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

```asm
<parser_copy_word>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  mov    (%rdi),%rbp
  jmp    L44
  nopl   0x0(%rax,%rax,1)
L25:
  cmpb   $0x0,0x0(%rbp,%rbx,1)
  je     L29
  add    $0x1,%rbx
L44:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isspace@plt>
  test   %eax,%eax
  je     L25
L29:
  xor    %r12d,%r12d
  test   %rbx,%rbx
  je     L49
  lea    0x1(%rbx),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L49
  add    %rbx,0x0(%r13)
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L49:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *parser_copy_word(char **cur) {
  char *value = *cur;
  size_t i;
  for (i = 0; !isspace(value[i]) && value[i] != '\0'; i++) {
  }
  if (i == 0) {
    return NULL;
  }
  char *result = malloc(i + 1);
  if (!result) {
    return NULL;
  }
  *cur += i;
  memcpy(result, value, i);
  result[i] = '\0';
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<virtual_machine_resource_pool_linked_feedback>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <linked_mem_pool_feedback@plt>
  leave
  ret
```

assistant: ```c
int virtual_machine_resource_pool_linked_feedback(void *pool_ptr) {
  struct linked_mem_pool *linked_mem_pool_ptr = pool_ptr;
  return linked_mem_pool_feedback(linked_mem_pool_ptr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<virtual_machine_resource_pool_linked_feedback>:
  endbr64
  sub    $0x8,%rsp
  call   <linked_mem_pool_feedback@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int virtual_machine_resource_pool_linked_feedback(void *pool_ptr) {
  struct linked_mem_pool *linked_mem_pool_ptr = pool_ptr;
  return linked_mem_pool_feedback(linked_mem_pool_ptr);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<virtual_machine_resource_pool_linked_feedback>:
  endbr64
  jmp    1050 <linked_mem_pool_feedback@plt>
```

assistant: ```c
int virtual_machine_resource_pool_linked_feedback(void *pool_ptr) {
  struct linked_mem_pool *linked_mem_pool_ptr = pool_ptr;
  return linked_mem_pool_feedback(linked_mem_pool_ptr);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<virtual_machine_resource_pool_linked_feedback>:
  endbr64
  jmp    1050 <linked_mem_pool_feedback@plt>
```

assistant: ```c
int virtual_machine_resource_pool_linked_feedback(void *pool_ptr) {
  struct linked_mem_pool *linked_mem_pool_ptr = pool_ptr;
  return linked_mem_pool_feedback(linked_mem_pool_ptr);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<consume_comment>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  jmp    L20
L37:
  cmpb   $0x2a,-0x1(%rbp)
  jne    L20
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0x2f,%al
  je     L26
L20:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x1(%rbp)
  mov    D3(%rip),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1(%rbp)
  jne    L37
  jmp    L6
L26:
  nop
L6:
  nop
  leave
  ret
```

assistant: ```c
void consume_comment() {
  char c;
  while ((c = getchar()) != EOF) {
    if (c == '*') {
      if (getchar() == '/') {
        break;
      }
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<consume_comment>:
  endbr64
  sub    $0x8,%rsp
L2:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D7(%rip),%rdx
  cmp    %al,(%rdx)
  je     L4
  cmp    $0x2a,%al
  jne    L2
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0x2f,%al
  jne    L2
L4:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void consume_comment() {
  char c;
  while ((c = getchar()) != EOF) {
    if (c == '*') {
      if (getchar() == '/') {
        break;
      }
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<consume_comment>:
  endbr64
  push   %rbx
  mov    D2(%rip),%rbx
  jmp    L27
  xchg   %ax,%ax
L9:
  cmp    $0x2a,%al
  je     L3
L27:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %al,(%rbx)
  jne    L9
  pop    %rbx
  ret
  nopl   0x0(%rax)
L3:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x2f,%al
  jne    L27
  pop    %rbx
  ret
```

assistant: ```c
void consume_comment() {
  char c;
  while ((c = getchar()) != EOF) {
    if (c == '*') {
      if (getchar() == '/') {
        break;
      }
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<consume_comment>:
  endbr64
  push   %rbx
  mov    D3(%rip),%rbx
  jmp    L18
  xchg   %ax,%ax
L21:
  cmp    $0x2a,%al
  je     L33
L18:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    %al,(%rbx)
  jne    L21
  pop    %rbx
  ret
  nopl   0x0(%rax)
L33:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0x2f,%al
  jne    L18
  pop    %rbx
  ret
```

assistant: ```c
void consume_comment() {
  char c;
  while ((c = getchar()) != EOF) {
    if (c == '*') {
      if (getchar() == '/') {
        break;
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<conv_de_bin>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%r8
  mov    %edx,%r11d
  test   %edx,%edx
  js     L8
  lea    0x1(%rdx),%eax
  lea    (%rsi,%rax,2),%rdi
L30:
  movw   $0x0,(%rsi)
  add    $0x2,%rsi
  cmp    %rdi,%rsi
  jne    L30
L55:
  movslq %r11d,%rax
  lea    0x2(%rax,%rax,1),%r9
  lea    (%r8,%r9,1),%rsi
  lea    -0x4(%rsi,%r9,1),%r10
  lea    -0x2(%rsi,%r9,1),%rbx
  add    %rsi,%r9
  lea    0x2(%r11,%r11,1),%eax
  sub    %eax,%ecx
  test   %ecx,%ecx
  jle    L9
  mov    %ecx,%ecx
  mov    $0x0,%edx
L20:
  movswl (%r8,%rdx,2),%eax
  movswl 0x2(%r8,%rdx,2),%ebp
  add    %ebp,%eax
  movswl 0x4(%r8,%rdx,2),%ebp
  add    %ebp,%eax
  movswl -0x2(%rsi,%rdx,2),%ebp
  add    %ebp,%eax
  movswl (%rsi,%rdx,2),%ebp
  lea    0x0(%rbp,%rbp,8),%ebp
  add    %ebp,%eax
  movswl 0x2(%rsi,%rdx,2),%ebp
  add    %ebp,%eax
  movswl (%r10,%rdx,2),%ebp
  add    %ebp,%eax
  movswl (%rbx,%rdx,2),%ebp
  add    %ebp,%eax
  movswl (%r9,%rdx,2),%ebp
  add    %ebp,%eax
  mov    %ax,(%rdi,%rdx,2)
  add    $0x1,%rdx
  cmp    %rdx,%rcx
  jne    L20
  lea    (%rdi,%rcx,2),%rdi
L9:
  test   %r11d,%r11d
  js     L68
  lea    0x1(%r11),%eax
  lea    (%rdi,%rax,2),%rax
L27:
  movw   $0x0,(%rdi)
  add    $0x2,%rdi
  cmp    %rdi,%rax
  jne    L27
L68:
  pop    %rbx
  pop    %rbp
  ret
L8:
  mov    %rsi,%rdi
  jmp    L55
```

assistant: ```c
void conv_de_bin(short int *src, short int *dst, int ncol, int npix) {
  short int *i0, *i1, *i2, *i3, *i4, *i5, *i6, *i7, *i8;
  int i, res;
  for (i = 0; i < ncol + 1; ++i) {
    *dst = 0;
    ++dst;
  }
  src += ncol + 1;
  i0 = src - ncol - 1;
  i1 = src - ncol;
  i2 = src - ncol + 1;
  i3 = src - 1;
  i4 = src;
  i5 = src + 1;
  i6 = src + ncol - 1;
  i7 = src + ncol;
  i8 = src + ncol + 1;
  npix -= 2 * (ncol + 1);
  for (i = 0; i < npix; ++i) {
    res = (*i0 + *i1 + *i2 + *i3 + *i4 * 9 + *i5 + *i6 + *i7 + *i8);
    *dst = res;
    dst++;
    i0++;
    i1++;
    i2++;
    i3++;
