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
<main>:
  endbr64
  push   %r12
  mov    $0xd,%esi
  mov    $0x3c,%edi
  call   <bitOr@plt>
  mov    $0x3d,%ecx
  mov    $0xd,%edx
  mov    $0x3c,%esi
  mov    %eax,%r12d
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%ecx
  mov    $0xd,%edx
  mov    $0x3c,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D10
type_guessed: string
name: unpack_data
arguments: data_tag: D9
type_guessed: string

tool: D10: %d|%d should be %d


tool: D9: bitOr(%d, %d) -> %d


assistant: ```c
int main() {
  int a = 60;
  int b = 13;
  int c = bitOr(a, b);
  printf("%d|%d should be %d\n", a, b, 61);
  printf("bitOr(%d, %d) -> %d\n", a, b, c);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <peek@plt>
  cmp    $0x2f,%al
  jle    L69
  mov    $0x0,%eax
  call   <peek@plt>
  cmp    $0x39,%al
  jg     L69
  mov    $0x0,%eax
  call   <number@plt>
  jmp    L41
L69:
  mov    $0x0,%eax
  call   <peek@plt>
  cmp    $0x28,%al
  jne    L14
  mov    $0x0,%eax
  call   <get@plt>
  mov    $0x0,%eax
  call   <expression@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  call   <get@plt>
  mov    -0x4(%rbp),%eax
  jmp    L41
L14:
  mov    $0x0,%eax
  call   <peek@plt>
  cmp    $0x2d,%al
  jne    L16
  mov    $0x0,%eax
  call   <get@plt>
  mov    $0x0,%eax
  call   <expression@plt>
  neg    %eax
  jmp    L41
L16:
  mov    $0x0,%eax
L41:
  leave
  ret
```

assistant: ```c
int factor() {
  if (peek() >= '0' && peek() <= '9')
    return number();
  else if (peek() == '(') {
    get();
    int result = expression();
    get();
    return result;
  } else if (peek() == '-') {
    get();
    return -expression();
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<factor>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <peek@plt>
  cmp    $0x2f,%al
  jle    L45
  mov    $0x0,%eax
  call   <peek@plt>
  cmp    $0x39,%al
  jle    L6
L45:
  mov    $0x0,%eax
  call   <peek@plt>
  cmp    $0x28,%al
  je     L17
  mov    $0x0,%eax
  call   <peek@plt>
  mov    $0x0,%ebx
  cmp    $0x2d,%al
  je     L56
L31:
  mov    %ebx,%eax
  pop    %rbx
  ret
L6:
  mov    $0x0,%eax
  call   <number@plt>
  mov    %eax,%ebx
  jmp    L31
L17:
  mov    $0x0,%eax
  call   <get@plt>
  mov    $0x0,%eax
  call   <expression@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <get@plt>
  jmp    L31
L56:
  mov    $0x0,%eax
  call   <get@plt>
  mov    $0x0,%eax
  call   <expression@plt>
  neg    %eax
  mov    %eax,%ebx
  jmp    L31
```

assistant: ```c
int factor() {
  if (peek() >= '0' && peek() <= '9')
    return number();
  else if (peek() == '(') {
    get();
    int result = expression();
    get();
    return result;
  } else if (peek() == '-') {
    get();
    return -expression();
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<factor>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <peek@plt>
  cmp    $0x2f,%al
  jle    L16
  xor    %eax,%eax
  call   <peek@plt>
  cmp    $0x39,%al
  jle    L8
L16:
  xor    %eax,%eax
  call   <peek@plt>
  cmp    $0x28,%al
  mov    $0x0,%eax
  je     L12
  call   <peek@plt>
  xor    %r12d,%r12d
  cmp    $0x2d,%al
  je     L48
L36:
  mov    %r12d,%eax
  pop    %r12
  ret
  xchg   %ax,%ax
L8:
  xor    %eax,%eax
  pop    %r12
  jmp    10a0 <number@plt>
  nopl   0x0(%rax)
L12:
  call   <get@plt>
  xor    %eax,%eax
  call   <expression@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <get@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L48:
  xor    %eax,%eax
  call   <get@plt>
  xor    %eax,%eax
  call   <expression@plt>
  neg    %eax
  mov    %eax,%r12d
  jmp    L36
```

assistant: ```c
int factor() {
  if (peek() >= '0' && peek() <= '9')
    return number();
  else if (peek() == '(') {
    get();
    int result = expression();
    get();
    return result;
  } else if (peek() == '-') {
    get();
    return -expression();
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<factor>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <peek@plt>
  cmp    $0x2f,%al
  jle    L10
  xor    %eax,%eax
  call   <peek@plt>
  cmp    $0x39,%al
  jle    L9
L10:
  xor    %eax,%eax
  call   <peek@plt>
  cmp    $0x28,%al
  mov    $0x0,%eax
  je     L25
  call   <peek@plt>
  xor    %r12d,%r12d
  cmp    $0x2d,%al
  je     L19
L46:
  mov    %r12d,%eax
  pop    %r12
  ret
  xchg   %ax,%ax
L9:
  xor    %eax,%eax
  pop    %r12
  jmp    10a0 <number@plt>
  nopl   0x0(%rax)
L25:
  call   <get@plt>
  xor    %eax,%eax
  call   <expression@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <get@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L19:
  xor    %eax,%eax
  call   <get@plt>
  xor    %eax,%eax
  call   <expression@plt>
  neg    %eax
  mov    %eax,%r12d
  jmp    L46
```

assistant: ```c
int factor() {
  if (peek() >= '0' && peek() <= '9')
    return number();
  else if (peek() == '(') {
    get();
    int result = expression();
    get();
    return result;
  } else if (peek() == '-') {
    get();
    return -expression();
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test_short>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0xffff,-0xd6(%rbp)
  movw   $0xffff,-0xd4(%rbp)
  movw   $0x0,-0xd8(%rbp)
  jmp    L22
L55:
  movzwl -0xd6(%rbp),%edx
  movzwl -0xd4(%rbp),%eax
  add    %edx,%eax
  mov    %ax,-0xd2(%rbp)
  movw   $0x0,-0xda(%rbp)
  jmp    L63
L11:
  movswl -0xd8(%rbp),%eax
  imul   $0xffffffec,%eax,%edx
  movswl -0xda(%rbp),%eax
  add    %edx,%eax
  cltq
  movzwl -0xd2(%rbp),%edx
  mov    %dx,-0xd0(%rbp,%rax,2)
  movzwl -0xda(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0xda(%rbp)
L63:
  cmpw   $0x13,-0xda(%rbp)
  jle    L11
  movswl -0xd8(%rbp),%edx
  mov    $0x5,%eax
  sub    %edx,%eax
  cltq
  movzwl -0xd2(%rbp),%edx
  mov    %dx,-0xd0(%rbp,%rax,2)
  movzwl -0xd8(%rbp),%eax
  sub    $0x1,%eax
  mov    %ax,-0xd8(%rbp)
L22:
  cmpw   $0xfffc,-0xd8(%rbp)
  jge    L55
  movw   $0x0,-0xda(%rbp)
  jmp    L1
L30:
  movswl -0xda(%rbp),%eax
  cltq
  movzwl -0xd0(%rbp,%rax,2),%eax
  cmp    $0xfffe,%ax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movzwl -0xda(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0xda(%rbp)
L1:
  cmpw   $0x63,-0xda(%rbp)
  jle    L30
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
void test_short() {
  signed short b[100];
  signed short i, j, x = -1, y = -1;
#pragma acc parallel loop copyout(b)
  for (j = 0; j > -5; j--) {
    short c = x + y;
#pragma acc loop vector
    for (i = 0; i < 20; i++)
      b[-j * 20 + i] = c;
    b[5 - j] = c;
  }
  for (i = 0; i < 100; i++)
    assert(b[i] == -2);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_short>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x28(%rsp),%rdx
  mov    %rbx,%rsi
  mov    $0x0,%ecx
L5:
  lea    -0x28(%rdx),%rax
L1:
  movw   $0xfffe,(%rax)
  add    $0x2,%rax
  cmp    %rdx,%rax
  jne    L1
  movw   $0xfffe,0xa(%rsi)
  add    $0x2,%rsi
  add    $0x14,%ecx
  add    $0x28,%rdx
  cmp    $0x64,%ecx
  jne    L5
  lea    0xc8(%rbx),%rbp
L30:
  cmpw   $0xfffe,(%rbx)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x2,%rbx
  cmp    %rbx,%rbp
  jne    L30
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0xd8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_short() {
  signed short b[100];
  signed short i, j, x = -1, y = -1;
#pragma acc parallel loop copyout(b)
  for (j = 0; j > -5; j--) {
    short c = x + y;
#pragma acc loop vector
    for (i = 0; i < 20; i++)
      b[-j * 20 + i] = c;
    b[5 - j] = c;
  }
  for (i = 0; i < 100; i++)
    assert(b[i] == -2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_short>:
  endbr64
  push   %rbp
  xor    %ecx,%ecx
  push   %rbx
  sub    $0xd8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x28(%rsp),%rdx
  mov    %rbx,%rsi
L39:
  lea    -0x28(%rdx),%rax
  nopl   0x0(%rax)
L3:
  mov    $0xfffffffe,%edi
  add    $0x2,%rax
  mov    %di,-0x2(%rax)
  cmp    %rdx,%rax
  jne    L3
  mov    $0xfffffffe,%edx
  add    $0x14,%ecx
  add    $0x2,%rsi
  mov    %dx,0x8(%rsi)
  lea    0x28(%rax),%rdx
  cmp    $0x64,%ecx
  jne    L39
  lea    0xc8(%rbx),%rbp
  nopw   0x0(%rax,%rax,1)
L38:
  xor    %edi,%edi
  cmpw   $0xfffe,(%rbx)
  sete   %dil
  add    $0x2,%rbx
  call   <assert@plt>
  cmp    %rbx,%rbp
  jne    L38
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0xd8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_short() {
  signed short b[100];
  signed short i, j, x = -1, y = -1;
#pragma acc parallel loop copyout(b)
  for (j = 0; j > -5; j--) {
    short c = x + y;
#pragma acc loop vector
    for (i = 0; i < 20; i++)
      b[-j * 20 + i] = c;
    b[5 - j] = c;
  }
  for (i = 0; i < 100; i++)
    assert(b[i] == -2);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test_short>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0xd8,%rsp
  movdqa D19(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  mov    D19(%rip),%rax
  mov    %rsp,%rbx
  lea    0xc8(%rsp),%rbp
  movaps %xmm0,(%rsp)
  mov    %rax,0x20(%rsp)
  mov    %rax,0x48(%rsp)
  mov    %rax,0x70(%rsp)
  mov    %rax,0x98(%rsp)
  mov    %rax,0xc0(%rsp)
  movaps %xmm0,0x10(%rsp)
  movups %xmm0,0x28(%rsp)
  movups %xmm0,0x38(%rsp)
  movaps %xmm0,0x50(%rsp)
  movaps %xmm0,0x60(%rsp)
  movups %xmm0,0x78(%rsp)
  movups %xmm0,0x88(%rsp)
  movaps %xmm0,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  nopl   0x0(%rax,%rax,1)
L8:
  xor    %edi,%edi
  cmpw   $0xfffe,(%rbx)
  sete   %dil
  add    $0x2,%rbx
  call   <assert@plt>
  cmp    %rbx,%rbp
  jne    L8
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0xd8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D19
type: byte[16]

