如何解决用于将const char *转换为char *的strdup
我为霍夫曼树设计了将二进制代码转换为较短的bin代码的方法。在main中,如果您调用Binary tree.init(q),则树将以key:频率和value:bin代码出现。问题是用char *转换const char *。我看过一些代码,在这里我使用strdup对其进行了转换。有时效果很好,但有时不起作用。所以我签出了该函数的参数。呼叫strdup或其他人有问题吗?
#pragma once
#include <stdio.h>
#include <queue>
#include <iostream>
#include "pch.h"
#include <string.h>
#include <string>
#define _CRT_SECURE_NO_WARNINGS
//this is a header file
using namespace std;
class Node
{
public:
//key : frequency,value : code
int f;
char* code;
Node* left;
Node* right;
int getFrequency()
{
return f;
}
char* getCode()
{
return code;
}
void init(int frequency,char* codestring)
{
f = frequency;
code = codestring;
}
Node* getLeft() {
return left;
}
Node* getRight()
{
return right;
}
void setLeft(Node* L)
{
left = L;
}
void setRight(Node* R)
{
right = R;
}
void setFrequency(int frequency)
{
f = frequency;
}
void setCode(char* string)
{
code = string;
}
};
class BinaryTree
{
public:
typedef priority_queue<int,vector<int>,greater<int>> pq;
pq q;
Node* proot;
int sizeofqueue;
void init(pq PriorityQueue)
{
q = PriorityQueue;
sizeofqueue = q.size();
N = 0;
int comparetimes = q.size() - 1;
for (int i = 0; i < comparetimes; i++)
{
if (i == 0)
{
put_first_two_nodes();
}
else
{
if (proot->getFrequency() <= q.top())
{
put_right_node();
}
else if (proot->getFrequency() > q.top())
{
put_left_node();
}
q.pop();
}
}
}
void put_first_two_nodes()
{
Node* pleft = new Node();
(*pleft).setFrequency(q.top());
(*pleft).setCode("0");
q.pop();
Node* pright = new Node();
(*pright).setFrequency(q.top());
(*pright).setCode("1");
put(pleft,pright);
q.pop();
}
void put_right_node()
{
Node* pright = new Node();
pright->setFrequency(q.top());
pright->setCode("1");
put(proot,pright);
appendcode(0);
}
void appendcode(int prefix)
{
string pre;
if (prefix == 1) pre = "1";
else pre = "0";
Node* targetNode = proot->getRight();
char* rcode = targetNode->getRight()->getCode();
char* lcode = targetNode->getLeft()->getCode();
string lefts = pre;
string rights = pre;
lefts.append(lcode);
rights.append(rcode);
char* leftstring = strdup(lefts.c_str());
char* rightstring = strdup(rights.c_str());
targetNode->getLeft()->setCode(leftstring);
targetNode->getRight()->setCode(rightstring);
free(leftstring);
free(rightstring);
}
void put_left_node()
{
Node* pleft = new Node();
pleft->setFrequency(q.top());
pleft->setCode("0");
put(pleft,proot);
appendcode(1);
}
char* get(int k)
{
return getItem(*proot,k);
}
char* getItem(Node root,int k)
{
//if there's no node
if (&root == nullptr) return "";
//if f or root > k,search left sibling
if (root.getFrequency() > k) return getItem(*(root.getLeft()),k);
//else,search right sibling
else if (root.getFrequency() < k) return getItem(*(root.getRight()),k);
//get it
else return root.getCode();
}
void put(Node* left,Node* right)
{
put_item(left,right);
}
void put_item(Node* left,Node* right)
{
//make new node that has sibling with left and right
Node* newnode = new Node();
newnode->setLeft(left);
newnode->setRight(right);
//exchange the new node and root without losing data
Node* temp;
temp = proot;
proot = newnode;
newnode = temp;
//proot's frequency : left f + right f
(*proot).setFrequency((*left).getFrequency() + (*right).getFrequency());
}
void printpost()
{
postorder(proot);
}
void postorder(Node* root)
{
if (root != nullptr)
{
if (root->getLeft() != nullptr) postorder(root->getLeft());
if (root->getRight() != nullptr) postorder(root->getRight());
printf("%d : %s ",root->getFrequency(),root->getCode());
}
}
private:
int N;
Node root;
};
解决方法
您不应该在c++中完全使用const char*
和char*
(除非有时要处理旧版或外部接口)。
切换您的代码以使用例如。改为使用std::string
或std::string_view
(c++17)(string_view需要更多的理解才能正确处理,并且可以说是const的-所以我会坚持不懈地进行字符串处理)。通过引用或必要时通过const引用传递std::string
。对于大多数程序来说,std::string
的开销是可以忽略的。
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