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

```asm
<newcopydim>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L25
  mov    (%rdi),%rbp
  mov    %rbp,%rdi
  call   <newdim@plt>
  test   %rax,%rax
  je     L31
  test   %rbp,%rbp
  js     L31
  lea    0x1(%rbp),%rsi
  mov    $0x0,%edx
L37:
  mov    (%rbx,%rdx,8),%rcx
  mov    %rcx,(%rax,%rdx,8)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L37
L31:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L25:
  mov    %rdi,%rax
  jmp    L31
```

assistant: ```c
long *newcopydim(const long *dim) {
  long *newdimension;
  long N, idim;
  if (!dim)
    return ((long *)NULL);
  N = dim[0];
  if (!(newdimension = newdim(N)))
    return ((long *)NULL);
  for (idim = 0; idim <= N; idim++)
    newdimension[idim] = dim[idim];
  return (newdimension);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<newcopydim>:
  endbr64
  test   %rdi,%rdi
  je     L17
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rbp
  mov    %rbp,%rdi
  call   <newdim@plt>
  test   %rax,%rax
  je     L1
  test   %rbp,%rbp
  js     L22
  lea    0x1(%rbp),%rsi
  xor    %edx,%edx
  nopl   (%rax)
L34:
  mov    (%rbx,%rdx,8),%rcx
  mov    %rcx,(%rax,%rdx,8)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L34
L22:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  ret
```

assistant: ```c
long *newcopydim(const long *dim) {
  long *newdimension;
  long N, idim;
  if (!dim)
    return ((long *)NULL);
  N = dim[0];
  if (!(newdimension = newdim(N)))
    return ((long *)NULL);
  for (idim = 0; idim <= N; idim++)
    newdimension[idim] = dim[idim];
  return (newdimension);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<newcopydim>:
  endbr64
  test   %rdi,%rdi
  je     L23
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%rbp
  mov    %rbp,%rdi
  call   <newdim@plt>
  test   %rax,%rax
  je     L64
  test   %rbp,%rbp
  js     L31
  lea    0x8(%rbx),%rdx
  cmp    %rdx,%rax
  je     L51
  cmp    $0x2,%rbp
  jbe    L51
  add    $0x1,%rbp
  xor    %edx,%edx
  mov    %rbp,%rcx
  shr    %rcx
  shl    $0x4,%rcx
  nopw   0x0(%rax,%rax,1)
L79:
  movdqu (%rbx,%rdx,1),%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %rcx,%rdx
  jne    L79
  mov    %rbp,%rdx
  and    $0xfffffffffffffffe,%rdx
  and    $0x1,%ebp
  je     L31
  mov    (%rbx,%rdx,8),%rcx
  mov    %rcx,(%rax,%rdx,8)
L31:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L64:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L51:
  add    $0x1,%rbp
  xor    %edx,%edx
  cs nopw 0x0(%rax,%rax,1)
L35:
  mov    (%rbx,%rdx,8),%rcx
  mov    %rcx,(%rax,%rdx,8)
  add    $0x1,%rdx
  cmp    %rdx,%rbp
  jne    L35
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  xor    %eax,%eax
  ret
```

assistant: ```c
long *newcopydim(const long *dim) {
  long *newdimension;
  long N, idim;
  if (!dim)
    return ((long *)NULL);
  N = dim[0];
  if (!(newdimension = newdim(N)))
    return ((long *)NULL);
  for (idim = 0; idim <= N; idim++)
    newdimension[idim] = dim[idim];
  return (newdimension);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strings_is_white_space>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strings_length@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L27
L36:
  mov    -0x8(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jle    L26
  mov    $0x0,%eax
  jmp    L15
L26:
  addl   $0x1,-0x8(%rbp)
L27:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jb     L36
  mov    $0x1,%eax
L15:
  leave
  ret
```

assistant: ```c
bool strings_is_white_space(char *string) {
  assert(string);
  unsigned int length = strings_length(string);
  for (unsigned int k = 0; k < length; k++) {
    if (string[k] > 0x20)
      return false;
  }
  return true;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strings_is_white_space>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <strings_length@plt>
  test   %eax,%eax
  je     L21
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rax
L37:
  cmpb   $0x20,(%rdx)
  jg     L22
  add    $0x1,%rdx
  cmp    %rax,%rdx
  jne    L37
  mov    $0x1,%eax
  jmp    L15
L21:
  mov    $0x1,%eax
  jmp    L15
L22:
  mov    $0x0,%eax
L15:
  pop    %rbx
  ret
```

