Use of double type in float subtraction vs precision
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From: LiloLilo on 26 Jun 2010 23:56 Hi, can double type rapresent exactly every result when subtracting two floats? Example: float a; float b; double result; result = (double)a - (double)b; Does result suffer of some rounding issue in this case or it would always rapresent the very correct result? Thank you all for help. -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Jens Schmidt on 27 Jun 2010 21:08 LiloLilo wrote: > can double type rapresent exactly every result when subtracting two > floats? > double result; > > result = (double)a - (double)b; > > Does result suffer of some rounding issue in this case or it would > always rapresent the very correct result? Because the Range of possible float values is usually larger than the number of digits in double, the answer is no. Example: use 1e+30 for a and 1e-30 for b. -- Greetings, Jens Schmidt [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Walter Bright on 27 Jun 2010 21:09 LiloLilo wrote: > can double type rapresent exactly every result when subtracting two > floats? Example: > > float a; > float b; > > double result; > > result = (double)a - (double)b; > > Does result suffer of some rounding issue in this case or it would > always rapresent the very correct result? If the difference in the binary exponents of a and b exceed the additional significand precision bits in a double (29), you'll lose a corresponding number of bits of precision. The reason is because in order to do the subtraction, the significands must be normalized, meaning shifted left or right until the exponents of a and b match. When significand bits get shifted off the end of the representation, they're gone. So, no, a double is not anywhere near a complete answer to float precision. --- Walter Bright, Digital Mars free C, C++, D programming language compilers http://www.digitalmars.com -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Pierre Asselin on 28 Jun 2010 05:39 LiloLilo <danilobrambilla(a)tiscali.it> wrote: > can double type rapresent exactly every result when subtracting two > floats? No. > Example: > float a; > float b; > double result; > result = (double)a - (double)b; > Does result suffer of some rounding issue in this case or it would > always rapresent the very correct result? float a= 1e20f; float b= 1e-20f; double result= (double)a - (double)b; -- pa at panix dot com [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ]
From: Keith H Duggar on 28 Jun 2010 06:14
On Jun 27, 10:56 am, LiloLilo <danilobrambi...(a)tiscali.it> wrote: > Hi, > > can double type rapresent exactly every result when subtracting two > floats? Example: > > float a; > float b; > > double result; > > result = (double)a - (double)b; > > Does result suffer of some rounding issue in this case or it would > always rapresent the very correct result? No. "What Every Scientist Should Know About Floating-Point Arithmetic" http://dlc.sun.com/pdf/800-7895/800-7895.pdf http://en.wikipedia.org/wiki/Kahan_summation_algorithm KHD -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] |