ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D13(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rbp
  call   <scanf@plt>
  xor    %eax,%eax
  mov    $0x32,%ecx
  mov    %rbp,%rdi
  rep stos %rax,%es:(%rdi)
  mov    0x8(%rsp),%eax
  movl   $0x0,(%rdi)
  test   %eax,%eax
  jle    L10
  xor    %ebx,%ebx
  lea    0xc(%rsp),%r13
  nop
L5:
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <scanf@plt>
  movslq 0xc(%rsp),%rax
  addl   $0x1,0x10(%rsp,%rax,4)
  cmp    %ebx,0x8(%rsp)
  jg     L5
L10:
  xor    %eax,%eax
  jmp    L53
  nopw   0x0(%rax,%rax,1)
L25:
  add    $0x1,%rax
  cmp    $0x65,%rax
  je     L17
L53:
  mov    %eax,%esi
  testb  $0x1,0x0(%rbp,%rax,4)
  je     L25
  lea    D1(%rip),%rdi
  call   <printf@plt>
L17:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x1b8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D13
type_guessed: string
name: unpack_data
arguments: data_label: D1
type_guessed: string

tool: D13: %d

tool: D1: %d


assistant: ```c
int main() {
  int n, a[101], i, s;
  scanf("%d", &n);
  for (i = 0; i < 101; ++i)
    a[i] = 0;
  for (i = 0; i < n; ++i) {
    scanf("%d", &s);
    a[s]++;
  }
  for (i = 0; i < 101; ++i)
    if (a[i] & 0x1) {
      printf("%d\n", i);
      break;
    }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<reservaMemoria>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L0
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L1
L0:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L1:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
type: string
name: unpack_data
arguments: tag: D12
type: string

tool: D1: 
Error en la reserva de memoria

tool: D12: 
Se ha reservado memoria para %i elementos


assistant: ```c
*reservaMemoria(int nEle) {
  int *aux;
  if ((aux = (int *)calloc(sizeof(int), nEle)) == NULL) {
    printf("\nError en la reserva de memoria\n");
    exit(-1);
  } else {
    printf("\nSe ha reservado memoria para %i elementos\n", nEle);
  }
  return aux;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reservaMemoria>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %edi,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L18
  lea    D7(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L15:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  mov    %ebp,%esi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L15
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
datatype: string
name: extract_symbolic_data
arguments: label: D18
datatype: string

tool: D7: 
Error en la reserva de memoria

tool: D18: 
Se ha reservado memoria para %i elementos


assistant: ```c
*reservaMemoria(int nEle) {
  int *aux;
  if ((aux = (int *)calloc(sizeof(int), nEle)) == NULL) {
    printf("\nError en la reserva de memoria\n");
    exit(-1);
  } else {
    printf("\nSe ha reservado memoria para %i elementos\n", nEle);
  }
  return aux;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<reservaMemoria>:
  endbr64
  push   %r12
  mov    %edi,%esi
  push   %rbp
  mov    %edi,%ebp
  mov    $0x4,%edi
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L6
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %ebp,%esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D10
type_infered: string
name: read_data
arguments: data_tag: D15
type_infered: string

tool: D10: 
Error en la reserva de memoria

tool: D15: 
Se ha reservado memoria para %i elementos


assistant: ```c
*reservaMemoria(int nEle) {
  int *aux;
  if ((aux = (int *)calloc(sizeof(int), nEle)) == NULL) {
    printf("\nError en la reserva de memoria\n");
    exit(-1);
  } else {
    printf("\nSe ha reservado memoria para %i elementos\n", nEle);
  }
  return aux;
}
```
user: What is the c source code of the assembly code below:

