ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rev>:
  endbr64
  test   %rdi,%rdi
  je     L23
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L24
  nopl   0x0(%rax)
L29:
  mov    %rbp,%rdi
  call   <strrev@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <free@plt>
L24:
  mov    %rbx,%rdi
  call   <getln@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L29
  xor    %edi,%edi
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L23:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int rev(FILE *in) {
  char *line = NULL;
  if (!in)
    return -1;
  while ((line = getln(in))) {
    strrev(line);
    puts(line);
    free(line);
  }
  free(line);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rev>:
  endbr64
  test   %rdi,%rdi
  je     L5
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L7
  nopl   0x0(%rax)
L9:
  mov    %rbp,%rdi
  call   <strrev@plt>
  mov    %rbp,%rdi
  call   <puts@plt>
  mov    %rbp,%rdi
  call   <free@plt>
L7:
  mov    %rbx,%rdi
  call   <getln@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L9
  xor    %edi,%edi
  call   <free@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L5:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int rev(FILE *in) {
  char *line = NULL;
  if (!in)
    return -1;
  while ((line = getln(in))) {
    strrev(line);
    puts(line);
    free(line);
  }
  free(line);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x10(%rbp)
  movl   $0x2,-0xc(%rbp)
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x14(%rbp),%eax
  test   %eax,%eax
  je     L55
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L37
L35:
  mov    -0xc(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L0
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L7
L0:
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L7:
  addl   $0x1,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L37:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jl     L35
L55:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L45
  call   <__stack_chk_fail@plt>
L45:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D13
type_guessed: string
name: retrieve_data
arguments: data_label: D32
type_guessed: string
name: retrieve_data
arguments: data_label: D45
type_guessed: string
name: retrieve_data
arguments: data_label: D7
type_guessed: string
name: retrieve_data
arguments: data_label: D47
type_guessed: string

tool: D13: Informe o limite: 

tool: D32: %d

tool: D45: 1

tool: D7: -1/%d


tool: D47: 1/%d


assistant: ```c
int main(void) {
  int n, i = 1, deno = 2;
  printf("Informe o limite: ");
  scanf("%d", &n);
  if (n != 0) {
    printf("1\n");
    while (i < n) {
      if (deno % 2 == 0)
        printf("-1/%d\n", deno);
      else
        printf("1/%d\n", deno);
      deno++;
      i++;
    }
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D30(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D43(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x4(%rsp)
  jne    L7
L68:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L56
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  lea    D14(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L68
  mov    $0x2,%ebx
  lea    D5(%rip),%r12
  lea    D42(%rip),%rbp
  jmp    L31
L23:
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L11:
  lea    0x1(%rbx),%eax
  cmp    %ebx,0x4(%rsp)
  jle    L68
  mov    %eax,%ebx
L31:
  test   $0x1,%bl
  jne    L23
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L11
L56:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D30
data_type: string
name: parse_data
arguments: data_label: D43
data_type: string
name: parse_data
arguments: data_label: D14
data_type: string
name: parse_data
arguments: data_label: D5
data_type: string
name: parse_data
arguments: data_label: D42
data_type: string

tool: D30: Informe o limite: 

tool: D43: %d

tool: D14: 1

tool: D5: 1/%d


tool: D42: -1/%d


assistant: ```c
int main(void) {
  int n, i = 1, deno = 2;
  printf("Informe o limite: ");
  scanf("%d", &n);
  if (n != 0) {
    printf("1\n");
    while (i < n) {
      if (deno % 2 == 0)
        printf("-1/%d\n", deno);
      else
        printf("1/%d\n", deno);
      deno++;
      i++;
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D34(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D33(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L46
L44:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L46:
  lea    D28(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L44
  mov    $0x2,%ebx
  lea    D16(%rip),%r12
  lea    D15(%rip),%rbp
  jmp    L50
  nopl   0x0(%rax)
L18:
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L23:
  lea    0x1(%rbx),%eax
  cmp    %ebx,0x4(%rsp)
  jle    L44
  mov    %eax,%ebx
L50:
  mov    %ebx,%esi
  test   $0x1,%bl
  je     L18
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L23
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D34
datatype: string
name: unpack_data
arguments: label: D33
datatype: string
name: unpack_data
arguments: label: D28
datatype: string
name: unpack_data
arguments: label: D16
datatype: string
name: unpack_data
arguments: label: D15
datatype: string

tool: D34: Informe o limite: 

