ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<warnx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L2
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L2:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <vwarnx@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  vwarnx(NULL, fmt, args);
  va_end(args);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<warnx>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <vwarnx@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  vwarnx(NULL, fmt, args);
  va_end(args);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<warnx>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <vwarnx@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  vwarnx(NULL, fmt, args);
  va_end(args);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<warnx>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  mov    %rdi,%rsi
  push   %rbp
  mov    %rdi,%rbp
  mov    %r12d,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %r12d,%edx
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <vwarnx@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void warnx(const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  vwarnx(NULL, fmt, args);
  va_end(args);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L83
L45:
  lea    -0x19(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movzbl -0x19(%rbp),%eax
  cmp    $0x55,%al
  jne    L25
  addl   $0x1,-0x14(%rbp)
  jmp    L22
L25:
  subl   $0x1,-0x14(%rbp)
L22:
  cmpl   $0x0,-0x14(%rbp)
  jns    L17
  cmpl   $0x0,-0xc(%rbp)
  jne    L17
  addl   $0x1,-0x10(%rbp)
  movl   $0x1,-0xc(%rbp)
L17:
  cmpl   $0x0,-0x14(%rbp)
  js     L58
  cmpl   $0x1,-0xc(%rbp)
  jne    L58
  movl   $0x0,-0xc(%rbp)
L58:
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x18(%rbp)
L83:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jg     L45
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L56
  call   <__stack_chk_fail@plt>
L56:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D25
data_type: string
name: extract_symbolic_data
arguments: data_tag: D14
data_type: string

tool: D25: %d

tool: D14:  %c

assistant: ```c
int main() {
  int n, l = 0, cnt = 0, e = 0;
  scanf("%d", &n);
  char i;
  for (; n > 0; n--) {
    scanf(" %c", &i);
    if (i == 'U')
      l++;
    else
      l--;
    if (l < 0 && e == 0) {
      cnt++;
      e = 1;
    }
    if (l >= 0 && e == 1)
      e = 0;
  }
  printf("%d", cnt);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L84
  mov    $0x0,%ebp
  mov    $0x0,%r12d
  mov    $0x0,%ebx
  lea    D11(%rip),%r13
  jmp    L89
L72:
  add    $0x1,%ebx
  jmp    L93
L71:
  test   %ebx,%ebx
  js     L17
  cmp    $0x1,%ebp
  mov    $0x0,%eax
  cmove  %eax,%ebp
L17:
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jle    L26
L89:
  lea    0x3(%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmpb   $0x55,0x3(%rsp)
  je     L72
  sub    $0x1,%ebx
L93:
  test   %ebx,%ebx
  jns    L71
  test   %ebp,%ebp
  jne    L71
  add    $0x1,%r12d
  mov    $0x1,%ebp
  jmp    L17
L84:
  mov    $0x0,%r12d
L26:
  mov    %r12d,%esi
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L70
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L70:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D24
type_guessed: string
name: extract_symbolic_data
arguments: label: D11
type_guessed: string

tool: D24: %d

tool: D11:  %c

assistant: ```c
int main() {
  int n, l = 0, cnt = 0, e = 0;
  scanf("%d", &n);
  char i;
  for (; n > 0; n--) {
    scanf(" %c", &i);
    if (i == 'U')
      l++;
    else
      l--;
    if (l < 0 && e == 0) {
      cnt++;
      e = 1;
    }
    if (l >= 0 && e == 1)
      e = 0;
  }
  printf("%d", cnt);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D11(%rip),%r14
  push   %r13
  mov    %r14,%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L116
  xor    %r13d,%r13d
  xor    %ebp,%ebp
  lea    0x3(%rsp),%r12
  xor    %r15d,%r15d
  lea    D14(%rip),%rbx
  jmp    L70
  nopl   0x0(%rax,%rax,1)
L2:
  sub    $0x1,%r15d
  test   %r15d,%r15d
  jns    L14
L78:
  test   $0x1,%r13b
  jne    L111
  add    $0x1,%ebp
  mov    $0x1,%r13d
L71:
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jle    L106
L70:
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpb   $0x55,0x3(%rsp)
  jne    L2
  add    $0x1,%r15d
  test   %r15d,%r15d
  js     L78
L14:
  xor    %eax,%eax
  test   $0x1,%r13b
  cmovne %eax,%r13d
L47:
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L70
L106:
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L111:
  test   %r15d,%r15d
  js     L71
  xor    %eax,%eax
  test   $0x1,%r13b
  cmovne %eax,%r13d
  jmp    L47
L116:
  xor    %ebp,%ebp
  jmp    L106
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D11
data_type: string
name: extract_symbolic_data
arguments: data_label: D14
data_type: string

