Float and Double - 5.1 divide by 3.0
Hi everyone,
I am quite new to java programming in term of using it. I wrote a test class to learn about float and double. However when I wrote the below code, I got a strange result. So, I wonder if java's float and double have any kind of exception I should be aware of. This won't happen if I just change from 3 to other numbers. Any suggestion? Thanks in advance.
public class MathTest {
public static void main (String args[]) {
float f1 = 5.1F;
float f2 = 3.0F;
double d1 = 5.1;
double d2 = 3.0;
System.out.println("Float result : " + f1/f2);
System.out.println("Double result : " + d1/d2);
Result:
Float result : 1.6999999
Double result : 1.7
This is due to the way that binary numbers convert to digital numbers and is expected. See these threads for explanations:
http://onesearch.sun.com/search/onesearch/index.jsp?qt=%2Bprecision+%2Bwrong+double&col=developer-forums&rf=0&chooseCat=allJava&subCat=siteid%3Ajava
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I have done the following program which is supposed to deal with temperature and scales.
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public class Temperature {
private float temp;
private char scale;
public static void main(String args[]) {
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public Temperature() {
temp = 0;
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scale = pScale;
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F:\programs\ch4>javac Temperature.java
Temperature.java:8: cannot find symbol
symbol : constructor Temperature(double,char)
location: class Temperature
Temperature t = new Temperature(40.5, 'F');
*^*
*1 error*
I don't understand why the number 40.5 is considered like a double and that I can't use Temperature(float, char);
If I'm not mistaken the compiler is asking for a constructor like Temperature(double, char) right ??
thanks.Can I second Mr TuringPest, and Sir J.Bloch... Do NOT use floats, unless of course you have a really really good reason. Internally, java does all it's floating point arithmetic in doubles anyways, so you're just shooting yourself in the accuracy-foot by forcing it to: promote to double to calculate, "narrow" result back to float... promote, narrow ... promote, narrow ... promote, narrow ... The net result can be astoundingly inaccurate... So just use doubles.
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void testDoit()
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why "double" is the ouput in the previous example!hummm
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Really, acording the "Java primitives data types"
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mathematically
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Hi, does anyone knows how can i get more precise calculations with floating point numbers? Let's see an example:
public class Teste {
public static void main(String args[]) {
float i = 0;
while (i<=10) {
System.out.println(i);
i=i+0.1f;
/* output
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.70000005
0.8000001
0.9000001
1.0000001
1.1000001
1.2000002
1.3000002
1.4000002
1.5000002
1.6000003
1.7000003
1.8000003
1.9000003
2.0000002
2.1000001
2.2
2.3
2.3999999
2.4999998
2.5999997
2.6999996
2.7999995
2.8999994
2.9999993
3.0999992
3.199999
3.299999
3.399999
3.4999988
3.5999987
3.6999986
3.7999985
3.8999984
3.9999983
4.0999985
4.1999984
4.2999983
4.399998
4.499998
4.599998
4.699998
4.799998
4.8999977
4.9999976
5.0999975
5.1999974
5.2999973
5.399997
5.499997
5.599997
5.699997
5.799997
5.8999968
5.9999967
6.0999966
6.1999965
6.2999964
6.3999963
6.499996
6.599996
6.699996
6.799996
6.899996
6.9999957
7.0999956
7.1999955
7.2999954
7.3999953
7.499995
7.599995
7.699995
7.799995
7.899995
7.9999948
8.099995
8.199995
8.299995
8.399996
8.499996
8.599997
8.699997
8.799997
8.899998
8.999998
9.099998
9.199999
9.299999
9.4
9.5
9.6
9.700001
9.800001
9.900002
*/http://forum.java.sun.com/thread.jsp?forum=31&thread=357174
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Thanks.Hi
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Use like this.
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Get bckif you are getting error still...
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<<KUDOS ARE WELCOME>>
ELECTRO SAM
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