user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x288,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x278(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebx
  lea    D19(%rip),%r12
L100:
  mov    %ebx,%ebp
  lea    0xf(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  movzbl 0xf(%rsp),%eax
  cmp    $0xa,%al
  je     L63
  cmp    $0x20,%al
  mov    $0x2d,%edx
  cmove  %edx,%eax
  mov    %al,0x1a0(%rsp,%rbx,1)
  add    $0x1,%rbx
  jmp    L100
L63:
  lea    0x10(%rsp),%rdi
  mov    $0x32,%ecx
  mov    $0x0,%eax
  rep stos %rax,%es:(%rdi)
  test   %ebx,%ebx
  jle    L102
  movslq %ebx,%rax
  lea    0x19f(%rsp,%rax,1),%rdx
  lea    0x210(%rsp),%rax
  lea    -0x1(%rbx),%ecx
  lea    0x211(%rsp,%rcx,1),%rsi
L121:
  movzbl (%rdx),%ecx
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L121
L102:
  movslq %ebp,%rbp
  movb   $0x2a,0x210(%rsp,%rbp,1)
  lea    0x10(%rsp),%rbp
  mov    %rbp,%r14
  mov    $0x0,%r13d
  mov    $0x0,%r15d
  jmp    L3
L24:
  mov    %r13d,%eax
  mov    %r15d,%r12d
L18:
  mov    %r12d,(%r14)
  cmp    $0x2d,%dil
  sete   %r13b
  movzbl %r13b,%r13d
  add    %eax,%r13d
  add    $0x4,%r14
  movslq %r13d,%rax
  cmpb   $0x2a,0x210(%rsp,%rax,1)
  je     L48
L3:
  movslq %r13d,%rax
  movzbl 0x210(%rsp,%rax,1),%edi
  cmp    $0x2a,%dil
  je     L24
  cmp    $0x2d,%dil
  je     L24
  lea    0x1(%r13),%ebx
  movslq %ebx,%rbx
  mov    %r15d,%r12d
L90:
  mov    %ebx,%esi
  sub    %r13d,%esi
  sub    $0x1,%esi
  movsbl %dil,%edi
  sub    $0x30,%edi
  call   <expo@plt>
  add    %eax,%r12d
  mov    %ebx,%eax
  movzbl 0x210(%rsp,%rbx,1),%edi
  add    $0x1,%rbx
  cmp    $0x2a,%dil
  je     L18
  cmp    $0x2d,%dil
  jne    L90
  jmp    L18
L48:
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rdi
  call   <stack@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    0x190(%rbp),%rax
  mov    $0x0,%esi
L64:
  add    0x0(%rbp),%esi
  add    $0x4,%rbp
  cmp    %rax,%rbp
  jne    L64
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x278(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L46
  mov    $0x0,%eax
  add    $0x288,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L46:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D19
type: string
name: extract_symbolic_data
arguments: label: D24
type: string
name: extract_symbolic_data
arguments: label: D4
type: string

tool: D19: %c

tool: D24: Your list: 

tool: D4: The sum of the integers is: %d


assistant: ```c
int main(void) {
  char list[100];
  int counter = 0;
  char temp;
  do {
    scanf("%c", &temp);
    if (temp == '\n') {
      break;
    }
    if (temp == ' ') {
      list[counter] = '-';
      counter++;
      continue;
    }
    list[counter] = temp;
    counter++;
  } while (1);
  int intlist[100] = {0};
  char revlist[100];
  int i;
  for (i = 0; i < counter; i++) {
    revlist[i] = list[counter - 1 - i];
  }
  revlist[counter] = '*';
  int tet = 0, sum, exp, cot = 0, temp1;
  do {
    exp = 0;
    sum = 0;
    while (revlist[tet] != '-') {
      if (revlist[tet] == '*') {
        break;
      }
      temp1 = revlist[tet] - '0';
      sum += expo(temp1, exp);
      exp++;
      tet++;
    }
    intlist[cot] = sum;
    cot++;
    if (revlist[tet] == '-') {
      tet++;
      continue;
    }
  } while (revlist[tet] != '*');
  printf("Your list: ");
  stack(intlist);
  printf("\n");
  int sum2 = 0;
  for (i = 0; i < 100; i++) {
    sum2 += intlist[i];
  }
  printf("The sum of the integers is: %d\n", sum2);
  return 0;
}
```