sqrt function in C++ issue

Question

#include <iostream>
#include <cmath>

using namespace std;

int main(){
    
    int n;
    cin >> n;
    
    int i = sqrt(1 + 2 * n * (n + 1)) - 1;
    
    cout << i;
}

I have written a simple program which utilizes the sqrt() function in C++. The above program prints out a negative value on the console for n = 32768 even though the input to sqrt() is positive. I tried changing the statement from int i = sqrt(1 + 2 * n * (n + 1)) - 1; to
double i = sqrt(1 + 2 * n * (n + 1)) - 1; but the error is not resolving.

Output:

32768
-2147483648

The above output is for int i = sqrt(1 + 2 * n * (n + 1)) - 1;

Please help!

`1 + 2 * n * (n + 1)` is an expression of type `int`. It seems to have overflowed. — njuffa, Sep 08 '20 at 17:46
If the result of calling `sqrt` (or any other function) doesn’t make sense, look at the argument(s) that it’s being called with. — Pete Becker, Sep 08 '20 at 19:18
Try calculating `1 + 2 * 32768 * (32768 + 1)` by hand you would quickly notice that it is `2147549185` and likely overflowed `int`. — Ranoiaetep, Sep 08 '20 at 19:21

score 6 · Accepted Answer · answered Sep 08 '20 at 17:47

6

Change int n to double n. Your calculation 1 + 2 * n * (n + 1) overflows the range of int which for 32bits is -2,147,483,648 to 2,147,483,647.

Side note: int may not be 32bit, it depends on the platform (however, usually most of the time it is 32bit)

answered Sep 08 '20 at 17:47

Martin Perry

9,232
8
46
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That's right, except overflow isn't guaranteed. Attempting to cast a double outside of representable range to int is UB. See SO post [Handling overflow when casting doubles to integers in C](https://stackoverflow.com/a/30424410/13223986). – Pascal Getreuer Sep 09 '20 at 00:50

sqrt function in C++ issue

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