Why is the integer comparision giving the wrong results?

Asked May 14 '19 at 11:38

Active May 14 '19 at 11:41

Viewed 59 times

I have been trying to solve a problem which requires me to write a function called mySqrt of the form int mySqrt(int x).

I am supposed to compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

I tried to implement a straightforward algorithm where I start from i=0 and then increment the value of i until $i*i<=x$ and then return i-1.

public int mySqrt(int x) {
    int i=0;
    while(i*i<=x){
        i++;
    }
    return (i-1);
}

I have an idea of what is happening but I really can't understand why. When i is large enough such that i^2 >2^31-1 the comparison in the while loop (i*i<=x) for x=2147395600 (which is less that 2^31-1) gives me a true. Why would that be? And as a consequence I am getting junk results. What is exactly happening when I try to evaluate (i^2<=x) when i^2 can no longer be stored in an int?

edited May 14 '19 at 11:41

Michael

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asked May 14 '19 at 11:38

user3503589

1

https://stackoverflow.com/questions/5131131/what-happens-when-you-increment-an-integer-beyond-its-max-value – Steve Smith May 14 '19 at 11:40
1

Hint: try logging the result of `i * i` in your loop. – Jon Skeet May 14 '19 at 11:41
1

Integer overflow. You wrap back to negative. – Michael May 14 '19 at 11:42
So i resolved it by typecasting i*i to a long. Thank you – user3503589 May 14 '19 at 11:50

Why is the integer comparision giving the wrong results?

0 Answers0