ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<int_from_string>:
  endbr64
  mov    (%rdi),%rdx
  movzbl (%rdx),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L14
  add    $0x1,%rdx
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L6:
  sub    $0x30,%eax
  lea    (%r8,%r8,4),%ecx
  mov    %rdx,(%rdi)
  add    $0x1,%rdx
  movsbl %al,%eax
  lea    (%rax,%rcx,2),%r8d
  movzbl -0x1(%rdx),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  jbe    L6
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L14:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int int_from_string(const char **str) {
  int ret = 0;
  while (**str >= '0' && **str <= '9') {
    ret *= 10;
    ret += **str - '0';
    (*str)++;
  }
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<int_from_string>:
  endbr64
  mov    (%rdi),%rdx
  movzbl (%rdx),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  ja     L5
  add    $0x1,%rdx
  xor    %r8d,%r8d
  nopl   0x0(%rax)
L18:
  sub    $0x30,%eax
  lea    (%r8,%r8,4),%ecx
  mov    %rdx,(%rdi)
  add    $0x1,%rdx
  movsbl %al,%eax
  lea    (%rax,%rcx,2),%r8d
  movzbl -0x1(%rdx),%eax
  lea    -0x30(%rax),%ecx
  cmp    $0x9,%cl
  jbe    L18
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L5:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int int_from_string(const char **str) {
  int ret = 0;
  while (**str >= '0' && **str <= '9') {
    ret *= 10;
    ret += **str - '0';
    (*str)++;
  }
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L1:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L1
```

assistant: ```c
int main(void) {
  for (;;) {
    sleep(1);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
L0:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L0
```

assistant: ```c
int main(void) {
  for (;;) {
    sleep(1);
  }
  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
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L9:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L9
```

assistant: ```c
int main(void) {
  for (;;) {
    sleep(1);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L5:
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L5
```

assistant: ```c
int main(void) {
  for (;;) {
    sleep(1);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<i_to_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpl   $0x0,-0x34(%rbp)
  jns    L35
  mov    -0x40(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  movb   $0x2d,(%rax)
  negl   -0x34(%rbp)
L35:
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L39:
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  imul   $0xfffffff6,%eax,%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  cmpl   $0x0,-0x1c(%rbp)
  je     L26
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x34(%rbp)
  jmp    L39
L26:
  nop
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  sub    -0x40(%rbp),%rax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L28