```asm
<reservaMemoria>:
  endbr64
  push   %r12
  mov    %edi,%esi
  push   %rbp
  mov    %edi,%ebp
  mov    $0x4,%edi
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L4
  lea    D5(%rip),%rdi
  call   <puts@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %ebp,%esi
  lea    D11(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D5
datatype: string
name: parse_data
arguments: label: D11
datatype: string

tool: D5: 
Error en la reserva de memoria

tool: D11: 
Se ha reservado memoria para %i elementos


assistant: ```c
*reservaMemoria(int nEle) {
  int *aux;
  if ((aux = (int *)calloc(sizeof(int), nEle)) == NULL) {
    printf("\nError en la reserva de memoria\n");
    exit(-1);
  } else {
    printf("\nSe ha reservado memoria para %i elementos\n", nEle);
  }
  return aux;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<in_set>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  cmpb   $0x0,-0x1c(%rbp)
  jne    L9
  mov    $0x0,%eax
  jmp    L54
L9:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L56
L10:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    $0x2d,%eax
  je     L96
  cmp    $0x4d,%eax
  jne    L66
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L84
  mov    $0x1,%eax
  jmp    L54
L84:
  subl   $0x2,-0x4(%rbp)
  jmp    L56
L96:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jl     L47
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jg     L47
  mov    $0x1,%eax
  jmp    L54
L47:
  subl   $0x3,-0x4(%rbp)
  jmp    L56
L66:
  mov    $0xfffffffa,%eax
  jmp    L54
L56:
  cmpl   $0x0,-0x4(%rbp)
  jg     L10
  mov    $0x0,%eax
L54:
  pop    %rbp
  ret
```

assistant: ```c
int in_set(set, c)
char *set;
char c;
{
  int n;
  if (c == 0)
    return (0);
  n = *set++;
  while (n > 0)
    switch (*set++) {
    case 'M':
      if (*set++ == c)
        return (1);
      n -= 2;
      continue;
    case '-':
      if (*set++ <= c && c <= *set++)
        return (1);
      n -= 3;
      continue;
    default:
      return (-6);
    }
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<in_set>:
  endbr64
  mov    %esi,%r8d
  mov    $0x0,%eax
  test   %sil,%sil
  je     L34
  lea    0x1(%rdi),%rax
  movsbl (%rdi),%ecx
  test   %ecx,%ecx
  jle    L41
L17:
  movzbl (%rax),%edx
  cmp    $0x2d,%dl
  je     L26
  cmp    $0x4d,%dl
  jne    L28
  cmp    %r8b,0x1(%rax)
  je     L32
  add    $0x2,%rax
  sub    $0x2,%ecx
  jmp    L44
L26:
  cmp    %r8b,0x1(%rax)
  jg     L43
  cmp    %r8b,0x2(%rax)
  jl     L50
  mov    $0x1,%eax
L34:
  ret
L43:
  add    $0x2,%rax
L13:
  sub    $0x3,%ecx
L44:
  test   %ecx,%ecx
  jg     L17
  mov    $0x0,%eax
  ret
L50:
  add    $0x3,%rax
  jmp    L13
L41:
  mov    $0x0,%eax
  ret
L28:
  mov    $0xfffffffa,%eax
  ret
L32:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int in_set(set, c)
char *set;
char c;
{
  int n;
  if (c == 0)
    return (0);
  n = *set++;
  while (n > 0)
    switch (*set++) {
    case 'M':
      if (*set++ == c)
        return (1);
      n -= 2;
      continue;
    case '-':
      if (*set++ <= c && c <= *set++)
        return (1);
      n -= 3;
      continue;
    default:
      return (-6);
    }
  return (0);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<in_set>:
  endbr64
  mov    %esi,%r9d
  xor    %r8d,%r8d
  test   %sil,%sil
  je     L56
  movsbl (%rdi),%ecx
  lea    0x1(%rdi),%rax
  test   %ecx,%ecx
  jle    L56
L33:
  movzbl (%rax),%edx
  cmp    $0x2d,%dl
  je     L58
  cmp    $0x4d,%dl
  jne    L15
  cmp    %r9b,0x1(%rax)
  je     L44
  add    $0x2,%rax
  sub    $0x2,%ecx
L85:
  test   %ecx,%ecx
  jg     L33
  xor    %r8d,%r8d
L56:
  mov    %r8d,%eax
  ret
  nop
L15:
  mov    $0xfffffffa,%r8d
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L58:
  cmp    %r9b,0x1(%rax)
  jg     L69
  cmp    %r9b,0x2(%rax)
  jl     L13
L44:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L13:
  add    $0x3,%rax
L23:
  sub    $0x3,%ecx
  jmp    L85
  nopl   0x0(%rax)
L69:
  add    $0x2,%rax
  jmp    L23
```

assistant: ```c
int in_set(set, c)
char *set;
char c;
{
  int n;
  if (c == 0)
    return (0);
  n = *set++;
  while (n > 0)
    switch (*set++) {
    case 'M':
      if (*set++ == c)
        return (1);
      n -= 2;
      continue;
    case '-':
      if (*set++ <= c && c <= *set++)
        return (1);
      n -= 3;
      continue;
    default:
      return (-6);
    }
  return (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<in_set>:
  endbr64
  mov    %esi,%r9d
  xor    %r8d,%r8d
  test   %sil,%sil
  je     L84
  movsbl (%rdi),%ecx
  lea    0x1(%rdi),%rax
  test   %ecx,%ecx
  jle    L84
L43:
  movzbl (%rax),%edx
  cmp    $0x2d,%dl
  je     L77
  cmp    $0x4d,%dl
  jne    L66
  cmp    %r9b,0x1(%rax)
  je     L9
  add    $0x2,%rax
  sub    $0x2,%ecx
L21:
  test   %ecx,%ecx
  jg     L43
  xor    %r8d,%r8d
L84:
  mov    %r8d,%eax
  ret
  nop
L66:
  mov    $0xfffffffa,%r8d
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L77:
  cmp    %r9b,0x1(%rax)
  jg     L45
  cmp    %r9b,0x2(%rax)
  jl     L56
L9:
  mov    $0x1,%r8d
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L56:
  add    $0x3,%rax
L40:
  sub    $0x3,%ecx
  jmp    L21
  nopl   0x0(%rax)
L45:
  add    $0x2,%rax
  jmp    L40
```

assistant: ```c
int in_set(set, c)
char *set;
char c;
{
  int n;
  if (c == 0)
    return (0);
  n = *set++;
  while (n > 0)
    switch (*set++) {
    case 'M':
      if (*set++ == c)
        return (1);
      n -= 2;
      continue;
    case '-':
      if (*set++ <= c && c <= *set++)
        return (1);
      n -= 3;
      continue;
    default:
      return (-6);
    }
  return (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<show_syms>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <plintf@plt>
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <plintf@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <plintf@plt>
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <plintf@plt>
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <plintf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D12
type: string
name: unpack_data
arguments: data_label: D41
type: string
name: unpack_data
arguments: data_label: D34
type: string
name: unpack_data
arguments: data_label: D17
type: string
name: unpack_data
arguments: data_label: D24
type: string

tool: D12: (    ,    )    +    -      *    ^    **    / 


tool: D41: sin  cos  tan  atan  atan2 acos asin


tool: D34: exp  ln   log  log10 tanh  cosh sinh 


tool: D17: max  min  heav flr   mod   sign sqrt 


tool: D24: t    pi   ran  


assistant: ```c
void show_syms(void) {
  plintf("(    ,    )    +    -      *    ^    **    / \n");
  plintf("sin  cos  tan  atan  atan2 acos asin\n");
  plintf("exp  ln   log  log10 tanh  cosh sinh \n");
  plintf("max  min  heav flr   mod   sign sqrt \n");
  plintf("t    pi   ran  \n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<show_syms>:
  endbr64
  sub    $0x8,%rsp
  lea    D29(%rip),%rdi
  call   <plintf@plt>
  lea    D0(%rip),%rdi
  call   <plintf@plt>
  lea    D20(%rip),%rdi
  call   <plintf@plt>
  lea    D1(%rip),%rdi
  call   <plintf@plt>
  lea    D4(%rip),%rdi
  call   <plintf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D29
type_guessed: string
name: unpack_data
arguments: label: D0
type_guessed: string
name: unpack_data
arguments: label: D20
type_guessed: string
name: unpack_data
arguments: label: D1
type_guessed: string
name: unpack_data
