user: What would the C source look like for the assembly code provided here?

```asm
<sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x20(%rbp)
  jmp    L26
L47:
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L49
L1:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jle    L20
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L20:
  addl   $0x1,-0x1c(%rbp)
L49:
  cmpl   $0x9,-0x1c(%rbp)
  jle    L1
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  addl   $0x1,-0x20(%rbp)
L26:
  cmpl   $0x8,-0x20(%rbp)
  jle    L47
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x20(%rbp)
  jmp    L43
L46:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x20(%rbp)
L43:
  cmpl   $0x9,-0x20(%rbp)
  jle    L46
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L53
  call   <__stack_chk_fail@plt>
L53:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D13
type_infered: string
name: unpack_data
arguments: tag: D11
type_infered: string

tool: D13: 
 result:

tool: D11: 
 %5d%10s

assistant: ```c
void sort(int num[], char name[10][8]) {
  int i, j, min, templ;
  char temp2[8];
  for (i = 0; i < 10 - 1; i++) {
    min = i;
    for (j = i; j < 10; j++)
      if (num[min] > num[j])
        min = j;
    templ = num[i];
    strcpy(temp2, name[i]);
    num[i] = num[min];
    strcpy(name[i], name[min]);
    num[min] = templ;
    strcpy(name[min], temp2);
  }
  printf("\n result:\n");
  for (i = 0; i < 10; i++)
    printf("\n %5d%10s", num[i], name[i]);
}
```