tool: D33: %d

tool: D28: 1

tool: D16: 1/%d


tool: D15: -1/%d


assistant: ```c
int main(void) {
  int n, i = 1, deno = 2;
  printf("Informe o limite: ");
  scanf("%d", &n);
  if (n != 0) {
    printf("1\n");
    while (i < n) {
      if (deno % 2 == 0)
        printf("-1/%d\n", deno);
      else
        printf("1/%d\n", deno);
      deno++;
      i++;
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    D17(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jne    L33
L41:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L33:
  lea    D4(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x1,0x4(%rsp)
  jle    L41
  mov    $0x2,%ebx
  lea    D30(%rip),%r12
  lea    D3(%rip),%rbp
  jmp    L40
  nopl   0x0(%rax)
L16:
  mov    %eax,%ebx
L40:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  test   $0x1,%bl
  je     L54
  mov    %r12,%rdi
L54:
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x1(%rbx),%eax
  cmp    %ebx,0x4(%rsp)
  jg     L16
  jmp    L41
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D17
datatype: string
name: retrieve_data
arguments: label: D26
datatype: string
name: retrieve_data
arguments: label: D4
datatype: string
name: retrieve_data
arguments: label: D30
datatype: string
name: retrieve_data
arguments: label: D3
datatype: string

tool: D17: Informe o limite: 

tool: D26: %d

tool: D4: 1

tool: D30: 1/%d


tool: D3: -1/%d


assistant: ```c
int main(void) {
  int n, i = 1, deno = 2;
  printf("Informe o limite: ");
  scanf("%d", &n);
  if (n != 0) {
    printf("1\n");
    while (i < n) {
      if (deno % 2 == 0)
        printf("-1/%d\n", deno);
      else
        printf("1/%d\n", deno);
      deno++;
      i++;
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<lmsc_U2l>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L19
L13:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    -0x8(%rbp),%rdx
  mov    %al,(%rdx)
  addq   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L13
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *lmsc_U2l(char *p) {
  char *s;
  for (s = p; *s; s++) {
    *s = tolower((int)*s);
  }
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<lmsc_U2l>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L11
  mov    %rbp,%rbx
L9:
  movsbl %dil,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L9
L11:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *lmsc_U2l(char *p) {
  char *s;
  for (s = p; *s; s++) {
    *s = tolower((int)*s);
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<lmsc_U2l>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L10
  mov    %r12,%rbx
  nopl   0x0(%rax)
L18:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L18
L10:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *lmsc_U2l(char *p) {
  char *s;
  for (s = p; *s; s++) {
    *s = tolower((int)*s);
  }
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<lmsc_U2l>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%edi
  test   %dil,%dil
  je     L13
  mov    %r12,%rbx
  nopl   0x0(%rax)
L4:
  call   <tolower@plt>
  add    $0x1,%rbx
  mov    %al,-0x1(%rbx)
  movsbl (%rbx),%edi
  test   %dil,%dil
  jne    L4
L13:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *lmsc_U2l(char *p) {
  char *s;
  for (s = p; *s; s++) {
    *s = tolower((int)*s);
  }
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<anon_addr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %edx,-0x48(%rbp)
  mov    %ecx,-0x54(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x50(%rbp),%rax
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0xffffffff,-0x34(%rbp)
  cmpl   $0x80,-0x44(%rbp)
  je     L3
  cmpl   $0x81,-0x44(%rbp)
  jne    L23
  movl   $0x81,-0x30(%rbp)
  mov    -0x48(%rbp),%eax
  mov    %eax,%edi
  call   <htons@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x54(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  lea    -0x30(%rbp),%rcx
  mov    -0x50(%rbp),%rax
  mov    $0x28,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0x28,-0x34(%rbp)
  jmp    L23
L3:
  movl   $0x80,-0x2c(%rbp)
  mov    -0x48(%rbp),%eax
  mov    %eax,%edi
  call   <htons@plt>
  mov    %rax,-0x20(%rbp)
  lea    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    $0x8,%edx
  mov    $0xffffffff,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x28(%rbp),%rax
  lea    0x10(%rax),%rdx
  mov    -0x54(%rbp),%eax
  mov    %eax,(%rdx)
  lea    -0x30(%rbp),%rcx
  mov    -0x50(%rbp),%rax
  mov    $0x28,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0x28,-0x34(%rbp)
  nop
L23:
  mov    -0x34(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