    i4++;
    i5++;
    i6++;
    i7++;
    i8++;
  }
  for (i = 0; i < ncol + 1; ++i) {
    *dst = 0;
    ++dst;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<conv_de_bin>:
  endbr64
  push   %r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  test   %edx,%edx
  js     L2
  lea    0x2(%r12,%r12,1),%r14
  xor    %esi,%esi
  mov    %r14,%rdx
  call   <memset@plt>
  mov    %rax,%rdi
  add    %r14,%rdi
L2:
  lea    0x2(%r12,%r12,1),%r8
  lea    0x2(%rbp,%rbp,1),%eax
  mov    %r13d,%esi
  lea    (%rbx,%r8,1),%rcx
  sub    %eax,%esi
  lea    -0x4(%rcx,%r8,1),%r9
  lea    -0x2(%rcx,%r8,1),%r10
  add    %rcx,%r8
  test   %esi,%esi
  jle    L32
  movslq %esi,%rsi
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L27:
  movswl 0x2(%rbx,%rdx,2),%r11d
  movswl (%rbx,%rdx,2),%eax
  add    %r11d,%eax
  movswl 0x4(%rbx,%rdx,2),%r11d
  add    %r11d,%eax
  movswl -0x2(%rcx,%rdx,2),%r11d
  add    %r11d,%eax
  movswl (%rcx,%rdx,2),%r11d
  lea    (%r11,%r11,8),%r11d
  add    %r11d,%eax
  movswl 0x2(%rcx,%rdx,2),%r11d
  add    %r11d,%eax
  movswl (%r9,%rdx,2),%r11d
  add    %r11d,%eax
  movswl (%r10,%rdx,2),%r11d
  add    %r11d,%eax
  movswl (%r8,%rdx,2),%r11d
  add    %r11d,%eax
  mov    %ax,(%rdi,%rdx,2)
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L27
  lea    (%rdi,%rsi,2),%rdi
L32:
  test   %ebp,%ebp
  js     L15
  pop    %rbx
  lea    0x2(%r12,%r12,1),%rdx
  pop    %rbp
  xor    %esi,%esi
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <memset@plt>
  nopl   (%rax)
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void conv_de_bin(short int *src, short int *dst, int ncol, int npix) {
  short int *i0, *i1, *i2, *i3, *i4, *i5, *i6, *i7, *i8;
  int i, res;
  for (i = 0; i < ncol + 1; ++i) {
    *dst = 0;
    ++dst;
  }
  src += ncol + 1;
  i0 = src - ncol - 1;
  i1 = src - ncol;
  i2 = src - ncol + 1;
  i3 = src - 1;
  i4 = src;
  i5 = src + 1;
  i6 = src + ncol - 1;
  i7 = src + ncol;
  i8 = src + ncol + 1;
  npix -= 2 * (ncol + 1);
  for (i = 0; i < npix; ++i) {
    res = (*i0 + *i1 + *i2 + *i3 + *i4 * 9 + *i5 + *i6 + *i7 + *i8);
    *dst = res;
    dst++;
    i0++;
    i1++;
    i2++;
    i3++;
    i4++;
    i5++;
    i6++;
    i7++;
    i8++;
  }
  for (i = 0; i < ncol + 1; ++i) {
    *dst = 0;
    ++dst;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<cd_play_from>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <cd_get_tracks@plt>
  test   %rax,%rax
  je     L4
  mov    $0xffffffff,%eax
  jmp    L6
L4:
  mov    -0xc(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <play@plt>
L6:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int cd_play_from(int track) {
  int last;
  if (cd_get_tracks(0, &last) != 0)
    return -1;
  return play(track, last);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<cd_play_from>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  mov    $0x0,%edi
  call   <cd_get_tracks@plt>
  test   %rax,%rax
  jne    L2
  mov    0x4(%rsp),%esi
  mov    %ebx,%edi
  call   <play@plt>
L11:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x10,%rsp
  pop    %rbx
  ret
L2:
  mov    $0xffffffff,%eax
  jmp    L11
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cd_play_from(int track) {
  int last;
  if (cd_get_tracks(0, &last) != 0)
    return -1;
  return play(track, last);
}
```
user: What is the c source code of the assembly code below?

