关于static修饰符

蒟蒻在做 P1919 【模板】A*B Problem 升级版（FFT 快速傅里叶变换） 这题时遇到一个关于static修饰符的问题：

在NTT时：

• 若在临时前加入static修饰符，则会无报错消息地编译失败：

  void NTT(int *g, bool fl, int n)
  {
  tsetup(n);
  static ull f[MAXN << 1], w[MAXN << 1] = {1};//就是这里
  for (int i = 0; i < n; i++)
      f[i] = (((ll)MOD << 5) + g[tr[i]]) % MOD;
  for (int len = 1; len < n; len <<= 1)
  {
      ull rt = fpow(fl ? _G : invG, (MOD - 1) / (len << 1));
      for (int i = 1; i < len; i++)
          w[i] = w[i - 1] * rt % MOD;
      for (int p = 0; p < n; p += (len << 1))
          for (int cur = 0; cur < len; cur++)
          {
              int t = w[cur] * f[p | len | cur] % MOD;
              f[p | len | cur] = f[p | cur] + MOD - t;
              f[p | cur] = f[p | cur] + t;
          }
      if (len == (1 << 10))
          for (int i = 0; i < n; i++)
              f[i] %= MOD;
  }
  if (fl)
      for (int i = 0; i < n; i++)
          g[i] = f[i] % MOD;
  else
  {
      ull invn = inv(n);
      for (int i  =0; i < n; i++)
          g[i] = f[i] % MOD * invn % MOD;
  }
  }

• 若去除修饰符，则可以愉快的AC：

  oid NTT(int *g, bool fl, int n)
  {
  tsetup(n);
  ull f[MAXN << 1], w[MAXN << 1] = {1};//还是这里
  for (int i = 0; i < n; i++)
      f[i] = (((ll)MOD << 5) + g[tr[i]]) % MOD;
  for (int len = 1; len < n; len <<= 1)
  {
      ull rt = fpow(fl ? _G : invG, (MOD - 1) / (len << 1));
      for (int i = 1; i < len; i++)
          w[i] = w[i - 1] * rt % MOD;
      for (int p = 0; p < n; p += (len << 1))
          for (int cur = 0; cur < len; cur++)
          {
              int t = w[cur] * f[p | len | cur] % MOD;
              f[p | len | cur] = f[p | cur] + MOD - t;
              f[p | cur] = f[p | cur] + t;
          }
      if (len == (1 << 10))
          for (int i = 0; i < n; i++)
              f[i] %= MOD;
  }
  if (fl)
      for (int i = 0; i < n; i++)
          g[i] = f[i] % MOD;
  else
  {
      ull invn = inv(n);
      for (int i  =0; i < n; i++)
          g[i] = f[i] % MOD * invn % MOD;
  }
  }

  有没有大佬知道这是什么原因？