如何在函数C编程中打印2D数组

Question

I'm new to 2D arrays, structs and have a limited knowledge about pointers. My problem is that the display function only displays addresses. i dont know which part of my code is wrong or missing. can you give me suggestions to fix it or ideas? Thanks!

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <assert.h>


struct Inventory
{
    char name[100];
    float latestCost;
    int stock;
    int sold;
};
struct Inventory *getInfo(void);
void display(struct Inventory *items[][4], int n);
main()
{
    char select;
    struct Inventory items[10][4];
    int i,n,j, *ptr;

    printf("\nEnter how many items in the inventory:\n");
    scanf("%d", &n);

     ptr = malloc(sizeof(struct Inventory));

    for(i= 0; i < n; i++)
    {
        items[i][4] = *getInfo();
    }
    display(&items,n);

    getch();
}
struct Inventory *getInfo(void)
{
    struct Inventory *items = malloc(sizeof(struct Inventory));
    assert(items != NULL);
    fflush(stdin);
    printf("\nName of the item: \n");
    gets(items->name);
    printf("\nCost:");
    scanf("%f", &items->latestCost);
    printf("\nStock:");
    scanf("%d", &items->stock);
    printf("\nTotal Sold:\n");
    scanf("%d", &items->sold);
    return items;
}
void display(struct Inventory *items[][4], int n)
{
    int i, j;

    for(i = 0; i < n; i++)
    {
        for(j = 0; j < 4; j++)
        {
            printf("%d\t", items[i][j]);
        }
        printf("\n");
    }

}

Answer 1

Here

void display(struct Inventory *items[][4], int n);

you want a pointer to an array of 4 Inventory s, not a 2d array of pointers to Inventory , change to

void display(struct Inventory items[][4], int n)

or

void display(struct Inventory (*items)[4], int n)

---

Here

display(&items,n);

you don't need to pass the address

display(items,n);

is enough

---

printf("%d\t", items[i][j]);

%d prints an integer, use some member of the struct:

printf("%d\t", items[i][j].stock);

or

printf("%d\t", items[i][j].sold);