传递带有结构的向量指针 C++

huangapple
go评论94阅读模式
英文:

Passing pointers of vectors with structures C++

问题

I've translated the code you provided as requested:

  1. void setCellInfo(CELL_MESH* target, int Global_ID, int node0, vector<NODE_MESH *>* NodeStore, vector<CELL_MESH *>* CellStore) {
  3.     CellStore->push_back(target); // No Errors
  4.     target->Global_ID = Global_ID; // No Errors
  6.     if (node0 != 0) {
  7.         target->node[0] = (*NodeStore)[static_cast<vector<NODE_MESH *>::size_type>(node0)]->ID; // ERROR 1
  8.         target->node_pointers[0] = (*NodeStore)[static_cast<vector<NODE_MESH *>::size_type>(node0)]; // ERROR 2
  9.     }
  10. }
  12. int main() {
  14.     int i = 0;
  16.     for (double y = 0; y < nodes_y; y++) {
  17.         for (double x = 0; x < nodes_x; x++) {
  18.             NODE_MESH* newNode = new NODE_MESH;
  19.             NodeStore.push_back(newNode);
  20.             newNode->ID = i;
  21.             i++;
  22.         }
  23.     }
  25.     i = 0;
  27.     for (int y = 0; y < cells_y; y++) { // nodes_y since horizontal faces are aligned with nodes horizontally (same y)
  28.         for (int x = 0; x < cells_x; x++) { // x coordinate for horizontal faces is in-between nodes, so 0.5 with a count for faces
  30.             CELL_MESH* newCell = new CELL_MESH;
  32.             setCellInfo(newCell, i, nodes_x * y + x, &NodeStore, &CellStore);
  34.             i++;
  35.         }
  36.     }
  38.     return 0;
  39. }

I've made the necessary translations, such as replacing &gt; with >, &amp; with &, and using static_cast to cast the index to the appropriate type.

英文:

I've declared the following function in C++

  1. void setCellInfo (CELL_MESH* target, int Global_ID, int node0,vector&lt;NODE_MESH *&gt;* NodeStore, vector&lt;CELL_MESH *&gt;* CellStore) {
  3.     CellStore-&gt;push_back(target); //No Errors
  4.     target-&gt;Global_ID = Global_ID; //No Errors
  5.     
  6.     if (node0 != 0) {
  7.         target-&gt;node[0] = NodeStore[vector&lt;NODE_MESH *&gt;::size_type(node0)]-&gt;ID; //ERROR 1
  8.         target-&gt;node_pointers[0] = NodeStore[vector&lt;NODE_MESH *&gt;::size_type(node0)]; //ERROR 2
  9.     }
  10. }

ERROR1: Gives me a "No member named 'ID' in 'std::vector<NODE_MESH *>'" for the target->node[] attributions although its the entities from the pointers within the vector that have this ID member. Since I'm trying to get a specific entity in the vector using NodeStore[value], I would think it would work.

ERROR2: Gives me "Assigning to 'NODE_MESH *' from incompatible type 'vector<NODE_MESH *>'" for all the target->node_pointers attributions. This seems to be the same problem but with pointers directly (without the ID member).

the NodeStore and CellStore vectors a defined as follows outside the function

  1. vector&lt;NODE_MESH*&gt; NodeStore;
  2. vector&lt;CELL_MESH*&gt; CellStore;

I then try to use the function like this, with 'i' being the int Global_ID and 'nodes_x*y+x' being some integer.

  1. CELL_MESH *newCell = new CELL_MESH;
  3. setCellInfo (&amp;newCell, i, nodes_x*y+x, &amp;NodeStore, &amp;CellStore);

I've tried many different alterations to pointers but can't get it work. Would you know how to ?

Here's a simplified complete version:

  1. #include &lt;vector&gt;
  3. using namespace std;
  5. typedef struct NODE_MESH{
  6.     int ID;
  7. }NODE_MESH;
  9. typedef struct CELL_MESH{
  10.     int Global_ID;
  11.     NODE_MESH* node_pointers[4];
  12.     int node[4];
  13. }CELL_MESH;
  15. vector&lt;NODE_MESH*&gt; NodeStore;
  16. vector&lt;CELL_MESH*&gt; CellStore;
  18. double nodes_y = 5;
  19. double nodes_x = 4;
  20. int cells_y = 4;
  21. int cells_x = 3;
  23. void setCellInfo (CELL_MESH* target, int Global_ID, int node0,vector&lt;NODE_MESH *&gt;* NodeStore, vector&lt;CELL_MESH *&gt;* CellStore) {
  25.     CellStore-&gt;push_back(target); //No Errors
  26.     target-&gt;Global_ID = Global_ID; //No Errors
  27.     
  28.     if (node0 != 0) {
  29.         target-&gt;node[0] = NodeStore[vector&lt;NODE_MESH *&gt;::size_type(node0)]-&gt;ID; //ERROR 1
  30.         target-&gt;node_pointers[0] = NodeStore[vector&lt;NODE_MESH *&gt;::size_type(node0)]; //ERROR 2
  31.     }
  32. }
  34. int main() {
  35.     
  36.     int i = 0;
  37.     
  38.     for (double y = 0; y &lt; nodes_y; y++) {
  39.         for (double x = 0; x &lt; nodes_x; x++) {
  40.             NODE_MESH *newNode = new NODE_MESH;
  41.             NodeStore.push_back(newNode);
  42.             newNode -&gt; ID = i;
  43.             i++;
  44.         }
  45.     }
  46.     
  47.     i = 0;
  48.     
  49.     for (int y = 0; y &lt; cells_y; y++) { //nodes_y since horizontal faces are aligned with nodes horizontaly (same y)
  50.         for (int x = 0; x &lt; cells_x; x++) { //x coordinate for horizontal faces is in-between nodes so 0.5 with count for faces
  51.             
  52.             CELL_MESH *newCell = new CELL_MESH;
  53.             
  54.             setCellInfo (newCell, i, nodes_x*y+x, &amp;NodeStore, &amp;CellStore);
  55.             
  56.             i++;
  57.             
  58.         }
  59.     }
  60.     
  61.     return 0;
  62. }

