ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  cmp    $0xb,%eax
  ja     L65
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D74(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D74(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D109(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D116(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D135(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D116(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D116(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L65
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
L65:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D28
type_infered: string
name: retrieve_data
arguments: label: D53
type_infered: string
name: retrieve_data
arguments: label: D109
type_infered: string
name: retrieve_data
arguments: label: D25
type_infered: string
name: retrieve_data
arguments: label: D50
type_infered: string
name: retrieve_data
arguments: label: D116
type_infered: string
name: retrieve_data
arguments: label: D60
type_infered: string
name: retrieve_data
arguments: label: D135
type_infered: string
name: retrieve_data
arguments: label: D15
type_infered: string

tool: D28: %d

tool: D53: 5


tool: D109: 20


tool: D25: 12


tool: D50: 2


tool: D116: 1


tool: D60: 4


tool: D135: 6


tool: D15: 0


assistant: ```c
int main(void) {
  int input;
  scanf("%d", &input);
  switch (input) {
  case 0:
    printf("5\n");
    break;
  case 1:
    printf("20\n");
    break;
  case 2:
    printf("12\n");
    break;
  case 3:
    printf("2\n");
    break;
  case 4:
    printf("1\n");
    break;
  case 5:
    printf("4\n");
    break;
  case 6:
    printf("6\n");
    break;
  case 7:
    printf("1\n");
    break;
  case 8:
    printf("4\n");
    break;
  case 9:
    printf("4\n");
    break;
  case 10:
    printf("1\n");
    break;
  case 11:
    printf("0\n");
    break;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D8(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0xb,0x4(%rsp)
  ja     L91
  mov    0x4(%rsp),%eax
  lea    D87(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D135(%rip),%rdi
  call   <printf@plt>
L91:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L80
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D48(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D100(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D32(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D123(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D111(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D32(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D123(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D123(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D32(%rip),%rdi
  call   <printf@plt>
  jmp    L91
  lea    D15(%rip),%rdi
  call   <printf@plt>
  jmp    L91
L80:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D8
type_guessed: string
name: read_data
arguments: tag: D135
type_guessed: string
name: read_data
arguments: tag: D24
type_guessed: string
name: read_data
arguments: tag: D48
type_guessed: string
name: read_data
arguments: tag: D100
type_guessed: string
name: read_data
arguments: tag: D32
type_guessed: string
name: read_data
arguments: tag: D123
type_guessed: string
name: read_data
arguments: tag: D111
type_guessed: string
name: read_data
arguments: tag: D15
type_guessed: string

tool: D8: %d

tool: D135: 5


tool: D24: 20


tool: D48: 12


tool: D100: 2


tool: D32: 1


tool: D123: 4


tool: D111: 6


tool: D15: 0


assistant: ```c
int main(void) {
  int input;
  scanf("%d", &input);
  switch (input) {
  case 0:
    printf("5\n");
    break;
  case 1:
    printf("20\n");
    break;
  case 2:
    printf("12\n");
    break;
  case 3:
    printf("2\n");
    break;
  case 4:
    printf("1\n");
    break;
  case 5:
    printf("4\n");
    break;
  case 6:
    printf("6\n");
    break;
  case 7:
    printf("1\n");
    break;
  case 8:
    printf("4\n");
    break;
  case 9:
    printf("4\n");
    break;
  case 10:
    printf("1\n");
    break;
  case 11:
    printf("0\n");
    break;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D56(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0xb,0x4(%rsp)
  ja     L88
  mov    0x4(%rsp),%eax
  lea    D71(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D95(%rip),%rdi
  call   <printf@plt>
L88:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  lea    D97(%rip),%rdi
  call   <printf@plt>
  jmp    L88
  lea    D41(%rip),%rdi
  call   <printf@plt>
  jmp    L88
  lea    D25(%rip),%rdi
  call   <printf@plt>
  jmp    L88
  lea    D34(%rip),%rdi
  call   <printf@plt>
  jmp    L88
  lea    D30(%rip),%rdi
  call   <printf@plt>
  jmp    L88
  lea    D79(%rip),%rdi
  call   <printf@plt>
  jmp    L88
  lea    D36(%rip),%rdi
  call   <printf@plt>
  jmp    L88
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D56
type: string
name: retrieve_data
arguments: data_tag: D95
type: string
name: retrieve_data
arguments: data_tag: D97
type: string
name: retrieve_data
arguments: data_tag: D41
type: string
name: retrieve_data
arguments: data_tag: D25
type: string
name: retrieve_data
arguments: data_tag: D34
type: string
name: retrieve_data
arguments: data_tag: D30
type: string
name: retrieve_data
arguments: data_tag: D79
type: string
name: retrieve_data
arguments: data_tag: D36
type: string