tool: D11: %d

tool: D14:  %c

assistant: ```c
int main() {
  int n, l = 0, cnt = 0, e = 0;
  scanf("%d", &n);
  char i;
  for (; n > 0; n--) {
    scanf(" %c", &i);
    if (i == 'U')
      l++;
    else
      l--;
    if (l < 0 && e == 0) {
      cnt++;
      e = 1;
    }
    if (l >= 0 && e == 1)
      e = 0;
  }
  printf("%d", cnt);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D14(%rip),%r14
  push   %r13
  mov    %r14,%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L111
  xor    %r13d,%r13d
  xor    %ebp,%ebp
  lea    0x3(%rsp),%r12
  xor    %r15d,%r15d
  lea    D16(%rip),%rbx
  jmp    L65
  nopl   0x0(%rax,%rax,1)
L5:
  sub    $0x1,%r15d
  test   %r15d,%r15d
  jns    L78
L46:
  test   $0x1,%r13b
  jne    L49
  add    $0x1,%ebp
  mov    $0x1,%r13d
L27:
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jle    L11
L65:
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpb   $0x55,0x3(%rsp)
  jne    L5
  add    $0x1,%r15d
  test   %r15d,%r15d
  js     L46
L78:
  xor    %eax,%eax
  test   $0x1,%r13b
  cmovne %eax,%r13d
L85:
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  test   %eax,%eax
  jg     L65
L11:
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L49:
  test   %r15d,%r15d
  js     L27
  xor    %eax,%eax
  test   $0x1,%r13b
  cmovne %eax,%r13d
  jmp    L85
L111:
  xor    %ebp,%ebp
  jmp    L11
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D14
data_type: string
name: read_data
arguments: label: D16
data_type: string

tool: D14: %d

tool: D16:  %c

assistant: ```c
int main() {
  int n, l = 0, cnt = 0, e = 0;
  scanf("%d", &n);
  char i;
  for (; n > 0; n--) {
    scanf(" %c", &i);
    if (i == 'U')
      l++;
    else
      l--;
    if (l < 0 && e == 0) {
      cnt++;
      e = 1;
    }
    if (l >= 0 && e == 1)
      e = 0;
  }
  printf("%d", cnt);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

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

assistant: ```c
int backtrace(void **buffer, int size) {
  void *ebp = __builtin_frame_address(0);
  return backtrace_np(ebp, buffer, size);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<backtrace>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %esi,%edx
  mov    %rdi,%rsi
  mov    %rbp,%rdi
  call   <backtrace_np@plt>
  pop    %rbp
  ret
```

assistant: ```c
int backtrace(void **buffer, int size) {
  void *ebp = __builtin_frame_address(0);
  return backtrace_np(ebp, buffer, size);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<backtrace>:
  endbr64
  push   %rbp
  mov    %esi,%edx
  mov    %rdi,%rsi
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1050 <backtrace_np@plt>
```

assistant: ```c
int backtrace(void **buffer, int size) {
  void *ebp = __builtin_frame_address(0);
  return backtrace_np(ebp, buffer, size);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<backtrace>:
  endbr64
  push   %rbp
  mov    %esi,%edx
  mov    %rdi,%rsi
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1050 <backtrace_np@plt>
```

assistant: ```c
int backtrace(void **buffer, int size) {
  void *ebp = __builtin_frame_address(0);
  return backtrace_np(ebp, buffer, size);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<product>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  cqto
  idivq  -0x8(%rbp)
  cmp    %rax,-0x10(%rbp)
  jle    L18
  mov    $0xffffffffffffffff,%rax
  jmp    L3
L18:
  mov    -0x8(%rbp),%rax
  imul   -0x10(%rbp),%rax
L3:
  pop    %rbp
  ret
```

assistant: ```c
long product(long a, long b) {
  if (b > LONG_MAX / a)
    return -1;
  return a * b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<product>:
  endbr64
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  cqto
  idiv   %rdi
  cmp    %rsi,%rax
  jl     L2
  mov    %rdi,%rax
  imul   %rsi,%rax
  ret
L2:
  mov    $0xffffffffffffffff,%rax
  ret
```

assistant: ```c
long product(long a, long b) {
  if (b > LONG_MAX / a)
    return -1;
  return a * b;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<product>:
  endbr64
  mov    D9(%rip),%rax
  mov    (%rax),%rax
  cqto
  idiv   %rdi
  cmp    %rsi,%rax
  jl     L0
  mov    %rdi,%rax
  imul   %rsi,%rax
  ret
L0:
  mov    $0xffffffffffffffff,%rax
  ret
```