L38:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x21(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x40(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x21(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x20(%rbp)
L28:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x20(%rbp)
  jl     L38
  mov    -0x10(%rbp),%rax
  sub    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int i_to_str(int val, char *str) {
  int len, i;
  char *s;
  char *begin = str;
  if (val < 0) {
    *str++ = '-';
    val = -val;
  }
  s = str;
  for (;;) {
    int ni = val / 10;
    int digit = val - ni * 10;
    *s++ = (char)('0' + digit);
    if (ni == 0) {
      break;
    }
    val = ni;
  }
  *s = '\0';
  len = (int)(s - str);
  for (i = 0; i < len / 2; ++i) {
    char ch = str[i];
    str[i] = str[len - 1 - i];
    str[len - 1 - i] = ch;
  }
  return (int)(s - begin);
}
```
user: Translate the below assembly code into C programming language:

```asm
<i_to_str>:
  endbr64
  mov    %rsi,%r9
  test   %edi,%edi
  js     L20
L2:
  mov    %rsi,%rax
L11:
  mov    %edi,%r8d
  movslq %edi,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  sar    $0x1f,%edi
  sub    %edi,%edx
  mov    %edx,%edi
  add    $0x1,%rax
  imul   $0xfffffff6,%edx,%ecx
  lea    0x30(%rcx,%r8,1),%ecx
  mov    %cl,-0x1(%rax)
  test   %edx,%edx
  jne    L11
  movb   $0x0,(%rax)
  mov    %rax,%rdx
  sub    %rsi,%rdx
  mov    %edx,%r10d
  shr    $0x1f,%r10d
  add    %edx,%r10d
  sar    %r10d
  cmp    $0x1,%edx
  jle    L37
  movslq %edx,%rdx
  lea    -0x1(%rsi,%rdx,1),%rcx
  mov    $0x0,%edx
L18:
  movzbl (%rsi,%rdx,1),%edi
  movzbl (%rcx),%r8d
  mov    %r8b,(%rsi,%rdx,1)
  mov    %dil,(%rcx)
  add    $0x1,%rdx
  sub    $0x1,%rcx
  cmp    %edx,%r10d
  jg     L18
L37:
  sub    %r9d,%eax
  ret
L20:
  lea    0x1(%rsi),%rsi
  movb   $0x2d,(%r9)
  neg    %edi
  jmp    L2
```

assistant: ```c
int i_to_str(int val, char *str) {
  int len, i;
  char *s;
  char *begin = str;
  if (val < 0) {
    *str++ = '-';
    val = -val;
  }
  s = str;
  for (;;) {
    int ni = val / 10;
    int digit = val - ni * 10;
    *s++ = (char)('0' + digit);
    if (ni == 0) {
      break;
    }
    val = ni;
  }
  *s = '\0';
  len = (int)(s - str);
  for (i = 0; i < len / 2; ++i) {
    char ch = str[i];
    str[i] = str[len - 1 - i];
    str[len - 1 - i] = ch;
  }
  return (int)(s - begin);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<i_to_str>:
  endbr64
  mov    %rsi,%r8
  test   %edi,%edi
  jns    L35
  movb   $0x2d,(%r8)
  lea    0x1(%rsi),%rsi
  neg    %edi
L35:
  mov    %rsi,%rax
  mov    $0xcccccccd,%r9d
  xchg   %ax,%ax
L4:
  mov    %edi,%edi
  add    $0x1,%rax
  mov    %rdi,%rcx
  imul   %r9,%rdi
  shr    $0x23,%rdi
  imul   $0xfffffff6,%edi,%edx
  lea    0x30(%rdx,%rcx,1),%edx
  mov    %dl,-0x1(%rax)
  test   %edi,%edi
  jne    L4
  mov    %rax,%rdx
  movb   $0x0,(%rax)
  sub    %rsi,%rdx
  mov    %edx,%r10d
  shr    $0x1f,%r10d
  add    %edx,%r10d
  sar    %r10d
  cmp    $0x1,%edx
  jle    L3
  sub    $0x1,%edx
  xor    %ecx,%ecx
  movslq %edx,%rdx
  add    %rsi,%rdx
  nopl   (%rax)
L10:
  movzbl (%rdx),%r9d
  movzbl (%rsi,%rcx,1),%edi
  sub    $0x1,%rdx
  mov    %r9b,(%rsi,%rcx,1)
  add    $0x1,%rcx
  mov    %dil,0x1(%rdx)
  cmp    %ecx,%r10d
  jg     L10
L3:
  sub    %r8d,%eax
  ret
```

assistant: ```c
int i_to_str(int val, char *str) {
  int len, i;
  char *s;
  char *begin = str;
  if (val < 0) {
    *str++ = '-';
    val = -val;
  }
  s = str;
  for (;;) {
    int ni = val / 10;
    int digit = val - ni * 10;
    *s++ = (char)('0' + digit);
    if (ni == 0) {
      break;
    }
    val = ni;
  }
  *s = '\0';
  len = (int)(s - str);
  for (i = 0; i < len / 2; ++i) {
    char ch = str[i];
    str[i] = str[len - 1 - i];
    str[len - 1 - i] = ch;
  }
  return (int)(s - begin);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<i_to_str>:
  endbr64
  mov    %rsi,%r8
  test   %edi,%edi
  jns    L23
  movb   $0x2d,(%r8)
  lea    0x1(%rsi),%rsi
  neg    %edi
L23:
  mov    %rsi,%rax
  mov    $0xcccccccd,%r9d
  xchg   %ax,%ax
L31:
  mov    %edi,%edi
  add    $0x1,%rax
  mov    %rdi,%rcx
  imul   %r9,%rdi
  shr    $0x23,%rdi
  imul   $0xfffffff6,%edi,%edx
  lea    0x30(%rdx,%rcx,1),%edx
  mov    %dl,-0x1(%rax)
  test   %edi,%edi
  jne    L31
  mov    %rax,%rdx
  movb   $0x0,(%rax)
  sub    %rsi,%rdx
  mov    %edx,%r10d
  shr    $0x1f,%r10d
  add    %edx,%r10d
  sar    %r10d
  cmp    $0x1,%edx
  jle    L7
  sub    $0x1,%edx
  xor    %ecx,%ecx
  movslq %edx,%rdx
  add    %rsi,%rdx
  nopl   (%rax)
L32:
  movzbl (%rdx),%r9d
  movzbl (%rsi,%rcx,1),%edi
  sub    $0x1,%rdx
  mov    %r9b,(%rsi,%rcx,1)
  add    $0x1,%rcx
  mov    %dil,0x1(%rdx)
  cmp    %ecx,%r10d
  jg     L32
L7:
  sub    %r8d,%eax
  ret
```

assistant: ```c
int i_to_str(int val, char *str) {
  int len, i;
  char *s;
  char *begin = str;
  if (val < 0) {
    *str++ = '-';
    val = -val;
  }
  s = str;
  for (;;) {
    int ni = val / 10;
    int digit = val - ni * 10;
    *s++ = (char)('0' + digit);
    if (ni == 0) {
      break;
    }
    val = ni;
  }
  *s = '\0';
  len = (int)(s - str);
  for (i = 0; i < len / 2; ++i) {
    char ch = str[i];
    str[i] = str[len - 1 - i];
    str[len - 1 - i] = ch;
  }
  return (int)(s - begin);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x748,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x2738(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <printf@plt>
  lea    0x1c(%rsp),%rsi
  lea    D44(%rip),%rdi
  call   <scanf@plt>
  mov    0x1c(%rsp),%ebx
  test   %ebx,%ebx
  js     L122
  lea    0x1(%rbx),%ebp
  mov    $0xc8,%edi
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  lea    -0xa8(%rsp),%r13
  mov    $0x0,%r12d
  mov    $0x1,%r10d
  jmp    L15
L36:
  add    $0x4,%rax
  mov    %esi,%ecx
L83:
  lea    (%rax,%r9,1),%rsi
  mov    %edx,(%rsi,%rdi,1)
  lea    0x1(%rcx),%esi
  cmp    %r8d,%ecx
  je     L57
  mov    %r10d,%edx
  test   %esi,%esi
  je     L36
  mov    (%rax),%edx
  add    0x4(%rax),%edx
  jmp    L36
L57:
  sub    $0xc8,%r9
  add    $0xc8,%rdi
  mov    %r11d,%r8d
L15:
  lea    0x1(%r8),%r11d
  cmp    %ebp,%r11d
  je     L52
  test   %r11d,%r11d
  jle    L57
