CODEWITHSUNDEEP
java
toffoo
toffoo

Reputation: 29

Add 0's to the end of an integer

for(i=0; i<array.length; i++){
  sum = 4 * 5;
}

What I'm trying to do is add ((array.length - 1) - i) 0's to the value of sum. For this example assume array length is 3. sum equals 20. So for the first iteration of the loop i want to add ((3 - 1) - 0) 0's to the value of sum, so sum would be 2000. The next iteration would be ((3 - 1) - 1) 0's. so sum would equal 200 and so on. I hope what I am trying to achieve is clear.

So my questions are:

Is it possible to just shift an int to add extra digits? My search thus far suggests it is not.

If not, how can i achieve my desired goal?

Thankyou for reading my question and any help would be greatly apreciated.

Upvotes: 0

Views: 4190

Answers (6)

corvec
corvec

Reputation: 288

You can add n zeroes to the end of a number, sum by multiplying sum by 10 * n.

int sum = 20;
for (int i = 0; i < ary.length; ++i) {
    int zeroesToAdd = ary.length - 1 - i
    sum *= (zeroesToAdd > 0) ? zeroesToAdd * 10 : 1
}
System.out.println("Sum after loop: " + sum);

Upvotes: 1

Mitesh Pathak
Mitesh Pathak

Reputation: 1211

For array of length = n; you will end up adding (n - 1) + (n - 2) + ... + 2 + 1 + 0 zeros for i = 0, 1, ... n-2, n-1 respectively.

Therefore, number of zeros to append (z) = n * (n-1) / 2

So the answer is sum * (10 ^ z)

[EDIT] The above can be used to find the answer after N iteration. (I miss read the question)

int n = array.length;
long sum = 20;
long pow = Math.pow(10, n-1); // for i = 0
for (int i = 0; i < n; i++) {
    System.out.println(sum*pow);
    pow /= 10;
}

Upvotes: 0

Elliott Frisch
Elliott Frisch

Reputation: 201537

Multiply by 10 instead, and use < (not >) like

int sum = 20;
int[] array = { 1, 2, 3 };
for (int i = 0; i < array.length; i++) {
    int powSum = sum;
    for (int j = array.length - 1; j > i; j--) {
        powSum *= 10;               
    }
    System.out.println(powSum);
}

Output is (as requested)

2000
200
20

Upvotes: 0

Victor
Victor

Reputation: 13

  1. You will want to check your for-loop condition: i>array.length. Since i starts at 0, this loop will not run unless the array's length is also 0 (an empty array). The correct condition is i < array.length.
  2. This "shift" you want can be achieved by creating a temporary variable inside the loop that is equal to the sum times 10i. In Java's Math library, there is a pow(a,b) function that computes ab. With that in mind, what you want is something like this:
int oldSum = 4 * 5;

for (int i = 0; i < array.length; i++) {
    int newSum = oldSum * Math.pow(10,i);
}

Upvotes: 0

brso05
brso05

Reputation: 13232

for(int i=array.length; i>0; i--){
  sum = 20;
  for(int j=0; j < (i - 1); j++)
  {
     sum *= 10;
  }
}

Use inner loop to multiply by 10 the number of times i is for that iteration. You would need to reset sum in your outer loop each time.

Upvotes: 0

SamTebbs33
SamTebbs33

Reputation: 5647

You can just multiply it by 10 however many times.

200 * 10 = 2000

etc

So in your case, you'd have to use a for loop until the end of the array and multiply sum every iteration. Be careful though, because the max value of an int is 2^31, so it of surpasses that, it will roll back to 0

Upvotes: 4

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