Array Elements Check

Applications Programming
1

Hi everyone, :slight_smile:

I'm trying to make a simple C program that scans an array of chars to see if its elements are similar.
I can't understand what's wrong. Could you help me to fix this? Here is the code.

Thanks!

#include<stdio.h>

int main() {
    int arr[10];
    int i, len;
    int flag = 0;

    printf("Enter a word: ");

    for (i = 0; i < 10; i++)
        scanf("%d", &arr);

    len = sizeof(arr);

    for (i = 0; i < len - 1; i++) {
        if (arr != arr[i + 1])
            flag = 1;
    }

    if (flag == 0)
        printf("The array has different elements.");
    else
        printf("The array has similar elements");

    return 0;

}
2

You didn't say what's wrong?

        if (arr != arr[i + 1])
            flag = 1;
    }

    if (flag == 0)
        printf("The array has different elements.");

Shouldn't this be if (flag == 1) ?

3

No. Unfortunately with

if (flag == 1)

it doesn't work.

4

Give this a try...

#include <stdio.h>

int main(void) {
    int arr[10];
    int i, x, len;
    int flag = 0;

    printf("Enter a word: ");

    for (i = 0; i < 10; i++) {
        scanf("%d", &x);
        arr=x;
    }
     // len = sizeof(arr);
    len = (int)( sizeof(arr) / sizeof(arr[0]) );

    for (i = 0; i < len - 1; i++) {
        if (arr != arr[i + 1]) {
            flag = 1;
        }
    }

    if (flag == 1) {
        printf("The array has different elements.");
    } else {
        printf("The array has similar elements");
    }
    return 0;
}
5

Thank you for your reply, but unfortunately it still not working. :frowning:

6

You are aware that this script only accepts numbers as input, right?

7

Yeah, of course. I can't understand why it doesn't work. :(:(:frowning:
I'm using Eclipse with Windows. I'm going mad! :confused:

8 
akshay@Aix:/tmp$ cat tmp.c 
#include <stdio.h>
#define N_ELEMENTS(array) (sizeof(array)/sizeof((array)[0]))


void main() {
    int i, x, arr[10];

    printf("Enter a word: ");

    for (i = 0; i < 10; i++) {
        scanf("%d", &x);
        arr=x;
    }
   
    for (i = 0; i < N_ELEMENTS(arr); i++) { printf("%d\n",arr);
        if (arr != arr[i + 1]) {
            printf("The array has different elements\n");
	    return ;
        }
    }

  
    printf("The array has similar elements\n");
}

akshay@Aix:/tmp$ gcc tmp.c 

akshay@Aix:/tmp$ ./a.out 
Enter a word: 1 1 1 1 1 1 1 1 1 1
1
1
1
1
1
1
1
1
1
1
The array has similar elements

akshay@Aix:/tmp$ ./a.out 
Enter a word: 1 2 3 4 5 6 7 8 9 10
1
The array has different elements
1 Like
9

I hear ya. I had issues like that using Geany in Linux. Might be the compiler. It compiles and runs well from a terminal though.

gcc tmp.c -o tmp
10

This is what's going wrong:

scanf("%d", &x);

The first time you run it, it sees a number and scans it.

The second time you run it, it hits whitespace of some sort -- a space or newline. This is not a digit, so it gives up.

The third time you run it, it hits the same whitespace. It's going to just keep slamming its head into the wall over and over again.

In summary, don't use scanf, it's a royal pain. Read lines yourself, tokenize them yourself, then use sscanf if you must. It will never pull this trick on you.

char buf[512];
int arr[10];

while(fgets(buf, 512, stdin))
{
        int n=0;
        char *tok=strtok(buf, " \r\n\t"); // Split on any whitespace, including newlines (which fgets includes).

        while(tok != NULL)
        {
                if(sscanf(tok, "%d", arr+n) != 0)
                {
                        n++;
                        tok=strtok(NULL, " \r\n\t"); // Keep splitting on the last string we gave it until done.
                        continue;
                }

                break;
        }

        printf("scanned %d numbers\n", n);
}
11

It works! Thank you for solving my problem. :slight_smile: