user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movl   $0x80000000,-0x24(%rbp)
  movl   $0xffffffff,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  mov    %rax,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x20,-0x1c(%rbp)
  movl   $0x1,-0x2c(%rbp)
  movl   $0x0,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%ecx
  shll   %cl,-0x2c(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L126
  negq   -0x18(%rbp)
L126:
  cmpl   $0x0,-0x20(%rbp)
  jns    L118
  negq   -0x10(%rbp)
L118:
  cmpq   $0x1,-0x10(%rbp)
  jne    L6
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,-0x24(%rbp)
  jns    L103
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L103:
  cmpl   $0x0,-0x20(%rbp)
  jns    L55
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L55:
  cmpl   $0x0,-0x24(%rbp)
  js     L30
  cmpl   $0x0,-0x20(%rbp)
  js     L32
L30:
  cmpl   $0x0,-0x24(%rbp)
  jns    L10
  cmpl   $0x0,-0x20(%rbp)
  js     L10
L32:
  negq   -0x8(%rbp)
L10:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  jmp    L88
L59:
  movl   $0xffffffff,-0x28(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %eax,-0x2c(%rbp)
  jmp    L100
L92:
  shll   -0x2c(%rbp)
  addl   $0x1,-0x28(%rbp)
L100:
  mov    -0x2c(%rbp),%eax
  cltq
  cmp    %rax,-0x18(%rbp)
  jge    L92
  mov    -0x2c(%rbp),%eax
  sar    %eax
  cltq
  sub    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  cltq
  add    %rax,-0x8(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jge    L59
  cmpl   $0x0,-0x24(%rbp)
  setg   %dl
  cmpl   $0x0,-0x20(%rbp)
  setg   %al
  xor    %edx,%eax
  test   %al,%al
  je     L96
  negq   -0x8(%rbp)
L96:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
L88:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D17
datatype: string
name: parse_data
arguments: data_label: D26
datatype: string
name: parse_data
arguments: data_label: D1
datatype: string
name: parse_data
arguments: data_label: D32
datatype: string

tool: D17: pre res=%ld


tool: D26: a<0,

tool: D1: b<0

tool: D32: res= %ld


assistant: ```c
int main() {
  int a = -2147483648;
  int b = -1;
  long long a1 = (long long)a;
  long long b1 = (long long)b;
  long long res = 0;
  int len = sizeof(int) * 8;
  int tmp = 1;
  int step = 0;
  tmp <<= len - 2;
  if (a < 0)
    a1 = 0 - a1;
  if (b < 0)
    b1 = 0 - b1;
  if (b1 == 1) {
    res = a1;
    printf("pre res=%ld\n", res);
    if (a < 0)
      printf("a<0,");
    if (b < 0)
      printf("b<0\n");
    if (((a >= 0) && (b < 0)) || ((a < 0) && (b >= 0)))
      res = 0 - res;
    printf("res= %ld\n", res);
    return res;
  }
  while (a1 >= b1) {
    step = -1;
    tmp = b1;
    while (a1 >= tmp) {
      tmp <<= 1;
      step++;
    }
    a1 -= tmp >> 1;
    res += 1 << step;
  }
  if ((a > 0) != (b > 0)) {
    res = 0 - res;
  }
  printf("res= %ld\n", res);
  return res;
}
```