```asm
<cd_play_from>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <cd_get_tracks@plt>
  test   %rax,%rax
  jne    L12
  mov    0x4(%rsp),%esi
  mov    %ebp,%edi
  call   <play@plt>
L4:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x10,%rsp
  pop    %rbp
  ret
  nopl   (%rax)
L12:
  mov    $0xffffffff,%eax
  jmp    L4
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cd_play_from(int track) {
  int last;
  if (cd_get_tracks(0, &last) != 0)
    return -1;
  return play(track, last);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<cd_play_from>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <cd_get_tracks@plt>
  test   %rax,%rax
  jne    L9
  mov    0x4(%rsp),%esi
  mov    %ebp,%edi
  call   <play@plt>
L22:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L11
  add    $0x10,%rsp
  pop    %rbp
  ret
  nopl   (%rax)
L9:
  mov    $0xffffffff,%eax
  jmp    L22
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cd_play_from(int track) {
  int last;
  if (cd_get_tracks(0, &last) != 0)
    return -1;
  return play(track, last);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strip_whitespace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L103
  mov    $0x0,%eax
  jmp    L34
L103:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L42
  mov    $0x0,%eax
  jmp    L34
L42:
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L57
L36:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L127
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  jne    L31
L127:
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L129
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  jne    L31
L129:
  addq   $0x1,-0x18(%rbp)
  jmp    L57
L31:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  jne    L118
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x20,(%rax)
  addq   $0x1,-0x18(%rbp)
  jmp    L57
L118:
  mov    -0x18(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L57:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L36
  mov    -0x8(%rbp),%rax
L34:
  leave
  ret
```

assistant: ```c
char *strip_whitespace(const char *input) {
  unsigned int n_chars;
  const char *read;
  char *write;
  char *output;
  if (input == NULL) {
    return NULL;
  }
  n_chars = strlen(input);
  if (n_chars == 0) {
    return NULL;
  }
  output = calloc(n_chars + 1, sizeof(char));
  read = input;
  write = output;
  while (*read != '\0') {
    if ((*read == ' ' || *read == '\t') && (*(read + 1) == ' ' || *(read + 1) == '\t')) {
      read++;
      continue;
    }
    if (*read == '\t') {
      *(write++) = ' ';
      read++;
    } else {
      *(write++) = *(read++);
    }
  }
  return output;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strip_whitespace>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L82
  call   <strlen@plt>
  mov    $0x0,%r9d
  test   %eax,%eax
  je     L60
  lea    0x1(%rax),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%r9
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L60
  lea    0x1(%rbx),%rcx
  mov    %rax,%rsi
  jmp    L66
L68:
  mov    %rcx,%rdi
  mov    %dl,(%rsi)
  lea    0x1(%rsi),%rsi
L10:
  movzbl (%rdi),%edx
  add    $0x1,%rcx
  test   %dl,%dl
  je     L60
L66:
  cmp    $0x20,%dl
  je     L11
  cmp    $0x9,%dl
  jne    L68
L11:
  mov    %rcx,%rdi
  movzbl (%rcx),%r8d
  cmp    $0x20,%r8b
  je     L10
  cmp    $0x9,%r8b
  je     L10
  cmp    $0x9,%dl
  jne    L68
  movb   $0x20,(%rsi)
  lea    0x1(%rsi),%rsi
  jmp    L10
L82:
  mov    %rdi,%r9
L60:
  mov    %r9,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *strip_whitespace(const char *input) {
  unsigned int n_chars;
  const char *read;
  char *write;
  char *output;
  if (input == NULL) {
    return NULL;
  }
  n_chars = strlen(input);
  if (n_chars == 0) {
    return NULL;
  }
  output = calloc(n_chars + 1, sizeof(char));
  read = input;
  write = output;
  while (*read != '\0') {
    if ((*read == ' ' || *read == '\t') && (*(read + 1) == ' ' || *(read + 1) == '\t')) {
      read++;
      continue;
    }
    if (*read == '\t') {
      *(write++) = ' ';
      read++;
    } else {
      *(write++) = *(read++);
    }
  }
  return output;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strip_whitespace>:
  endbr64
  test   %rdi,%rdi
  je     L61
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  je     L88
  lea    0x1(%rax),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L85
  mov    %rax,%rsi
  jmp    L36
  nop
L9:
  mov    %dl,(%rsi)
  mov    %rcx,%rsi
  movzbl (%rbx),%ecx
L82:
  mov    %ecx,%edx
  test   %cl,%cl
  je     L85
L36:
  add    $0x1,%rbx
  cmp    $0x20,%dl
  je     L74
  lea    0x1(%rsi),%rcx
  cmp    $0x9,%dl
  jne    L9
L74:
  movzbl (%rbx),%ecx
  cmp    $0x20,%cl
  je     L82
  cmp    $0x9,%cl
  je     L82
  lea    0x1(%rsi),%rcx
  cmp    $0x9,%dl
  jne    L9
  movb   $0x20,(%rsi)
  mov    %rcx,%rsi
  movzbl (%rbx),%ecx
  mov    %ecx,%edx
  test   %cl,%cl
  jne    L36
L85:
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L88:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L61:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *strip_whitespace(const char *input) {
  unsigned int n_chars;
  const char *read;
  char *write;
  char *output;
  if (input == NULL) {
    return NULL;
  }
  n_chars = strlen(input);
  if (n_chars == 0) {
    return NULL;
  }
  output = calloc(n_chars + 1, sizeof(char));
  read = input;
  write = output;
  while (*read != '\0') {
    if ((*read == ' ' || *read == '\t') && (*(read + 1) == ' ' || *(read + 1) == '\t')) {
      read++;
      continue;
    }
    if (*read == '\t') {
      *(write++) = ' ';
      read++;
    } else {
      *(write++) = *(read++);
    }
  }
  return output;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strip_whitespace>:
  endbr64
  test   %rdi,%rdi
  je     L81
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  je     L1
  lea    0x1(%rax),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  movzbl (%rbx),%edx
  test   %dl,%dl
  je     L14
  mov    %rax,%rsi
  jmp    L71
  nop
L38:
  mov    %dl,(%rsi)
  mov    %rcx,%rsi
  movzbl (%rbx),%ecx
L49:
  mov    %ecx,%edx
  test   %cl,%cl
  je     L14
L71:
  add    $0x1,%rbx
  cmp    $0x20,%dl
  je     L105
  lea    0x1(%rsi),%rcx
  cmp    $0x9,%dl
  jne    L38
L105:
  movzbl (%rbx),%ecx
  cmp    $0x20,%cl
  je     L49
  cmp    $0x9,%cl
  je     L49
  lea    0x1(%rsi),%rcx
  cmp    $0x9,%dl
  jne    L38
  movb   $0x20,(%rsi)
  mov    %rcx,%rsi
  movzbl (%rbx),%ecx
  mov    %ecx,%edx
  test   %cl,%cl
  jne    L71
L14:
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L81:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *strip_whitespace(const char *input) {
  unsigned int n_chars;
  const char *read;
  char *write;
  char *output;
  if (input == NULL) {
    return NULL;
  }
  n_chars = strlen(input);
  if (n_chars == 0) {
    return NULL;
  }
  output = calloc(n_chars + 1, sizeof(char));
  read = input;
  write = output;
  while (*read != '\0') {
    if ((*read == ' ' || *read == '\t') && (*(read + 1) == ' ' || *(read + 1) == '\t')) {
      read++;
      continue;
    }
    if (*read == '\t') {
      *(write++) = ' ';
      read++;
    } else {
      *(write++) = *(read++);
    }
  }
  return output;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<EEPROMBlockUnlock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  mov    $0x400af004,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    $0x400af020,%eax
  movl   $0xffffffff,(%rax)
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x2,%rax
  sub    $0x4,%rax
  add    %rax,-0x10(%rbp)
  jmp    L7
L1:
  mov    -0x10(%rbp),%rax
  lea    -0x4(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    $0x400af020,%edx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x8(%rbp)
L7:
  cmpl   $0x0,-0x8(%rbp)
  jne    L1
  mov    $0x400af020,%eax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t EEPROMBlockUnlock(uint32_t ui32Block, uint32_t *pui32Password, uint32_t ui32Count) {
  ;
  ;
  ;
  (*((volatile uint32_t *)(0x400AF004))) = ui32Block;
  (*((volatile uint32_t *)(0x400AF020))) = 0xFFFFFFFF;
  pui32Password += (ui32Count - 1);
  while (ui32Count) {
    (*((volatile uint32_t *)(0x400AF020))) = *pui32Password--;
    ui32Count--;
  }
  return ((*((volatile uint32_t *)(0x400AF020))));
}
```
user: What is the c source code of the assembly code below?

