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Topic: Why is this an infinte loop in C++ ?**Question:**
I' trying to calculate the number of steps required to get to 1 from a number by using the floor function and sqrt function. I have to get to 1 by calculating the floor of square root of number.
for example for number 4 the output should be 2 because
sqrt 4 = 2
sqrt 2 = 1.4
and floor of 1.4 = 1
my code is like:
int n ;
cout << "Please enter a number"<< endl;
cin >> n;
int counter = 0;
double val = n; //
int steps = floor (sqrt (val));
while (steps >= 1)
{
steps = floor (sqrt (val));
counter++;
}
cout << "It takes " <

June 18, 2019 / By Florence

Steps is always equal to floor(sqrt(val)) so it never falls to 1. Even if you had val = 1, the square root would still be 1 so the while loop would never exit. I think what you intended was steps = floor(sqrt(steps)); inside the while loop. then change the while test to just greater than 1. You may want to change how steps is initialized as well. Shadow Wolf

👍 228 | 👎 10

Did you like the answer? Because the loop's condition is that steps >= 1, and the value of steps never changes. In the loop, you set steps equal to floor(sqrt(val)), but since val never changes, steps never changes.

👍 90 | 👎 1

1st: while (steps>=1),if steps reachs 1 while loop is still continued so your loop will not have exit way. 2nd: steps=floor(sqrt(val)); But the value of val is never change so your loop will be never exit. Fix: while (steps>1) { val=sqrt(val); steps=floor(val); counter++; }

👍 81 | 👎 -8

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