user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x30,%rsp
  mov    %rsi,%rbx
  cmp    $0x2,%edi
  je     L64
  mov    (%rsi),%rsi
  lea    D36(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L46:
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  ret
L64:
  mov    0x8(%rsi),%rax
  movsbl (%rax),%edi
  call   <isdigit@plt>
  test   %eax,%eax
  jne    L40
  lea    D25(%rip),%rdi
  call   <puts@plt>
  jmp    L46
L40:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  fld    %st(0)
  fstpt  (%rsp)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,%rbx
  add    $0x10,%rsp
  mov    %rax,%rdx
  mov    $0x0,%eax
  fldz
  fldt   (%rsp)
  fcomip %st(1),%st
  fstp   %st(0)
  jbe    L94
L110:
  movl   $0x1,(%rdx)
  add    $0x1,%rax
  add    $0x4,%rdx
  mov    %rax,0x18(%rsp)
  fildll 0x18(%rsp)
  fldt   (%rsp)
  fcomip %st(1),%st
  fstp   %st(0)
  ja     L110
L94:
  flds   0xe03(%rip)        # 2074 <_fini+0xd08>
  fldt   (%rsp)
  fcomip %st(1),%st
  jbe    L63
  lea    0x4(%rbx),%rsi
  lea    0xc(%rbx),%r9
  fldz
  fstpt  0x20(%rsp)
  mov    $0x0,%r10d
  mov    $0x2,%ecx
  jmp    L125
L28:
  add    $0x1,%r10d
  fldt   0x20(%rsp)
  faddp  %st,%st(1)
  fstpt  0x20(%rsp)
  jmp    L86
L142:
  fstp   %st(0)
L86:
  add    $0x1,%rcx
  mov    %rcx,0x18(%rsp)
  fildll 0x18(%rsp)
  add    $0x4,%rsi
  add    $0x8,%r9
  fldt   (%rsp)
  fcomip %st(1),%st
  jbe    L121
L125:
  cmpl   $0x0,(%rsi)
  je     L142
  mov    %rcx,%rdi
  lea    (%rcx,%rcx,1),%rdx
  mov    %rdx,%rax
  mov    %rdx,0x18(%rsp)
  fildll 0x18(%rsp)
  fldt   (%rsp)
  fcomip %st(1),%st
  fstp   %st(0)
  jbe    L28
  lea    0x0(,%rcx,4),%r8
  mov    %r9,%rdx
L16:
  movl   $0x0,(%rdx)
  add    %rdi,%rax
  add    %r8,%rdx
  mov    %rax,0x18(%rsp)
  fildll 0x18(%rsp)
  fldt   (%rsp)
  fcomip %st(1),%st
  fstp   %st(0)
  ja     L16
  jmp    L28
L63:
  fstp   %st(0)
  fldz
  fstpt  0x20(%rsp)
  mov    $0x0,%r10d
  jmp    L115
L121:
  fstp   %st(0)
L115:
  push   0x8(%rsp)
  push   0x8(%rsp)
  mov    %r10d,%esi
  lea    D28(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  push   0x38(%rsp)
  push   0x38(%rsp)
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x20,%rsp
  mov    %rbx,%rdi
  call   <free@plt>
  jmp    L46
```

tool_calls: name: extract_symbolic_data
arguments: tag: D36
type: string
name: extract_symbolic_data
arguments: tag: D25
type: string
name: extract_symbolic_data
arguments: tag: D28
type: string
name: extract_symbolic_data
arguments: tag: D17
type: string

tool: D36: Usage: %s [RANGE]...


tool: D25: Range must be a number.

tool: D28: There are %d primes below %LG.


tool: D17: The sum of these primes is %LG.


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc != 2)
    printf("Usage: %s [RANGE]...\n", argv[0]);
  else if (!isdigit(argv[1][0]))
    printf("Range must be a number.\n");
  else {
    const long double max = strtol(argv[1], NULL, 10);
    int *set = (int *)calloc(max, sizeof(int));
    long i = 0;
    for (i = 0; i < max; i++)
      set[i] = 1;
    long pos = 0;
    int count = 0;
    long double sum = 0;
    for (pos = 2; pos < max; pos++) {
      if (set[pos - 1] != 0) {
        for (i = 2; (i * pos) < max; i++)
          set[(i * pos - 1)] = 0;
        count++;
        sum = sum + pos;
      }
    }
    printf("There are %d primes below %LG.\n", count, max);
    printf("The sum of these primes is %LG.\n", sum);
    free(set);
  }
  return 0;
}
```