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

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x1,%edx
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,0x4(%rsp)
  mov    %rsi,0x8(%rsp)
L49:
  call   <get_next_tab_stop@plt>
  mov    $0x2,%r15d
  xor    %r13d,%r13d
  mov    %eax,%r14d
L100:
  xor    %ebx,%ebx
  xor    %ebp,%ebp
L185:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%r12d
  mov    D2(%rip),%rax
  cmp    %r12d,(%rax)
  je     L51
  cmp    %r14d,%r13d
  je     L38
  add    $0x1,%r13d
L157:
  cmp    $0x20,%r12d
  jne    L64
  add    $0x1,%ebx
  cmp    %r14d,%r13d
  jne    L185
  mov    0x8(%rsp),%rsi
  mov    0x4(%rsp),%edi
  mov    %r15d,%edx
  lea    0x1(%r15),%ebx
  mov    %ebx,%r15d
  add    $0x1,%ebp
  xor    %r13d,%r13d
  xor    %ebx,%ebx
  call   <get_next_tab_stop@plt>
  mov    %eax,%r14d
  jmp    L185
  nop
L64:
  movsbl %r12b,%eax
  mov    %eax,(%rsp)
  cmp    $0x9,%r12d
  je     L135
  test   %ebp,%ebp
  je     L110
  nopl   (%rax)
L32:
  mov    $0x9,%edi
  call   <putchar@plt>
  sub    $0x1,%ebp
  jne    L32
L110:
  test   %ebx,%ebx
  je     L184
  nopl   0x0(%rax,%rax,1)
L23:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L23
L184:
  mov    (%rsp),%edi
  call   <putchar@plt>
  cmp    $0xa,%r12d
  jne    L100
  mov    0x8(%rsp),%rsi
  mov    0x4(%rsp),%edi
  mov    $0x1,%edx
  jmp    L49
  nopl   0x0(%rax)
L38:
  mov    0x8(%rsp),%rsi
  mov    0x4(%rsp),%edi
  mov    %r15d,%edx
  lea    0x1(%r15),%r13d
  mov    %r13d,%r15d
  mov    $0x1,%r13d
  call   <get_next_tab_stop@plt>
  mov    %eax,%r14d
  jmp    L157
  xchg   %ax,%ax
L51:
  test   %ebp,%ebp
  je     L83
  nopl   0x0(%rax)
L40:
  mov    $0x9,%edi
  call   <putchar@plt>
  sub    $0x1,%ebp
  jne    L40
L83:
  test   %ebx,%ebx
  je     L22
  nopl   0x0(%rax,%rax,1)
L111:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L111
L22:
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L135:
  mov    0x8(%rsp),%rsi
  mov    0x4(%rsp),%edi
  mov    %r15d,%edx
  lea    0x1(%r15),%ebx
  call   <get_next_tab_stop@plt>
  mov    %eax,%r14d
  test   %ebp,%ebp
  je     L144
  mov    %ebx,%r15d
  xor    %r13d,%r13d
  xor    %ebx,%ebx
  jmp    L32
L144:
  mov    (%rsp),%edi
  mov    %ebx,%r15d
  xor    %r13d,%r13d
  xor    %ebx,%ebx
  call   <putchar@plt>
  jmp    L185
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c, t, s, pos, next_stop, index;
  t = s = pos = 0;
  index = 1;
  next_stop = get_next_tab_stop(argc, argv, index++);
  while ((c = getchar()) != EOF) {
    if (pos == next_stop) {
      pos = 1;
      next_stop = get_next_tab_stop(argc, argv, index++);
    } else {
      pos++;
    }
    if (c == ' ') {
      if (pos == next_stop) {
        t++;
        pos = 0;
        s = 0;
        next_stop = get_next_tab_stop(argc, argv, index++);
      } else {
        s++;
      }
    } else {
      if (c == '\t') {
        s = 0;
        pos = 0;
        next_stop = get_next_tab_stop(argc, argv, index++);
      }
      {
        for (; t > 0; t--)
          putchar('\t');
        for (; s > 0; s--)
          putchar(' ');
      };
      putchar(c);
      if (c == '\n') {
        t = s = pos = 0;
        index = 1;
        next_stop = get_next_tab_stop(argc, argv, index++);
      }
    }
  }
  {
    for (; t > 0; t--)
      putchar('\t');
    for (; s > 0; s--)
      putchar(' ');
  };
  return 0;
}
```