ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ENGINE_register_all_pkey_asn1_meths>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <ENGINE_get_first@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L15
L10:
  mov    %rbx,%rdi
  call   <ENGINE_register_pkey_asn1_meths@plt>
  mov    %rbx,%rdi
  call   <ENGINE_get_next@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L10
L15:
  pop    %rbx
  ret
```

assistant: ```c
void ENGINE_register_all_pkey_asn1_meths(void) {
  ENGINE *e;
  for (e = ENGINE_get_first(); e; e = ENGINE_get_next(e))
    ENGINE_register_pkey_asn1_meths(e);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ENGINE_register_all_pkey_asn1_meths>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <ENGINE_get_first@plt>
  test   %rax,%rax
  je     L15
  mov    %rax,%rbp
  nopl   0x0(%rax)
L10:
  mov    %rbp,%rdi
  call   <ENGINE_register_pkey_asn1_meths@plt>
  mov    %rbp,%rdi
  call   <ENGINE_get_next@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L10
L15:
  pop    %rbp
  ret
```

assistant: ```c
void ENGINE_register_all_pkey_asn1_meths(void) {
  ENGINE *e;
  for (e = ENGINE_get_first(); e; e = ENGINE_get_next(e))
    ENGINE_register_pkey_asn1_meths(e);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ENGINE_register_all_pkey_asn1_meths>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <ENGINE_get_first@plt>
  test   %rax,%rax
  je     L1
  mov    %rax,%rbp
  nopl   0x0(%rax)
L19:
  mov    %rbp,%rdi
  call   <ENGINE_register_pkey_asn1_meths@plt>
  mov    %rbp,%rdi
  call   <ENGINE_get_next@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L19
L1:
  pop    %rbp
  ret
```

assistant: ```c
void ENGINE_register_all_pkey_asn1_meths(void) {
  ENGINE *e;
  for (e = ENGINE_get_first(); e; e = ENGINE_get_next(e))
    ENGINE_register_pkey_asn1_meths(e);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<trio_equal_case_max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  cmpq   $0x0,-0x8(%rbp)
  je     L3
  cmpq   $0x0,-0x18(%rbp)
  je     L3
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L16
L3:
  mov    $0x0,%eax
L16:
  leave
  ret
```

assistant: ```c
int trio_equal_case_max(const char *first, size_t max, const char *second) {
  assert(first);
  assert(second);
  if ((first != 0) && (second != 0)) {
    return (0 == strncmp(first, second, max));
  }
  return (1 == 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<trio_equal_case_max>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%rbp
  call   <assert@plt>
  mov    %rbp,%rdi
  call   <assert@plt>
  test   %rbx,%rbx
  je     L23
  test   %rbp,%rbp
  jne    L3
L23:
  mov    $0x0,%eax
L14:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L14
```

assistant: ```c
int trio_equal_case_max(const char *first, size_t max, const char *second) {
  assert(first);
  assert(second);
  if ((first != 0) && (second != 0)) {
    return (0 == strncmp(first, second, max));
  }
  return (1 == 0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<trio_equal_case_max>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <assert@plt>
  mov    %r12,%rdi
  call   <assert@plt>
  test   %rbp,%rbp
  je     L0
  test   %r12,%r12
  jne    L9
L0:
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L9:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  pop    %rbp
  pop    %r12
  test   %rax,%rax
  pop    %r13
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int trio_equal_case_max(const char *first, size_t max, const char *second) {
  assert(first);
  assert(second);
  if ((first != 0) && (second != 0)) {
    return (0 == strncmp(first, second, max));
  }
  return (1 == 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<trio_equal_case_max>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <assert@plt>
  mov    %r12,%rdi