tool: D56: %d

tool: D95: 1


tool: D97: 4


tool: D41: 2


tool: D25: 6


tool: D34: 0


tool: D30: 5


tool: D79: 20


tool: D36: 12


assistant: ```c
int main(void) {
  int input;
  scanf("%d", &input);
  switch (input) {
  case 0:
    printf("5\n");
    break;
  case 1:
    printf("20\n");
    break;
  case 2:
    printf("12\n");
    break;
  case 3:
    printf("2\n");
    break;
  case 4:
    printf("1\n");
    break;
  case 5:
    printf("4\n");
    break;
  case 6:
    printf("6\n");
    break;
  case 7:
    printf("1\n");
    break;
  case 8:
    printf("4\n");
    break;
  case 9:
    printf("4\n");
    break;
  case 10:
    printf("1\n");
    break;
  case 11:
    printf("0\n");
    break;
  }
  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
  sub    $0x18,%rsp
  lea    D30(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  cmpl   $0xb,0x4(%rsp)
  ja     L36
  mov    0x4(%rsp),%eax
  lea    D38(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D80(%rip),%rdi
  call   <printf@plt>
L36:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  lea    D51(%rip),%rdi
  call   <printf@plt>
  jmp    L36
  lea    D29(%rip),%rdi
  call   <printf@plt>
  jmp    L36
  lea    D43(%rip),%rdi
  call   <printf@plt>
  jmp    L36
  lea    D73(%rip),%rdi
  call   <printf@plt>
  jmp    L36
  lea    D22(%rip),%rdi
  call   <printf@plt>
  jmp    L36
  lea    D70(%rip),%rdi
  call   <printf@plt>
  jmp    L36
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L36
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D30
data_type: string
name: unpack_data
arguments: data_tag: D80
data_type: string
name: unpack_data
arguments: data_tag: D51
data_type: string
name: unpack_data
arguments: data_tag: D29
data_type: string
name: unpack_data
arguments: data_tag: D43
data_type: string
name: unpack_data
arguments: data_tag: D73
data_type: string
name: unpack_data
arguments: data_tag: D22
data_type: string
name: unpack_data
arguments: data_tag: D70
data_type: string
name: unpack_data
arguments: data_tag: D6
data_type: string

tool: D30: %d

tool: D80: 1


tool: D51: 4


tool: D29: 2


tool: D43: 6


tool: D73: 0


tool: D22: 5


tool: D70: 20


tool: D6: 12


assistant: ```c
int main(void) {
  int input;
  scanf("%d", &input);
  switch (input) {
  case 0:
    printf("5\n");
    break;
  case 1:
    printf("20\n");
    break;
  case 2:
    printf("12\n");
    break;
  case 3:
    printf("2\n");
    break;
  case 4:
    printf("1\n");
    break;
  case 5:
    printf("4\n");
    break;
  case 6:
    printf("6\n");
    break;
  case 7:
    printf("1\n");
    break;
  case 8:
    printf("4\n");
    break;
  case 9:
    printf("4\n");
    break;
  case 10:
    printf("1\n");
    break;
  case 11:
    printf("0\n");
    break;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%edi
  call   <count_to_hundred@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <print_num_reverse@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <recurse@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D19
type: string

tool: D19: 


assistant: ```c
int main() {
  count_to_hundred(1);
  printf("\n");
  print_num_reverse(0);
  printf("\n");
  recurse(0);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x1,%edi
  call   <count_to_hundred@plt>
  lea    D5(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <print_num_reverse@plt>
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <recurse@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_label: D5
type_infered: string

