problem related to floats

[Abhishek Reddy]

float i=0.7f;
if(i<0.7)
System.out.println("hello");
else
System.out.println("world");

the output of this code is hello
why not world???

[Shaan Shar]

Originally posted by Abhishek Reddy Chepyala:
float i=0.7f;
if(i<0.7)
System.out.println("hello");
else
System.out.println("world");

the output of this code is hello
why not world???

By default 0.7 is double. So the precision of Double is more then Float.

Try out this program.

public class Sample {
public static void main(String[] args) {
double d = 0.7;
if (d < 0.7)
System.out.println("hello");
else
System.out.println("world");
}
}

You will have a clear view of the problem.

Hope this helps you out.

[Barry Gaunt]

float i=0.7f; // in this case i holds an approximation to the literal 0.7f to the precision of a float

i<0.7 // The 0.7 here is a double literal and the comparison has to be done with doubles.

When the float value in i is promoted to a double no extra precision can be added to the value.

So you have (something like):

0.69999990000000 < 0.69999999999999 which is true.

My explanation is only an approximation to what happens, but I think it conveys an idea of why you get the "wrong" answer.
[ October 03, 2006: Message edited by: Barry Gaunt ]

[Shaan Shar]

So My explanations was correct Barry??

[Barry Gaunt]

Originally posted by Ankur Sharma:
So My explanations was correct Barry??

Your statement that a double has more precision than a float is correct. I think we were answering the question at the same time (I am a very slow two finger typer )