assistant: ```c
bool strings_is_white_space(char *string) {
  assert(string);
  unsigned int length = strings_length(string);
  for (unsigned int k = 0; k < length; k++) {
    if (string[k] > 0x20)
      return false;
  }
  return true;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strings_is_white_space>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <strings_length@plt>
  test   %eax,%eax
  je     L14
  sub    $0x1,%eax
  mov    %rbx,%rdi
  lea    0x1(%rbx,%rax,1),%rax
  jmp    L21
  cs nopw 0x0(%rax,%rax,1)
L28:
  add    $0x1,%rdi
  cmp    %rax,%rdi
  je     L14
L21:
  cmpb   $0x20,(%rdi)
  jle    L28
  xor    %eax,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
bool strings_is_white_space(char *string) {
  assert(string);
  unsigned int length = strings_length(string);
  for (unsigned int k = 0; k < length; k++) {
    if (string[k] > 0x20)
      return false;
  }
  return true;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strings_is_white_space>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <assert@plt>
  mov    %rbx,%rdi
  call   <strings_length@plt>
  test   %eax,%eax
  je     L30
  sub    $0x1,%eax
  mov    %rbx,%rdi
  lea    0x1(%rbx,%rax,1),%rax
  jmp    L25
  cs nopw 0x0(%rax,%rax,1)
L33:
  add    $0x1,%rdi
  cmp    %rax,%rdi
  je     L30
L25:
  cmpb   $0x20,(%rdi)
  jle    L33
  xor    %eax,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
bool strings_is_white_space(char *string) {
  assert(string);
  unsigned int length = strings_length(string);
  for (unsigned int k = 0; k < length; k++) {
    if (string[k] > 0x20)
      return false;
  }
  return true;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rotateNibbles>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L14:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <rotateLeft@plt>
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L3:
  cmpl   $0xf,-0x4(%rbp)
  jle    L14
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
uint64_t rotateNibbles(uint64_t state) {
  int i;
  for (i = 0; i < 16; i++) {
    state = rotateLeft(state);
  }
  return state;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rotateNibbles>:
  endbr64
  push   %rbx
  mov    $0x10,%ebx
L2:
  call   <rotateLeft@plt>
  mov    %eax,%edi
  sub    $0x1,%ebx
  jne    L2
  pop    %rbx
  ret
```

assistant: ```c
uint64_t rotateNibbles(uint64_t state) {
  int i;
  for (i = 0; i < 16; i++) {
    state = rotateLeft(state);
  }
  return state;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rotateNibbles>:
  endbr64
  push   %rbx
  mov    $0x10,%ebx
  nopw   0x0(%rax,%rax,1)
L7:
  call   <rotateLeft@plt>
  mov    %eax,%edi
  sub    $0x1,%ebx
  jne    L7
  pop    %rbx
  ret
```

assistant: ```c
uint64_t rotateNibbles(uint64_t state) {
  int i;
  for (i = 0; i < 16; i++) {
    state = rotateLeft(state);
  }
  return state;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rotateNibbles>:
  endbr64
  push   %rbx
  mov    $0x10,%ebx
  nopw   0x0(%rax,%rax,1)
L3:
  call   <rotateLeft@plt>
  mov    %eax,%edi
  sub    $0x1,%ebx
  jne    L3
  pop    %rbx
  ret
```

assistant: ```c
uint64_t rotateNibbles(uint64_t state) {
  int i;
  for (i = 0; i < 16; i++) {
    state = rotateLeft(state);
  }
  return state;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L12
L29:
  lea    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <length@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    -0x10(%rbp),%rcx
  mov    %rax,%rdi
  call   <recurFunct@plt>
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L12:
  mov    -0x30(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  test   %rax,%rax
  jne    L29
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D19
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D21
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D18
type_guessed: string

tool: D19: %lld

tool: D21: %lld %lld %lld

tool: D18: %lld


assistant: ```c
int main() {
  long long int t, a, n, m, p, q, r, s, l;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld %lld %lld", &a, &n, &m);
    l = length(a);
    printf("%lld\n", recurFunct(a, n, m, l));
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```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
  lea    0x8(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L27
  lea    0x20(%rsp),%r13
  lea    0x18(%rsp),%r12
  lea    D25(%rip),%rbp
  lea    D10(%rip),%rbx
L18:
  lea    0x10(%rsp),%rsi
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rdi
  call   <length@plt>
  mov    %rax,%rcx
  mov    0x20(%rsp),%rdx
  mov    0x18(%rsp),%rsi
  mov    0x10(%rsp),%rdi
  call   <recurFunct@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  jne    L18
L27:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
datatype: string
name: unpack_data
arguments: data_label: D25
datatype: string
name: unpack_data
arguments: data_label: D10
datatype: string

tool: D3: %lld

tool: D25: %lld %lld %lld

tool: D10: %lld


assistant: ```c
int main() {
  long long int t, a, n, m, p, q, r, s, l;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld %lld %lld", &a, &n, &m);
    l = length(a);
    printf("%lld\n", recurFunct(a, n, m, l));
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r14
  lea    D2(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L23
  lea    0x20(%rsp),%r14
  lea    0x18(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D20(%rip),%rbp
  lea    D19(%rip),%rbx
  nopl   0x0(%rax)
L12:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rdi
  call   <length@plt>
  mov    0x20(%rsp),%rdx
  mov    0x18(%rsp),%rsi
  mov    0x10(%rsp),%rdi
  mov    %rax,%rcx
  call   <recurFunct@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  jne    L12
L23:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D2
type_guessed: string
name: disassemble_data
arguments: tag: D20
type_guessed: string
name: disassemble_data
arguments: tag: D19
type_guessed: string

tool: D2: %lld

tool: D20: %lld %lld %lld

tool: D19: %lld


assistant: ```c
int main() {
  long long int t, a, n, m, p, q, r, s, l;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld %lld %lld", &a, &n, &m);
    l = length(a);
    printf("%lld\n", recurFunct(a, n, m, l));
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D27(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  je     L25
  lea    0x20(%rsp),%r14
  lea    0x18(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D3(%rip),%rbp
  lea    D7(%rip),%rbx
  nopl   0x0(%rax)
L26:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rdi
  call   <length@plt>
  mov    0x20(%rsp),%rdx
  mov    0x18(%rsp),%rsi
  mov    0x10(%rsp),%rdi
  mov    %rax,%rcx
  call   <recurFunct@plt>
  mov    %rbx,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  jne    L26
L25:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D27
type_infered: string
name: retrieve_data
arguments: label: D3
type_infered: string
name: retrieve_data
arguments: label: D7
type_infered: string

tool: D27: %lld

tool: D3: %lld %lld %lld

tool: D7: %lld


assistant: ```c
int main() {
  long long int t, a, n, m, p, q, r, s, l;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld %lld %lld", &a, &n, &m);
    l = length(a);
    printf("%lld\n", recurFunct(a, n, m, l));
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cmpstrptr>:
  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,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  leave
  ret
```

