c++ negative square root

Question

My goal is to print *** if the square root is negative. But I can't think of a way to change default nan text to ***

for(int i=x1;i<=x2;i++){
   double y = sqrt(pow(i,2)+3*i-500);
    if(y = ?){
        outFile << "***";
    }

So, what should I write in the if statement to make it possible? Or maybe there is another way to check if the y is nan then print *

possible duplicate of [Checking if a double (or float) is nan in C++](http://stackoverflow.com/questions/570669/checking-if-a-double-or-float-is-nan-in-c) — Alok Save, Jan 29 '12 at 12:13

score 8 · Accepted Answer · answered Jan 29 '12 at 12:15

8

How about checking for a negative input to the square root function?

for (int i = x1; i <= x2; ++i)
{
    double x = pow(i, 2) + 3*i - 500;
    if (x < 0)
    {
        outFile << "***";
    }
    else
    {
        outFile << sqrt(x);
    }
}

answered Jan 29 '12 at 12:15

fredoverflow

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score 3 · Answer 2 · edited May 23 '17 at 11:50

3

Testing for NaN in C++ is tricky. Just use an if statement to avoid evaluating the sqrt if its argument is negative.

edited May 23 '17 at 11:50

Community

1
1

answered Jan 29 '12 at 12:14

Thomas

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score 2 · Answer 3 · answered Jan 29 '12 at 12:13

2

A nan number isn't equal to anything, even to itself.
You could simple test to see if if( y != y ).

answered Jan 29 '12 at 12:13

mmodahl

176
4

@RnD: This is not 100% reliable, check Alf's answer in the linked duplicate for more details. – Alok Save Jan 29 '12 at 12:18

score 0 · Answer 4 · answered Feb 06 '12 at 12:55

My goal is to print * if the square root is negative. But I can't think of a way to change default nan text to *

A square root never is negative. But it may be complex. See https://en.wikipedia.org/wiki/Complex_number

The idea is to expand the set of numbers into the so called complex plane, which contains a special number i for which is defined i² = -1. This allows us to generalize square roots:

sqrt(a b) = sqrt(a) sqrt(b)

So we can break down sqrt(-a) into sqrt(-1) sqrt(a) = i sqrt(a)

This allows us to change your program into

for(int i=x1;i<=x2;i++){
    double X = pow(i,2)+3*i-500;
    double y = sqrt(abs(x));
    if(X < 0){
        outFile << y << "i";
    } else {
        outFile << y;
    }
}

c++ negative square root

4 Answers4

Linked