tool: D5: 


assistant: ```c
int main() {
  count_to_hundred(1);
  printf("\n");
  print_num_reverse(0);
  printf("\n");
  recurse(0);
  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    $0x1,%edi
  lea    D5(%rip),%rbp
  call   <count_to_hundred@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <print_num_reverse@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <recurse@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
data_type: string

tool: D5: 


assistant: ```c
int main() {
  count_to_hundred(1);
  printf("\n");
  print_num_reverse(0);
  printf("\n");
  recurse(0);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x1,%edi
  lea    D6(%rip),%rbp
  call   <count_to_hundred@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <print_num_reverse@plt>
  mov    %rbp,%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <recurse@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

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

tool: D6: 


assistant: ```c
int main() {
  count_to_hundred(1);
  printf("\n");
  print_num_reverse(0);
  printf("\n");
  recurse(0);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<MatrixInver>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <zeros@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L2
L7:
  movl   $0x0,-0xc(%rbp)
  jmp    L30
L22:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rcx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L30:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L22
  addl   $0x1,-0x10(%rbp)
L2:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L7
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
double **MatrixInver(double **a, int m, int n) {
  int i, j;
  double **p;
  p = zeros(n, m);
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++)
      p[j][i] = a[i][j];
  }
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<MatrixInver>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    %edx,%ebx
  mov    %edx,%edi
  call   <zeros@plt>
  test   %ebp,%ebp
  jle    L4
  mov    %r12,%rsi
  lea    -0x1(%rbp),%edx
  lea    0x8(%r12,%rdx,8),%r9
  lea    -0x1(%rbx),%edx
  lea    0x8(,%rdx,8),%r8
  mov    $0x0,%edi
  jmp    L1
L17:
  mov    (%rsi),%rcx
  movsd  (%rcx,%rdx,1),%xmm0
  mov    (%rax,%rdx,1),%rcx
  movsd  %xmm0,(%rcx,%rdi,1)
  add    $0x8,%rdx
  cmp    %r8,%rdx
  jne    L17
L3:
  add    $0x8,%rsi
  add    $0x8,%rdi
  cmp    %r9,%rsi
  je     L4
L1:
  mov    $0x0,%edx
  test   %ebx,%ebx
  jg     L17
  jmp    L3
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
double **MatrixInver(double **a, int m, int n) {
  int i, j;
  double **p;
  p = zeros(n, m);
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++)
      p[j][i] = a[i][j];
  }
  return p;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<MatrixInver>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    %edx,%edi
  push   %rbx
  mov    %esi,%ebx
  call   <zeros@plt>
  test   %ebx,%ebx
  jle    L38
  movslq %ebx,%r10
  xor    %edi,%edi
  movslq %r12d,%r9
  shl    $0x3,%r10
  nopw   0x0(%rax,%rax,1)
L13:
  test   %r12d,%r12d
  jle    L22
  mov    0x0(%rbp,%rdi,1),%r8
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L21:
  movsd  (%r8,%rcx,8),%xmm0
  mov    (%rax,%rcx,8),%rsi
  add    $0x1,%rcx
  movsd  %xmm0,(%rsi,%rdi,1)
  cmp    %rcx,%r9
  jne    L21
L22:
  add    $0x8,%rdi
  cmp    %r10,%rdi
  jne    L13
L38:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
double **MatrixInver(double **a, int m, int n) {
