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

```asm
<main>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  lea    D58(%rip),%rsi
  lea    D30(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movaps %xmm0,0x10(%rsp)
  call   <fopen@plt>
  test   %rax,%rax
  je     L78
  lea    0xc(%rsp),%r13
  mov    %rax,%rdi
  mov    %rax,%rbx
  xor    %ebp,%ebp
  lea    D21(%rip),%r12
  mov    %r13,%rdx
  lea    D24(%rip),%r14
  mov    %r12,%rsi
  call   <fscanf@plt>
  mov    $0xff,%esi
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L10
L28:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <fscanf@plt>
L10:
  mov    %rbx,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L70
  mov    %ebp,%esi
  mov    0xc(%rsp),%r8d
  add    $0x1,%ebp
  mov    %r14,%rdi
  and    $0x3,%esi
  mov    0x10(%rsp,%rsi,4),%edx
  add    %r8d,%edx
  mov    %edx,%eax
  movzbl %dl,%edx
  shr    $0x8,%eax
  add    %edx,%eax
  mov    %r8d,%edx
  mov    %eax,%ecx
  shr    $0x8,%ecx
  add    %eax,%ecx
  xor    %eax,%eax
  mov    %ecx,0x10(%rsp,%rsi,4)
  mov    %ebp,%esi
  call   <printf@plt>
  cmp    $0x2d,%ebp
  je     L14
  imul   $0xcccccccd,%ebp,%eax
  ror    %eax
  cmp    $0x19999999,%eax
  ja     L28
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L28
L14:
  mov    $0x2c,%ebp
L70:
  mov    0x14(%rsp),%ecx
  mov    0x10(%rsp),%edx
  mov    %ebp,%esi
  xor    %eax,%eax
  mov    0x1c(%rsp),%r9d
  mov    0x18(%rsp),%r8d
  lea    D65(%rip),%rdi
  not    %ecx
  not    %edx
  not    %r9d
  not    %r8d
  call   <printf@plt>
  xor    %eax,%eax
L62:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L59
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L78:
  lea    D28(%rip),%rdi
  call   <puts@plt>
  or     $0xffffffff,%eax
  jmp    L62
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D58
datatype: string
name: disassemble_data
arguments: data_tag: D30
datatype: string
name: disassemble_data
arguments: data_tag: D21
datatype: string
name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D2
datatype: string
name: disassemble_data
arguments: data_tag: D65
datatype: string
name: disassemble_data
arguments: data_tag: D28
datatype: string

tool: D58: r

tool: D30: header.txt

tool: D21: %x

tool: D24: %02x: %02x sum(%x) 

tool: D2: test %d


tool: D65: 
hlen = %d,sum[0-3] %02x-%02x-%02x-%02x 


tool: D28: error opening header.txt

assistant: ```c
int main() {
  unsigned int sum[4] = {0};
  unsigned int hdata, hlen = 0;
  FILE *fp = fopen("header.txt", "r");
  if (fp == NULL) {
    printf("error opening header.txt\n");
    return -1;
  }
  fscanf(fp, "%x", &hdata);
  printf("test %d\n", 0 + 0xff);
  while (!feof(fp)) {
    sum[hlen % 4] += hdata;
    sum[hlen % 4] = (sum[hlen % 4] >> 8) + (sum[hlen % 4] & 0xff);
    sum[hlen % 4] += (sum[hlen % 4] >> 8);
    printf("%02x: %02x sum(%x) ", (hlen + 1), hdata, sum[hlen % 4]);
    if (hlen == 44)
      break;
    hlen++;
    if (hlen % 10 == 0)
      printf("\n");
    fscanf(fp, "%x", &hdata);
  }
  printf("\nhlen = %d,sum[0-3] %02x-%02x-%02x-%02x \n", hlen, ~sum[0], ~sum[1], ~sum[2], ~sum[3]);
  return 0;
}
```