```asm
<EEPROMBlockUnlock>:
  endbr64
  mov    %edi,0x400af004
  movl   $0xffffffff,0x400af020
  movslq %edx,%rax
  lea    -0x4(%rsi,%rax,4),%rcx
  test   %edx,%edx
  je     L7
  mov    %rcx,%rax
  lea    -0x1(%rdx),%edx
  shl    $0x2,%rdx
  sub    %rdx,%rcx
L19:
  mov    (%rax),%edx
  mov    %edx,0x400af020
  mov    %rax,%rdx
  sub    $0x4,%rax
  cmp    %rcx,%rdx
  jne    L19
L7:
  mov    0x400af020,%eax
  ret
```

assistant: ```c
uint32_t EEPROMBlockUnlock(uint32_t ui32Block, uint32_t *pui32Password, uint32_t ui32Count) {
  ;
  ;
  ;
  (*((volatile uint32_t *)(0x400AF004))) = ui32Block;
  (*((volatile uint32_t *)(0x400AF020))) = 0xFFFFFFFF;
  pui32Password += (ui32Count - 1);
  while (ui32Count) {
    (*((volatile uint32_t *)(0x400AF020))) = *pui32Password--;
    ui32Count--;
  }
  return ((*((volatile uint32_t *)(0x400AF020))));
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<EEPROMBlockUnlock>:
  endbr64
  movslq %edx,%rax
  mov    %edi,0x400af004
  movl   $0xffffffff,0x400af020
  lea    -0x4(%rsi,%rax,4),%rax
  test   %edx,%edx
  je     L15
  sub    $0x1,%edx
  mov    %rax,%rcx
  shl    $0x2,%rdx
  sub    %rdx,%rcx
  nop
L19:
  mov    (%rax),%edx
  mov    %edx,0x400af020
  mov    %rax,%rdx
  sub    $0x4,%rax
  cmp    %rcx,%rdx
  jne    L19
L15:
  mov    0x400af020,%eax
  ret
```

assistant: ```c
uint32_t EEPROMBlockUnlock(uint32_t ui32Block, uint32_t *pui32Password, uint32_t ui32Count) {
  ;
  ;
  ;
  (*((volatile uint32_t *)(0x400AF004))) = ui32Block;
  (*((volatile uint32_t *)(0x400AF020))) = 0xFFFFFFFF;
  pui32Password += (ui32Count - 1);
  while (ui32Count) {
    (*((volatile uint32_t *)(0x400AF020))) = *pui32Password--;
    ui32Count--;
  }
  return ((*((volatile uint32_t *)(0x400AF020))));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<EEPROMBlockUnlock>:
  endbr64
  movslq %edx,%rax
  mov    %edi,0x400af004
  movl   $0xffffffff,0x400af020
  lea    -0x4(%rsi,%rax,4),%rax
  test   %edx,%edx
  je     L12
  sub    $0x1,%edx
  mov    %rax,%rcx
  shl    $0x2,%rdx
  sub    %rdx,%rcx
  nop
L0:
  mov    (%rax),%edx
  mov    %edx,0x400af020
  mov    %rax,%rdx
  sub    $0x4,%rax
  cmp    %rcx,%rdx
  jne    L0
L12:
  mov    0x400af020,%eax
  ret
```

assistant: ```c
uint32_t EEPROMBlockUnlock(uint32_t ui32Block, uint32_t *pui32Password, uint32_t ui32Count) {
  ;
  ;
  ;
  (*((volatile uint32_t *)(0x400AF004))) = ui32Block;
  (*((volatile uint32_t *)(0x400AF020))) = 0xFFFFFFFF;
  pui32Password += (ui32Count - 1);
  while (ui32Count) {
    (*((volatile uint32_t *)(0x400AF020))) = *pui32Password--;
    ui32Count--;
  }
  return ((*((volatile uint32_t *)(0x400AF020))));
}
```
user: Translate the below assembly code into C programming language:

```asm
<str_trim_end>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <xstrdup@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <str_trim_end_inplace@plt>
  leave
  ret
```

assistant: ```c
char *str_trim_end(const char *str, const char *set) {
  char *copy = xstrdup(str);
  return str_trim_end_inplace(copy, set);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<str_trim_end>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  call   <xstrdup@plt>
  mov    %rax,%rdi
  mov    %rbx,%rsi
  call   <str_trim_end_inplace@plt>
  pop    %rbx
  ret
```

assistant: ```c
char *str_trim_end(const char *str, const char *set) {
  char *copy = xstrdup(str);
  return str_trim_end_inplace(copy, set);
}
```