assistant: ```c
socklen_t anon_addr(int domain, struct sockaddr_storage *sst, unsigned int id, unsigned int id2) {
  memset(sst, 0, sizeof(*sst));
  socklen_t ret = -1;
  switch (domain) {
  case AF_INET:;
    struct sockaddr_in sin;
    sin.sin_family = AF_INET;
    sin.sin_port = htons(id);
    sin.sin_addr.s_addr = id2;
    memcpy(sst, &sin, sizeof(sin));
    ret = sizeof(sin);
    break;
  case AF_INET6:;
    struct sockaddr_in6 sin6;
    sin6.sin6_family = AF_INET6;
    sin6.sin6_port = htons(id);
    memset(&sin6.sin6_addr, -1, sizeof(sin6.sin6_addr));
    sin6.sin6_addr.s6_addr16[4] = id2;
    memcpy(sst, &sin6, sizeof(sin6));
    ret = sizeof(sin6);
    break;
  }
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<anon_addr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,%ebx
  mov    %rsi,%rbp
  mov    %edx,%r13d
  mov    %ecx,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  cmp    $0x80,%ebx
  je     L18
  mov    $0xffffffff,%eax
  cmp    $0x81,%ebx
  jne    L37
  movl   $0x81,(%rsp)
  mov    %r13d,%edi
  call   <htons@plt>
  mov    %rax,0x20(%rsp)
  mov    %r12d,0x18(%rsp)
  mov    %rsp,%rsi
  mov    $0x28,%edx
  mov    %rbp,%rdi
  call   <memcpy@plt>
L38:
  mov    $0x28,%eax
L37:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  movl   $0x80,0x4(%rsp)
  mov    %r13d,%edi
  call   <htons@plt>
  mov    %rax,0x10(%rsp)
  mov    %rsp,%rbx
  lea    0x8(%rsp),%rdi
  mov    $0x8,%edx
  mov    $0xffffffff,%esi
  call   <memset@plt>
  mov    0x8(%rsp),%rax
  mov    %r12d,0x10(%rax)
  mov    $0x28,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  jmp    L38
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
socklen_t anon_addr(int domain, struct sockaddr_storage *sst, unsigned int id, unsigned int id2) {
  memset(sst, 0, sizeof(*sst));
  socklen_t ret = -1;
  switch (domain) {
  case AF_INET:;
    struct sockaddr_in sin;
    sin.sin_family = AF_INET;
    sin.sin_port = htons(id);
    sin.sin_addr.s_addr = id2;
    memcpy(sst, &sin, sizeof(sin));
    ret = sizeof(sin);
    break;
  case AF_INET6:;
    struct sockaddr_in6 sin6;
    sin6.sin6_family = AF_INET6;
    sin6.sin6_port = htons(id);
    memset(&sin6.sin6_addr, -1, sizeof(sin6.sin6_addr));
    sin6.sin6_addr.s6_addr16[4] = id2;
    memcpy(sst, &sin6, sizeof(sin6));
    ret = sizeof(sin6);
    break;
  }
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<anon_addr>:
  endbr64
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  mov    $0x8,%edx
  push   %rbp
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %edi,%ebx
  mov    %rbp,%rdi
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <memset@plt>
  cmp    $0x80,%ebx
  je     L24
  mov    $0xffffffff,%eax
  cmp    $0x81,%ebx
  jne    L7
  mov    %r12d,%edi
  movl   $0x81,(%rsp)
  call   <htons@plt>
  mov    %rsp,%rsi
  mov    $0x28,%edx
  mov    %rbp,%rdi
  mov    %rax,0x20(%rsp)
  mov    %r13d,0x18(%rsp)
  call   <memcpy@plt>
L9:
  mov    $0x28,%eax
L7:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L24:
  mov    %r12d,%edi
  movl   $0x80,0x4(%rsp)
  mov    %rsp,%r12
  call   <htons@plt>
  lea    0x8(%rsp),%rdi
  mov    $0x8,%edx
  mov    $0xffffffff,%esi
  mov    %rax,0x10(%rsp)
  call   <memset@plt>
  mov    0x8(%rsp),%rax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x28,%edx
  mov    %r13d,0x10(%rax)
  call   <memcpy@plt>
  jmp    L9
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
socklen_t anon_addr(int domain, struct sockaddr_storage *sst, unsigned int id, unsigned int id2) {
