2

我的insert函数出现段错误：

current->isWord = true;

一切编译正常，没有警告或错误 ( g++ -Wall -Wextra)。我的main函数只调用insert一次函数，它不会工作。这是我的代码；它是我.h和.cpp文件之间的混合体：

const int alphabetSize = 26;

struct Node
{
    bool isWord;
    Node* child[alphabetSize];
};

Dictionary::Dictionary()
{
    initNode(head); //Node* head; is defined in my .h file under private:
}

bool Dictionary::isPrefix(string s)
{
    Node* current = endOfString(s, false);
    if (current == NULL)
    {
        return false;
    }
    else
    {
        return true;
    }
}

bool Dictionary::isWord(string s)
{
    Node* current = endOfString(s, false);
    if (current == NULL)
    {
        return false;
    }
    else
    {
        return current->isWord;
    }
}

void Dictionary::insert(string s)
{
    Node* current = endOfString(s, true);
    current->isWord = true; //segfault here
}

//initializes a new Node
void Dictionary::initNode(Node* current)
{
    current = new Node;
    current->isWord = false;
    for (int i = 0; i < alphabetSize; i++)
    {
       current->child[i] = NULL;
    }
}

//returns a pointer to the Node of the last character in the string
//isInsert tells it whether it needs to initialize new Nodes
Node* Dictionary::endOfString(string s, bool isInsert)
{
    Node* current = head;
    Node* next = head;
    for (unsigned int i = 0; i < s.length(); i++)
    {
        if (isalpha(s[i]) == true)
        {
            int letter = (tolower(s[i]) - 'a');
            next = current->child[letter];
            if (next == NULL)
            {
                if (isInsert == false)
                {
                    return NULL;
                }

                initNode(next);
                current->child[letter] = next;
            }
            current = current->child[letter];
        }
    }

    return current;
}
4

2 回答 2

4

initNode创建一个新的Node并对其进行初始化，但随后将其丢弃。因为current是按值传递的，所以在函数内部修改时，更改不会传播到initNode. 直接的解决方法是让它通过引用传递：

void Dictionary::initNode(Node*& current)
于 2013-04-09T03:55:20.943 回答
1

问题在这里：

//initializes a new Node
void Dictionary::initNode(Node* current)
{
    current = new Node;
    current->isWord = false;
    for (int i = 0; i < alphabetSize; i++)
    {
       current->child[i] = NULL;
    }
}

current是按值传入的，因此当您更改current方法时，您正在更改传入内容的副本，而不是外部变量。尝试传入Node** current一个指向您的指针的指针，以便您可以编辑原始变量。你会这样称呼它；initNode(&next);并且在该方法中，您将取消引用 current 以便能够编辑原始变量。

于 2013-04-09T03:55:02.447 回答