  lea    0x0(%r13,%rdi,1),%rax
  mov    %r12d,%ecx
  mov    %r10d,%edx
  jmp    L83
L52:
  mov    $0x0,%r15d
  mov    $0x0,%r14d
  lea    D16(%rip),%r13
  lea    0x20(%rsp),%rax
  mov    %rax,0x8(%rsp)
  lea    D33(%rip),%r12
L40:
  sub    %r14d,%ebx
  test   %ebx,%ebx
  jle    L20
L59:
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L59
L20:
  test   %r14d,%r14d
  jle    L64
  mov    0x8(%rsp),%rax
  lea    (%rax,%r15,4),%rbx
  lea    -0x1(%r14),%eax
  add    %r15,%rax
  lea    0x24(%rsp,%rax,4),%rbp
L85:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L85
L64:
  lea    D11(%rip),%rdi
  call   <puts@plt>
  add    $0x1,%r14d
  mov    0x1c(%rsp),%ebx
  add    $0x32,%r15
  cmp    %r14d,%ebx
  jge    L40
L122:
  mov    0x2738(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x2748,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D14
type_infered: string
name: parse_data
arguments: data_label: D44
type_infered: string
name: parse_data
arguments: data_label: D16
type_infered: string
name: parse_data
arguments: data_label: D33
type_infered: string
name: parse_data
arguments: data_label: D11
type_infered: string

tool: D14: Enter the number of rows: 

tool: D44: %d

tool: D16:    

tool: D33: %6d

tool: D11: 


assistant: ```c
int main() {
  int a[50][50], x, i, j;
  printf("Enter the number of rows: ");
  scanf("%d", &x);
  for (i = 0; i <= x; i++) {
    for (j = 0; j < i; j++) {
      if (j == 0 || j == i) {
        a[i][j] = 1;
      } else {
        a[i][j] = a[i - 1][j] + a[i - 1][j - 1];
      }
    }
  }
  for (i = 0; i <= x; i++) {
    for (j = x - i; j > 0; j--)
      printf("   ");
    for (j = 0; j < i; j++)
      printf("%6d", a[i][j]);
    printf("\n\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x748,%rsp
  lea    D2(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x2738(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x1c(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  mov    0x1c(%rsp),%ebx
  test   %ebx,%ebx
  js     L124
  je     L45
  lea    0x20(%rsp),%r8
  lea    0x1(%rbx),%r11d
  mov    $0xc8,%r10d
  xor    %r9d,%r9d
  mov    $0x1,%esi
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %r8,%rax
  xor    %edx,%edx
  jmp    L3
  nop
L110:
  mov    $0x1,%edx
L112:
  mov    -0x4(%rax),%ecx
  lea    (%rax,%r9,1),%rdi
  add    (%rax),%ecx
  add    $0x1,%edx
  mov    %ecx,(%rdi,%r10,1)
  add    $0x4,%rax
  cmp    %esi,%edx
  je     L95
L3:
  test   %edx,%edx
  jne    L112
  movl   $0x1,0xc8(%r8)
  add    $0x4,%rax
  cmp    $0x1,%esi
  jne    L110
L95:
  add    $0x1,%esi
  add    $0xc8,%r8
  sub    $0xc8,%r9
  add    $0xc8,%r10
  cmp    %r11d,%esi
  jne    L8
L45:
  lea    0x20(%rsp),%rax
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  mov    %rax,(%rsp)
  lea    0x24(%rsp),%rax
  lea    D18(%rip),%r12
  mov    %rax,0x8(%rsp)
  lea    D21(%rip),%rbp
  nopl   0x0(%rax)
L26:
  sub    %r13d,%ebx
  test   %ebx,%ebx
  jle    L9
  nopw   0x0(%rax,%rax,1)
L39:
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L39
L9:
  test   %r13d,%r13d
