user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  lea    D1(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D48(%rip),%rbp
  push   %rbx
  sub    $0x98,%rsp
  movdqa D65(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  mov    D17(%rip),%rax
  mov    %rsp,%rbx
  movaps %xmm0,(%rsp)
  movdqa D21(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <puts@plt>
  cs nopw 0x0(%rax,%rax,1)
L29:
  mov    (%rbx,%r12,4),%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r12
  call   <printf@plt>
  cmp    $0xa,%r12
  jne    L29
  xor    %r12d,%r12d
  lea    0x30(%rsp),%r14
  lea    0x60(%rsp),%r13
L26:
  xor    %r15d,%r15d
  nopl   0x0(%rax,%rax,1)
L32:
  mov    (%rbx,%r15,4),%edi
  mov    %r12,%rsi
  call   <getVal@plt>
  mov    %eax,(%r14,%r15,4)
  add    $0x1,%r15
  cmp    $0xa,%r15
  jne    L32
  mov    %r12,%rsi
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  xor    %r15d,%r15d
  call   <printf@plt>
  mov    $0xa,%ecx
  mov    %r13,%rdx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <counting_sort@plt>
  nopl   0x0(%rax)
L19:
  mov    0x0(%r13,%r15,4),%edx
  mov    %r15,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  mov    %edx,(%rbx,%r15,4)
  add    $0x1,%r15
  call   <printf@plt>
  cmp    $0xa,%r15
  jne    L19
  add    $0x1,%r12
  cmp    $0x3,%r12
  jne    L26
  lea    D64(%rip),%rdi
  xor    %r12d,%r12d
  call   <puts@plt>
  nopl   0x0(%rax)
L7:
  mov    (%rbx,%r12,4),%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r12
  call   <printf@plt>
  cmp    $0xa,%r12
  jne    L7
  mov    $0x3,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L49
  add    $0x98,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L49:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D1
type: string
name: unpack_data
arguments: data_label: D48
type: string
name: unpack_data
arguments: data_label: D65
type: byte[16]
name: unpack_data
arguments: data_label: D17
type: byte[8]
name: unpack_data
arguments: data_label: D21
type: byte[16]
name: unpack_data
arguments: data_label: D37
type: string
name: unpack_data
arguments: data_label: D64
type: string
name: unpack_data
arguments: data_label: D7
type: string

tool: D1: The test data is:

tool: D48: %zu	,%d,


tool: {"D65": [0x46,0x03,0x00,0x00,0xca,0x01,0x00,0x00,0xd1,0x03,0x00,0x00,0x1e,0x01,0x00,0x00]}

tool: {"D17": [0x2d,0x02,0x00,0x00,0x63,0x02,0x00,0x00]}

tool: {"D21": [0x4a,0x03,0x00,0x00,0x72,0x03,0x00,0x00,0x31,0x03,0x00,0x00,0xb2,0x01,0x00,0x00]}

tool: D37: For the %zuth digit sorting:


tool: D64: For the final sorted data:

tool: D7: ....%d


assistant: ```c
int main(void) {
  int srcdata[10] = {838, 458, 977, 286, 842, 882, 817, 434, 557, 611};
  int testdata[10];
  int rtndata[10];
  int digit = 3;
  size_t idx, jdx;
  printf("The test data is:\n");
  for (idx = 0; idx < 10; idx++) {
    printf("%zu\t,%d,\n", idx, srcdata[idx]);
  }
  for (jdx = 0; jdx < digit; jdx++) {
    for (idx = 0; idx < 10; idx++) {
      testdata[idx] = getVal(srcdata[idx], jdx);
    }
    printf("For the %zuth digit sorting:\n", jdx);
    counting_sort(srcdata, testdata, rtndata, 10);
    for (idx = 0; idx < 10; idx++) {
      srcdata[idx] = rtndata[idx];
      printf("%zu\t,%d,\n", idx, rtndata[idx]);
    }
  }
  printf("For the final sorted data:\n");
  for (idx = 0; idx < 10; idx++) {
    printf("%zu\t,%d,\n", idx, srcdata[idx]);
  }
  printf("....%d\n", (int)pow(10, 3));
  return 0;
}
```