ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D35(%rip),%rsi
  lea    D2(%rip),%rdi
  push   %r13
  lea    D38(%rip),%r14
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <fopen@plt>
  mov    %rax,%r13
  jmp    L63
  nopl   0x0(%rax,%rax,1)
L46:
  add    $0x1,%r12
  cmp    $0x3e8,%r12
  je     L26
L63:
  mov    %r13,%rdi
  lea    (%rbx,%r12,4),%rdx
  mov    %r14,%rsi
  mov    %r12d,%ebp
  call   <fscanf@plt>
  mov    %r13,%rdi
  call   <feof@plt>
  test   %rax,%rax
  je     L46
  lea    -0x1(%r12),%esi
  test   %r12d,%r12d
  je     L1
L103:
  movslq %esi,%rax
  lea    (%rbx,%rax,4),%r8
  nopl   0x0(%rax)
L80:
  mov    (%r8),%r10d
  lea    -0x1(%rsi),%edi
  xor    %eax,%eax
  mov    %rdi,%r9
  mov    %r10d,%ecx
  test   %esi,%esi
  jg     L36
  jmp    L68
  nopl   0x0(%rax)
L27:
  mov    %rdx,%rax
L36:
  mov    (%rbx,%rax,4),%edx
  cmp    %edx,%ecx
  jge    L89
  mov    %edx,%ecx
  mov    %eax,%esi
L89:
  lea    0x1(%rax),%rdx
  cmp    %rax,%rdi
  jne    L27
  movslq %esi,%rsi
  mov    %ecx,(%r8)
  mov    %r10d,(%rsp,%rsi,4)
L15:
  sub    $0x4,%r8
  mov    %r9d,%esi
  jmp    L80
  nopw   0x0(%rax,%rax,1)
L68:
  cmp    $0xffffffff,%r9d
  jne    L15
  xor    %r12d,%r12d
  lea    D22(%rip),%r13
L41:
  mov    (%rbx,%r12,4),%esi
  mov    %r13,%rdi
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12d,%ebp
  jg     L41
L1:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L57
  add    $0xfb0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L26:
  mov    $0x3e8,%ebp
  mov    $0x3e7,%esi
  jmp    L103
L57:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D35
data_type: string
name: read_data
arguments: data_label: D2
data_type: string
name: read_data
arguments: data_label: D38
data_type: string
name: read_data
arguments: data_label: D22
data_type: string

tool: D35: r

tool: D2: res/input.txt

tool: D38: %d

tool: D22: %d


assistant: ```c
int main(void) {
  FILE *fp = fopen("res/input.txt", "r");
  int data[1000], size;
  for (size = 0; size < 1000; ++size) {
    fscanf(fp, "%d", &data[size]);
    if (feof(fp))
      break;
  }
  for (int i = size - 1; i >= 0; --i) {
    int max = i;
    for (int j = 0; j < i; ++j)
      if (data[j] > data[max])
        max = j;
    int tmp = data[i];
    data[i] = data[max];
    data[max] = tmp;
  }
  for (int i = 0; i < size; ++i)
    printf("%d\n", data[i]);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L21:
  lea    -0x9(%rbp),%rax
  mov    $0x1,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <read@plt>
  cmp    $0x1,%eax
  je     L24
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
  jmp    L19
L24:
  movzbl -0x9(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  movzbl -0x9(%rbp),%eax
  cmp    $0xa,%al
  jne    L21
  mov    $0x0,%eax
L19:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L38
  call   <__stack_chk_fail@plt>
L38:
  leave
  ret
```

assistant: ```c
main(argc, argv) char **argv;
{
  char c;
  for (;;) {
    if (read(0, &c, 1) != 1) {
      putchar('\n');
      return 1;
    }
    putchar(c);
    if (c == '\n')
      return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rbx
L24:
  mov    $0x1,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <read@plt>
  cmp    $0x1,%eax
  jne    L30
  movsbl 0x7(%rsp),%edi
  call   <putchar@plt>
  cmpb   $0xa,0x7(%rsp)
  jne    L24
  mov    $0x0,%eax
  jmp    L27
L30:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
L27:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x10,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
main(argc, argv) char **argv;
{
  char c;
  for (;;) {
    if (read(0, &c, 1) != 1) {
      putchar('\n');
      return 1;
    }
    putchar(c);
    if (c == '\n')
      return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rbx
  jmp    L14
L13:
  movsbl 0x7(%rsp),%edi
  call   <putchar@plt>
  cmpb   $0xa,0x7(%rsp)
  je     L33
L14:
  xor    %edi,%edi
  xor    %eax,%eax
  mov    $0x1,%edx
  mov    %rbx,%rsi
  call   <read@plt>
  cmp    $0x1,%eax
  je     L13
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
L26:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x10,%rsp
  pop    %rbx
  ret
L33:
  xor    %eax,%eax
  jmp    L26
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
main(argc, argv) char **argv;
{
  char c;
  for (;;) {
    if (read(0, &c, 1) != 1) {
      putchar('\n');
      return 1;
    }
    putchar(c);
    if (c == '\n')
      return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%rbx
  jmp    L12
L43:
  movsbl 0x7(%rsp),%edi
  call   <putchar@plt>
  cmpb   $0xa,0x7(%rsp)
  je     L34
L12:
  xor    %edi,%edi
  xor    %eax,%eax
  mov    $0x1,%edx
  mov    %rbx,%rsi
  call   <read@plt>
  cmp    $0x1,%eax
  je     L43
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x1,%eax
L37:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x10,%rsp
  pop    %rbx
  ret
L34:
  xor    %eax,%eax
  jmp    L37
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
main(argc, argv) char **argv;
{
  char c;
  for (;;) {
    if (read(0, &c, 1) != 1) {
      putchar('\n');
      return 1;
    }
    putchar(c);
    if (c == '\n')
      return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<merge>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  mov    %r8d,-0x3c(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L77
L28:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x30(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x14(%rbp)
L77:
  mov    -0x14(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jle    L28
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x38(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L23
L29:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jg     L49
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  jmp    L23
L49:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
L23:
  mov    -0x10(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jg     L53
  mov    -0xc(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jle    L29
L53:
  mov    -0x38(%rbp),%eax
  sub    -0x10(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L79
L83:
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x14(%rbp)
L79:
  mov    -0x14(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L83
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void merge(int array[], int my[], int start, int middle, int end) {
  int i;
  int left, right;
  int current;
  int remaining;
  for (i = start; i <= end; i++)
    my[i] = array[i];
  current = start;
  left = start;
  right = middle + 1;
  while (left <= middle && right <= end) {
    if (my[left] <= my[right]) {
      array[current++] = my[left++];
    } else {
      array[current++] = my[right++];
    }
  }
  remaining = middle - left;
  for (i = 0; i <= remaining; i++)
    array[current + i] = my[left + i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<merge>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%r11
  mov    %ecx,%ebx
  cmp    %r8d,%edx
  jg     L13
  movslq %edx,%rax
L34:
  mov    (%r11,%rax,4),%ecx
  mov    %ecx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jge    L34
L13:
  lea    0x1(%rbx),%eax
  cmp    %ebx,%edx
  jg     L95
  cmp    %eax,%r8d
  jl     L95
  lea    0x1(%rdx),%edi
  movslq %edx,%rcx
  lea    (%r11,%rcx,4),%rcx
  jmp    L54
L17:
  add    $0x1,%eax
  mov    %edi,%ebp
  mov    %r9d,(%rcx)
L101:
  add    $0x1,%edi
  add    $0x4,%rcx
  cmp    %ebx,%edx
  jg     L47
  cmp    %r8d,%eax
  jg     L47
L54:
  movslq %edx,%r9
  mov    (%rsi,%r9,4),%r10d
  movslq %eax,%r9
  mov    (%rsi,%r9,4),%r9d
  cmp    %r9d,%r10d
  jg     L17
  add    $0x1,%edx
  mov    %edi,%ebp
  mov    %r10d,(%rcx)
  jmp    L101
L95:
  mov    %edx,%ebp
L47:
  sub    %edx,%ebx
  js     L100
  lea    0x1(%rbx),%ecx
  shl    $0x2,%rcx
  movslq %edx,%rdx
  lea    (%rsi,%rdx,4),%rdi
  movslq %ebp,%rbp
  lea    (%r11,%rbp,4),%rsi
  mov    $0x0,%eax
L45:
  mov    (%rdi,%rax,1),%edx
  mov    %edx,(%rsi,%rax,1)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L45
L100:
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void merge(int array[], int my[], int start, int middle, int end) {
  int i;
  int left, right;
  int current;
  int remaining;
  for (i = start; i <= end; i++)
    my[i] = array[i];
  current = start;
  left = start;
  right = middle + 1;
  while (left <= middle && right <= end) {
    if (my[left] <= my[right]) {
      array[current++] = my[left++];
    } else {
      array[current++] = my[right++];
    }
  }
  remaining = middle - left;
  for (i = 0; i <= remaining; i++)
    array[current + i] = my[left + i];
}
```
user: Translate the below assembly code into C programming language:

```asm
<merge>:
  endbr64
  push   %rbx
  mov    %rdi,%r10
  mov    %ecx,%r11d
  movslq %edx,%rax
  cmp    %r8d,%edx
  jg     L107
  nopl   0x0(%rax,%rax,1)
L76:
  mov    (%r10,%rax,4),%ecx
  mov    %ecx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r8d
  jge    L76
L107:
  lea    0x1(%r11),%edi
  cmp    %r11d,%edx
  jg     L81
  cmp    %edi,%r8d
  lea    0x1(%rdx),%eax
  setge  %cl
  cltq
  test   %cl,%cl
  jne    L99
  jmp    L81
  nopl   0x0(%rax)
L74:
  add    $0x1,%edx
  mov    %ecx,-0x4(%r10,%rax,4)
  add    $0x1,%rax
  cmp    %r11d,%edx
  jg     L32
L43:
  cmp    %r8d,%edi
  jg     L32
L99:
  movslq %edx,%rcx
  movslq %edi,%r9
  movslq %eax,%rbx
  mov    (%rsi,%rcx,4),%ecx
  mov    (%rsi,%r9,4),%r9d
  cmp    %r9d,%ecx
  jle    L74
  mov    %r9d,%ecx
  add    $0x1,%edi
  mov    %ecx,-0x4(%r10,%rax,4)
  add    $0x1,%rax
  cmp    %r11d,%edx
  jle    L43
L32:
  sub    %edx,%r11d
  js     L119
  movslq %r11d,%r11
  movslq %edx,%rdx
  lea    (%r10,%rbx,4),%rcx
  xor    %eax,%eax
  lea    0x4(,%r11,4),%rdi
  lea    (%rsi,%rdx,4),%rsi
  nopl   0x0(%rax)
L33:
  mov    (%rsi,%rax,1),%edx
  mov    %edx,(%rcx,%rax,1)
  add    $0x4,%rax
  cmp    %rdi,%rax
  jne    L33
L119:
  pop    %rbx
  ret
  nopl   0x0(%rax)
L81:
  movslq %edx,%rbx
  jmp    L32
```