  je     L72
  mov    (%rsp),%rax
  mov    0x8(%rsp),%rcx
  lea    (%rax,%r14,4),%r15
  lea    -0x1(%r13),%eax
  add    %r14,%rax
  lea    (%rcx,%rax,4),%rbx
  nopl   0x0(%rax)
L104:
  mov    (%r15),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%r15
  call   <printf@plt>
  cmp    %rbx,%r15
  jne    L104
L72:
  lea    D14(%rip),%rdi
  add    $0x1,%r13d
  add    $0x32,%r14
  call   <puts@plt>
  mov    0x1c(%rsp),%ebx
  cmp    %r13d,%ebx
  jge    L26
L124:
  mov    0x2738(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L123
  add    $0x2748,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L123:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D2
datatype: string
name: disassemble_data
arguments: label: D6
datatype: string
name: disassemble_data
arguments: label: D18
datatype: string
name: disassemble_data
arguments: label: D21
datatype: string
name: disassemble_data
arguments: label: D14
datatype: string

tool: D2: Enter the number of rows: 

tool: D6: %d

tool: D18:    

tool: D21: %6d

tool: D14: 


assistant: ```c
int main() {
  int a[50][50], x, i, j;
  printf("Enter the number of rows: ");
  scanf("%d", &x);
  for (i = 0; i <= x; i++) {
    for (j = 0; j < i; j++) {
      if (j == 0 || j == i) {
        a[i][j] = 1;
      } else {
        a[i][j] = a[i - 1][j] + a[i - 1][j - 1];
      }
    }
  }
  for (i = 0; i <= x; i++) {
    for (j = x - i; j > 0; j--)
      printf("   ");
    for (j = 0; j < i; j++)
      printf("%6d", a[i][j]);
    printf("\n\n");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sulong_qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  cmp    %rcx,%rdx
  jl     L10
L12:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L10:
  mov    %rcx,%r13
  mov    %r8,%r14
  mov    %r9,%r12
  mov    %rdx,%rbx
  mov    %rdx,%r15
  imul   %r9,%r15
  add    %rsi,%r15
  lea    (%rdx,%rcx,1),%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rax,%rdx
  sar    %rdx
  imul   %r9,%rdx
  mov    %rsi,0x8(%rsp)
  add    %rsi,%rdx
  mov    %r9,%rcx
  mov    %r15,%rsi
  call   <sulong_swap@plt>
  mov    %ebx,0x4(%rsp)
  lea    0x1(%rbx),%ebp
  movslq %ebp,%rax
  cmp    %rax,%r13
  jl     L40
  mov    %rax,%rbp
  mov    %r12,%rbx
  imul   %rax,%rbx
  add    0x8(%rsp),%rbx
  jmp    L8
L22:
  add    %r12,%rbx
  add    $0x1,%rbp
  cmp    %rbp,%r13
  jl     L40
L8:
  mov    %r15,%rsi
  mov    %rbx,%rdi
  call   *%r14
  test   %eax,%eax
  jns    L22
  addl   $0x1,0x4(%rsp)
  mov    0x4(%rsp),%eax
  movslq %eax,%rsi
  imul   %r12,%rsi
  add    0x8(%rsp),%rsi
  mov    %r12,%rcx
  mov    %rbx,%rdx
  mov    0x10(%rsp),%rdi
  call   <sulong_swap@plt>
  jmp    L22
L40:
  mov    0x4(%rsp),%ebx
  movslq %ebx,%rdx
  imul   %r12,%rdx
  mov    0x8(%rsp),%rbp
  add    %rbp,%rdx
  mov    %r12,%rcx
  mov    %r15,%rsi
  mov    0x10(%rsp),%r15
  mov    %r15,%rdi
  call   <sulong_swap@plt>
  lea    -0x1(%rbx),%ecx
  movslq %ecx,%rcx
  mov    %r12,%r9
  mov    %r14,%r8
  mov    0x18(%rsp),%rdx
  mov    %rbp,%rsi
  mov    %r15,%rdi
  call   <sulong_qsort>
  lea    0x1(%rbx),%edx