assistant: ```c
int cmpstrptr(const void *_x, const void *_y) {
  const char *x = *(const char **)_x;
  const char *y = *(const char **)_y;
  return strcmp(x, y);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<cmpstrptr>:
  endbr64
  sub    $0x8,%rsp
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  call   <strcmp@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int cmpstrptr(const void *_x, const void *_y) {
  const char *x = *(const char **)_x;
  const char *y = *(const char **)_y;
  return strcmp(x, y);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cmpstrptr>:
  endbr64
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  jmp    1050 <strcmp@plt>
```

assistant: ```c
int cmpstrptr(const void *_x, const void *_y) {
  const char *x = *(const char **)_x;
  const char *y = *(const char **)_y;
  return strcmp(x, y);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<cmpstrptr>:
  endbr64
  mov    (%rsi),%rsi
  mov    (%rdi),%rdi
  jmp    1050 <strcmp@plt>
```

assistant: ```c
int cmpstrptr(const void *_x, const void *_y) {
  const char *x = *(const char **)_x;
  const char *y = *(const char **)_y;
  return strcmp(x, y);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<compare_digest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L19
L26:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rdx),%edx
  movzbl %dl,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L38
  mov    -0x4(%rbp),%eax
  jmp    L12
L38:
  addl   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  cmp    %eax,-0x24(%rbp)
  ja     L26
  mov    $0x0,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int compare_digest(unsigned char *digest_1, unsigned char *digest_2, unsigned int len) {
  int i, c;
  for (i = 0; i < len; ++i) {
    c = digest_1[i] - digest_2[i];
    if (c)
      return c;
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<compare_digest>:
  endbr64
  test   %edx,%edx
  je     L17
  mov    %edx,%edx
  mov    $0x0,%ecx
L26:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%r8d
  sub    %r8d,%eax
  jne    L16
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L26
  ret
L17:
  mov    $0x0,%eax
L16:
  ret
```

assistant: ```c
int compare_digest(unsigned char *digest_1, unsigned char *digest_2, unsigned int len) {
  int i, c;
  for (i = 0; i < len; ++i) {
    c = digest_1[i] - digest_2[i];
    if (c)
      return c;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<compare_digest>:
  endbr64
  test   %edx,%edx
  je     L24
  mov    %edx,%edx
  xor    %ecx,%ecx
  jmp    L22
  xchg   %ax,%ax
L14:
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L16
L22:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%r8d
  sub    %r8d,%eax
  je     L14
  ret
  nopl   0x0(%rax,%rax,1)
L16:
  ret
L24:
  xor    %eax,%eax
  ret
```