arguments: label: D4
type_guessed: string

tool: D29: (    ,    )    +    -      *    ^    **    / 


tool: D0: sin  cos  tan  atan  atan2 acos asin


tool: D20: exp  ln   log  log10 tanh  cosh sinh 


tool: D1: max  min  heav flr   mod   sign sqrt 


tool: D4: t    pi   ran  


assistant: ```c
void show_syms(void) {
  plintf("(    ,    )    +    -      *    ^    **    / \n");
  plintf("sin  cos  tan  atan  atan2 acos asin\n");
  plintf("exp  ln   log  log10 tanh  cosh sinh \n");
  plintf("max  min  heav flr   mod   sign sqrt \n");
  plintf("t    pi   ran  \n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<show_syms>:
  endbr64
  sub    $0x8,%rsp
  lea    D33(%rip),%rdi
  call   <plintf@plt>
  lea    D40(%rip),%rdi
  call   <plintf@plt>
  lea    D43(%rip),%rdi
  call   <plintf@plt>
  lea    D13(%rip),%rdi
  call   <plintf@plt>
  lea    D2(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <plintf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D33
data_type: string
name: retrieve_data
arguments: data_label: D40
data_type: string
name: retrieve_data
arguments: data_label: D43
data_type: string
name: retrieve_data
arguments: data_label: D13
data_type: string
name: retrieve_data
arguments: data_label: D2
data_type: string

tool: D33: (    ,    )    +    -      *    ^    **    / 


tool: D40: sin  cos  tan  atan  atan2 acos asin


tool: D43: exp  ln   log  log10 tanh  cosh sinh 


tool: D13: max  min  heav flr   mod   sign sqrt 


tool: D2: t    pi   ran  


assistant: ```c
void show_syms(void) {
  plintf("(    ,    )    +    -      *    ^    **    / \n");
  plintf("sin  cos  tan  atan  atan2 acos asin\n");
  plintf("exp  ln   log  log10 tanh  cosh sinh \n");
  plintf("max  min  heav flr   mod   sign sqrt \n");
  plintf("t    pi   ran  \n");
}
```
user: What is the c source code of the assembly code below?

```asm
<show_syms>:
  endbr64
  sub    $0x8,%rsp
  lea    D31(%rip),%rdi
  call   <plintf@plt>
  lea    D4(%rip),%rdi
  call   <plintf@plt>
  lea    D40(%rip),%rdi
  call   <plintf@plt>
  lea    D38(%rip),%rdi
  call   <plintf@plt>
  lea    D2(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <plintf@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D31
type_infered: string
name: disassemble_data
arguments: tag: D4
type_infered: string
name: disassemble_data
arguments: tag: D40
type_infered: string
name: disassemble_data
arguments: tag: D38
type_infered: string
name: disassemble_data
arguments: tag: D2
type_infered: string

tool: D31: (    ,    )    +    -      *    ^    **    / 


tool: D4: sin  cos  tan  atan  atan2 acos asin


tool: D40: exp  ln   log  log10 tanh  cosh sinh 


tool: D38: max  min  heav flr   mod   sign sqrt 


tool: D2: t    pi   ran  


assistant: ```c
void show_syms(void) {
  plintf("(    ,    )    +    -      *    ^    **    / \n");
  plintf("sin  cos  tan  atan  atan2 acos asin\n");
  plintf("exp  ln   log  log10 tanh  cosh sinh \n");
  plintf("max  min  heav flr   mod   sign sqrt \n");
  plintf("t    pi   ran  \n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<s_memcpy4orgb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8d,-0x30(%rbp)
  mov    %r9d,-0x44(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    $0x8,%eax
  sub    -0x44(%rbp),%eax
  mov    %eax,-0x44(%rbp)
  mov    $0x8,%eax
  sub    0x10(%rbp),%eax
  mov    %eax,0x10(%rbp)
  mov    $0x8,%eax
  sub    0x18(%rbp),%eax
  mov    %eax,0x18(%rbp)