  movslq %edx,%rdx
  mov    %r12,%r9
  mov    %r14,%r8
  mov    %r13,%rcx
  mov    %rbp,%rsi
  mov    %r15,%rdi
  call   <sulong_qsort>
  jmp    L12
```

assistant: ```c
void sulong_qsort(char *buffer, char *v, long left, long right, int (*comp)(const void *, const void *), size_t size) {
  int i, last;
  if (left >= right) {
    return;
  }
  sulong_swap(buffer, &v[left * size], &v[((left + right) / 2) * size], size);
  last = left;
  for (i = left + 1; i <= right; i++) {
    if (comp(&(v[i * size]), &(v[left * size])) < 0) {
      last++;
      sulong_swap(buffer, &(v[last * size]), &(v[i * size]), size);
    }
  }
  sulong_swap(buffer, &(v[left * size]), &(v[last * size]), size);
  sulong_qsort(buffer, v, left, last - 1, comp, size);
  sulong_qsort(buffer, v, last + 1, right, comp, size);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sulong_qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  mov    %rsi,0x8(%rsp)
  mov    %rdx,0x18(%rsp)
  cmp    %rcx,%rdx
  jge    L22
  mov    %rcx,%r12
  mov    %r8,%r14
  mov    %r9,%rbx
L64:
  mov    0x18(%rsp),%r15
  mov    0x8(%rsp),%rcx
  mov    0x10(%rsp),%rdi
  lea    (%r12,%r15,1),%rax
  mov    %r15,%rbp
  mov    %rax,%rdx
  imul   %rbx,%rbp
  shr    $0x3f,%rdx
  add    %rax,%rdx
  sar    %rdx
  add    %rcx,%rbp
  imul   %rbx,%rdx
  mov    %rbp,%rsi
  add    %rcx,%rdx
  mov    %rbx,%rcx
  call   <sulong_swap@plt>
  lea    0x1(%r15),%r8d
  mov    %r15d,0x4(%rsp)
  movslq %r8d,%r8
  cmp    %r8,%r12
  jl     L24
  mov    %rbx,%rdx
  mov    0x8(%rsp),%rcx
  mov    %r8,%r13
  imul   %r8,%rdx
  lea    (%rcx,%rdx,1),%r15
  jmp    L27
  nopl   (%rax)
L26:
  add    $0x1,%r13
  add    %rbx,%r15
  cmp    %r13,%r12
  jl     L24
L27:
  mov    %rbp,%rsi
  mov    %r15,%rdi
  call   *%r14
  test   %eax,%eax
  jns    L26
  addl   $0x1,0x4(%rsp)
  movslq 0x4(%rsp),%rsi
  mov    %r15,%rdx
  mov    %rbx,%rcx
  mov    0x10(%rsp),%rdi
  add    $0x1,%r13
  add    %rbx,%r15
  imul   %rbx,%rsi
  add    0x8(%rsp),%rsi
  call   <sulong_swap@plt>
  cmp    %r13,%r12
  jge    L27
L24:
  movslq 0x4(%rsp),%rdx
  mov    0x8(%rsp),%r13
  mov    %rbp,%rsi
  mov    %rbx,%rcx
  mov    0x10(%rsp),%rbp
  mov    %rdx,%r15
  imul   %rbx,%rdx
  mov    %rbp,%rdi
  add    %r13,%rdx
  call   <sulong_swap@plt>
  mov    0x18(%rsp),%rdx
  mov    %r13,%rsi
  mov    %rbx,%r9
  lea    -0x1(%r15),%ecx
  mov    %r14,%r8
  lea    0x1(%r15),%r13d
  mov    %rbp,%rdi
  movslq %ecx,%rcx
  call   <sulong_qsort>
  movslq %r13d,%rax
  mov    %rax,0x18(%rsp)
  cmp    %r12,%rax
  jl     L64
L22:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void sulong_qsort(char *buffer, char *v, long left, long right, int (*comp)(const void *, const void *), size_t size) {
  int i, last;
  if (left >= right) {
    return;
  }
  sulong_swap(buffer, &v[left * size], &v[((left + right) / 2) * size], size);
  last = left;
  for (i = left + 1; i <= right; i++) {
    if (comp(&(v[i * size]), &(v[left * size])) < 0) {
      last++;
      sulong_swap(buffer, &(v[last * size]), &(v[i * size]), size);
    }