  call   <assert@plt>
  test   %rbp,%rbp
  je     L3
  test   %r12,%r12
  jne    L5
L3:
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L5:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  pop    %rbp
  pop    %r12
  test   %rax,%rax
  pop    %r13
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int trio_equal_case_max(const char *first, size_t max, const char *second) {
  assert(first);
  assert(second);
  if ((first != 0) && (second != 0)) {
    return (0 == strncmp(first, second, max));
  }
  return (1 == 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
<outBuffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8d,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L34
L24:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpl   $0x0,-0x30(%rbp)
  je     L15
  cmpb   $0x60,-0x9(%rbp)
  jle    L15
  cmpb   $0x7a,-0x9(%rbp)
  jg     L15
  movzbl -0x9(%rbp),%eax
  sub    $0x20,%eax
  mov    %al,-0x9(%rbp)
L15:
  movsbl -0x9(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rcx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   *%rcx
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L34:
  mov    -0x4(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L24
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
unsigned outBuffer(void (*myoutchar)(void *arg, char), void *arg, const char *buffer, int len, unsigned toupper) {
  unsigned count = 0;
  int i;
  char c;
  for (i = 0; i < len; i++) {
    c = buffer[i];
    if (toupper && (c >= 'a' && c <= 'z')) {
      c = (char)(c - ('a' - 'A'));
    }
    (*myoutchar)(arg, c);
    count++;
  }
  return count;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<outBuffer>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %ecx,%ecx
  jle    L7
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %r8d,%r13d
  mov    %rdx,%rbx
  mov    %ecx,%r15d
  lea    -0x1(%rcx),%eax
  lea    0x1(%rdx,%rax,1),%r14
  jmp    L10
L4:
  movsbl %sil,%esi
  mov    %r12,%rdi
  call   *%rbp
  add    $0x1,%rbx
  cmp    %r14,%rbx
  je     L26
L10:
  movzbl (%rbx),%esi
  test   %r13d,%r13d
  je     L4
  lea    -0x61(%rsi),%edx
  lea    -0x20(%rsi),%eax
  cmp    $0x1a,%dl
  cmovb  %eax,%esi
  jmp    L4
L7:
  mov    $0x0,%r15d
L26:
  mov    %r15d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned outBuffer(void (*myoutchar)(void *arg, char), void *arg, const char *buffer, int len, unsigned toupper) {
  unsigned count = 0;
  int i;
  char c;
  for (i = 0; i < len; i++) {
    c = buffer[i];
    if (toupper && (c >= 'a' && c <= 'z')) {
      c = (char)(c - ('a' - 'A'));
    }
    (*myoutchar)(arg, c);
    count++;
  }
  return count;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<outBuffer>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %ecx,%ecx
  jle    L8
  lea    -0x1(%rcx),%eax
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %ecx,%r14d
  mov    %r8d,%r12d
  mov    %rdx,%r15
  lea    0x1(%rdx,%rax,1),%r13
  nopl   (%rax)
L25:
  movzbl (%r15),%esi
  test   %r12d,%r12d
  je     L16
  lea    -0x61(%rsi),%edx
  lea    -0x20(%rsi),%eax
  cmp    $0x1a,%dl
  cmovb  %eax,%esi
L16:
  add    $0x1,%r15
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   *%rbx
  cmp    %r13,%r15
  jne    L25
L13:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L8:
  xor    %r14d,%r14d
  jmp    L13
```

assistant: ```c
unsigned outBuffer(void (*myoutchar)(void *arg, char), void *arg, const char *buffer, int len, unsigned toupper) {
  unsigned count = 0;
  int i;
  char c;
  for (i = 0; i < len; i++) {
    c = buffer[i];
    if (toupper && (c >= 'a' && c <= 'z')) {
      c = (char)(c - ('a' - 'A'));
    }
    (*myoutchar)(arg, c);
    count++;
  }
  return count;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<outBuffer>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %ecx,%ecx