  memset(sst, 0, sizeof(*sst));
  socklen_t ret = -1;
  switch (domain) {
  case AF_INET:;
    struct sockaddr_in sin;
    sin.sin_family = AF_INET;
    sin.sin_port = htons(id);
    sin.sin_addr.s_addr = id2;
    memcpy(sst, &sin, sizeof(sin));
    ret = sizeof(sin);
    break;
  case AF_INET6:;
    struct sockaddr_in6 sin6;
    sin6.sin6_family = AF_INET6;
    sin6.sin6_port = htons(id);
    memset(&sin6.sin6_addr, -1, sizeof(sin6.sin6_addr));
    sin6.sin6_addr.s6_addr16[4] = id2;
    memcpy(sst, &sin6, sizeof(sin6));
    ret = sizeof(sin6);
    break;
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<anon_addr>:
  endbr64
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  mov    $0x8,%edx
  push   %rbp
  mov    %rsi,%rbp
  xor    %esi,%esi
  push   %rbx
  mov    %edi,%ebx
  mov    %rbp,%rdi
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <memset@plt>
  cmp    $0x80,%ebx
  je     L22
  mov    $0xffffffff,%eax
  cmp    $0x81,%ebx
  jne    L11
  mov    %r12d,%edi
  movl   $0x81,(%rsp)
  call   <htons@plt>
  mov    %rsp,%rsi
  mov    $0x28,%edx
  mov    %rbp,%rdi
  mov    %rax,0x20(%rsp)
  mov    %r13d,0x18(%rsp)
  call   <memcpy@plt>
L3:
  mov    $0x28,%eax
L11:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L30
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L22:
  mov    %r12d,%edi
  movl   $0x80,0x4(%rsp)
  mov    %rsp,%r12
  call   <htons@plt>
  lea    0x8(%rsp),%rdi
  mov    $0x8,%edx
  mov    $0xffffffff,%esi
  mov    %rax,0x10(%rsp)
  call   <memset@plt>
  mov    0x8(%rsp),%rax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x28,%edx
  mov    %r13d,0x10(%rax)
  call   <memcpy@plt>
  jmp    L3
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
socklen_t anon_addr(int domain, struct sockaddr_storage *sst, unsigned int id, unsigned int id2) {
  memset(sst, 0, sizeof(*sst));
  socklen_t ret = -1;
  switch (domain) {
  case AF_INET:;
    struct sockaddr_in sin;
    sin.sin_family = AF_INET;
    sin.sin_port = htons(id);
    sin.sin_addr.s_addr = id2;
    memcpy(sst, &sin, sizeof(sin));
    ret = sizeof(sin);
    break;
  case AF_INET6:;
    struct sockaddr_in6 sin6;
    sin6.sin6_family = AF_INET6;
    sin6.sin6_port = htons(id);
    memset(&sin6.sin6_addr, -1, sizeof(sin6.sin6_addr));
    sin6.sin6_addr.s6_addr16[4] = id2;
    memcpy(sst, &sin6, sizeof(sin6));
    ret = sizeof(sin6);
    break;
  }
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memcpy_crc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  movb   $0x0,-0x12(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L3
L16:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  movzbl -0x11(%rbp),%eax
  add    %al,-0x12(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl -0x11(%rbp),%edx
  mov    %dl,(%rax)
L3:
  mov    -0x38(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x38(%rbp)
  test   %rax,%rax
  jne    L16
  movzbl -0x12(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char memcpy_crc(void *dest, const void *src, size_t n) {
  unsigned char sum = 0;
  unsigned char tmp;
  char *cs = (char *)src;
  char *cd = (char *)dest;
  while (n--) {
    tmp = *cs++;
    sum += tmp;
    *cd++ = tmp;
  }
  return sum;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<memcpy_crc>:
  endbr64
  test   %rdx,%rdx
  je     L25
  mov    $0x0,%eax
  mov    $0x0,%r8d
L21:
  movzbl (%rsi,%rax,1),%ecx
  add    %ecx,%r8d
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L21
