How to Show all elements of an array declared in a C struct?

Question

In that code:

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

struct cadastro {
   char nome[50];
   int idade;
   char rua[50];
   int numero;

   };
int main()
{
struct cadastro c[4]; //Array[4] de estruturas
for(int i=0; i<4; i++) {
    gets(c[i].nome);
    scanf("%d", &c[i].idade);
    gets(c[i].rua);
    scanf("%d", &c[i].numero);
}
 system("pause");
 return 0;
}

How to print all the elements of the struct, without doing it manually with a printf. For he will exhibit four times.

Answer 1

Hello, I think it’s unclear what you meant to ask.

First, there is no method more recommended for output values in C than with printf.

To print the names, just follow the same logic you used to read them:

for (i=0;i<4;i++){
    printf("%s\n",c[i].nome);
}

Answer 2

To print all elements at once, with a single simple instruction, use 1 function!

void cadastro_print(struct cadastro *p) {
    printf("nome: %s\n", p->nome);
    printf("idade: %d\n", p->idade);
    printf("rua: %s\n", p->rua);
    printf("numero: %d\n", p->numero);
    printf("\n");
}

and, in main(), calls this function

    for (int i = 0; i < 4; i++) cadastro_print(c + i);