将typdef结构作为参数传递给函数

Question

When i run this code i got this error : [Error] subscripted value is neither array nor pointer nor vector

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/*Defined a data type named User with typedef*/
typedef struct User{
    char firstName[50];
    char lastName[50];
    int phonenumber; 
}user;


int main(int argc, char *argv[]) {

    user users[2];/*users defined "user" type*/

    strcpy(users[0].firstName,"furkan");

    strcpy(users[1].lastName,"xxxx");

    users[0].phonenumber = 1;

    users[1].phonenumber = 2 ;

    print_users(users);

    return 0;
}

/*Function for printing user type users values*/
void print_users(user usr)
{
    int j=0;

    for(j=0;j<10;j++)
    {
        printf("%-10s%-20s%7d\n",usr[j].firstName,usr[j].lastName,usr[j].phonenumber);
    }
}

I can make this function without typedef but i wonder if there is a way to make this happen

Answer 1

void print_users(user *usr)

this should be the parameters that your function receive, because inside your function you're acessing usr[j], so that means that usr need to be a pointer and not a structure itself.

ah, just to say, your for goes from 0 to 9 (10 positions), and your only allocated 2 positions.

Answer 2

The function parameter

void print_users(user usr);
                 ^^^^^^^

is a scalar object. You may not apply the subscript operator for a scalar object.

If you want that the function deals with an array then you should declare the function at least like

void print_users(user usr[]);
                 ^^^^^^^^^^

Take into account that it is not clear why the function uses magic number 10.

for(j=0;j<10;j++)
        ^^^^^

At the same time in the main you declared an array of only two elements

user users[2];

Thus it will be correctly to declare the function like

void print_users(user usr[], size_t n );

and to use the variable n in the loop

for(j=0;j < n;j++)
        ^^^^^

Correspondingly the function can be called like

print_users( users, 2 );