L17:
  mov    %r8d,%eax
  ret
L25:
  mov    $0x0,%r8d
  jmp    L17
```

assistant: ```c
unsigned char memcpy_crc(void *dest, const void *src, size_t n) {
  unsigned char sum = 0;
  unsigned char tmp;
  char *cs = (char *)src;
  char *cd = (char *)dest;
  while (n--) {
    tmp = *cs++;
    sum += tmp;
    *cd++ = tmp;
  }
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<memcpy_crc>:
  endbr64
  test   %rdx,%rdx
  je     L17
  xor    %eax,%eax
  xor    %r8d,%r8d
  xchg   %ax,%ax
L7:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  add    %ecx,%r8d
  cmp    %rdx,%rax
  jne    L7
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned char memcpy_crc(void *dest, const void *src, size_t n) {
  unsigned char sum = 0;
  unsigned char tmp;
  char *cs = (char *)src;
  char *cd = (char *)dest;
  while (n--) {
    tmp = *cs++;
    sum += tmp;
    *cd++ = tmp;
  }
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<memcpy_crc>:
  endbr64
  mov    %rdi,%r9
  mov    %rsi,%rcx
  lea    -0x1(%rdx),%r10
  test   %rdx,%rdx
  je     L136
  lea    0x1(%rsi),%rsi
  mov    %rdi,%rax
  sub    %rsi,%rax
  cmp    $0xe,%rax
  jbe    L125
  cmp    $0x6,%r10
  jbe    L125
  cmp    $0xe,%r10
  jbe    L53
  mov    %rdx,%rsi
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  and    $0xfffffffffffffff0,%rsi
  nopl   0x0(%rax)
L37:
  movdqu (%rcx,%rax,1),%xmm2
  movups %xmm2,(%r9,%rax,1)
  add    $0x10,%rax
  paddb  %xmm2,%xmm0
  cmp    %rsi,%rax
  jne    L37
  movdqa %xmm0,%xmm1
  mov    %rdx,%rsi
  psrldq $0x8,%xmm1
  and    $0xfffffffffffffff0,%rsi
  paddb  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  lea    (%rcx,%rsi,1),%r8
  sub    %rsi,%r10
  psadbw %xmm1,%xmm0
  lea    (%r9,%rsi,1),%rdi
  movd   %xmm0,%eax
  cmp    %rsi,%rdx
  je     L142
  sub    %rsi,%rdx
  lea    -0x1(%rdx),%r11
  cmp    $0x6,%r11
  jbe    L130
L57:
  movq   (%rcx,%rsi,1),%xmm0
  pxor   %xmm1,%xmm1
  movdqa %xmm0,%xmm4
  movq   %xmm0,(%r9,%rsi,1)
  psadbw %xmm1,%xmm4
  movq   %xmm4,%rcx
  add    %ecx,%eax
  mov    %rdx,%rcx
  and    $0xfffffffffffffff8,%rcx
  add    %rcx,%r8
  add    %rcx,%rdi
  sub    %rcx,%r10
  cmp    %rcx,%rdx
  je     L142
L130:
  movzbl (%r8),%edx
  mov    %dl,(%rdi)
  add    %edx,%eax
  test   %r10,%r10
  je     L142
  movzbl 0x1(%r8),%edx
  mov    %dl,0x1(%rdi)
  add    %edx,%eax
  cmp    $0x1,%r10
  je     L142
  movzbl 0x2(%r8),%edx
  mov    %dl,0x2(%rdi)
  add    %edx,%eax
  cmp    $0x2,%r10
  je     L142
  movzbl 0x3(%r8),%edx
  mov    %dl,0x3(%rdi)
  add    %edx,%eax
  cmp    $0x3,%r10
  je     L142
  movzbl 0x4(%r8),%edx
  mov    %dl,0x4(%rdi)
  add    %edx,%eax
  cmp    $0x4,%r10
  je     L142
  movzbl 0x5(%r8),%edx
  mov    %dl,0x5(%rdi)
  add    %edx,%eax
  cmp    $0x5,%r10
  je     L142
  movzbl 0x6(%r8),%edx
  mov    %dl,0x6(%rdi)
  add    %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L136:
  xor    %eax,%eax
L142:
  ret
  nopl   0x0(%rax,%rax,1)
L125:
  xor    %esi,%esi
  xor    %eax,%eax
  nopl   0x0(%rax)
L46:
  movzbl (%rcx,%rsi,1),%edi
  mov    %dil,(%r9,%rsi,1)
  add    $0x1,%rsi
  add    %edi,%eax
  cmp    %rsi,%rdx
  jne    L46
  ret
L53:
  mov    %rcx,%r8
  xor    %esi,%esi
  xor    %eax,%eax
  jmp    L57
```

assistant: ```c
unsigned char memcpy_crc(void *dest, const void *src, size_t n) {
  unsigned char sum = 0;
  unsigned char tmp;
  char *cs = (char *)src;
  char *cd = (char *)dest;
  while (n--) {
    tmp = *cs++;
    sum += tmp;
    *cd++ = tmp;
  }
  return sum;
}
```