assistant: ```c
long product(long a, long b) {
  if (b > LONG_MAX / a)
    return -1;
  return a * b;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<product>:
  endbr64
  mov    D8(%rip),%rax
  mov    (%rax),%rax
  cqto
  idiv   %rdi
  cmp    %rsi,%rax
  jl     L3
  mov    %rdi,%rax
  imul   %rsi,%rax
  ret
L3:
  mov    $0xffffffffffffffff,%rax
  ret
```

assistant: ```c
long product(long a, long b) {
  if (b > LONG_MAX / a)
    return -1;
  return a * b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test34>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L7
  mov    $0x3,%eax
  jmp    L18
L7:
  mov    $0x5,%eax
L18:
  pop    %rbp
  ret
```

assistant: ```c
int test34(int x) {
  if (x == 1) {
    return 3;
  } else if (x == 2 || 1) {
    return 5;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test34>:
  endbr64
  cmp    $0x1,%edi
  setne  %al
  movzbl %al,%eax
  lea    0x3(%rax,%rax,1),%eax
  ret
```

assistant: ```c
int test34(int x) {
  if (x == 1) {
    return 3;
  } else if (x == 2 || 1) {
    return 5;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<test34>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%edi
  setne  %al
  lea    0x3(%rax,%rax,1),%eax
  ret
```

assistant: ```c
int test34(int x) {
  if (x == 1) {
    return 3;
  } else if (x == 2 || 1) {
    return 5;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<test34>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%edi
  setne  %al
  lea    0x3(%rax,%rax,1),%eax
  ret
```

assistant: ```c
int test34(int x) {
  if (x == 1) {
    return 3;
  } else if (x == 2 || 1) {
    return 5;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<max_heapify>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  add    %eax,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jg     L33
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L33
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L51
L33:
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L51:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jg     L42
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L42
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L42:
  mov    -0x1c(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  je     L12
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%edx
  mov    -0x10(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <max_heapify>
L12:
  nop
  leave
  ret
```

assistant: ```c
void max_heapify(int data[], int i, int heap_size) {
  int l = (2 * i), r = (2 * i + 1), largest;
  if (l <= heap_size && data[l] > data[i]) {
    largest = l;
  } else {
    largest = i;
  }
  if (r <= heap_size && data[r] > data[largest]) {
    largest = r;
  }
  if (i != largest) {
    int temp = data[i];
    data[i] = data[largest];
    data[largest] = temp;
    max_heapify(data, largest, heap_size);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<max_heapify>:
  endbr64
  lea    (%rsi,%rsi,1),%eax
  lea    0x1(%rax),%r8d
  cmp    %edx,%eax
  jg     L64
  movslq %eax,%rcx
  movslq %esi,%r9
  mov    (%rdi,%r9,4),%r10d
  cmp    %r10d,(%rdi,%rcx,4)
  jle    L64
  cmp    %edx,%r8d
  jg     L11
  movslq %r8d,%r9
  mov    (%rdi,%rcx,4),%ecx
  cmp    %ecx,(%rdi,%r9,4)
  cmovle %eax,%r8d
  jmp    L10
L11:
  mov    %eax,%r8d
  jmp    L10
L64:
  cmp    %edx,%r8d
  jg     L61
  movslq %r8d,%rax
  movslq %esi,%rcx
  mov    (%rdi,%rcx,4),%ecx
  cmp    %ecx,(%rdi,%rax,4)
  jle    L34
L10:
  cmp    %esi,%r8d
  je     L61
  sub    $0x8,%rsp
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  mov    (%rcx),%esi
  movslq %r8d,%rax
  lea    (%rdi,%rax,4),%rax
  mov    (%rax),%r9d
  mov    %r9d,(%rcx)
  mov    %esi,(%rax)
  mov    %r8d,%esi
  call   <max_heapify>
  add    $0x8,%rsp
  ret
L34:
  ret
L61:
  ret
```

assistant: ```c
void max_heapify(int data[], int i, int heap_size) {
  int l = (2 * i), r = (2 * i + 1), largest;
  if (l <= heap_size && data[l] > data[i]) {
    largest = l;
  } else {
    largest = i;
  }
  if (r <= heap_size && data[r] > data[largest]) {
    largest = r;
  }
  if (i != largest) {
    int temp = data[i];
    data[i] = data[largest];
    data[largest] = temp;
    max_heapify(data, largest, heap_size);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<max_heapify>:
  endbr64
  mov    %rdi,%rcx
  jmp    L24
  nopl   0x0(%rax)
L45:
  movslq %esi,%r8
  movslq %edi,%r9
  mov    (%rcx,%r8,4),%r11d
  cmp    %r11d,(%rcx,%r9,4)
  jle    L7
  cmp    %edx,%eax
  jle    L19
  mov    %edi,%eax
L36:
  cmp    %esi,%eax
  je     L48
  movslq %eax,%rsi
  lea    (%rcx,%r8,4),%rdi
  lea    (%rcx,%rsi,4),%rsi
  mov    (%rdi),%r8d
  mov    (%rsi),%r9d
  mov    %r9d,(%rdi)
  mov    %r8d,(%rsi)
  mov    %eax,%esi
L24:
  lea    (%rsi,%rsi,1),%edi
  lea    0x1(%rdi),%eax
  cmp    %edx,%edi
  jle    L45
L7:
  cmp    %edx,%eax
  jg     L48
  movslq %esi,%r8
  movslq %eax,%rdi
  mov    (%rcx,%r8,4),%r10d
  cmp    %r10d,(%rcx,%rdi,4)
  jg     L36
L48:
  ret
L19:
  movslq %eax,%r10
  mov    (%rcx,%r9,4),%r11d
  cmp    %r11d,(%rcx,%r10,4)
  cmovle %edi,%eax
  jmp    L36
```

