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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x30(%rbp)
  movl   $0x7,-0x2c(%rbp)
  movl   $0x9,-0x28(%rbp)
  movl   $0xb,-0x24(%rbp)
  movl   $0x0,-0x20(%rbp)
  movl   $0x6,-0x1c(%rbp)
  movl   $0x7,-0x18(%rbp)
  movl   $0x5,-0x14(%rbp)
  movl   $0x4,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L5
L43:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L5:
  cmpl   $0x9,-0x34(%rbp)
  jle    L43
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    -0x30(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <inv@plt>
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x34(%rbp)
  jmp    L10
L18:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L10:
  cmpl   $0x9,-0x34(%rbp)
  jle    L18
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D27
type_guessed: string
name: read_data
arguments: label: D38
type_guessed: string
name: read_data
arguments: label: D26
type_guessed: string

tool: D27: The original array:

tool: D38: %5d

tool: D26: the reveredArray is :

assistant: ```c
int main() {
  void inv(int x[], int n);
  int i, a[10] = {3, 7, 9, 11, 0, 6, 7, 5, 4, 2};
  printf("The original array:\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\n");
  inv(a, 10);
  printf("the reveredArray is :\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x3,(%rsp)
  movl   $0x7,0x4(%rsp)
  movl   $0x9,0x8(%rsp)
  movl   $0xb,0xc(%rsp)
  movl   $0x0,0x10(%rsp)
  movl   $0x6,0x14(%rsp)
  movl   $0x7,0x18(%rsp)
  movl   $0x5,0x1c(%rsp)
  movl   $0x4,0x20(%rsp)
  movl   $0x2,0x24(%rsp)
  lea    D13(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  mov    %rbx,%rbp
  lea    D19(%rip),%r13
L7:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbp
  cmp    %r12,%rbp
  jne    L7
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rsp,%rdi
  mov    $0xa,%esi
  call   <inv@plt>
  lea    D34(%rip),%rdi
  call   <puts@plt>
  lea    D19(%rip),%rbp
L27:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L27
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D13
type_infered: string
name: retrieve_data
arguments: data_label: D19
type_infered: string
name: retrieve_data
arguments: data_label: D34
type_infered: string

tool: D13: The original array:

tool: D19: %5d

tool: D34: the reveredArray is :

assistant: ```c
int main() {
  void inv(int x[], int n);
  int i, a[10] = {3, 7, 9, 11, 0, 6, 7, 5, 4, 2};
  printf("The original array:\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\n");
  inv(a, 10);
  printf("the reveredArray is :\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D10(%rip),%rdi
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x700000003,%rax
  mov    %rsp,%r14
  lea    0x28(%rsp),%r12
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %rax,(%rsp)
  movabs $0xb00000009,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x600000000,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x500000007,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x200000004,%rax
  mov    %rax,0x20(%rsp)
  call   <puts@plt>
  nopl   0x0(%rax)
L16:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L16
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r14,%rdi
  mov    $0xa,%esi
  call   <inv@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  nopl   0x0(%rax)
L13:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L13
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D10
type_infered: string
name: disassemble_data
arguments: tag: D14
type_infered: string
name: disassemble_data
arguments: tag: D13
type_infered: string

tool: D10: The original array:

tool: D14: %5d

tool: D13: the reveredArray is :

assistant: ```c
int main() {
  void inv(int x[], int n);
  int i, a[10] = {3, 7, 9, 11, 0, 6, 7, 5, 4, 2};
  printf("The original array:\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\n");
  inv(a, 10);
  printf("the reveredArray is :\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\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   %r14
  lea    D50(%rip),%rdi
  push   %r13
  lea    D5(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D57(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D24(%rip),%rax
  mov    %rsp,%r14
  lea    0x28(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D58(%rip),%xmm0
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    0x0(%rbp),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbp
  call   <printf@plt>
  cmp    %r12,%rbp
  jne    L16
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r14,%rdi
  mov    $0xa,%esi
  call   <inv@plt>
  lea    D42(%rip),%rdi
  call   <puts@plt>
  nopl   0x0(%rax)
L4:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L4
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D50
type_guessed: string
name: disassemble_data
arguments: tag: D5
type_guessed: string
name: disassemble_data
arguments: tag: D57
type_guessed: byte[16]
name: disassemble_data
arguments: tag: D24
type_guessed: byte[8]
name: disassemble_data
arguments: tag: D58
type_guessed: byte[16]
name: disassemble_data
arguments: tag: D42
type_guessed: string

tool: D50: The original array:

tool: D5: %5d

tool: {"D57": [0x03,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x0b,0x00,0x00,0x00]}