assistant: ```c
void merge(int array[], int my[], int start, int middle, int end) {
  int i;
  int left, right;
  int current;
  int remaining;
  for (i = start; i <= end; i++)
    my[i] = array[i];
  current = start;
  left = start;
  right = middle + 1;
  while (left <= middle && right <= end) {
    if (my[left] <= my[right]) {
      array[current++] = my[left++];
    } else {
      array[current++] = my[right++];
    }
  }
  remaining = middle - left;
  for (i = 0; i <= remaining; i++)
    array[current + i] = my[left + i];
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<mm_round_epi32_sse2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_add_epi32@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <_mm_srai_epi32@plt>
  leave
  ret
```

assistant: ```c
__m128i mm_round_epi32_sse2(const __m128i *const src, const __m128i *const half_depth, const int depth) {
  const __m128i nearest_src = _mm_add_epi32(*src, *half_depth);
  return _mm_srai_epi32(nearest_src, depth);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mm_round_epi32_sse2>:
  endbr64
  push   %rbx
  mov    %edx,%ebx
  mov    (%rsi),%esi
  mov    (%rdi),%edi
  call   <_mm_add_epi32@plt>
  mov    %eax,%edi
  mov    %ebx,%esi
  call   <_mm_srai_epi32@plt>
  pop    %rbx
  ret
```

assistant: ```c
__m128i mm_round_epi32_sse2(const __m128i *const src, const __m128i *const half_depth, const int depth) {
  const __m128i nearest_src = _mm_add_epi32(*src, *half_depth);
  return _mm_srai_epi32(nearest_src, depth);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mm_round_epi32_sse2>:
  endbr64
  push   %rbp
  mov    (%rsi),%esi
  mov    %edx,%ebp
  mov    (%rdi),%edi
  call   <_mm_add_epi32@plt>
  mov    %ebp,%esi
  pop    %rbp
  mov    %eax,%edi
  jmp    1070 <_mm_srai_epi32@plt>
```

assistant: ```c
__m128i mm_round_epi32_sse2(const __m128i *const src, const __m128i *const half_depth, const int depth) {
  const __m128i nearest_src = _mm_add_epi32(*src, *half_depth);
  return _mm_srai_epi32(nearest_src, depth);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mm_round_epi32_sse2>:
  endbr64
  push   %rbp
  mov    (%rsi),%esi
  mov    %edx,%ebp
  mov    (%rdi),%edi
  call   <_mm_add_epi32@plt>
  mov    %ebp,%esi
  pop    %rbp
  mov    %eax,%edi
  jmp    1070 <_mm_srai_epi32@plt>
```

assistant: ```c
__m128i mm_round_epi32_sse2(const __m128i *const src, const __m128i *const half_depth, const int depth) {
  const __m128i nearest_src = _mm_add_epi32(*src, *half_depth);
  return _mm_srai_epi32(nearest_src, depth);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<quick>:
  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
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L28
L31:
  addl   $0x1,-0x10(%rbp)
L64:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jg     L31
  jmp    L67
L61:
  subl   $0x1,-0xc(%rbp)
L67:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L61
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L28
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L28:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L64
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L22
  mov    -0xc(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick>
L22:
  mov    -0x10(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L75
  mov    -0x20(%rbp),%edx
  mov    -0x10(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick>
  nop
L75:
  nop
  leave
  ret
```

assistant: ```c
void quick(int *v, int inicio, int fim) {
  int i, j, pivo, aux;
  i = inicio;
  j = fim;
  pivo = v[(inicio + fim) / 2];
  while (i < j) {
    while (v[i] < pivo)
      i++;
    while (v[j] > pivo)
      j--;
    if (i <= j) {
      aux = v[i];
      v[i++] = v[j];
      v[j--] = aux;
    }
  }
  if (j > inicio)
    quick(v, inicio, j);
  if (i < fim)
    quick(v, i, fim);
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<quick>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r10d
  mov    %edx,%r12d
  lea    (%rsi,%rdx,1),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  cltq
  mov    (%rdi,%rax,4),%esi
  mov    %r12d,%edx
  mov    %r10d,%ebx
  jmp    L12
L130:
  add    $0x1,%ebx
  mov    %edi,(%r8)
  sub    $0x1,%edx
  mov    %ecx,(%r9)
  jmp    L12
L73:
  cmp    %r10d,%edx
  jg     L40
L14:
  cmp    %r12d,%ebx
  jl     L69
L59:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L40:
  mov    %r10d,%esi
  mov    %rbp,%rdi
  call   <quick>
  jmp    L14
L69:
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <quick>
  jmp    L59
L15:
  movslq %edx,%rax
  lea    0x0(%rbp,%rax,4),%r9
  mov    (%r9),%edi
  cmp    %edi,%esi
  jge    L130
L67:
  movslq %edx,%rax
  lea    -0x4(%rbp,%rax,4),%rax
L42:
  sub    $0x1,%edx
  mov    %rax,%r9
  sub    $0x4,%rax
  mov    0x4(%rax),%edi
  cmp    %esi,%edi
  jg     L42
L94:
  cmp    %ebx,%edx
  jge    L130
L12:
  cmp    %edx,%ebx
  jge    L73
  movslq %ebx,%rax
  shl    $0x2,%rax
  lea    0x0(%rbp,%rax,1),%r8
  mov    (%r8),%ecx
  lea    0x4(%rbp,%rax,1),%rax
  cmp    %ecx,%esi
  jle    L15
L101:
  add    $0x1,%ebx
  mov    %rax,%r8
  add    $0x4,%rax
  mov    -0x4(%rax),%ecx
  cmp    %esi,%ecx
  jl     L101
  movslq %edx,%rax
  lea    0x0(%rbp,%rax,4),%r9
  mov    (%r9),%edi
  cmp    %edi,%esi
  jl     L67
  jmp    L94
```