tool: {"D19": [0xfe,0xff,0xfe,0xff,0xfe,0xff,0xfe,0xff,0xfe,0xff,0xfe,0xff,0xfe,0xff,0xfe,0xff]}

assistant: ```c
void test_short() {
  signed short b[100];
  signed short i, j, x = -1, y = -1;
#pragma acc parallel loop copyout(b)
  for (j = 0; j > -5; j--) {
    short c = x + y;
#pragma acc loop vector
    for (i = 0; i < 20; i++)
      b[-j * 20 + i] = c;
    b[5 - j] = c;
  }
  for (i = 0; i < 100; i++)
    assert(b[i] == -2);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rBve99009253050923_br_build_path>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L30
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L30
  mov    -0x28(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x1,-0x1c(%rbp)
  jmp    L13
L30:
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
L13:
  mov    -0x30(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %rax,-0x8(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L31
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L31:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
datatype: string

tool: D8: /

assistant: ```c
char *rBve99009253050923_br_build_path(const char *dir, const char *file) {
  char *dir2, *result;
  size_t len;
  int must_free = 0;
  len = strlen(dir);
  if (len > 0 && dir[len - 1] != '/') {
    dir2 = rBve99009253050923_br_strcat(dir, "/");
    must_free = 1;
  } else
    dir2 = (char *)dir;
  result = rBve99009253050923_br_strcat(dir2, file);
  if (must_free)
    free(dir2);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rBve99009253050923_br_build_path>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  test   %rax,%rax
  je     L27
  cmpb   $0x2f,-0x1(%rbx,%rax,1)
  jne    L2
L27:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %rax,%rbx
L6:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  lea    D1(%rip),%rsi
  mov    %rbx,%rdi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %rax,%r12
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %rax,%rbx
  mov    %r12,%rdi
  call   <free@plt>
  jmp    L6
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type_infered: string

