%rcx)
  add    $0x10,%rcx
  cmp    %rsi,%rcx
  jne    L18
  mov    %edx,%ecx
  and    $0xfffffffc,%ecx
  test   $0x3,%dl
  je     L16
L46:
  movslq %ecx,%rsi
  mov    %r10d,(%rdi,%rsi,4)
  lea    0x0(,%rsi,4),%r11
  lea    0x1(%rcx),%esi
  cmp    %esi,%edx
  jle    L16
  add    $0x2,%ecx
  mov    %r10d,0x4(%rdi,%r11,1)
  cmp    %ecx,%edx
  jle    L16
  mov    %r10d,0x8(%rdi,%r11,1)
L16:
  cmp    %edx,%r9d
  jle    L110
  mov    %r9d,%r10d
  sub    %edx,%r10d
  lea    -0x1(%r10),%ecx
  cmp    $0x2,%ecx
  jbe    L36
  movslq %edx,%rcx
  movd   %eax,%xmm2
  lea    (%rdi,%rcx,4),%rsi
  mov    %r10d,%ecx
  pshufd $0x0,%xmm2,%xmm0
  shr    $0x2,%ecx
  lea    -0x1(%rcx),%r8d
  lea    0x10(%rsi),%rcx
  shl    $0x4,%r8
  add    %rcx,%r8
  jmp    L59
  nopl   0x0(%rax)
L15:
  add    $0x10,%rcx
L59:
  movups %xmm0,(%rsi)
  mov    %rcx,%rsi
  cmp    %rcx,%r8
  jne    L15
  mov    %r10d,%ecx
  and    $0xfffffffc,%ecx
  lea    (%rdx,%rcx,1),%r8d
  cmp    %ecx,%r10d
  je     L65
L36:
  movslq %r8d,%rdx
  mov    %eax,(%rdi,%rdx,4)
  lea    0x0(,%rdx,4),%rcx
  lea    0x1(%r8),%edx
  cmp    %edx,%r9d
  jle    L65
  add    $0x2,%r8d
  mov    %eax,0x4(%rdi,%rcx,1)
  cmp    %r8d,%r9d
  jle    L65
  mov    %eax,0x8(%rdi,%rcx,1)
L65:
  ret
  nop
L110:
  ret
L123:
  xor    %ecx,%ecx
  jmp    L46
```

assistant: ```c
void splitLen(int *lens, int threads, int n) {
  int i, l, m;
  l = n / threads;
  m = n % threads;
  for (i = 0; i < m; ++i)
    lens[i] = l + 1;
  for (i = m; i < threads; ++i)
    lens[i] = l;
}
```