user: 这段汇编代码用C语言表示是什么样的?

```asm
<qsufsort>:
  endbr64
  push   %r15
  mov    %rdx,%r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %ecx,%ebp
  mov    $0x80,%ecx
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  rep stos %rax,%es:(%rdi)
  test   %ebp,%ebp
  jle    L113
  mov    %rdx,%rax
  lea    -0x1(%rbp),%edx
  lea    0x8(%r15,%rdx,8),%rcx
L35:
  mov    (%rax),%rdx
  add    $0x8,%rax
  addl   $0x1,(%rsp,%rdx,4)
  cmp    %rax,%rcx
  jne    L35
L113:
  lea    0x4(%rsp),%rdi
  lea    0x400(%rsp),%rcx
  mov    %rdi,%rbx
  mov    %rdi,%rax
  nopl   0x0(%rax,%rax,1)
L81:
  mov    -0x4(%rax),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L81
  mov    $0x3fc,%edx
  mov    %r14,%rsi
  call   <memmove@plt>
  movl   $0x0,(%rsp)
  test   %ebp,%ebp
  jle    L135
  movslq %ebp,%rsi
  xor    %edx,%edx
  nopw   0x0(%rax,%rax,1)
L16:
  mov    (%r15,%rdx,8),%rcx
  mov    (%rsp,%rcx,4),%eax
  add    $0x1,%eax
  mov    %eax,(%rsp,%rcx,4)
  cltq
  mov    %edx,(%r12,%rax,4)
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L16
  xor    %eax,%eax
  xchg   %ax,%ax
L155:
  mov    (%r15,%rax,8),%rdx
  mov    (%rsp,%rdx,4),%edx
  mov    %edx,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L155
  mov    (%rsp),%eax
L111:
  movl   $0x0,0x0(%r13,%rsi,4)
  lea    0x3fc(%r14),%rcx
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L131:
  add    $0x4,%rbx
  mov    %edx,%eax
L6:
  mov    (%rbx),%edx
  add    $0x1,%eax
  cmp    %eax,%edx
  jne    L51
  movslq %edx,%rax
  movl   $0xffffffff,(%r12,%rax,4)
L51:
  cmp    %rcx,%rbx
  jne    L131
  mov    %ebp,%ebx
  movl   $0xffffffff,(%r12)
  mov    $0xffffffffffffffff,%rax
  mov    $0x1,%r14d
  not    %ebx
  test   %ebp,%ebp
  je     L158
L70:
  xor    %ecx,%ecx
  xor    %edx,%edx
  test   %ebp,%ebp
  jns    L38
  jmp    L64
  nopw   0x0(%rax,%rax,1)
L107:
  test   %ecx,%ecx
  je     L73
  mov    %edx,%eax
  sub    %ecx,%eax
  neg    %ecx
  cltq
  mov    %ecx,(%r12,%rax,4)
  movslq %edx,%rax
  movslq (%r12,%rax,4),%rax
L73:
  mov    0x0(%r13,%rax,4),%r15d
  mov    %r14d,%r8d
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x1,%r15d
  mov    %r15d,%ecx
  sub    %edx,%ecx
  call   <split@plt>
  mov    %r15d,%edx
  xor    %ecx,%ecx
  cmp    %ebp,%edx
  jg     L124
L3:
  movslq %edx,%rax
  movslq (%r12,%rax,4),%rax
L38:
  test   %eax,%eax
  jns    L107
  sub    %eax,%edx
  sub    %eax,%ecx
  cmp    %ebp,%edx
  jle    L3
L124:
  test   %ecx,%ecx
  je     L11
  sub    %ecx,%edx
  neg    %ecx
  movslq %edx,%rdx
  mov    %ecx,(%r12,%rdx,4)
L11:
  movslq (%r12),%rax
L64:
  add    %r14d,%r14d
  cmp    %eax,%ebx
  jne    L70
  xor    %eax,%eax
  test   %ebp,%ebp
  js     L165
  nop
L197:
  movslq 0x0(%r13,%rax,4),%rdx
  mov    %eax,(%r12,%rdx,4)
  add    $0x1,%rax
  cmp    %eax,%ebp
  jge    L197
L165:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L140
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L135:
  xor    %eax,%eax
  movslq %ebp,%rsi
  jmp    L111
L158:
  movslq 0x0(%r13),%rax
  movl   $0x0,(%r12,%rax,4)
  jmp    L165
L140:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void qsufsort(int32_t *I, int32_t *V, u_char *old, int32_t oldsize) {
  int32_t buckets[256];
  int32_t i, h, len;
  for (i = 0; i < 256; i++)
    buckets[i] = 0;
  for (i = 0; i < oldsize; i++)
    buckets[old[i]]++;
  for (i = 1; i < 256; i++)
    buckets[i] += buckets[i - 1];
  for (i = 255; i > 0; i--)
    buckets[i] = buckets[i - 1];
  buckets[0] = 0;
  for (i = 0; i < oldsize; i++)
    I[++buckets[old[i]]] = i;
  I[0] = oldsize;
  for (i = 0; i < oldsize; i++)
    V[i] = buckets[old[i]];
  V[oldsize] = 0;
  for (i = 1; i < 256; i++)
    if (buckets[i] == buckets[i - 1] + 1)
      I[buckets[i]] = -1;
  I[0] = -1;
  for (h = 1; I[0] != -(oldsize + 1); h += h) {
    len = 0;
    for (i = 0; i < oldsize + 1;) {
      if (I[i] < 0) {
        len -= I[i];
        i -= I[i];
      } else {
        if (len)
          I[i - len] = -len;
        len = V[I[i]] + 1 - i;
        split(I, V, i, len, h);
        i += len;
        len = 0;
      };
    };
    if (len)
      I[i - len] = -len;
  };
  for (i = 0; i < oldsize + 1; i++)
    I[V[i]] = i;
}
```