  jle    L0
  lea    -0x1(%rcx),%eax
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%rbx
  mov    %ecx,%r13d
  lea    0x1(%rdx,%rax,1),%r14
  test   %r8d,%r8d
  je     L45
  nopl   0x0(%rax)
L43:
  movsbl (%rbx),%esi
  lea    -0x61(%rsi),%eax
  cmp    $0x19,%al
  ja     L8
  sub    $0x20,%esi
  movzbl %sil,%esi
L8:
  add    $0x1,%rbx
  mov    %r12,%rdi
  call   *%rbp
  cmp    %r14,%rbx
  jne    L43
L3:
  pop    %rbx
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L45:
  movsbl (%rbx),%esi
  add    $0x1,%rbx
  mov    %r12,%rdi
  call   *%rbp
  cmp    %r14,%rbx
  jne    L45
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L0:
  xor    %r13d,%r13d
  pop    %rbx
  pop    %rbp
  mov    %r13d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
unsigned outBuffer(void (*myoutchar)(void *arg, char), void *arg, const char *buffer, int len, unsigned toupper) {
  unsigned count = 0;
  int i;
  char c;
  for (i = 0; i < len; i++) {
    c = buffer[i];
    if (toupper && (c >= 'a' && c <= 'z')) {
      c = (char)(c - ('a' - 'A'));
    }
    (*myoutchar)(arg, c);
    count++;
  }
  return count;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bbm_com_hostif_select>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  movzbl -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <bbm_hostif_select@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int bbm_com_hostif_select(void *handle, unsigned char hostif) {
  int res;
  res = bbm_hostif_select(handle, hostif);
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bbm_com_hostif_select>:
  endbr64
  sub    $0x8,%rsp
  movzbl %sil,%esi
  call   <bbm_hostif_select@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int bbm_com_hostif_select(void *handle, unsigned char hostif) {
  int res;
  res = bbm_hostif_select(handle, hostif);
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bbm_com_hostif_select>:
  endbr64
  movzbl %sil,%esi
  jmp    1050 <bbm_hostif_select@plt>
```

assistant: ```c
int bbm_com_hostif_select(void *handle, unsigned char hostif) {
  int res;
  res = bbm_hostif_select(handle, hostif);
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bbm_com_hostif_select>:
  endbr64
  movzbl %sil,%esi
  jmp    1050 <bbm_hostif_select@plt>
```

assistant: ```c
int bbm_com_hostif_select(void *handle, unsigned char hostif) {
  int res;
  res = bbm_hostif_select(handle, hostif);
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<LLdiv10>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x3,-0x20(%rbp)
  jmp    L23
L15:
  mov    -0x1c(%rbp),%eax
  shl    $0x10,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  cltq
  movzwl -0x10(%rbp,%rax,2),%eax
  movzwl %ax,%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%edx
  mov    %edx,%ecx
  mov    $0xcccccccd,%eax
  imul   %rcx,%rax
  shr    $0x20,%rax
  mov    %eax,%ecx
  shr    $0x3,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%eax
  cltq
  mov    %dx,-0x10(%rbp,%rax,2)
  subl   $0x1,-0x20(%rbp)
L23:
  cmpl   $0x0,-0x20(%rbp)
  jns    L15
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x1c(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
void LLdiv10(unsigned long long a, unsigned long long *q, int *r) {
  union {
    long long value;
    unsigned short s[4];
  } LL;
  int i;
  unsigned int cur, sq, sr = 0;
  LL.value = a;
  for (i = 3; i >= 0; i--) {
    cur = sr * 0x10000 + LL.s[i];
    sq = cur / 10;
    sr = cur % 10;
    LL.s[i] = sq;
  }
  *q = LL.value;
  *r = sr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<LLdiv10>:
  endbr64
  sub    $0x18,%rsp
  mov    %rsi,%r11
  mov    %rdx,%r10
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,(%rsp)
  mov    %rsp,%rdx
  lea    -0x8(%rsp),%r9
  mov    $0xcccccccd,%r8d
L14:
  shl    $0x10,%eax
  movzwl 0x6(%rdx),%ecx
