C++ c如何传递对 struct 数组的引用，而不是设置 struct 的副本

struct数组包含struct个元素，而不是对struct个元素的引用.相反，您需要有一个指针数组，如

Student **students;

并在重新锁定时以这种方式继续，因此temp将具有相同的类型.然后你可以这样做:

temp[size-1] = student;

因为temp将是一个由Student个指针组成的数组，因此您将能够将引用存储为它的元素.

欲了解更多信息，请阅读https://www.geeksforgeeks.org/how-to-declare-and-initialize-an-array-of-pointers-to-a-structure-in-c/

它将其指定为此类动态 struct 指针数组的一般形式:

< structure_name >  ** < array_name > = <structure_name **> malloc ( sizeof( <structure_name> )* size ) ;

同一页提供了以下示例代码以供说明:

// C Program to implement
// Dynamic Array of structure
// pointer
#include <stdio.h>
#include <stdlib.h>
 
typedef struct node {
    int number;
    char character;
} node;
 
int main(void)
{
    // First pointer to structure
    node* structure_ptr1 = (node*)malloc(sizeof(node));
    // Second pointer to structure
    node* structure_ptr2 = (node*)malloc(sizeof(node));
    // Third pointer to structure
    node* structure_ptr3 = (node*)malloc(sizeof(node));
    // Fourth pointer to structure
    node* structure_ptr4 = (node*)malloc(sizeof(node));
 
    // Initializing structure pointers
 
    structure_ptr1->number = 100;
    structure_ptr1->character = 'a';
 
    structure_ptr2->number = 200;
    structure_ptr2->character = 'b';
 
    structure_ptr3->number = 300;
    structure_ptr3->character = 'c';
 
    structure_ptr4->number = 400;
    structure_ptr4->character = 'd';
 
    // Declaring the array dynamically for structure
    // pointers
 
    // Note that double pointer is
    // used for pointer to pointer
    node** structure_array
        = (node**)malloc(sizeof(node) * 4);
 
    // Initializing the array of structure pointers
 
    structure_array[0] = structure_ptr1;
    structure_array[1] = structure_ptr2;
    structure_array[2] = structure_ptr3;
    structure_array[3] = structure_ptr4;
 
    // Accessing dynamic 1D arrays - just like static 1D
    // arrays
    printf("Data:\n");
    for (int i = 0; i < 4; i++) {
        printf("%c - ", structure_array[i]->character);
        printf("%d\t\t", structure_array[i]->number);
        printf("\n");
    }
}