  jmp    L20
L23:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  cmp    %eax,-0x2c(%rbp)
  jne    L15
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  shr    $0x18,%eax
  mov    %al,-0x13(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  shr    $0x10,%eax
  mov    %al,-0x12(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  shr    $0x8,%eax
  mov    %al,-0x11(%rbp)
  movzbl -0x13(%rbp),%edx
  mov    -0x44(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    0x20(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  movzbl -0x12(%rbp),%edx
  mov    0x10(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    0x28(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,%esi
  movzbl -0x11(%rbp),%edx
  mov    0x18(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    0x30(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %esi,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
L15:
  addq   $0x4,-0x10(%rbp)
  addq   $0x4,-0x8(%rbp)
  addq   $0x1,-0x28(%rbp)
L20:
  mov    -0x30(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x30(%rbp)
  test   %eax,%eax
  jne    L23
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void s_memcpy4orgb(unsigned char *m, int id, void *dest, unsigned int *rgb, int n, int rl, int gl, int bl, int ro, int go, int bo) {
  unsigned char dr;
  unsigned char dg;
  unsigned char db;
  unsigned int *d = (unsigned int *)dest;
  unsigned int *s = (unsigned int *)rgb;
  rl = 8 - rl;
  gl = 8 - gl;
  bl = 8 - bl;
  while (n--) {
    if (*m == id) {
      dr = *s >> 0x18;
      dg = *s >> 0x10;
      db = *s >> 0x08;
      *d = ((dr >> rl) << ro) | ((dg >> gl) << go) | ((db >> bl) << bo);
    }
    d++;
    s++;
    m++;
  };
}
```
user: Translate the below assembly code into C programming language:

```asm
<s_memcpy4orgb>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r10d
  mov    %rdx,%rsi
  mov    0x48(%rsp),%r12d
  mov    0x50(%rsp),%r13d
  mov    0x58(%rsp),%r14d
  mov    $0x8,%edx
  mov    %edx,%r11d
  sub    %r9d,%r11d
  mov    %edx,%ebp
  sub    0x38(%rsp),%ebp
  sub    0x40(%rsp),%edx
  test   %r8d,%r8d
  je     L23
  mov    %rdi,%rax
  mov    %rcx,%rdi
  mov    %edx,%r9d
  mov    %r8d,%r8d
  add    %rax,%r8
  jmp    L1
L33:
  add    $0x4,%rsi
  add    $0x4,%rdi
  add    $0x1,%rax
  cmp    %rax,%r8
  je     L23
L1:
  movzbl (%rax),%edx
  cmp    %r10d,%edx
  jne    L33
  mov    (%rdi),%ebx
  mov    %ebx,%edx
  shr    $0x10,%edx
  movzbl %dl,%edx
  mov    %ebp,%ecx
  sar    %cl,%edx
  mov    %r13d,%ecx
  shl    %cl,%edx
  movzbl %bh,%ecx
  mov    %ecx,%r15d
  mov    %r9d,%ecx
  sar    %cl,%r15d
  mov    %r14d,%ecx
  shl    %cl,%r15d
  or     %r15d,%edx
  shr    $0x18,%ebx
  mov    %r11d,%ecx
  sar    %cl,%ebx
  mov    %r12d,%ecx
  shl    %cl,%ebx
  or     %ebx,%edx
  mov    %edx,(%rsi)
  jmp    L33
L23:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void s_memcpy4orgb(unsigned char *m, int id, void *dest, unsigned int *rgb, int n, int rl, int gl, int bl, int ro, int go, int bo) {
  unsigned char dr;
  unsigned char dg;
  unsigned char db;
  unsigned int *d = (unsigned int *)dest;
  unsigned int *s = (unsigned int *)rgb;
  rl = 8 - rl;
  gl = 8 - gl;
  bl = 8 - bl;
  while (n--) {
    if (*m == id) {
      dr = *s >> 0x18;
      dg = *s >> 0x10;
      db = *s >> 0x08;
      *d = ((dr >> rl) << ro) | ((dg >> gl) << go) | ((db >> bl) << bo);
    }
    d++;
    s++;
    m++;
  };
