Question

    struct AdjListNode
    {
        int dest;
        struct AdjListNode* next;
    };

    // A structure to represent an adjacency list
    struct AdjList
    {
        struct AdjListNode *head;  // pointer to head node of list
    };

    // A structure to represent a graph. A graph is an array of adjacency lists.
    // Size of array will be V (number of vertices in graph)
    struct Graph
    {
        int V;
        struct AdjList* array;
    };


    struct Graph* createGraph(int V)
    {
        struct Graph* graph = (struct Graph*) malloc(sizeof(struct Graph));
        graph->V = V;

        // Create an array of adjacency lists.  Size of array will be V
        graph->array = (struct AdjList*) malloc(V * sizeof(struct AdjList));

         // Initialize each adjacency list as empty by making head as NULL
        int i;
        for (i = 0; i < V; ++i)
            graph->array[i].head = NULL;

        return graph;
    }

    // Adds an edge to an undirected graph
    void addEdge(struct Graph* graph, int src, int dest)
    {
        // Add an edge from src to dest.  A new node is added to the adjacency
        // list of src.  The node is added at the begining
        struct AdjListNode* newNode = newAdjListNode(dest);
        newNode->next = graph->array[src].head;
        graph->array[src].head = newNode;

        // Since graph is undirected, add an edge from dest to src also
        newNode = newAdjListNode(src);
        newNode->next = grap`enter code here`h->array[dest].head;
        graph->array[dest].head = newNode;
    }

特别是

graph->array[i].head = NULL;

好的，所以图形是Graph
因此使用 - ＆gt;访问数组，它是AdjList类型的图形对象有一个 AdjListNode类型的指针，它也是一个指针。

因为Graph中的数组成员是指向AdjList的指针，所以为什么是

graph->array[i]->head =  NULL;

没用过？

我知道当我们访问不是指针的struct成员时会使用.运算符。

somestructpointer -> itsmember

基本上是糖衣

(*somestructpointer).itsmember

我不明白发生了什么。 HELP。

Answer 1

AdjList* array是指向数组数组的指针。基本上，写array[i]等同于写*(array + i)。

所以我们有：

graph->array[i].head

变为

(*(graph->array + i)).head

作为一个有趣的奖金并且让你更加困惑，你也可以写i[array]，因为它基本上也是*(i + array)。

Answer 2

graph是一个指针 graph->array取消引用上面的指针以获取数组变量 graph->array[i]使用数组表示法来访问位置i的成员。

我们到达的成员是struct AdjList类型，它不是指针。这是一个struct。所以点（.）是合适的。

为什么指针值被用作。而不是 - ＆gt;在此

2 个答案: