[英]Why does the following code failed to compile (C++ lambda problem)
Well I'm coding a function in which I used recursive lambda in C++ (GNU C++14).好吧,我正在编写 function,其中我在 C++(GNU C++14)中使用了递归 lambda。 So I tried the
y_combinator
method, beginning with a declaration outside the main()
function (I've used using namespace std;
before this).所以我尝试了
y_combinator
方法,从main()
function 之外的声明开始(在此之前我使用过using namespace std;
)。
#include <bits/stdc++.h>
using namespace std;
template<class Fun>
class y_combinator_result {
Fun fun_;
public:
template<class T>
explicit y_combinator_result(T &&fun): fun_(forward<T>(fun)) {}
template<class ...Args>
decltype(auto) operator()(Args &&...args) {
return fun_(ref(*this), forward<Args>(args)...);
}
};
template<class Fun>
decltype(auto) y_combinator(Fun &&fun) {
return y_combinator_result<decay_t<Fun>>(forward<Fun>(fun));
}
int main () {
cin.tie(0)->sync_with_stdio(0);
int N; cin >> N;
vector<int> adj[N+1];
for (int i = 0; i < N-1; i++) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
vector<int> d(N+1);
auto dfs = y_combinator([&](auto dfs, int u, int p, int &r) {
if (u == p) d[u] = 0;
if (d[u] > d[r]) r = u;
for (auto &v : adj[u]) {
if (v != p) {
d[v] = d[u]+1;
dfs(v, u, r);
}
}
});
}
Now I have a simple, easy to understand question for you guys: What make this lambda function failed to even compile (this happen inside the main function)?现在我有一个简单易懂的问题给你们:是什么让这个 lambda function 甚至无法编译(这发生在主函数内部)?
The thing is, when I'm trying to compile it, I received this catastrophic message:问题是,当我试图编译它时,我收到了这个灾难性的消息:
test1.cpp: In function 'int main()':
test1.cpp:36:29: error: use of deleted function 'main()::<lambda(auto:1, int, int, int&)>::~<lambda>()'
36 | auto dfs = y_combinator([&](auto dfs, int u, int p, int &r) {
| ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
37 | if (u == p) d[u] = 0;
| ~~~~~~~~~~~~~~~~~~~~~
38 | if (d[u] > d[r]) r = u;
| ~~~~~~~~~~~~~~~~~~~~~~~
39 | for (auto &v : adj[u]) {
| ~~~~~~~~~~~~~~~~~~~~~~~~
40 | if (v != p) {
| ~~~~~~~~~~~~~
41 | d[v] = d[u]+1;
| ~~~~~~~~~~~~~~
42 | dfs(v, u, r);
| ~~~~~~~~~~~~~
43 | }
| ~
44 | }
| ~
45 | });
| ~
test1.cpp:36:31: note: 'main()::<lambda(auto:1, int, int, int&)>::~<lambda>()' is implicitly deleted because the default definition would be ill-formed:
36 | auto dfs = y_combinator([&](auto dfs, int u, int p, int &r) {
| ^
So I try a small fix by capturing 2 vector name d
and adj
所以我通过捕获 2 个向量名称
d
和adj
来尝试一个小修复
#include <bits/stdc++.h>
using namespace std;
template<class Fun>
class y_combinator_result {
Fun fun_;
public:
template<class T>
explicit y_combinator_result(T &&fun): fun_(forward<T>(fun)) {}
template<class ...Args>
decltype(auto) operator()(Args &&...args) {
return fun_(ref(*this), forward<Args>(args)...);
}
};
template<class Fun>
decltype(auto) y_combinator(Fun &&fun) {
return y_combinator_result<decay_t<Fun>>(forward<Fun>(fun));
}
int main () {
cin.tie(0)->sync_with_stdio(0);
int N; cin >> N;
vector<int> adj[N+1];
for (int i = 0; i < N-1; i++) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
vector<int> d(N+1);
auto dfs = y_combinator([&d, &adj](auto dfs, int u, int p, int &r) { // I changed this line
if (u == p) d[u] = 0;
if (d[u] > d[r]) r = u;
for (auto &v : adj[u]) {
if (v != p) {
d[v] = d[u]+1;
dfs(v, u, r);
}
}
});
}
and the program compiled as intended... Obviously there is a very big question from my perspective: Why did the first piece of code that I used [&]
to capture everything in side the main()
function failed, but the second one work?并且程序按预期编译...显然从我的角度来看有一个非常大的问题:为什么我使用
[&]
捕获main()
function 中的所有内容的第一段代码失败了,但第二段代码有效? Is there any way around this, cuz I don't want to spend time capturing variables...有什么办法可以解决这个问题,因为我不想花时间捕捉变量......
P/s: forgive me for my bad english. P/s:原谅我的英语不好。 I'm not a native speaker.
我不是母语人士。
The C++ does not have VLA vector<int> adj[N+1];
C++ 没有 VLA
vector<int> adj[N+1];
. . Using this VLA causes the compiler error.
使用此 VLA 会导致编译器错误。 See Variable Length Array (VLA) in C++ compilers .
请参阅C++ 编译器中的可变长度数组 (VLA) 。
Making vector<int> adj[N+1];
制作
vector<int> adj[N+1];
-> vector<vector<int>> adj(N+1);
->
vector<vector<int>> adj(N+1);
makes the example compliable.使示例兼容。
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