How to get sequence of numbers and then print the last 5?

Question

Im trying to make a program that will get sequence from the user that end with 0, and then i want to print the last 5 numbers (not including the 0).

I can assume that the user will input all the numbers in one line and will end it with 0.

I wrote that code but something is wrong with it, I think its something about the scanf line.

Input:

1 6 9 5 2 1 4 3 0

Output: no output

#include <stdio.h>
#define N 5

int main()
{
    int arr[N] = {0};
    int last_input, j;
    
    printf("please enter more than %d number and than enter 0: \n", N);
    
    last_input = 0;
    while (last_input<N) {
       scanf(" %d", &j);
       if (j == '0') {
          last_input = N;
          break;
       }
       else {
          arr[last_input] = j;
       }
       if (last_input==(N-1)) {
          last_input=-1;
       }
       ++last_input;
   }
    
    
    printf("The last %d numbers u entered are:\n", N); 
    
    for (j=(last_input+1); j<N; ++j) {
       printf(" %d", arr[j]);    
    }

    for (j=0; j<last_input; ++j) {
       printf(" %d", arr[j]);  
    }

    return 0;
}

Answer 1

This comparison

if (j == '0') {

does not make a sense because the user will try to enter the integer value 0 instead of the value (for example ASCII 30h or EBCDIC F0h) for the character '0'.

You need to write at least

if (j == 0) {

Due to these sub-statements of the if statement

  last_input = N;
  break;

this for loop

for (j=(last_input+1); j<N; ++j) {
   printf(" %d", arr[j]);    
}

is never executed and does not make a sense.

This statement

last_input=-1;

results in breaking the order of the N last elements in its output. And moreover the result value of the variable last_input will be incorrect.

You need to move elements of the array one position left. For this purpose you can use a loop of standard C function memmove.

The program can look the following way.

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


int main( void ) 
{
    enum { N = 5 };
    int arr[N];

    printf( "Please enter at least not less than %d numbers (0 - stop): ", N );

    size_t count = 0;

    for (int num; scanf( "%d", &num ) == 1 && num != 0; )
    {
        if (count != N)
        {
            arr[count++] = num;
        }
        else
        {
            memmove( arr, arr + 1, ( N - 1 ) * sizeof( int ) );
            arr[N - 1] = num;
        }
    }

    if (count != 0)
    {
        printf( "The last %zu numbers u entered are: ", count );
        for (size_t i = 0; i < count; i++)
        {
            printf( "%d ", arr[i] );
        }
        putchar( '\n' );
    }
    else
    {
        puts( "There are no entered numbers." );
    }
}

The program output might look like

Please enter at least not less than 5 numbers (0 - stop): 1 2 3 4 5 6 7 8 9 0
The last 5 numbers u entered are: 5 6 7 8 9

Answer 2

I made some changes based on ur comments and now its work fine!

#include <stdio.h>
#define N 5

int main()
{
    int arr[N] = {0};
    int last_input, j;
    
    printf("please enter more than %d number and than enter 0: \n", N);
    
    last_input = 0;
    while (last_input<N) {
       scanf("%d", &j);
       if (j == 0) {
          break;
       }
       else {
          arr[last_input] = j;
       }
       if (last_input==(N-1)) {
          last_input=-1;
       }
       ++last_input;
   }
    
    
    printf("The last %d numbers u entered are:\n", N); 
    
    for (j=(last_input); j<N; ++j) {
       printf("%d ", arr[j]);    
    }

    for (j=0; j<last_input; ++j) {
       printf("%d ", arr[j]);  
    }

    return 0;
}

thank u guys <3.