  int i, j;
  double **p;
  p = zeros(n, m);
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++)
      p[j][i] = a[i][j];
  }
  return p;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<MatrixInver>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    %edx,%edi
  push   %rbx
  mov    %edx,%ebx
  call   <zeros@plt>
  test   %r12d,%r12d
  jle    L9
  test   %ebx,%ebx
  jle    L9
  movslq %r12d,%r9
  xor    %esi,%esi
  movslq %ebx,%r8
  shl    $0x3,%r9
  nopl   0x0(%rax)
L15:
  mov    0x0(%rbp,%rsi,1),%rdi
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L8:
  movsd  (%rdi,%rdx,8),%xmm0
  mov    (%rax,%rdx,8),%rcx
  add    $0x1,%rdx
  movsd  %xmm0,(%rcx,%rsi,1)
  cmp    %r8,%rdx
  jne    L8
  add    $0x8,%rsi
  cmp    %rsi,%r9
  jne    L15
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
double **MatrixInver(double **a, int m, int n) {
  int i, j;
  double **p;
  p = zeros(n, m);
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++)
      p[j][i] = a[i][j];
  }
  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
<remove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L44
L13:
  addl   $0x1,-0x4(%rbp)
L44:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L13
  subl   $0x1,-0x4(%rbp)
  jmp    L23
L34:
  subl   $0x1,-0x4(%rbp)
L23:
  cmpl   $0x0,-0x4(%rbp)
  js     L37
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L34
L37:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L34
  cmpl   $0x0,-0x4(%rbp)
  js     L57
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0xa,(%rax)
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L57:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int remove(char s[]) {
  int i;
  i = 0;
  while (s[i] != '\n') {
    ++i;
  }
  --i;
  while (i >= 0 && s[i] == '\t' || s[i] == ' ') {
    --i;
  }
  if (i >= 0) {
    ++i;
    s[i] = '\n';
    ++i;
    s[i] = '\0';
  }
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<remove>:
  endbr64
  cmpb   $0xa,(%rdi)
  je     L8
  mov    $0x1,%edx
L80:
  mov    %edx,%eax
  add    $0x1,%rdx
  cmpb   $0xa,-0x1(%rdi,%rdx,1)
  jne    L80
L61:
  sub    $0x1,%eax
  cltq
  jmp    L3
L8:
  mov    $0x0,%eax
  jmp    L61
L0:
  cmpb   $0x20,(%rdi,%rax,1)
  jne    L41
L6:
  sub    $0x1,%rax
L3:
  mov    %eax,%edx
  test   %eax,%eax
  js     L0
  cmpb   $0x9,(%rdi,%rax,1)
  jne    L0
  jmp    L6
L41:
  test   %edx,%edx
  js     L63
  lea    0x1(%rdx),%eax
  cltq
  movb   $0xa,(%rdi,%rax,1)
  add    $0x2,%edx
  movslq %edx,%rax
  movb   $0x0,(%rdi,%rax,1)
L63:
  mov    %edx,%eax
  ret
```

assistant: ```c
int remove(char s[]) {
  int i;
  i = 0;
  while (s[i] != '\n') {
    ++i;
  }
  --i;
  while (i >= 0 && s[i] == '\t' || s[i] == ' ') {
    --i;
  }
  if (i >= 0) {
    ++i;
    s[i] = '\n';
    ++i;
    s[i] = '\0';
  }
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<remove>:
  endbr64
  cmpb   $0xa,(%rdi)
  je     L72
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L1:
  movslq %edx,%rax
  add    $0x1,%rdx
  cmpb   $0xa,(%rdi,%rdx,1)
  jne    L1
  jmp    L5
  nop
L71:
  sub    $0x1,%rax
L5:
  movzbl (%rdi,%rax,1),%edx
  mov    %eax,%r8d
  test   %eax,%eax
  js     L31
  cmp    $0x9,%dl
  je     L71
L31:
  cmp    $0x20,%dl
  je     L71
  test   %r8d,%r8d
  js     L66
  lea    0x1(%r8),%eax
  add    $0x2,%r8d
  cltq
  movb   $0xa,(%rdi,%rax,1)
  movslq %r8d,%rax
  movb   $0x0,(%rdi,%rax,1)
L66:
  mov    %r8d,%eax
  ret
L72:
  mov    $0xffffffffffffffff,%rax
  jmp    L5
```

