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Java Forum / General / January 2007Java is only calculating to once decimal place
I'm doing a calculation with a variable defined as a float. (e.g. take 5% and add 1) I should get 1.05 When I print the variable that should hold 1.05, I only get 1.0 Needless to say when I use the variable that should have 1.05 stored, in the rest of my calculation, I'm getting errounious results (i.e. cannot divide by 0). I am inputting a figure (integer) from the that needs to be calculated as a percent. I store the result in a variable as a float int entry; float answer; answer = 1+(entry/100); thus if I enter 5... answer = 1+(5/100) answer = 1.05 when I have java display answer (System.out.println(answer);) I have 1.0 which thows the rest of my calculation that the above formula is in, out of whack I understood floats are suppose to have more than one significant digit? > I'm doing a calculation with a variable defined as a float. > (e.g. take 5% and add 1) > I should get 1.05 > > When I print the variable that should hold 1.05, I only get 1.0 [...] > I understood floats are suppose to have more than one significant digit? There is nothing wrong with the float. The problem is that you are using integer arithmetic to calculate the 5%, which results in 0. Instead of dividing 5/100, try dividing 5/100.0 or 5/(float)100, etc. /gordon  Signature[ don't email me support questions or followups ] g o r d o n + n e w s @ b a l d e r 1 3 . s e > I'm doing a calculation with a variable defined as a float. > (e.g. take 5% and add 1) > I should get 1.05 ... > int entry; > float answer; > > answer = 1+(entry/100); No, you are not doing a calculation with a variable defined as a float, and that is precisely your problem: 5/100 == 0 but 5.0/100 would be quite close to the 0.05 that you expected. (Your will see that some floating point calculations give results like 0.049999999999998 instead of the precise 0.05; that's why I avoid saying simply 5.0/100 == 0.05 above.) On Jan 26, 5:19 am, Jussi Piitulainen <jpiit...@ling.helsinki.fi> wrote: > rler...@telus.net writes: > > I'm doing a calculation with a variable defined as a float. [quoted text clipped - 7 lines] > float, and that is precisely your problem: 5/100 == 0 but 5.0/100 > would be quite close to the 0.05 that you expected. > (Your will see that some floating point calculations give results > like 0.049999999999998 instead of the precise 0.05; that's why I > avoid saying simply 5.0/100 == 0.05 above.) Since 1/20 is not representable as an IEEE floating point number, the literal 0.05 is not exactly 1/20 either. In fact (5.0/100 == 0.05) is always true in IEEE, though neither is exactly 1/20. > I'm doing a calculation with a variable defined as a float. > (e.g. take 5% and add 1) > I should get 1.05 > > When I print the variable that should hold 1.05, I only get 1.0 It helps a /lot/ if you show /exactly/ what code shows the problem. Otherwise we have to guess. Fortuately in this case we have an easy-peasy answer (or two). > int entry; > float answer; [quoted text clipped - 4 lines] > > answer = 1+(5/100) 5 is an integer literal. 100 is an integer literal. So the / is an integer divide. So the result is 0. So answer becomes 1(.0). BOOM. Try using `answer = 1 + entry/100.0;`. But note the following: (a) floats /are not/ decimal. The answer won't be exactly 1.05, but a binary approximation to it. (b) if you're playing fast-and-loose with floating point values, you might be better off using doubles, not floats. (c) Depending on exactly what you're doing with your percentages, you might want to consider using decimals or scaled integers. SignatureChris "electric hedgehog" Dollin "Who do you serve, and who do you trust?" /Crusade/ > I'm doing a calculation with a variable defined as a float. > (e.g. take 5% and add 1) [quoted text clipped - 13 lines] > > answer = 1+(entry/100); That is a purely integer calculation. 5/100 as an integer is 0. > thus if I enter 5... > > answer = 1+(5/100) > answer = 1.05 Arithmetically yes, but computationally the answer is 0, as above. > when I have java display answer (System.out.println(answer);) I have > 1.0 which thows the rest of my calculation that the above formula is > in, out of whack > > I understood floats are suppose to have more than one significant digit? Floats do, integers don't have any decimal places. You need to ensure that your calculation is performed as a floating point operation: answer = 1.0f+(entry/100.0f); SignatureNigel Wade, System Administrator, Space Plasma Physics Group, University of Leicester, Leicester, LE1 7RH, UK E-mail : nmw@ion.le.ac.uk Phone : +44 (0)116 2523548, Fax : +44 (0)116 2523555 Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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