传递指针数组功能 [英] Passing pointers to arrays to functions

问题描述

我存储指针结构数组我的信息。换句话说，阵列的每个元素是一个指向一个链表。

I am storing my information in an array of pointers to structs. In other words, each element of the array is a pointer to a linked list.

我不知道数组应该有多长等等，而不是初始化数组中的我的main（）函数，而不是我的intialize双指针

I don't know how long the array should be, so instead of initializing the array in my main() function, I instead intialize the double pointer

struct graph** graph_array;

然后，一旦我获得该阵列的长度，我尝试使用函数GraphInitialize初始化graph_array的每个元素：

Then once I obtain the length of the array, I try to initialize each element of graph_array using a function GraphInitialize:

int GraphInitialize(struct graph* *graph_array,int vertices)
{ 
  struct graph* graph_array2[vertices+1];
  graph_array = graph_array2;
  int i;

  for (i=0;i<vertices+1;i++)
  {
    graph_array[i] = NULL;
  }

  return 0;
}

但出于某种原因，这是不返回更新graph_array主（）。基本上，这个功能被本地更新graph_array，并没有改变正在取得。因此，任何时候，我尝试访问的graph_array它赛格故障，因为它未初始化的元素。我在做什么错了？

But for some reason this is not returning the updated graph_array to main(). Basically, this function is updating graph_array locally, and no change is being made. As a result, any time I try to access an element of graph_array it seg faults because it is not initialized. What am I doing wrong?

编辑：在与汤姆的啊康沃我要补充别的东西，使这个更加混乱

Following the convo with Tom Ahh I should add something else that makes this more confusing.

我不直接从主调用GraphIntialize（）。相反，我打电话从主要的getData（），并传递一个指向graph_array到GETDATA如下图所示。

I don't call GraphIntialize directly from main(). Instead, I call getdata() from main, and pass a pointer to graph_array to getdata as shown below.

    getdata(argc, argv, vertpt, edgept, &graph_array)

    int getdata(int argc, char *argv[], int *verts, int *edges, struct graph* **graph_array)

接着的getData会从我的输入文件的顶点的数目，并使用该调用GraphInitialize：

Then getdata gets the number of vertices from my input file, and uses that to call GraphInitialize:

    if ((GraphInitialize(&graph_array, *verts)) == -1)
    {
      printf("GraphCreate failed");
      return 0;
    }

这导致错误：预计'结构图3ASTERISKS'，但参数的类型为结构图4ASTERISKS

This results in an error: "expected 'struct graph 3ASTERISKS' but argument is of type 'struct graph 4ASTERISKS'.

推荐答案

在您指定的东西 graph_array ，您只需将其分配给它的本地副本。在功能它所做的更改不会被看到，可以通过调用者。您需要通过指针的值传递它能够改变它的值。你的函数原型更改为 INT GraphInitialize（图结构*** graph_array，INT顶点）键，当你调用它，使用 GraphInitialize（安培; graph_array， 42）。

When you assign something to graph_array, you simply assign it to its local copy. The changes made to it in the function will not be see-able by the caller. You need to pass it by pointer value to be able to change its value. Change your function prototype to int GraphInitialize(struct graph ***graph_array,int vertices) and when you call it, use GraphInitialize(&graph_array, 42).

当您创建 graph_array2 在code第二个问题是，您声明它是您的本地 GraphInitialize（）的功能。因此，退出你的函数时，graph_array2被破坏，即使你把它分配给 * graph_array 。 （星解引用指针将其分配到它所指向的值）。

Second problem in your code is when you create graph_array2, you declare it to be local to your GraphInitialize() function. Thus, when exiting your function, graph_array2 is destroyed, even if you assigned it to *graph_array. (the star dereferences the pointer to assign it to the value it points to).

更改为分配 * graph_array =的malloc（sizeof的（* graph_array）*顶点）; ，你应该罚款

change your assignation to *graph_array = malloc(sizeof(*graph_array) * vertices); and you should be fine.

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