c++ - 如何访问由另一个函数作为指针返回的对象的函数。

Question

又是那个在制作令人难以置信的模拟器的人!这次我遇到了一个问题，该函数由于语法原因被迫返回指向对象的指针。我想访问所述指针指向的对象的函数。我该怎么办？我的代码涉及的功能如下。

boggleNode * nextNode(int adj){
    return adjacency[adj];              //For switching between nodes. 
}
bool getUsed(){
    return isUsed;
}
private:
boggleNode * adjacency[8];
char letter;
bool isUsed;
};

最后，这里是包含上述函数的函数:

int sift(boggleNode bog, string word, string matcher){
int endofnodes;
string matchee;
if (bog.getData() == 'q')
    matchee = matcher + bog.getData() + 'u';
else
    matchee = matcher + bog.getData();
if (compare(word, matcher) == 0){
    cout << word << endl;
    return 5;                                                 //If it finds the word, escape the loop. 
}
if (word.find(matcher) == 0){
bog.makeUsed();
for (int j = 0; j < 8; j++){
    if (bog.nextNode(j) != NULL){
        if ((bog.nextNode(j)).getUsed() == 0){
            endofnodes = sift(bog.nextNode(j), word, matchee);
            if (endofnodes == 5)
                return endofnodes;
            }
        }
    bog.reset();
    return 0;
    //If it doesn't find anything, move onto next adjacency. 
    /*any words share a common starting letter sequence with matcher*/
    //Sift using an adjacent node, and add the old string to the new one. 
    }
}
}

其中我特别询问这一行:

if ((bog.nextNode(j)).getUsed() == 0)

当我尝试编译此代码时，出现“错误:成员请求getUsed' in (&bog)->boggleNode::nextNode(j)'，非类输入 `boggleNode*' "

非常感谢任何帮助。提前致谢!

Answer 1

您应该使用 -> 而不是 。:

if ((bog.nextNode(j))->getUsed() == 0)

在这种情况下，这是简写:

if ((*(bog.nextNode(j))).getUsed() == 0)