  }
  sulong_swap(buffer, &(v[left * size]), &(v[last * size]), size);
  sulong_qsort(buffer, v, left, last - 1, comp, size);
  sulong_qsort(buffer, v, last + 1, right, comp, size);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sulong_qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x10(%rsp)
  mov    %rsi,0x8(%rsp)
  mov    %rdx,0x18(%rsp)
  cmp    %rcx,%rdx
  jge    L35
  mov    %rcx,%r12
  mov    %r8,%r14
  mov    %r9,%rbx
L41:
  mov    0x18(%rsp),%r15
  mov    0x8(%rsp),%rcx
  mov    0x10(%rsp),%rdi
  lea    (%r12,%r15,1),%rax
  mov    %r15,%rbp
  mov    %rax,%rdx
  imul   %rbx,%rbp
  shr    $0x3f,%rdx
  add    %rax,%rdx
  sar    %rdx
  add    %rcx,%rbp
  imul   %rbx,%rdx
  mov    %rbp,%rsi
  add    %rcx,%rdx
  mov    %rbx,%rcx
  call   <sulong_swap@plt>
  lea    0x1(%r15),%r8d
  mov    %r15d,0x4(%rsp)
  movslq %r8d,%r8
  cmp    %r8,%r12
  jl     L4
  mov    %rbx,%rdx
  mov    0x8(%rsp),%rcx
  mov    %r8,%r13
  imul   %r8,%rdx
  lea    (%rcx,%rdx,1),%r15
  jmp    L67
  nopl   (%rax)
L66:
  add    $0x1,%r13
  add    %rbx,%r15
  cmp    %r13,%r12
  jl     L4
L67:
  mov    %rbp,%rsi
  mov    %r15,%rdi
  call   *%r14
  test   %eax,%eax
  jns    L66
  addl   $0x1,0x4(%rsp)
  movslq 0x4(%rsp),%rsi
  mov    %r15,%rdx
  mov    %rbx,%rcx
  mov    0x10(%rsp),%rdi
  add    $0x1,%r13
  add    %rbx,%r15
  imul   %rbx,%rsi
  add    0x8(%rsp),%rsi
  call   <sulong_swap@plt>
  cmp    %r13,%r12
  jge    L67
L4:
  movslq 0x4(%rsp),%rdx
  mov    0x8(%rsp),%r13
  mov    %rbp,%rsi
  mov    %rbx,%rcx
  mov    0x10(%rsp),%rbp
  mov    %rdx,%r15
  imul   %rbx,%rdx
  mov    %rbp,%rdi
  add    %r13,%rdx
  call   <sulong_swap@plt>
  mov    0x18(%rsp),%rdx
  mov    %r13,%rsi
  mov    %rbx,%r9
  lea    -0x1(%r15),%ecx
  mov    %r14,%r8
  lea    0x1(%r15),%r13d
  mov    %rbp,%rdi
  movslq %ecx,%rcx
  call   <sulong_qsort>
  movslq %r13d,%rax
  mov    %rax,0x18(%rsp)
  cmp    %r12,%rax
  jl     L41
L35:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void sulong_qsort(char *buffer, char *v, long left, long right, int (*comp)(const void *, const void *), size_t size) {
  int i, last;
  if (left >= right) {
    return;
  }
  sulong_swap(buffer, &v[left * size], &v[((left + right) / 2) * size], size);
  last = left;
  for (i = left + 1; i <= right; i++) {
    if (comp(&(v[i * size]), &(v[left * size])) < 0) {
      last++;
      sulong_swap(buffer, &(v[last * size]), &(v[i * size]), size);
    }
  }
  sulong_swap(buffer, &(v[left * size]), &(v[last * size]), size);
  sulong_qsort(buffer, v, left, last - 1, comp, size);
  sulong_qsort(buffer, v, last + 1, right, comp, size);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vprint>:
  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     L7
  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)
L7:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <vfprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

assistant: ```c
void vprint(const char *fmt, ...) {
  va_list ap;
  va_start(ap, fmt);
  vfprintf(stdout, fmt, ap);
  va_end(ap);
}
```