assistant: ```c
int remove(char s[]) {
  int i;
  i = 0;
  while (s[i] != '\n') {
    ++i;
  }
  --i;
  while (i >= 0 && s[i] == '\t' || s[i] == ' ') {
    --i;
  }
  if (i >= 0) {
    ++i;
    s[i] = '\n';
    ++i;
    s[i] = '\0';
  }
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<remove>:
  endbr64
  cmpb   $0xa,(%rdi)
  je     L42
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L4:
  movslq %edx,%rax
  add    $0x1,%rdx
  cmpb   $0xa,(%rdi,%rdx,1)
  jne    L4
  jmp    L34
  nop
L38:
  sub    $0x1,%rax
L34:
  mov    %eax,%r8d
  test   %eax,%eax
  js     L65
  cmpb   $0x9,(%rdi,%rax,1)
  je     L38
L65:
  cmpb   $0x20,(%rdi,%rax,1)
  je     L38
  test   %r8d,%r8d
  js     L23
  lea    0x1(%r8),%eax
  add    $0x2,%r8d
  cltq
  movb   $0xa,(%rdi,%rax,1)
  movslq %r8d,%rax
  movb   $0x0,(%rdi,%rax,1)
L23:
  mov    %r8d,%eax
  ret
L42:
  mov    $0xffffffffffffffff,%rax
  jmp    L34
```

assistant: ```c
int remove(char s[]) {
  int i;
  i = 0;
  while (s[i] != '\n') {
    ++i;
  }
  --i;
  while (i >= 0 && s[i] == '\t' || s[i] == ' ') {
    --i;
  }
  if (i >= 0) {
    ++i;
    s[i] = '\n';
    ++i;
    s[i] = '\0';
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<BufferReturn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L16
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <Free@plt>
  mov    -0x10(%rbp),%rax
  jmp    L12
L16:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
L12:
  leave
  ret
```

assistant: ```c
char *BufferReturn(char **Buffer, char *UserBuffer) {
  if (UserBuffer != NULL) {
    strcpy(UserBuffer, *Buffer);
    Free((void **)Buffer);
    return (UserBuffer);
  } else {
    return (*Buffer);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<BufferReturn>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  test   %rsi,%rsi
  je     L4
  mov    %rsi,%rbx
  mov    (%rdi),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <Free@plt>
  mov    %rbx,%rax
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  mov    (%rdi),%rax
  jmp    L8
```

assistant: ```c
char *BufferReturn(char **Buffer, char *UserBuffer) {
  if (UserBuffer != NULL) {
    strcpy(UserBuffer, *Buffer);
    Free((void **)Buffer);
    return (UserBuffer);
  } else {
    return (*Buffer);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<BufferReturn>:
  endbr64
  mov    (%rdi),%rax
  test   %rsi,%rsi
  je     L9
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rax,%rsi
  mov    %rbx,%rdi
  sub    $0x8,%rsp
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <Free@plt>
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L9:
  ret
```

assistant: ```c
char *BufferReturn(char **Buffer, char *UserBuffer) {
  if (UserBuffer != NULL) {
    strcpy(UserBuffer, *Buffer);
    Free((void **)Buffer);
    return (UserBuffer);
  } else {
    return (*Buffer);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<BufferReturn>:
  endbr64
  mov    (%rdi),%rax
  test   %rsi,%rsi
  je     L4
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rax,%rsi
  mov    %rbx,%rdi
  sub    $0x8,%rsp
  call   <strcpy@plt>
  mov    %rbp,%rdi
  call   <Free@plt>
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L4:
  ret
```

assistant: ```c
char *BufferReturn(char **Buffer, char *UserBuffer) {
  if (UserBuffer != NULL) {
    strcpy(UserBuffer, *Buffer);
    Free((void **)Buffer);
    return (UserBuffer);
  } else {
    return (*Buffer);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ordinate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jle    L0
  subl   $0x1,-0x4(%rbp)
  jmp    L13
L0:
  movl   $0x0,-0x4(%rbp)
L13:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int ordinate(const char *s) {
  int result = atoi(s);
  if (result > 0)
    --result;
  else
    result = 0;
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ordinate>:
  endbr64
  sub    $0x8,%rsp
  call   <atoi@plt>
  lea    -0x1(%rax),%edx
  test   %eax,%eax
  mov    $0x0,%eax
  cmovg  %edx,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int ordinate(const char *s) {
  int result = atoi(s);
  if (result > 0)
    --result;
  else
    result = 0;
  return result;
}
```