相同的代码在使用不同版本的C++时会产生不同的输出。

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英文:

The same code gives different output when using different versions of C++

问题

我正在解决这个问题,然后我在不同版本的C++之间遇到了奇怪的行为。当我使用C++17时,代码输出正确,但当我切换到C++14时,输出完全改变了。我使用的编译器是g++ (x86_64-posix-seh-rev1, Built by MinGW-Builds project) 13.1.0。如果有人能指出发生了什么,我将不胜感激。

这是我对该问题的解决方案。

#include <bits/stdc++.h>

#define nl '\n'

using namespace std;

using i64 = long long;

struct PersistentSegmentTree {
  struct Node {
    int value, left, right;

    Node() : value(0), left(0), right(0) {}
  };
  vector<Node> T = {Node()};

  int update(int root, int l, int r, int p, int v) {
    int id = T.size();
    T.emplace_back();
    T[id] = T[root];
    if (l == r) {
      T[id].value += v;
      return id;
    }
    
    int mid = (l + r) / 2;
    if (p <= mid) {
      T[id].left = update(T[root].left, l, mid, p, v);
    } else {
      T[id].right = update(T[root].right, mid + 1, r, p, v);
    }
    
    T[id].value = T[T[id].left].value + T[T[id].right].value;
    return id;
  }

  int query(int from, int to, int l, int r, int k) {
    if (l == r) {
      return l;
    }
    int mid = (l + r) / 2;
    int left_count = T[T[to].left].value - T[T[from].left].value;
    if (left_count >= k) {
      return query(T[from].left, T[to].left, l, mid, k);
    } else {
      return query(T[from].right, T[to].right, mid + 1, r, k - left_count);
    }
  }
} tree;

signed main() {
  cin.tie(0), ios::sync_with_stdio(0);

  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i]; 
  }

  vector<int> roots;
  roots.push_back(0);
  
  auto vals = a;
  sort(vals.begin(), vals.end());
  vals.erase(unique(vals.begin(), vals.end()), vals.end());

  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(vals.begin(), vals.end(), a[i]) - vals.begin();
    roots.push_back(tree.update(roots.back(), 0, vals.size() - 1, a[i], 1));
  }

  for (int qi = 0; qi < q; qi++) {
    int l, r, k;
    cin >> l >> r >> k;
    cout << vals[tree.query(roots[l - 1], roots[r], 0, vals.size() - 1, k)] << nl;    
  }
}

测试用例是:

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

C++17 输出(正确):

5 
6 
3

C++14 输出(错误):

7
7
4

提前感谢。

英文:

I was solving this problem and then I encounter this weird behaviour between different versions of C++. When I use C++17, the code gives correct output, but when I switch to C++14, the output changes completely. I was compiling with g++ (x86_64-posix-seh-rev1, Built by MinGW-Builds project) 13.1.0. I would appreciate it if someone would point out what is happening.

This is my solution to that problem.

#include &lt;bits/stdc++.h&gt;

#define nl &#39;\n&#39;

using namespace std;

using i64 = long long;

struct PersistentSegmentTree {
  struct Node {
    int value, left, right;

    Node() : value(0), left(0), right(0) {}
  };
  vector&lt;Node&gt; T = {Node()};

  int update(int root, int l, int r, int p, int v) {
    int id = T.size();
    T.emplace_back();
    T[id] = T[root];
    if (l == r) {
      T[id].value += v;
      return id;
    }
    
    int mid = (l + r) / 2;
    if (p &lt;= mid) {
      T[id].left = update(T[root].left, l, mid, p, v);
    } else {
      T[id].right = update(T[root].right, mid + 1, r, p, v);
    }
    
    T[id].value = T[T[id].left].value + T[T[id].right].value;
    return id;
  }

  int query(int from, int to, int l, int r, int k) {
    if (l == r) {
      return l;
    }
    int mid = (l + r) / 2;
    int left_count = T[T[to].left].value - T[T[from].left].value;
    if (left_count &gt;= k) {
      return query(T[from].left, T[to].left, l, mid, k);
    } else {
      return query(T[from].right, T[to].right, mid + 1, r, k - left_count);
    }
  }
} tree;

signed main() {
  cin.tie(0), ios::sync_with_stdio(0);

  int n, q;
  cin &gt;&gt; n &gt;&gt; q;
  vector&lt;int&gt; a(n);
  for (int i = 0; i &lt; n; i++) {
    cin &gt;&gt; a[i]; 
  }

  vector&lt;int&gt; roots;
  roots.push_back(0);
  
  auto vals = a;
  sort(vals.begin(), vals.end());
  vals.erase(unique(vals.begin(), vals.end()), vals.end());

  for (int i = 0; i &lt; n; i++) {
    a[i] = lower_bound(vals.begin(), vals.end(), a[i]) - vals.begin();
    roots.push_back(tree.update(roots.back(), 0, vals.size() - 1, a[i], 1));
  }

  for (int qi = 0; qi &lt; q; qi++) {
    int l, r, k;
    cin &gt;&gt; l &gt;&gt; r &gt;&gt; k;
    cout &lt;&lt; vals[tree.query(roots[l - 1], roots[r], 0, vals.size() - 1, k)] &lt;&lt; nl;    
  }
}

The test case is:

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

C++17 output (correct):

5 
6 
3

C++14 output (incorrect):

7
7
4

Thanks in advance.

答案1

得分: 3

你在所有的作业中都有问题,比如T.at(id).left = update(...

在C++17之前,对赋值语句的左右两边的求值顺序是未指定的。T.at(id).left似乎首先被求值,其结果类型是Node&amp;。然后调用update()函数,该函数向向量T.emplace_back()添加一个新元素,这会使所有引用失效,因此首先求值的引用T.at(id).left现在是一个指向堆中已释放内存的悬空引用。在update()完成后,将结果值赋给悬空指针会导致堆使用已释放内存的问题。

修复方法非常简单:

  const auto left = update(T.at(root).left, l, mid, p, v);
T.at(id).left = left;

希望你能自己修复其他的赋值语句。

英文:

You have issues in all assignments like T.at(id).left = update(....

The evaluation of the left and right sides of the assignment is unspecified until C++17. T.at(id).left seems to be evaluated first, its result type is Node&amp;. Then update() is called, that adds a new element to the vector T.emplace_back() - this invalidates all references, thus the first evaluated reference T.at(id).left is now a dangling reference to a freed memory in the heap. After uppdate() has finished, the assignent of the result value to the dangling pointer results in the heap-use-after-free.

The fix is quite trivial:

  const auto left = update(T.at(root).left, l, mid, p, v);
T.at(id).left = left;

I hope you will fix the other assignment yourself.

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  • 本文由 发表于 2023年8月9日 10:40:39
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