tool: {"D24": [0x04,0x00,0x00,0x00,0x02,0x00,0x00,0x00]}

tool: {"D58": [0x00,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0x05,0x00,0x00,0x00]}

tool: D42: the reveredArray is :

assistant: ```c
int main() {
  void inv(int x[], int n);
  int i, a[10] = {3, 7, 9, 11, 0, 6, 7, 5, 4, 2};
  printf("The original array:\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\n");
  inv(a, 10);
  printf("the reveredArray is :\n");
  for (i = 0; i < 10; i++)
    printf("%5d", a[i]);
  printf("\n");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Ftestdirempty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <opendir@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L35
  mov    $0x1,%eax
  jmp    L16
L56:
  addl   $0x1,-0x14(%rbp)
  cmpl   $0x2,-0x14(%rbp)
  jg     L29
L35:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <readdir@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L56
  jmp    L11
L29:
  nop
L11:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <closedir@plt>
  cmpl   $0x2,-0x14(%rbp)
  jg     L5
  mov    $0x1,%eax
  jmp    L16
L5:
  mov    $0x0,%eax
L16:
  leave
  ret
```

assistant: ```c
int Ftestdirempty(char *dirname) {
  int n = 0;
  struct dirent *d;
  DIR *dir = opendir(dirname);
  if (dir == NULL)
    return 1;
  while ((d = readdir(dir)) != NULL)
    if (++n > 2)
      break;
  closedir(dir);
  if (n <= 2)
    return 1;
  else
    return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Ftestdirempty>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <opendir@plt>
  test   %rax,%rax
  je     L0
  mov    %rax,%rbp
  mov    $0x0,%ebx
L20:
  mov    %rbp,%rdi
  call   <readdir@plt>
  test   %rax,%rax
  je     L15
  add    $0x1,%ebx
  cmp    $0x3,%ebx
  jne    L20
L15:
  mov    %rbp,%rdi
  call   <closedir@plt>
  cmp    $0x2,%ebx
  setle  %al
  movzbl %al,%eax
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  mov    $0x1,%eax
  jmp    L17
```

assistant: ```c
int Ftestdirempty(char *dirname) {
  int n = 0;
  struct dirent *d;
  DIR *dir = opendir(dirname);
  if (dir == NULL)
    return 1;
  while ((d = readdir(dir)) != NULL)
    if (++n > 2)
      break;
  closedir(dir);
  if (n <= 2)
    return 1;
  else
    return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Ftestdirempty>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <opendir@plt>
  test   %rax,%rax
  je     L12
  mov    %rax,%rbp
  mov    $0x3,%ebx
L5:
  mov    %rbp,%rdi
  call   <readdir@plt>
  test   %rax,%rax
  je     L21
  sub    $0x1,%ebx
  jne    L5
  mov    %rbp,%rdi
  call   <closedir@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nop
L21:
  mov    %rbp,%rdi
  call   <closedir@plt>
L12:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int Ftestdirempty(char *dirname) {
  int n = 0;
  struct dirent *d;
  DIR *dir = opendir(dirname);
  if (dir == NULL)
    return 1;
  while ((d = readdir(dir)) != NULL)
    if (++n > 2)
      break;
  closedir(dir);
  if (n <= 2)
    return 1;
  else
    return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Ftestdirempty>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <opendir@plt>
  test   %rax,%rax
  je     L16
  mov    %rax,%rbp
  mov    $0x3,%ebx
L29:
  mov    %rbp,%rdi
  call   <readdir@plt>
  test   %rax,%rax
  je     L20
  sub    $0x1,%ebx
  jne    L29
  mov    %rbp,%rdi
  call   <closedir@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nop
L20:
  mov    %rbp,%rdi
  call   <closedir@plt>
L16:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int Ftestdirempty(char *dirname) {
  int n = 0;
  struct dirent *d;
  DIR *dir = opendir(dirname);
  if (dir == NULL)
    return 1;
  while ((d = readdir(dir)) != NULL)
    if (++n > 2)
      break;
  closedir(dir);
  if (n <= 2)
    return 1;
  else
    return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strsep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L29
  mov    $0x0,%eax
  jmp    L1
L29:
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcspn@plt>
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L7
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x0,(%rax)
  jmp    L32
L7:
  movq   $0x0,-0x10(%rbp)
L32:
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
L1:
  leave
  ret
```

assistant: ```c
char *strsep(char **str, const char *sep) {
  char *s = *str, *end;
  if (!s)
    return NULL;
  end = s + strcspn(s, sep);
  if (*end)
    *end++ = 0;
  else
    end = 0;
  *str = end;
  return s;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strsep>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rbp
  test   %rbp,%rbp
  je     L8
  mov    %rdi,%rbx
  mov    %rbp,%rdi
  call   <strcspn@plt>
  cltq
  add    %rbp,%rax
  mov    $0x0,%edx
  cmpb   $0x0,(%rax)
  je     L13
  lea    0x1(%rax),%rdx
  movb   $0x0,(%rax)
L13:
  mov    %rdx,(%rbx)
L8:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *strsep(char **str, const char *sep) {
  char *s = *str, *end;
  if (!s)
    return NULL;
  end = s + strcspn(s, sep);
  if (*end)
    *end++ = 0;
  else
    end = 0;
  *str = end;
  return s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strsep>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r12
  test   %r12,%r12
  je     L23
  mov    %rdi,%rbx
  mov    %r12,%rdi
  call   <strcspn@plt>
  xor    %edx,%edx
  cltq
  add    %r12,%rax
  cmpb   $0x0,(%rax)
  jne    L4
L6:
  mov    %rdx,(%rbx)
L23:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  movb   $0x0,(%rax)
  lea    0x1(%rax),%rdx
  jmp    L6
```