assistant: ```c
void max_heapify(int data[], int i, int heap_size) {
  int l = (2 * i), r = (2 * i + 1), largest;
  if (l <= heap_size && data[l] > data[i]) {
    largest = l;
  } else {
    largest = i;
  }
  if (r <= heap_size && data[r] > data[largest]) {
    largest = r;
  }
  if (i != largest) {
    int temp = data[i];
    data[i] = data[largest];
    data[largest] = temp;
    max_heapify(data, largest, heap_size);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<max_heapify>:
  endbr64
  mov    %rdi,%rcx
  jmp    L31
  nopl   0x0(%rax)
L47:
  movslq %eax,%r10
  movslq %esi,%r11
  lea    (%rcx,%r10,4),%rdi
  mov    (%rdi),%r8d
  cmp    (%rcx,%r11,4),%r8d
  jle    L57
  cmp    %edx,%r9d
  jle    L60
L78:
  cmp    %esi,%eax
  je     L5
  movslq %esi,%rsi
  lea    (%rcx,%rsi,4),%rsi
  mov    (%rsi),%r9d
  mov    %r8d,(%rsi)
  mov    %eax,%esi
  mov    %r9d,(%rdi)
L31:
  lea    (%rsi,%rsi,1),%eax
  lea    0x1(%rax),%r9d
  cmp    %edx,%eax
  jle    L47
L57:
  cmp    %edx,%r9d
  jg     L5
  mov    %esi,%eax
  movslq %esi,%r10
L60:
  movslq %r9d,%rdi
  lea    (%rcx,%r10,4),%r10
  lea    (%rcx,%rdi,4),%rdi
  mov    (%r10),%r11d
  mov    (%rdi),%r8d
  cmp    %r11d,%r8d
  jg     L14
  mov    %r11d,%r8d
  mov    %r10,%rdi
  jmp    L78
  nopl   0x0(%rax)
L14:
  mov    %r9d,%eax
  jmp    L78
  nopl   (%rax)
L5:
  ret
```

assistant: ```c
void max_heapify(int data[], int i, int heap_size) {
  int l = (2 * i), r = (2 * i + 1), largest;
  if (l <= heap_size && data[l] > data[i]) {
    largest = l;
  } else {
    largest = i;
  }
  if (r <= heap_size && data[r] > data[largest]) {
    largest = r;
  }
  if (i != largest) {
    int temp = data[i];
    data[i] = data[largest];
    data[largest] = temp;
    max_heapify(data, largest, heap_size);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mmul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    D15(%rip),%eax
  cmp    %eax,-0x24(%rbp)
  jle    L12
  mov    D46(%rip),%rax
  test   %rax,%rax
  je     L37
  mov    D46(%rip),%rax
  mov    %rax,%rdi
  call   <MemFree@plt>
L37:
  mov    -0x24(%rbp),%eax
  mov    %eax,D43(%rip)
  mov    D43(%rip),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <MemAlloc@plt>
  mov    %rax,D43(%rip)
L12:
  movl   $0x0,-0x8(%rbp)
  jmp    L11
L62:
  mov    D43(%rip),%rdx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x0,-0x4(%rbp)
  jmp    L47
L29:
  mov    D43(%rip),%rdx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x8(%rbp),%eax
  imul   -0x24(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm2,%xmm0
  mov    D43(%rip),%rdx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L47:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L29
  addl   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L62
  movl   $0x0,-0x8(%rbp)
  jmp    L54
L55:
  mov    D43(%rip),%rdx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L54:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L55
  nop
  nop
  leave
  ret
```

assistant: ```c
void mmul(double *a, double *b, int n) {
  int i, j;
  static double *r = 0;
  static int rsize = 0;
  if (rsize < n) {
    if (r)
      MemFree(r);
    rsize = n;
    r = (double *)MemAlloc(sizeof(double) * rsize);
  }
  for (i = 0; i < n; i++) {
    r[i] = 0.0;
    for (j = 0; j < n; j++)
      r[i] += a[i * n + j] * b[j];
  }
  for (i = 0; i < n; i++)
    b[i] = r[i];
}
```