assistant: ```c
void quick(int *v, int inicio, int fim) {
  int i, j, pivo, aux;
  i = inicio;
  j = fim;
  pivo = v[(inicio + fim) / 2];
  while (i < j) {
    while (v[i] < pivo)
      i++;
    while (v[j] > pivo)
      j--;
    if (i <= j) {
      aux = v[i];
      v[i++] = v[j];
      v[j--] = aux;
    }
  }
  if (j > inicio)
    quick(v, inicio, j);
  if (i < fim)
    quick(v, i, fim);
  return;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<quick>:
  endbr64
  push   %r12
  mov    %esi,%r10d
  mov    %rdi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
L112:
  lea    (%r10,%rbp,1),%edx
  mov    %r10d,%ebx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  mov    %ebp,%edx
  sar    %eax
  cltq
  mov    (%r12,%rax,4),%edi
  nopl   0x0(%rax,%rax,1)
L104:
  cmp    %edx,%ebx
  jge    L103
L101:
  movslq %ebx,%rax
  movslq %edx,%r11
  shl    $0x2,%rax
  lea    (%r12,%r11,4),%r9
  lea    (%r12,%rax,1),%rcx
  mov    (%r9),%r8d
  lea    0x4(%r12,%rax,1),%rax
  mov    (%rcx),%esi
  cmp    %edi,%esi
  jge    L44
  nopl   0x0(%rax)
L7:
  mov    %rax,%rcx
  mov    (%rax),%esi
  lea    0x4(%rax),%rax
  add    $0x1,%ebx
  cmp    %edi,%esi
  jl     L7
  cmp    %r8d,%edi
  jge    L67
L78:
  lea    -0x4(%r12,%r11,4),%rax
  nopw   0x0(%rax,%rax,1)
L48:
  mov    %rax,%r9
  mov    (%rax),%r8d
  sub    $0x4,%rax
  sub    $0x1,%edx
  cmp    %edi,%r8d
  jg     L48
L67:
  cmp    %ebx,%edx
  jl     L104
  mov    %r8d,(%rcx)
  add    $0x1,%ebx
  sub    $0x1,%edx
  mov    %esi,(%r9)
L54:
  cmp    %edx,%ebx
  jl     L101
L103:
  cmp    %r10d,%edx
  jg     L38
L64:
  cmp    %ebp,%ebx
  jge    L31
  mov    %ebx,%r10d
  jmp    L112
  nop
L44:
  cmp    %r8d,%edi
  jl     L78
  mov    %r8d,(%rcx)
  add    $0x1,%ebx
  sub    $0x1,%edx
  mov    %esi,(%r9)
  jmp    L54
L31:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L38:
  mov    %r10d,%esi
  mov    %r12,%rdi
  call   <quick>
  jmp    L64
```

assistant: ```c
void quick(int *v, int inicio, int fim) {
  int i, j, pivo, aux;
  i = inicio;
  j = fim;
  pivo = v[(inicio + fim) / 2];
  while (i < j) {
    while (v[i] < pivo)
      i++;
    while (v[j] > pivo)
      j--;
    if (i <= j) {
      aux = v[i];
      v[i++] = v[j];
      v[j--] = aux;
    }
  }
  if (j > inicio)
    quick(v, inicio, j);
  if (i < fim)
    quick(v, i, fim);
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L51
  cmpl   $0x2,-0x14(%rbp)
  je     L51
  mov    $0x0,%eax
  jmp    L18
L51:
  movl   $0x3,-0x8(%rbp)
  jmp    L49
L61:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L1
  mov    $0x0,%eax
  jmp    L18
L1:
  addl   $0x2,-0x8(%rbp)
L49:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L61
  mov    $0x1,%eax
L18:
  leave
  ret
```

assistant: ```c
int is_prime(int n) {
  int i, s = sqrt(n) + 1;
  if (n % 2 == 0 && n != 2)
    return 0;
  for (i = 3; i < s; i += 2)
    if (n % i == 0)
      return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<is_prime>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <sqrt@plt>
  mov    %eax,%esi
  test   $0x1,%bl
  jne    L29
  mov    $0x0,%eax
  cmp    $0x2,%ebx
  jne    L17
L29:
  cmp    $0x2,%esi
  jle    L75
  movslq %ebx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  je     L17
  mov    $0x3,%ecx
L10:
  add    $0x2,%ecx
  cmp    %ecx,%esi
  jl     L40
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %edx,%edx
  jne    L10
  jmp    L17
L40:
  mov    $0x1,%eax
L17:
  pop    %rbx
  ret
L75:
  mov    $0x1,%eax
  jmp    L17
```

assistant: ```c
int is_prime(int n) {
  int i, s = sqrt(n) + 1;
  if (n % 2 == 0 && n != 2)
    return 0;
  for (i = 3; i < s; i += 2)
    if (n % i == 0)
      return 0;
  return 1;
}
```