  add    %eax,%ecx
  mov    %ecx,%eax
  imul   %r8,%rax
  shr    $0x23,%rax
  mov    %eax,%edi
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  mov    %ecx,%eax
  sub    %esi,%eax
  mov    %di,0x6(%rdx)
  sub    $0x2,%rdx
  cmp    %r9,%rdx
  jne    L14
  mov    (%rsp),%rdx
  mov    %rdx,(%r11)
  mov    %eax,(%r10)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void LLdiv10(unsigned long long a, unsigned long long *q, int *r) {
  union {
    long long value;
    unsigned short s[4];
  } LL;
  int i;
  unsigned int cur, sq, sr = 0;
  LL.value = a;
  for (i = 3; i >= 0; i--) {
    cur = sr * 0x10000 + LL.s[i];
    sq = cur / 10;
    sr = cur % 10;
    LL.s[i] = sq;
  }
  *q = LL.value;
  *r = sr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<LLdiv10>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdx,%r10
  mov    %rsi,%r11
  mov    $0xcccccccd,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,(%rsp)
  mov    %rsp,%rdx
  lea    -0x8(%rsp),%r9
L15:
  movzwl 0x6(%rdx),%ecx
  shl    $0x10,%eax
  sub    $0x2,%rdx
  add    %ecx,%eax
  mov    %rax,%rcx
  imul   %r8,%rax
  shr    $0x23,%rax
  mov    %eax,%esi
  lea    (%rax,%rax,4),%edi
  mov    %ecx,%eax
  add    %edi,%edi
  mov    %si,0x8(%rdx)
  sub    %edi,%eax
  cmp    %rdx,%r9
  jne    L15
  mov    (%rsp),%rdx
  mov    %rdx,(%r11)
  mov    %eax,(%r10)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x18,%rsp
  ret
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void LLdiv10(unsigned long long a, unsigned long long *q, int *r) {
  union {
    long long value;
    unsigned short s[4];
  } LL;
  int i;
  unsigned int cur, sq, sr = 0;
  LL.value = a;
  for (i = 3; i >= 0; i--) {
    cur = sr * 0x10000 + LL.s[i];
    sq = cur / 10;
    sr = cur % 10;
    LL.s[i] = sq;
  }
  *q = LL.value;
  *r = sr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<LLdiv10>:
  endbr64
  mov    %rdi,%rax
  mov    $0xcccccccd,%r10d
  mov    %rdx,%r8
  mov    %rdi,%rcx
  shr    $0x30,%rax
  mov    %rax,%rdx
  mov    %eax,%eax
  imul   %r10,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edi
  movzwl %ax,%eax
  add    %edi,%edi
  shl    $0x10,%rax
  sub    %edi,%edx
  mov    %rcx,%rdi
  shr    $0x20,%rdi
  shl    $0x10,%edx
  movzwl %di,%edi
  add    %edx,%edi
  mov    %rdi,%rdx
  imul   %r10,%rdi
  shr    $0x23,%rdi
  lea    (%rdi,%rdi,4),%r9d
  movzwl %di,%edi
  add    %r9d,%r9d
  or     %rdi,%rax
  sub    %r9d,%edx
  mov    %ecx,%r9d
  movzwl %cx,%ecx
  shl    $0x10,%rax
  shr    $0x10,%r9d
  shl    $0x10,%edx
  add    %r9d,%edx
  mov    %rdx,%r9
  imul   %r10,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%r11d
  movzwl %dx,%edx
  add    %r11d,%r11d
  or     %rdx,%rax
  sub    %r11d,%r9d
  shl    $0x10,%rax
  shl    $0x10,%r9d
  add    %ecx,%r9d
  mov    %r9,%rcx
  imul   %r10,%r9
  shr    $0x23,%r9
  movzwl %r9w,%edx
  or     %rdx,%rax
  mov    %rax,(%rsi)
  lea    (%r9,%r9,4),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,(%r8)
  ret
```

assistant: ```c
void LLdiv10(unsigned long long a, unsigned long long *q, int *r) {
  union {
    long long value;
    unsigned short s[4];
  } LL;
  int i;
  unsigned int cur, sq, sr = 0;
  LL.value = a;
  for (i = 3; i >= 0; i--) {
    cur = sr * 0x10000 + LL.s[i];
    sq = cur / 10;
    sr = cur % 10;
    LL.s[i] = sq;
  }
  *q = LL.value;
  *r = sr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <is_valid_password@plt>
  test   %rax,%rax
  je     L18
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L25
L18:
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L25:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
type: string
name: extract_symbolic_data
arguments: tag: D23
type: string
name: extract_symbolic_data
arguments: tag: D7
type: string

tool: D15: %s 

tool: D23: is a valid password.

