user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    $0x0,%eax
  mov    %eax,-0x24(%rbp)
  mov    $0x10,%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x28(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rcx
  mov    $0x0,%ebx
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%ebx
  mov    $0x0,%edx
  div    %rbx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L36:
  cmp    %rdx,%rsp
  je     L8
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L36
L8:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L6
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L6:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x30(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,%r10
  mov    $0x21,%edi
  call   <add.0>
  lea    -0x30(%rbp),%rax
  mov    %rax,%r10
  mov    $0x1b,%edi
  call   <add.0>
  lea    -0x30(%rbp),%rax
  mov    %rax,%r10
  mov    $0x39,%edi
  call   <add.0>
  lea    -0x30(%rbp),%rax
  mov    %rax,%r10
  mov    $0x43,%edi
  call   <add.0>
  lea    -0x30(%rbp),%rax
  mov    %rax,%r10
  mov    $0x11,%edi
  call   <add.0>
  lea    -0x30(%rbp),%rax
  mov    %rax,%r10
  mov    $0x57,%edi
  call   <add.0>
  mov    $0x5b,%edi
  call   <putchar@plt>
  movl   $0x1,-0x3c(%rbp)
  jmp    L11
L49:
  mov    -0x30(%rbp),%rax
  mov    -0x3c(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x3c(%rbp)
L11:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x3c(%rbp)
  jle    L49
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%edi
  call   <exit@plt>
```

tool_calls: name: read_data
arguments: data_label: D1
type_infered: string
name: read_data
arguments: data_label: D12
type_infered: string

tool: D1: %d, 

tool: D12: ]


assistant: ```c
int main(int argc, char *argv[]) {
  int heapItems = 0;
  int heapSize = 16;
  int minHeap[heapSize];
  int upHeap(int x) {
    if (x == 1) {
      return 0;
    }
    if (minHeap[parent(x)] <= minHeap[x]) {
      return 0;
    }
    int temp = minHeap[x];
    minHeap[x] = minHeap[parent(x)];
    minHeap[parent(x)] = temp;
    upHeap(parent(x));
    return 0;
  }
  int add(int value) {
    if (heapItems == 0) {
      minHeap[1] = value;
      ++heapItems;
      return 0;
    }
    if ((heapItems + 1) == heapSize) {
    }
    ++heapItems;
    minHeap[heapItems] = value;
    upHeap(heapItems);
    return 0;
  }
  add(33);
  add(27);
  add(57);
  add(67);
  add(17);
  add(87);
  printf("[");
  int i = 1;
  for (; i <= heapItems; ++i) {
    printf("%d, ", minHeap[i]);
  }
  printf("]\n\n");
  exit(0);
}
```