Comments on: 1 – 0.8 – 0.2 = ? http://anil.wordpress.com/2005/12/16/1-08-02/ All the things I am Interested in. Tue, 03 Nov 2009 08:55:34 +0000 http://wordpress.com/ hourly 1 By: Anil http://anil.wordpress.com/2005/12/16/1-08-02/#comment-9 Anil Wed, 01 Feb 2006 22:52:28 +0000 http://anil.wordpress.com/2005/12/16/1-08-02/#comment-9 Wow, Google corrected the problem. Wow, Google corrected the problem.

]]> By: Peter http://anil.wordpress.com/2005/12/16/1-08-02/#comment-8 Peter Wed, 01 Feb 2006 19:24:53 +0000 http://anil.wordpress.com/2005/12/16/1-08-02/#comment-8 The problem is that there is no binary number (using IEEE floating point representation, as in use by all modern computer architectures) that is exactly equal to either 0.8 or 0.2. The question you have asked the calculator (or programming language) to perform is "subtract from 1 the binary numbers that are closest to 0.8 and 0.2. Then print the result in some format." All agree on the result, but there is some leeway in how it is printed. The problem is that there is no binary number (using IEEE floating point representation, as in use by all modern computer architectures) that is exactly equal to either 0.8 or 0.2. The question you have asked the calculator (or programming language) to perform is “subtract from 1 the binary numbers that are closest to 0.8 and 0.2. Then print the result in some format.” All agree on the result, but there is some leeway in how it is printed.

]]> By: DYA http://anil.wordpress.com/2005/12/16/1-08-02/#comment-7 DYA Sat, 21 Jan 2006 04:15:03 +0000 http://anil.wordpress.com/2005/12/16/1-08-02/#comment-7 I have found why pearl and Java hade made the same mistake and the C gived ya the right one. When you use %f in the function printf the value of the variable 'c' The value of 'c' is rounded cause you cant print out the exponetial part. That is why you get a -0 but use the parameter %e and you'll see the same anwser as Java and pearl did. So Pearl and Java use the %e parameter when hey print out there Double. And why -5.551115×10^-17 ? Simple, there is no real 0 in float just a close one. Why no real 0 ? Simple, a computer doesnt understand what is a fractional number so they trick the machine into coding the number inside another one. So float number are only logical number and not real number for the computer. If you get hard time reading this, that mean my english havent improved. I have found why pearl and Java hade made the same mistake and the C gived ya the right one.

When you use %f in the function printf the value of the variable ‘c’ The value of ‘c’ is rounded cause you cant print out the exponetial part. That is why you get a -0 but use the parameter %e and you’ll see the same anwser as Java and pearl did.

So Pearl and Java use the %e parameter when hey print out there Double.

And why -5.551115×10^-17 ? Simple, there is no real 0 in float just a close one. Why no real 0 ? Simple, a computer doesnt understand what is a fractional number so they trick the machine into coding the number inside another one. So float number are only logical number and not real number for the computer.

If you get hard time reading this, that mean my english havent improved.

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