答案1

得分: 0

给定一个变量 T* t，语法 t[x] 等同于 *(t+x)，这就是引起混淆的原因。具体来说，NodeStore[vector&lt;NODE_MESH *&gt;::size_type(node0)] 的类型是 vector&lt;NODE_MESH *&gt;&amp;，而不是你期望的 NodeStore 的一个元素。

将你的代码改为通过引用传递变量：

  1. void setCellInfo (CELL_MESH* target, int Global_ID, int node0, vector&lt;NODE_MESH *&gt;&amp; NodeStore, vector&lt;CELL_MESH *&gt;&amp; CellStore) {
  3.     CellStore-&gt;push_back(target); //没有错误
  4.     target-&gt;Global_ID = Global_ID; //没有错误
  5.     
  6.     if (node0 != 0) {
  7.         target-&gt;node[0] = NodeStore[node0]-&gt;ID;
  8.         target-&gt;node_pointers[0] = NodeStore[node0];
  9.     }
  10. }

然后调用如下：

  1. setCellInfo(&amp;newCell, i, nodes_x*y+x, NodeStore, CellStore);

或者，你需要在索引之前解引用指针：

  1. (*NodeStore)[node0]-&gt;ID
英文:

Given a variable T* t, the syntax t[x] is equivalent to *(t+x), which is the cause of this confusion. Concretely, NodeStore[vector&lt;NODE_MESH *&gt;::size_type(node0)] is of type vector&lt;NODE_MESH *&gt;&amp; instead of an element of the NodeStore as you expected.

Change your code to take variables by reference instead:

  1. void setCellInfo (CELL_MESH* target, int Global_ID, int node0,vector&lt;NODE_MESH *&gt;&amp; NodeStore, vector&lt;CELL_MESH *&gt;&amp; CellStore) {
  3.     CellStore-&gt;push_back(target); //No Errors
  4.     target-&gt;Global_ID = Global_ID; //No Errors
  5.     
  6.     if (node0 != 0) {
  7.         target-&gt;node[0] = NodeStore[node0]-&gt;ID;
  8.         target-&gt;node_pointers[0] = NodeStore[node0];
  9.     }
  10. }

The call is then simply

  1. setCellInfo (&amp;newCell, i, nodes_x*y+x, NodeStore, CellStore);

Alternatively, you will need to dereference the pointer before indexing:

  1. (*NodeStore)[node0]-&gt;ID

答案2

得分: 0

最终解决方案：

  1. void setCellInfo(CELL_MESH* target, int Global_ID, int node0, vector<NODE_MESH *>& NodeStore, vector<CELL_MESH *>& CellStore) {
  3.     CellStore.push_back(target);
  4.     target->Global_ID = Global_ID;
  6.     if (node0 != 0) {
  7.         target->node[0] = NodeStore[node0]->ID;
  8.         target->node_pointers[0] = NodeStore[node0];
  9.     }
  10. }

在使用时：

  1. CELL_MESH *newCell = new CELL_MESH;
  3. setCellInfo(newCell, i, nodes_x * y + x, NodeStore, CellStore);
英文:

Final Solution:

  1. void setCellInfo (CELL_MESH* target, int Global_ID, int node0,vector&lt;NODE_MESH *&gt;&amp; NodeStore, vector&lt;CELL_MESH *&gt;&amp; CellStore) {
  3.     CellStore.push_back(target);
  4.     target-&gt;Global_ID = Global_ID;
  5.     
  6.     if (node0 != 0) {
  7.         target-&gt;node[0] = NodeStore[node0]-&gt;ID;
  8.         target-&gt;node_pointers[0] = NodeStore[node0];
  9.     }
  10. }

With this when using it:

  1. CELL_MESH *newCell = new CELL_MESH;
  2.         
  3. setCellInfo (newCell, i, nodes_x*y+x, NodeStore, CellStore);

huangapple
  • 本文由 发表于 2023年2月6日 19:29:31
  • 转载请务必保留本文链接：https://go.coder-hub.com/75360730.html
  • c++
  • function
  • function-pointers
  • structure
  • vector
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