Be seen. Boost your question’s priority for more expert views and faster solutions

Hello

I have built a calculator that is pretty much complete but for the inability to round numbers

what I want to do is say input a number x (double) press the "round" key, enter another number(y double and converted to int) and that number will be rounded to that number (y).

Example

9.4524 is entered (x)

Press Round Button

2 is entered (y)

result = 9.45 (9.4524 to y decimal places)

I have been trying to implement this but with out success

Here is some "example Code"

public double Rounded(Double d, int i){

BigDecimal rnd = new BigDecimal(d); /* the constructor is the only way I could find to convert */

rnd.setScale(i); /* a double to BigDecimal format */

return rnd.doubleValue();

}

Cheers

Murray S

PS I have little in the way of knowlege re: BigDecimal so quick tutorial would be good

I have built a calculator that is pretty much complete but for the inability to round numbers

what I want to do is say input a number x (double) press the "round" key, enter another number(y double and converted to int) and that number will be rounded to that number (y).

Example

9.4524 is entered (x)

Press Round Button

2 is entered (y)

result = 9.45 (9.4524 to y decimal places)

I have been trying to implement this but with out success

Here is some "example Code"

public double Rounded(Double d, int i){

BigDecimal rnd = new BigDecimal(d); /* the constructor is the only way I could find to convert */

rnd.setScale(i); /* a double to BigDecimal format */

return rnd.doubleValue();

}

Cheers

Murray S

PS I have little in the way of knowlege re: BigDecimal so quick tutorial would be good

double round(double d, int i) {

return ((int)((d + 5*Math.pow(10,-(i+1))) * Math.pow(10,i))/Math.pow(1

}

But could you explain it to me please, how did you come up with this algorithm &c.

Cheers

Murray

take the following number:

1.45643

If you want 2 decimal places, you can multiply the number by 100 to get:

145.643

now if you cast that to an integer, you will get

145

Last of all, wel divide by 100 to get

1.45

That gives us the original number to two decimal places. You also notice that first we add .005 to the number we are going to round, that way it rounds to the nearest integer when you cast it. So if the original number is 1.45643, and you are rounding to two decimal places, you get 1.46 instead of 1.45.

isn´t it to steps and intensive processing to get the result?

I want to use it to round the mouse move and place a component on the proper grid position. It wont slow down the operation?

If you knew the number of decimal places to round to before-hand, then you could make the algorithm much simpler by just doing:

d = Math.round(d * 100.0) / 100.0; // This does two decimal places

holdX and holdY are offsets from component origin to mouse positon.

public void setSnapLocation( int x, int y, int gridX, int gridY ) {

double ox, oy;

double gx, gy;

gx = gridX * 1.0;

gy = gridY * 1.0;

ox = Math.round( ( ( x - holdX ) * 1.0 + gx / 2 ) / gx ) * gx;

oy = Math.round( ( ( y - holdY ) * 1.0 + gy / 2 ) / gy ) * gy;

super.setLocation( (int)ox-GRABBERMARGIN, (int)oy-GRABBERMARGIN );

}

it´s a rubber move from mouse catch to every sides.

maybe I can give you some points to :)

thanks

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.

double d = 1.234567;

int i = 4;

d = d + 5*10^(-i - 1); // that would give us d = d + .00005, or d = 1.234617;

d = d * 10^i; // that gives us d = d * 10000, or d = 12346.17;

d = (int)d; // d = 12346

d = d / 10^i; // d = d / 10000, or d = 1.2346; That is the answer we want!