tool: D1: /

assistant: ```c
char *rBve99009253050923_br_build_path(const char *dir, const char *file) {
  char *dir2, *result;
  size_t len;
  int must_free = 0;
  len = strlen(dir);
  if (len > 0 && dir[len - 1] != '/') {
    dir2 = rBve99009253050923_br_strcat(dir, "/");
    must_free = 1;
  } else
    dir2 = (char *)dir;
  result = rBve99009253050923_br_strcat(dir2, file);
  if (must_free)
    free(dir2);
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rBve99009253050923_br_build_path>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %rax,%rax
  je     L14
  cmpb   $0x2f,-0x1(%rbp,%rax,1)
  jne    L16
L14:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <rBve99009253050923_br_strcat@plt>
  nopl   0x0(%rax)
L16:
  mov    %rbp,%rdi
  lea    D0(%rip),%rsi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D0
type_guessed: string

tool: D0: /

assistant: ```c
char *rBve99009253050923_br_build_path(const char *dir, const char *file) {
  char *dir2, *result;
  size_t len;
  int must_free = 0;
  len = strlen(dir);
  if (len > 0 && dir[len - 1] != '/') {
    dir2 = rBve99009253050923_br_strcat(dir, "/");
    must_free = 1;
  } else
    dir2 = (char *)dir;
  result = rBve99009253050923_br_strcat(dir2, file);
  if (must_free)
    free(dir2);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rBve99009253050923_br_build_path>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %rax,%rax
  je     L14
  cmpb   $0x2f,-0x1(%rbp,%rax,1)
  jne    L18
L14:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1090 <rBve99009253050923_br_strcat@plt>
  nopl   0x0(%rax)
L18:
  mov    %rbp,%rdi
  lea    D3(%rip),%rsi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %r12,%rsi
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <rBve99009253050923_br_strcat@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: label: D3
type_guessed: string

tool: D3: /

assistant: ```c
char *rBve99009253050923_br_build_path(const char *dir, const char *file) {
  char *dir2, *result;
  size_t len;
  int must_free = 0;
  len = strlen(dir);
  if (len > 0 && dir[len - 1] != '/') {
    dir2 = rBve99009253050923_br_strcat(dir, "/");
    must_free = 1;
  } else
    dir2 = (char *)dir;
  result = rBve99009253050923_br_strcat(dir2, file);
  if (must_free)
    free(dir2);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sockaddr_in_new>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sockaddr_in_set@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
struct sockaddr_in *sockaddr_in_new(const char *ipaddr, int port) {
  struct sockaddr_in *sa = malloc(sizeof(*sa));
  sa = sockaddr_in_set(sa, ipaddr, port);
  return sa;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sockaddr_in_new>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %ebp,%edx
  mov    %rbx,%rsi
  call   <sockaddr_in_set@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
struct sockaddr_in *sockaddr_in_new(const char *ipaddr, int port) {
  struct sockaddr_in *sa = malloc(sizeof(*sa));
  sa = sockaddr_in_set(sa, ipaddr, port);
  return sa;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sockaddr_in_new>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x4,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  add    $0x8,%rsp
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <sockaddr_in_set@plt>
```

assistant: ```c
struct sockaddr_in *sockaddr_in_new(const char *ipaddr, int port) {
  struct sockaddr_in *sa = malloc(sizeof(*sa));
  sa = sockaddr_in_set(sa, ipaddr, port);
  return sa;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sockaddr_in_new>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x4,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  add    $0x8,%rsp
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <sockaddr_in_set@plt>
```

assistant: ```c
struct sockaddr_in *sockaddr_in_new(const char *ipaddr, int port) {
  struct sockaddr_in *sa = malloc(sizeof(*sa));
  sa = sockaddr_in_set(sa, ipaddr, port);
  return sa;
}
```
user: Translate the below assembly code into C programming language:

```asm
<max_heapify>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
L24:
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  add    %eax,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L3
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L3
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L10
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L10
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L10:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  je     L22
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L24
L22:
  nop
  nop
  leave
  ret
```

assistant: ```c
void max_heapify(int A[], int index, int heapsize) {
  int iMax, iLeft, iRight;
  while (1) {
    iMax = index;
    iLeft = (2 * index + 1);
    iRight = 2 * (index + 1);
    if (iLeft < heapsize && A[iLeft] > A[iMax])
      iMax = iLeft;
    if (iRight < heapsize && A[iRight] > A[iMax])
      iMax = iRight;
    if (iMax != index) {
      swap(&A[iMax], &A[index]);
      index = iMax;
    } else
      break;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<max_heapify>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%eax
  mov    %edx,%r12d
  jmp    L77
L65:
  movslq %ebx,%rsi
  mov    0x0(%rbp,%rcx,4),%edi
  cmp    %edi,0x0(%rbp,%rsi,4)
  cmovle %edx,%ebx
  jmp    L64
L41:
  cmp    %r12d,%ebx
  jge    L86
  movslq %ebx,%rdx
  movslq %eax,%rcx
  mov    0x0(%rbp,%rcx,4),%ecx
  cmp    %ecx,0x0(%rbp,%rdx,4)
  jle    L86
L64:
  cmp    %ebx,%eax
  je     L86
  cltq
  lea    0x0(%rbp,%rax,4),%rsi
  movslq %ebx,%rax
  lea    0x0(%rbp,%rax,4),%rdi
  call   <swap@plt>
  mov    %ebx,%eax
L77:
  lea    (%rax,%rax,1),%ebx
  lea    0x1(%rbx),%edx
  add    $0x2,%ebx
  cmp    %r12d,%edx
  jge    L41
  movslq %edx,%rcx
  movslq %eax,%rsi
  mov    0x0(%rbp,%rsi,4),%edi
  cmp    %edi,0x0(%rbp,%rcx,4)
  jle    L41
  cmp    %r12d,%ebx
  jl     L65
  mov    %edx,%ebx
  jmp    L64
L86:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void max_heapify(int A[], int index, int heapsize) {
  int iMax, iLeft, iRight;
  while (1) {
    iMax = index;
    iLeft = (2 * index + 1);
    iRight = 2 * (index + 1);
    if (iLeft < heapsize && A[iLeft] > A[iMax])
      iMax = iLeft;
    if (iRight < heapsize && A[iRight] > A[iMax])
      iMax = iRight;
    if (iMax != index) {
      swap(&A[iMax], &A[index]);
      index = iMax;
    } else
      break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<max_heapify>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  jmp    L46
L69:
  movslq %esi,%rax
  movslq %edx,%rcx
  mov    0x0(%rbp,%rax,4),%edi
  cmp    %edi,0x0(%rbp,%rcx,4)
  jle    L61
  cmp    %r12d,%ebx
  jl     L32
  mov    %edx,%ebx
L39:
  cmp    %ebx,%esi
  je     L65
  lea    0x0(%rbp,%rax,4),%rsi
  movslq %ebx,%rax
  lea    0x0(%rbp,%rax,4),%rdi
  call   <swap@plt>
  mov    %ebx,%esi
L46:
  lea    (%rsi,%rsi,1),%ebx
  lea    0x1(%rbx),%edx
  add    $0x2,%ebx
  cmp    %r12d,%edx
  jl     L69
L61:
  cmp    %r12d,%ebx
  jge    L65
  movslq %esi,%rax
  movslq %ebx,%rdx
  mov    0x0(%rbp,%rax,4),%ecx
  cmp    %ecx,0x0(%rbp,%rdx,4)
  jg     L39
L65:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L32:
  movslq %ebx,%rdi
  mov    0x0(%rbp,%rcx,4),%ecx
  cmp    %ecx,0x0(%rbp,%rdi,4)
  cmovle %edx,%ebx
  jmp    L39
```

assistant: ```c
void max_heapify(int A[], int index, int heapsize) {
  int iMax, iLeft, iRight;
  while (1) {
    iMax = index;
    iLeft = (2 * index + 1);
    iRight = 2 * (index + 1);
    if (iLeft < heapsize && A[iLeft] > A[iMax])
      iMax = iLeft;
    if (iRight < heapsize && A[iRight] > A[iMax])
      iMax = iRight;
    if (iMax != index) {
      swap(&A[iMax], &A[index]);
      index = iMax;
    } else
      break;
  }
}
```