issue with if statement

[tony1420]

hello with this code the if statement get completely ignored, why is that? If i was not in increments of .1 and was just 1 while still being double it works but in this form when it approaches 9 the if statement does not work.

thank you for your time

  double a = .1;
	
	cout << "Thinking: ";
	for(double i = 0 ; i < 10; i+= .1){
		cout << i << endl;
		if((pow(a,2) + pow(i,2)) == 400)
			cout << a << " + " << i << endl;
		if(i == 9){
			cout << " i reset" << endl;
			i = 0;
		a = a + .1;
		}
	}

[Mats]

Make i an integer. Create a different double and assign the value of i to that double at the start of the for loop, before the if statements.

[tony1420]

i believe i am not understanding what you said. If i make a new double and assign the value of i which is now an integer and can't be in decimal form to it won't the double just be the same i value?

[giblit]

Basically the thing is you cant use the comparison operator ( == ) like that for any floating-point number. THey are not stored the same as integers.
9 could really be equal to 9.000000000000000000001 or even 8.999999999999999999 to compare you would have to put a range

like i < 9.0000000001 && i > 8.99999999999
or something along those lines.

basically what mats is saying is to do something like this I am guessing

for( int i = 0; i < 10; ++i )
{
    for( double j = 0.0; j < 1; j += .1 )
    {
        cout << j << endl;
        if( (pow( a , 2 ) + pow( j , 2 ) ) == 400 )
            cout << a << " + " << j << endl;
     }
     if( i == 9 )
     {
         cout << "i reset" << endl;
         i = 0;
         a += .1;
     }
}

though to be honest I am not sure why you are resetting i when it is equal to 9 if you have the loop running till 10.

[tony1420]

sorry i guess i should of explained the logic.

The idea is to find two numbers that when they are squared then added they will equal to 400. My thinking was to have a number that would scroll through testing possible answers example 5.2^2 + .1^2, 5.3^2 + .1^2, 5.4^2 + .1^2.... when at 9 the .1 would change to .2 and the cycle would continue over 5.2^2 + .2^2

[Mats]

Why not just do this -> i * i <- to square the number? Or even better, just square root 400? Also, if you finding the squares of only square numbers, you only need integers and not any other type.

[giblit]

hes tyring to find
x² + y² = 20²

basically it could have a trillion of results.

x = 0
y = 20

pretty much you can plug and chug this formula

a² + b² = c²
( Pythagorean theorem )

which means you pick a number for b

and do

a = sqrt( 20 - b );

[tony1420]

essentially pow and i*i is the same thing as i am still learning c++ i like to try different things. Does pow take more time to process then i * i ? of course on this level the difference is irrelevant but would be good to know. Also about just using integers yes it would make things simpler but same thing i would like to try different things.

[giblit]

pow uses a very complex formula it is something like using log's and natural logs. its not like i * i....doesn't work that way with rational numbers.

[tony1420]

interesting i will look that up i was under the impression it was the same thing. I knew about the a = sqrt approach but i was thinking of a different way because there is more to what i have to do but that might actually be the better solution since working with double using pow or I*i i don't get a solution. i wish i could of gotten it to work lol.