Array form of integer-CodePudding

I was trying to convert the array to integer sum=999999999999 (twelve 9) , when i am limiting the array to less than ten 9s it is giving the result but when i am giving the array of more than ten 9s it is giving an unexpected result , please explain it will be really helpful for me

int[] arr={9,9,9,9,9,9,9,9,9,9,9,9};
int p=arr.length-1;
int m;
int num=0;
for (int i = 0; i <= p; i  ) {
    m=(int) Math.pow(10, p-i);
    num  = arr[i]*m;           // it is executing like: 900 90 9=999
}

CodePudding user response：

this happens because you're exceeding the Integer.MAX_VALUE.

You can read about it here.

You can use instead of int a long, to store large values, and if that is not enough for you, you can use - BigInteger

BigInteger num = BigInteger.valueOf(0);
for (int i = 0; i <= p; i  ) {
    BigInteger m = BigInteger.valueOf((int) Math.pow(10, p-i));
    BigInteger next =  BigInteger.valueOf(arr[i]).multiply(m));
    num = num.add(BigInteger.valueOf(arr[i]*m));
}

CodePudding user response：

It is because an int is coded on 4 byte so technically you can only go from -2,147,483,648 to 2,147,483,647. Consider using the long type.

CodePudding user response：

A couple of things.

You don't need to use Math.pow.
for up to 18 digits, you can use a long to do the computation.
I added some extra digits to demonstrate

int[] arr={9,9,9,9,9,9,9,9,9,9,9,9,1,1,2,3,4};

long sum = 0;             // or BigInteger sum = BigInteger.ZERO;
for (int val : arr) {
    sum = sum * 10   val; // or sum.multiply(BigInteger.TEN).add(BigInteger.valueOf(val));
}
System.out.println(sum);

prints

99999999999911234

Here is the sequence for 1,2,3,4 so you can see what is happening.

- sum = 0
- sum = sum(0) * 10   1 (sum is now 1)
- sum = sum(1) * 10   2 (sum is now 12)
- sum = sum(12)* 10   3 (sum is now 123)
- sum = sum(123)*10   4 (sum is now 1234)

CodePudding user response：

Try using long instead of int.

CodePudding user response：

because it overflows integer boundry