Array comparison for integer

Question

Here I am comparing two array element , If array element are equal SETPASSWORD=1; else SETPASSWORD=0; But Here It is always printing password set status 1 even array are not equal

#include <stdio.h>
#define ARY_SIZE 4
int password_set;
int main (void)
{
  //Local Declarations
  int numbersA[ARY_SIZE];
  int numbersB[ARY_SIZE];
  int i;
  int j;

  //Statements
  printf("Please Enter 10 Integers For Array A\n");
  for (int i = 0; i < ARY_SIZE; i++)
    scanf("%d", &numbersA[i]);

  printf("\nPlease Enter 10 Integers For Array B\n");
  for (int j = 0; j < ARY_SIZE; j++)
    scanf("%d", &numbersB[j]);

  for (int i = 0; i < ARY_SIZE; i++)
  {
    for (int j = 0; j < ARY_SIZE; j++)
    {
      if (numbersA[i] == numbersB[j])
        password_set=1;
      else
        password_set=0;
    }
  }
  printf(" password setstaus =%d",password_set);
  return 0;
}

You set password_set every loop iteration so the result is going to be telling you whether the 10th numbers equal or not. — M.M
– M.M, Commented Mar 13, 2014 at 7:41
Using C++: password_set = std::equal(numbersA, numbersA + ARY_SIZE, numbersB); — juanchopanza
– juanchopanza, Commented Mar 13, 2014 at 8:02

Jabberwocky · Accepted Answer · 2014-03-13 07:38:18Z

4

Your logic is wrong. You must exit the loop as soon as one pair of numbers is non equal.

And you also need only one loop :

password_set = 1 ;

for (int j = 0; j < ARY_SIZE; j++)
{
  if (numbersA[j] != numbersB[j])
  {
    password_set = 0;
    break ;
  }
}

answered Mar 13, 2014 at 7:38

Jabberwocky

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2 Comments

RKNAYAK Over a year ago

Then where to set password_set = 1; if i set globally as 1 it will be always 1. if array are equalto 1 then only set to 1 else set to zero

Jabberwocky Over a year ago

You set password_set = 1 just before the loop as shown in the answer. If the two arrays are different password_set will be set to 0, otherwise it will stay 1.

hrkz · Accepted Answer · 2014-03-13 07:40:55Z

1

Your compare each value of the array in your for (if (numbersA[i] == numbersB[j])

It resets the value of password_set at each iterations. It means that the result printed will be the last index of your array.

And btw you need only one loop

answered Mar 13, 2014 at 7:40

hrkz

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Comments

sad1e · Accepted Answer · 2019-06-04 09:18:45Z

0

Actually, you comparison is wrong. Because you want to test two arrays is equal or not, you just test each number of these two arrays.

like this:

    password_set = 1;

    for (int i = 0; i < ARY_SIZE; i++) {
        if (numbersA[i] != numbersB[i]) {
            password_set=0;
            break;
        }
    }

edited Jun 4, 2019 at 9:18

answered Mar 13, 2014 at 7:43

sad1e

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2 Comments

Peter - Reinstate Monica Over a year ago

The password_set flag should really not be set back to 1 if just one pair happens to be equal (that would overwrite an unequality found before). You either initialize it to 1 and exit when the first inequality is found, or for god's sake continue to the end of the loop, but don't overwrite the 0 later with a 1 in the loop.

Peter - Reinstate Monica Over a year ago

Perhaps apple has a job offer for you if you are interested in safety critical software ;-) (scnr)

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