assistant: ```c
char *strsep(char **str, const char *sep) {
  char *s = *str, *end;
  if (!s)
    return NULL;
  end = s + strcspn(s, sep);
  if (*end)
    *end++ = 0;
  else
    end = 0;
  *str = end;
  return s;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strsep>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r12
  test   %r12,%r12
  je     L10
  mov    %rdi,%rbx
  mov    %r12,%rdi
  call   <strcspn@plt>
  xor    %edx,%edx
  cltq
  add    %r12,%rax
  cmpb   $0x0,(%rax)
  jne    L28
L22:
  mov    %rdx,(%rbx)
L10:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L28:
  movb   $0x0,(%rax)
  lea    0x1(%rax),%rdx
  jmp    L22
```

assistant: ```c
char *strsep(char **str, const char *sep) {
  char *s = *str, *end;
  if (!s)
    return NULL;
  end = s + strcspn(s, sep);
  if (*end)
    *end++ = 0;
  else
    end = 0;
  *str = end;
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rtrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L9
L28:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  je     L18
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L5
L18:
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L5
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L5:
  addq   $0x1,-0x8(%rbp)
L9:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  ja     L6
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x24(%rbp),%eax
  cltq
  cmp    %rax,%rdx
  jle    L28
L6:
  mov    -0x10(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jae    L8
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L8
  addq   $0x1,-0x10(%rbp)
L8:
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *rtrim(char *a, char *c, int d) {
  char *b, *x;
  b = c;
  for (x = a + 1; x <= c && x - a <= d; x++)
    if ((x == c || isspace(x[0])) && !isspace(x[-1]))
      b = x;
  if (b < c && !isspace(b[0]))
    b++;
  return (b);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rtrim>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  lea    0x1(%rdi),%rbx
  cmp    %rbx,%rsi
  jb     L13
  mov    %rdi,%r14
  movslq %edx,%r13
  test   %r13,%r13
  jle    L113
  lea    0x1(%rsi),%r15
  mov    %rsi,%r12
  jmp    L106
L93:
  movsbl -0x1(%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  cmove  %rbx,%r12
L82:
  add    $0x1,%rbx
  cmp    %r15,%rbx
  je     L109
  mov    %rbx,%rax
  sub    %r14,%rax
  cmp    %r13,%rax
  jg     L109
L106:
  cmp    %rbx,%rbp
  je     L93
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L82
  jmp    L93
L109:
  cmp    %rbp,%r12
  jb     L89
L61:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L89:
  movsbl (%r12),%edi
  call   <isspace@plt>
  cmp    $0x1,%rax
  adc    $0x0,%r12
  jmp    L61
L13:
  mov    %rsi,%r12
  jmp    L61
L113:
  mov    %rsi,%r12
  jmp    L61
```

assistant: ```c
char *rtrim(char *a, char *c, int d) {
  char *b, *x;
  b = c;
  for (x = a + 1; x <= c && x - a <= d; x++)
    if ((x == c || isspace(x[0])) && !isspace(x[-1]))
      b = x;
  if (b < c && !isspace(b[0]))
    b++;
  return (b);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rtrim>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  lea    0x1(%rdi),%rbx
  sub    $0x8,%rsp
  cmp    %rbx,%rsi
  jb     L43
  movslq %edx,%r13
  test   %r13,%r13
  jle    L43
  mov    %rdi,%r14
  lea    0x1(%rsi),%r15
  mov    %rsi,%r12
  jmp    L83
  cs nopw 0x0(%rax,%rax,1)
L18:
  mov    %rbx,%rax
  sub    %r14,%rax
  cmp    %r13,%rax
  jg     L46
L83:
  cmp    %rbx,%rbp
  je     L14
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L3
L14:
  movsbl -0x1(%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  cmove  %rbx,%r12
L3:
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L18
L46:
  cmp    %rbp,%r12
  jae    L24
  movsbl (%r12),%edi
  call   <isspace@plt>
  cmp    $0x1,%rax
  adc    $0x0,%r12
L24:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L43:
  mov    %rbp,%r12
  jmp    L24
```

assistant: ```c
char *rtrim(char *a, char *c, int d) {
  char *b, *x;
  b = c;
  for (x = a + 1; x <= c && x - a <= d; x++)
    if ((x == c || isspace(x[0])) && !isspace(x[-1]))
      b = x;
  if (b < c && !isspace(b[0]))
    b++;
  return (b);
}
```
user: What is the c source code of the assembly code below?

```asm
<rtrim>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  lea    0x1(%rdi),%rbx
  sub    $0x8,%rsp
  cmp    %rbx,%rsi
  jb     L5
  movslq %edx,%r13
  test   %r13,%r13
  jle    L5
  mov    %rdi,%r14
  lea    0x1(%rsi),%r15
  mov    %rsi,%r12
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L89:
  mov    %rbx,%rax
  sub    %r14,%rax
  cmp    %r13,%rax
  jg     L48
L16:
  cmp    %rbx,%rbp
  je     L63
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L71
L63:
  movsbl -0x1(%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  cmove  %rbx,%r12
L71:
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L89
L48:
  cmp    %rbp,%r12
  jae    L59
  movsbl (%r12),%edi
  call   <isspace@plt>
  cmp    $0x1,%rax
  adc    $0x0,%r12
L59:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L5:
  mov    %rbp,%r12
  jmp    L59
```

assistant: ```c
char *rtrim(char *a, char *c, int d) {
  char *b, *x;
  b = c;
  for (x = a + 1; x <= c && x - a <= d; x++)
    if ((x == c || isspace(x[0])) && !isspace(x[-1]))
      b = x;
  if (b < c && !isspace(b[0]))
    b++;
  return (b);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  jmp    L30
L43:
  cmpl   $0x20,-0x4(%rbp)
  jne    L16
  addl   $0x1,-0x8(%rbp)
  jmp    L22
L16:
  movl   $0x0,-0x8(%rbp)
L22:
  cmpl   $0x1,-0x8(%rbp)
  jg     L30
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L30:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L43
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int c;
  int n = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ') {
      ++n;
    } else {
      n = 0;
    }
    if (n < 2) {
      putchar(c);
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%ebx
  jmp    L8
L31:
  mov    $0x0,%ebx
L0:
  mov    %eax,%edi
  call   <putchar@plt>
L8:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D8(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L25
  cmp    $0x20,%eax
  jne    L31
  add    $0x1,%ebx
  cmp    $0x1,%ebx
  jg     L8
  jmp    L0
L25:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  int c;
  int n = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ') {
      ++n;
    } else {
      n = 0;
    }
    if (n < 2) {
      putchar(c);
    }
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  mov    D0(%rip),%rbp
L33:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,0x0(%rbp)
  je     L9
L35:
  cmp    $0x20,%edi
  jne    L24
  add    $0x1,%ebx
  cmp    $0x1,%ebx
  jne    L33
  call   <putchar@plt>
L26:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,0x0(%rbp)
  jne    L35
L9:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L24:
  xor    %ebx,%ebx
  call   <putchar@plt>
  jmp    L26
```

assistant: ```c
int main() {
  int c;
  int n = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ') {
      ++n;
    } else {
      n = 0;
    }
    if (n < 2) {
      putchar(c);
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  mov    D9(%rip),%rbp
L47:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,0x0(%rbp)
  je     L35
L10:
  cmp    $0x20,%edi
  jne    L19
  add    $0x1,%ebx
  cmp    $0x1,%ebx
  jne    L47
  call   <putchar@plt>
L36:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,0x0(%rbp)
  jne    L10
L35:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  xor    %ebx,%ebx
  call   <putchar@plt>
  jmp    L36
```

assistant: ```c
int main() {
  int c;
  int n = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ') {
      ++n;
    } else {
      n = 0;
    }
    if (n < 2) {
      putchar(c);
    }
  }
  return 0;
}
```