user: 这段汇编代码用C语言表示是什么样的？

```asm
<countSort>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  lea    0x0(,%r14,4),%edi
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  call   <malloc@plt>
  mov    $0x28,%edi
  mov    %rax,%r13
  call   <malloc@plt>
  pxor   %xmm0,%xmm0
  lea    -0x1(%r14),%ecx
  movq   $0x0,0x20(%rax)
  mov    %rax,%rbp
  movups %xmm0,(%rax)
  movups %xmm0,0x10(%rax)
  test   %r14d,%r14d
  jle    L95
  lea    -0x1(%r14),%eax
  mov    %r12,%rsi
  mov    %rax,%rcx
  lea    0x4(%r12,%rax,4),%rdi
  nopl   0x0(%rax)
L43:
  mov    (%rsi),%eax
  add    $0x4,%rsi
  cltd
  idiv   %ebx
  movslq %eax,%rdx
  mov    %eax,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r8d
  sar    $0x22,%rdx
  sub    %r8d,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  addl   $0x1,0x0(%rbp,%rax,4)
  cmp    %rdi,%rsi
  jne    L43
L95:
  mov    0x4(%rbp),%eax
  add    0x0(%rbp),%eax
  mov    %eax,0x4(%rbp)
  add    0x8(%rbp),%eax
  mov    %eax,0x8(%rbp)
  add    0xc(%rbp),%eax
  mov    %eax,0xc(%rbp)
  add    0x10(%rbp),%eax
  mov    %eax,0x10(%rbp)
  add    0x14(%rbp),%eax
  mov    %eax,0x14(%rbp)
  add    0x18(%rbp),%eax
  mov    %eax,0x18(%rbp)
  add    0x1c(%rbp),%eax
  mov    %eax,0x1c(%rbp)
  add    0x20(%rbp),%eax
  mov    %eax,0x20(%rbp)
  add    %eax,0x24(%rbp)
  test   %ecx,%ecx
  js     L14
  movslq %ecx,%rsi
  nopl   0x0(%rax)
L17:
  mov    (%r12,%rsi,4),%edi
  sub    $0x1,%rsi
  mov    %edi,%eax
  cltd
  idiv   %ebx
  movslq %eax,%rdx
  mov    %eax,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r8d
  sar    $0x22,%rdx
  sub    %r8d,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cltq
  lea    0x0(%rbp,%rax,4),%rax
  movslq (%rax),%rdx
  mov    %edi,-0x4(%r13,%rdx,4)
  subl   $0x1,(%rax)
  test   %esi,%esi
  jns    L17
L14:
  test   %r14d,%r14d
  jle    L97
  lea    0x4(%r13),%rdx
  mov    %r12,%rax
  sub    %rdx,%rax
  cmp    $0x8,%rax
  jbe    L77
  cmp    $0x2,%ecx
  jbe    L77
  mov    %r14d,%edx
  mov    %r12,%rcx
  mov    %r13,%rax
  shr    $0x2,%edx
  sub    %r13,%rcx
  shl    $0x4,%rdx
  add    %r13,%rdx
  cs nopw 0x0(%rax,%rax,1)
L64:
  movdqu (%rax),%xmm1
  movups %xmm1,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L64
  mov    %r14d,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%r14b
  je     L97
  mov    %eax,%ecx
  mov    0x0(%r13,%rcx,4),%esi
  lea    0x0(,%rcx,4),%rdx
  mov    %esi,(%r12,%rcx,4)
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r14d
  jle    L97
  mov    0x4(%r13,%rdx,1),%ecx
  add    $0x2,%eax
  mov    %ecx,0x4(%r12,%rdx,1)
  cmp    %eax,%r14d
  jle    L97
  mov    0x8(%r13,%rdx,1),%eax
  mov    %eax,0x8(%r12,%rdx,1)
L97:
  mov    %r13,%rdi
  call   <free@plt>
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <free@plt>
  nopl   0x0(%rax)
L77:
  lea    0x0(%r13,%r14,4),%rcx
  mov    %r13,%rax
  sub    %r13,%r12
  nopl   0x0(%rax,%rax,1)
L104:
  mov    (%rax),%edx
  mov    %edx,(%r12,%rax,1)
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L104
  jmp    L97
```

assistant: ```c
void countSort(int *dizi, int boyut, int taban) {
  int *output = (int *)malloc(sizeof(int) * boyut);
  int *count = (int *)malloc(sizeof(int) * 10);
  int i, mod;
  for (i = 0; i < 10; i++) {
    count[i] = 0;
  }
  for (i = 0; i < boyut; i++) {
    mod = (dizi[i] / taban) % 10;
    count[mod]++;
  }
  for (i = 1; i < 10; i++) {
    count[i] += count[i - 1];
  }
  for (i = boyut - 1; i >= 0; i--) {
    mod = (dizi[i] / taban) % 10;
    output[count[mod] - 1] = dizi[i];
    count[mod]--;
  }
  for (i = 0; i < boyut; i++) {
    dizi[i] = output[i];
  }
  free(output);
  free(count);
}
```