user: 这段汇编代码用C语言表示是什么样的？

```asm
<QuicksortAsc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rdx
  mov    %rdx,-0x38(%rbp)
  xor    %edx,%edx
  cmp    $0x1,%esi
  jbe    L201
  mov    %rdi,%r12
  mov    %esi,%eax
  cmp    $0x2,%esi
  je     L196
  shr    $0x1f,%esi
  mov    %rsp,-0x48(%rbp)
  add    %eax,%esi
  sar    %esi
  movslq %esi,%rdx
  mov    (%rdi,%rdx,4),%ebx
  movslq %eax,%rdx
  mov    %rsp,%rdi
  lea    0xf(,%rdx,4),%rdx
  mov    %rdx,%r9
  mov    %rdx,%rcx
  and    $0xfffffffffffff000,%r9
  and    $0xfffffffffffffff0,%rcx
  sub    %r9,%rdi
  cmp    %rdi,%rsp
  je     L181
L128:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdi,%rsp
  jne    L128
L181:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  je     L45
  orq    $0x0,-0x8(%rsp,%rcx,1)
L45:
  mov    %rdx,%rcx
  mov    %rsp,%rdi
  and    $0xfffffffffffff000,%rdx
  mov    %rsp,%r14
  sub    %rdx,%rdi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rdi,%rsp
  je     L161
L94:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rdi,%rsp
  jne    L94
L161:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  je     L148
  orq    $0x0,-0x8(%rsp,%rcx,1)
L148:
  mov    %rsp,%r9
  test   %eax,%eax
  jle    L130
  mov    %eax,%ecx
  xor    %r15d,%r15d
  xor    %eax,%eax
  xor    %r13d,%r13d
  jmp    L10
  nopl   0x0(%rax,%rax,1)
L144:
  movslq %r15d,%rdi
  add    $0x1,%r15d
  mov    %edx,(%r9,%rdi,4)
L75:
  add    $0x1,%rax
  cmp    %rax,%rcx
  je     L164
L10:
  cmp    %eax,%esi
  je     L75
  mov    (%r12,%rax,4),%edx
  cmp    %ebx,%edx
  jl     L144
  movslq %r13d,%rdi
  add    $0x1,%rax
  add    $0x1,%r13d
  mov    %edx,(%r14,%rdi,4)
  cmp    %rax,%rcx
  jne    L10
L164:
  mov    %r9,%rdi
  mov    %r15d,%esi
  mov    %r9,-0x50(%rbp)
  call   <QuicksortAsc>
  mov    %r13d,%esi
  mov    %r14,%rdi
  call   <QuicksortAsc>
  test   %r15d,%r15d
  mov    -0x50(%rbp),%r9
  je     L110
  movslq %r15d,%r15
  mov    %r9,%rsi
  mov    %r12,%rdi
  lea    0x0(,%r15,4),%rdx
  call   <memcpy@plt>
L105:
  mov    %ebx,(%r12,%r15,4)
  lea    0x0(,%r15,4),%rax
  test   %r13d,%r13d
  je     L51
  movslq %r13d,%rdx
  lea    0x4(%r12,%rax,1),%rdi
  mov    %r14,%rsi
  shl    $0x2,%rdx
  call   <memcpy@plt>
L51:
  mov    -0x48(%rbp),%rsp
L201:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L166
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L196:
  mov    (%rdi),%eax
  mov    0x4(%rdi),%edx
  cmp    %edx,%eax
  jle    L201
  mov    %edx,(%rdi)
  mov    %eax,0x4(%rdi)
  jmp    L201
L130:
  mov    %r9,%rdi
  xor    %esi,%esi
  call   <QuicksortAsc>
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <QuicksortAsc>
  mov    %ebx,(%r12)
  jmp    L51
L110:
  xor    %r15d,%r15d
  jmp    L105
L166:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void QuicksortAsc(int *B, int len) {
  int i;
  if ((len == 1) || (len == 0)) {
    return;
  } else if (len == 2) {
    if (B[0] > B[1]) {
      int tmp = B[0];
      B[0] = B[1];
      B[1] = tmp;
    }
    return;
  } else {
    int pivot = len / 2;
    int pivotElement = B[pivot];
    int larger[len];
    int smaller[len];
    int lCount = 0, sCount = 0;
    for (i = 0; i < len; i++) {
      if (i == pivot) {
      } else if (B[i] < B[pivot]) {
        smaller[sCount++] = B[i];
      } else {
        larger[lCount++] = B[i];
      }
    }
    QuicksortAsc(smaller, sCount);
    QuicksortAsc(larger, lCount);
    for (i = 0; i < sCount; i++) {
      B[i] = smaller[i];
    }
    B[sCount] = pivotElement;
    for (i = 0; i < lCount; i++) {
      B[i + sCount + 1] = larger[i];
    }
    return;
  }
}
```