assistant: ```c
int compare_digest(unsigned char *digest_1, unsigned char *digest_2, unsigned int len) {
  int i, c;
  for (i = 0; i < len; ++i) {
    c = digest_1[i] - digest_2[i];
    if (c)
      return c;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<compare_digest>:
  endbr64
  test   %edx,%edx
  je     L31
  mov    %edx,%edx
  xor    %ecx,%ecx
  jmp    L0
  xchg   %ax,%ax
L14:
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L26
L0:
  movzbl (%rdi,%rcx,1),%eax
  movzbl (%rsi,%rcx,1),%r8d
  sub    %r8d,%eax
  je     L14
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  ret
L31:
  xor    %eax,%eax
  ret
```

assistant: ```c
int compare_digest(unsigned char *digest_1, unsigned char *digest_2, unsigned int len) {
  int i, c;
  for (i = 0; i < len; ++i) {
    c = digest_1[i] - digest_2[i];
    if (c)
      return c;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_decrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    $0x10009000,%eax
  movl   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  shl    $0x10,%rax
  mov    $0x10009004,%edx
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  or     $0x83c00c00,%eax
  mov    $0x10009000,%edx
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <aes_fifos@plt>
  nop
  leave
  ret
```

assistant: ```c
void _decrypt(uint32_t value, void *inbuf, void *outbuf, size_t blocks) {
  *((volatile uint32_t *)0x10009000) = 0;
  *((volatile uint32_t *)0x10009004) = blocks << 16;
  *((volatile uint32_t *)0x10009000) = value | 0x80000000 | 0x02000000 | 0x01000000 | 0x00800000 | 0x00400000 | 0x00000800 | 0x00000400;
  aes_fifos(inbuf, outbuf, blocks);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_decrypt>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%eax
  mov    %rsi,%rdi
  mov    %rdx,%rsi
  mov    %rcx,%rdx
  movl   $0x0,0x10009000
  shl    $0x10,%rcx
  mov    %ecx,0x10009004
  or     $0x83c00c00,%eax
  mov    %eax,0x10009000
  call   <aes_fifos@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void _decrypt(uint32_t value, void *inbuf, void *outbuf, size_t blocks) {
  *((volatile uint32_t *)0x10009000) = 0;
  *((volatile uint32_t *)0x10009004) = blocks << 16;
  *((volatile uint32_t *)0x10009000) = value | 0x80000000 | 0x02000000 | 0x01000000 | 0x00800000 | 0x00400000 | 0x00000800 | 0x00000400;
  aes_fifos(inbuf, outbuf, blocks);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_decrypt>:
  endbr64
  mov    %edi,%r8d
  mov    %rcx,%rax
  movl   $0x0,0x10009000
  mov    %rsi,%rdi
  shl    $0x10,%rax
  or     $0x83c00c00,%r8d
  mov    %rdx,%rsi
  mov    %rcx,%rdx
  mov    %eax,0x10009004
  mov    %r8d,0x10009000
  jmp    1050 <aes_fifos@plt>
```

assistant: ```c
void _decrypt(uint32_t value, void *inbuf, void *outbuf, size_t blocks) {
  *((volatile uint32_t *)0x10009000) = 0;
  *((volatile uint32_t *)0x10009004) = blocks << 16;
  *((volatile uint32_t *)0x10009000) = value | 0x80000000 | 0x02000000 | 0x01000000 | 0x00800000 | 0x00400000 | 0x00000800 | 0x00000400;
  aes_fifos(inbuf, outbuf, blocks);
}
```
user: Translate the below assembly code into C programming language:

```asm
<_decrypt>:
  endbr64
  mov    %edi,%r8d
  mov    %rcx,%rax
  movl   $0x0,0x10009000
  mov    %rsi,%rdi
  shl    $0x10,%rax
  or     $0x83c00c00,%r8d
  mov    %rdx,%rsi
  mov    %rcx,%rdx
  mov    %eax,0x10009004
  mov    %r8d,0x10009000
  jmp    1050 <aes_fifos@plt>
```

assistant: ```c
void _decrypt(uint32_t value, void *inbuf, void *outbuf, size_t blocks) {
  *((volatile uint32_t *)0x10009000) = 0;
  *((volatile uint32_t *)0x10009004) = blocks << 16;
  *((volatile uint32_t *)0x10009000) = value | 0x80000000 | 0x02000000 | 0x01000000 | 0x00800000 | 0x00400000 | 0x00000800 | 0x00000400;
  aes_fifos(inbuf, outbuf, blocks);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xrop3_217_1_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x18(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x14(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  and    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  or     -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  xor    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  notl   -0x4(%rbp)
  mov    -0x14(%rbp),%eax
  not    %eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%eax
  not    %eax
  and    %eax,%edx
  mov    -0x18(%rbp),%eax
  and    -0x14(%rbp),%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%eax
  and    -0x10(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_217_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  unsigned stk3;
  stk3 = T & S;
  stk2 = D | stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```