我正在嘗試在C++中爲編程任務實現後綴trie。現在我認爲我有正確的想法,但是我一直在發生分段錯誤,而且我一直無法找到造成它的原因。使用OOP/C++實現後綴Trie
對於這個任務,我們鼓勵使用VIM /其他一些基本的文本編輯器,並從控制檯編譯程序。儘管如此,我已經下載了CLion來嘗試和調試代碼,所以我可以找到錯誤。
現在,在克利翁運行時,我得到的消息
terminate called after throwing an instance of 'std::bad_alloc'
what(): std::bad_alloc
試圖運行調試器會的消息
Error during pretty printers setup:
Undefined info command: "pretty-printer". Try "help info".
Some features and performance optimizations will not be available.
我是新來的克利翁,我不知道該怎麼辦關於這個(我使用的唯一的JetBrains IDE是Pycharm)。你能幫我解決這個問題嗎?
現在程序本身由三個類組成,Trie
,Edge
和Node
,其實現可以在下面看到。實施Trie背後的主要思想是在Trie.cpp
的構造函數中。
該代碼詳述如下。我感謝任何幫助。
Main.cpp的
#include <iostream>
using namespace std;
#include "Trie.hpp"
int main(){
string s = "Stef";
Trie trie(s);
return 0;
}
Trie.hpp
#ifndef TRIE_HPP
#define TRIE_HPP
#include <string>
#include "Node.hpp"
#include "Edge.hpp"
using namespace std;
class Trie{
private:
string T;
vector<Node> nodes;
void addWord(Node*, string);
public:
Trie(string);
};
#endif
Trie.cpp
#include <iostream>
#include <cstring>
#include "Trie.hpp"
using namespace std;
Trie::Trie(string T){
T += "#"; //terminating character
this->T = T;
vector<string> suffix; //array of suffixes
for(unsigned int i = 0; i < T.length(); i++)
suffix.push_back(T.substr(i, T.length()-i));
//Create the Root, and start from it
nodes.push_back(Node("")); //root has blank label
Node* currentNode = &nodes[0];
//While there are words in the array of suffixes
while(!suffix.empty()){
//If the character under consideration already has an edge, then this will be its index. Otherwise, it's -1.
int edgeIndex = currentNode->childLoc(suffix[0].at(0));
//If there is no such edge, add the rest of the word
if(edgeIndex == -1){
addWord(currentNode, suffix[0]); //add rest of word
suffix.erase(suffix.begin()); //erase the suffix from the suffix array
break; //break from the for loop
}
//if there is
else{
currentNode = (currentNode->getEdge(edgeIndex))->getTo(); //current Node is the next Node
suffix[0] = suffix[0].substr(1, suffix[0].length()); //remove first character
}
}
}
//This function adds the rest of a word
void Trie::addWord(Node* parent, string word){
for(unsigned int i = 0; i < word.length(); i++){ //For each remaining letter
nodes.push_back(Node(parent->getLabel()+word.at(i))); //Add a node with label of parent + label of edge
Edge e(word.at(i), parent, &nodes.back()); //Create an edge joining the parent to the node we just added
parent->addEdge(e); //Join the two with this edge
}
}
Node.hpp
#ifndef NODE_HPP
#define NODE_HPP
#include <string>
#include <vector>
#include "Edge.hpp"
using namespace std;
class Node{
private:
string label;
vector<Edge> outgoing_edges;
public:
Node();
Node(string);
string getLabel();
int childLoc(char);
void addEdge(Edge);
Edge* getEdge(int);
};
#endif
Node.cpp
#include "Node.hpp"
using namespace std;
Node::Node(){
}
Node::Node(string label){
this->label = label;
}
string Node::getLabel(){
return label;
}
//This function returns the edge matching the given label, returning -1 if there is no such edge.
int Node::childLoc(char label){
int loc = -1;
for(unsigned int i = 0; i < outgoing_edges.size(); i++)
if(outgoing_edges[i].getLabel() == label)
loc = i;
return loc;
}
void Node::addEdge(Edge e){
outgoing_edges.push_back(e);
}
Edge* Node::getEdge(int n){
return &outgoing_edges[n];
}
Edge.hpp
#ifndef EDGE_HPP
#define EDGE_HPP
#include <string>
using namespace std;
class Node; //Forward definition
class Edge{
private:
char label;
Node* from;
Node* to;
public:
Edge(char, Node*, Node*);
char getLabel();
Node* getTo();
Node* getFrom();
};
#endif
Edge.cpp
#include "Edge.hpp"
using namespace std;
Edge::Edge(char label, Node* from, Node* to){
this->label = label;
this->from = from;
this->to = to;
}
char Edge::getLabel(){
return label;
}
Node* Edge::getFrom(){
return from;
}
Node* Edge::getTo(){
return to;
}
感謝您的回覆。我不確定我是否理解 - 你的意思是*底層存儲被重新定位*?如果不是'&nodes [0]',我會如何設置'currentPointer'指向'nodes'的第一個元素? –
@LukeCollins向量將其元素存儲在動態分配的數組中。當你添加元素到它時,這個數組可能會被移動和展開。 (這在任何體面的入門書中都有介紹。)第二個問題的答案是,你不應該;你應該想出一個不同的方法來識別節點。 (我會使用節點的索引而不是地址。) – molbdnilo