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<s_memcpy4orgb>:
  endbr64
  mov    %esi,%r10d
  mov    %rdx,%rsi
  mov    $0x8,%edx
  push   %r15
  mov    %edx,%r11d
  push   %r14
  sub    %r9d,%r11d
  push   %r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    0x48(%rsp),%r12d
  mov    0x50(%rsp),%r13d
  mov    0x58(%rsp),%r14d
  sub    0x38(%rsp),%ebp
  sub    0x40(%rsp),%edx
  test   %r8d,%r8d
  je     L16
  mov    %rdi,%rax
  mov    %r8d,%r8d
  mov    %rcx,%rdi
  mov    %edx,%r9d
  add    %rax,%r8
  nopl   0x0(%rax)
L19:
  movzbl (%rax),%edx
  cmp    %r10d,%edx
  jne    L20
  mov    (%rdi),%ebx
  mov    %ebp,%ecx
  mov    %ebx,%edx
  shr    $0x10,%edx
  movzbl %dl,%edx
  sar    %cl,%edx
  mov    %r13d,%ecx
  shl    %cl,%edx
  movzbl %bh,%ecx
  shr    $0x18,%ebx
  mov    %ecx,%r15d
  mov    %r9d,%ecx
  sar    %cl,%r15d
  mov    %r14d,%ecx
  shl    %cl,%r15d
  mov    %r11d,%ecx
  sar    %cl,%ebx
  mov    %r12d,%ecx
  or     %r15d,%edx
  shl    %cl,%ebx
  or     %ebx,%edx
  mov    %edx,(%rsi)
L20:
  add    $0x1,%rax
  add    $0x4,%rsi
  add    $0x4,%rdi
  cmp    %rax,%r8
  jne    L19
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void s_memcpy4orgb(unsigned char *m, int id, void *dest, unsigned int *rgb, int n, int rl, int gl, int bl, int ro, int go, int bo) {
  unsigned char dr;
  unsigned char dg;
  unsigned char db;
  unsigned int *d = (unsigned int *)dest;
  unsigned int *s = (unsigned int *)rgb;
  rl = 8 - rl;
  gl = 8 - gl;
  bl = 8 - bl;
  while (n--) {
    if (*m == id) {
      dr = *s >> 0x18;
      dg = *s >> 0x10;
      db = *s >> 0x08;
      *d = ((dr >> rl) << ro) | ((dg >> gl) << go) | ((db >> bl) << bo);
    }
    d++;
    s++;
    m++;
  };
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<s_memcpy4orgb>:
  endbr64
  mov    %esi,%r10d
  mov    %rdx,%rsi
  mov    $0x8,%edx
  push   %r15
  mov    %edx,%r11d
  push   %r14
  sub    %r9d,%r11d
  push   %r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    0x48(%rsp),%r12d
  mov    0x50(%rsp),%r13d
  mov    0x58(%rsp),%r14d
  sub    0x38(%rsp),%ebp
  sub    0x40(%rsp),%edx
  test   %r8d,%r8d
  je     L8
  mov    %rdi,%rax
  mov    %r8d,%r8d
  mov    %rcx,%rdi
  mov    %edx,%r9d
  add    %rax,%r8
  nopl   0x0(%rax)
L26:
  movzbl (%rax),%edx
  cmp    %r10d,%edx
  jne    L2
  mov    (%rdi),%ebx
  mov    %ebp,%ecx
  mov    %ebx,%edx
  shr    $0x10,%edx
  movzbl %dl,%edx
  sar    %cl,%edx
  mov    %r13d,%ecx
  shl    %cl,%edx
  movzbl %bh,%ecx
  shr    $0x18,%ebx
  mov    %ecx,%r15d
  mov    %r9d,%ecx
  sar    %cl,%r15d
  mov    %r14d,%ecx
  shl    %cl,%r15d
  mov    %r11d,%ecx
  sar    %cl,%ebx
  mov    %r12d,%ecx
  or     %r15d,%edx
  shl    %cl,%ebx
  or     %ebx,%edx
  mov    %edx,(%rsi)
L2:
  add    $0x1,%rax
  add    $0x4,%rsi
  add    $0x4,%rdi
  cmp    %r8,%rax
  jne    L26
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void s_memcpy4orgb(unsigned char *m, int id, void *dest, unsigned int *rgb, int n, int rl, int gl, int bl, int ro, int go, int bo) {
  unsigned char dr;
  unsigned char dg;
  unsigned char db;
  unsigned int *d = (unsigned int *)dest;
  unsigned int *s = (unsigned int *)rgb;
  rl = 8 - rl;
  gl = 8 - gl;
  bl = 8 - bl;
  while (n--) {
    if (*m == id) {
      dr = *s >> 0x18;
      dg = *s >> 0x10;
      db = *s >> 0x08;
      *d = ((dr >> rl) << ro) | ((dg >> gl) << go) | ((db >> bl) << bo);
    }
    d++;
    s++;
    m++;
  };
}
```