tool: D7: is NOT a valid password.

assistant: ```c
int main() {
  char s[100];
  gets(s);
  printf("%s ", s);
  if (is_valid_password(s))
    printf("is a valid password.");
  else
    printf("is NOT a valid password.");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <gets@plt>
  mov    %rbx,%rsi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <is_valid_password@plt>
  test   %rax,%rax
  je     L18
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L9:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x70,%rsp
  pop    %rbx
  ret
L18:
  lea    D27(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L9
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D5
datatype: string
name: retrieve_data
arguments: label: D3
datatype: string
name: retrieve_data
arguments: label: D27
datatype: string

tool: D5: %s 

tool: D3: is a valid password.

tool: D27: is NOT a valid password.

assistant: ```c
int main() {
  char s[100];
  gets(s);
  printf("%s ", s);
  if (is_valid_password(s))
    printf("is a valid password.");
  else
    printf("is NOT a valid password.");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <gets@plt>
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  mov    %rbp,%rsi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <is_valid_password@plt>
  test   %rax,%rax
  je     L5
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L10:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L10
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D3
data_type: string
name: parse_data
arguments: tag: D14
data_type: string
name: parse_data
arguments: tag: D1
data_type: string

tool: D3: %s 

tool: D14: is a valid password.

tool: D1: is NOT a valid password.

assistant: ```c
int main() {
  char s[100];
  gets(s);
  printf("%s ", s);
  if (is_valid_password(s))
    printf("is a valid password.");
  else
    printf("is NOT a valid password.");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <gets@plt>
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  mov    %rbp,%rsi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <is_valid_password@plt>
  test   %rax,%rax
  je     L17
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L13:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L17:
  lea    D25(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L13
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D13
data_type: string
name: retrieve_data
arguments: tag: D29
data_type: string
name: retrieve_data
arguments: tag: D25
data_type: string

tool: D13: %s 

tool: D29: is a valid password.

tool: D25: is NOT a valid password.

assistant: ```c
int main() {
  char s[100];
  gets(s);
  printf("%s ", s);
  if (is_valid_password(s))
    printf("is a valid password.");
  else
    printf("is NOT a valid password.");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<residual_jacobi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x1,-0x14(%rbp)
  jmp    L6
L11:
  movl   $0x1,-0x18(%rbp)
  jmp    L16
L23:
  mov    -0x14(%rbp),%eax
  imul   -0x38(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x14(%rbp),%eax
  imul   -0x38(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x18(%rbp)
L16:
  mov    -0x38(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x18(%rbp)
  jb     L23
  addl   $0x1,-0x14(%rbp)
L6:
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jb     L11
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double residual_jacobi(double *u, double *utmp, unsigned sizex, unsigned sizey) {
  unsigned i, j;
  double unew, diff, sum = 0.0;
  for (j = 1; j < sizex - 1; j++) {
    for (i = 1; i < sizey - 1; i++) {
      diff = utmp[i + j * sizey] - u[i + j * sizey];
      sum += diff * diff;
    }
  }
  return sum;
}
```