precision

问题 As I know Single-precision floating-point number has 1 bit for sign, 8 bits for exponent and 23 bits for mantissa. I can understand that 7 digit integers fit 23 bit mantissa and don't loose precision but can't understand how a number like 1234567000000000 fits without loose "1,2,3,4,5,6,7" digits, what is the math behind this? 回答1: The IEEE-754 basic 32-bit binary floating-point format only guarantees that six significant decimal digits will survive a round-trip conversion, not seven.

问题 Is there a way to set the "minimum" number of decimal places that a std::ostream will output? For example, say I have two unknown double variables that I want to print (values added here for the sake of illustration): double a = 0; double b = 0.123456789; I can set my maximum decimal precision so that I output b exactly std::cout << std::setprecision(9) << b << std::endl; >>> 0.123456789 Is there a way to set a "minimum" precision (a minimum number of decimal places), while retaining the

问题 Is there a way to set the "minimum" number of decimal places that a std::ostream will output? For example, say I have two unknown double variables that I want to print (values added here for the sake of illustration): double a = 0; double b = 0.123456789; I can set my maximum decimal precision so that I output b exactly std::cout << std::setprecision(9) << b << std::endl; >>> 0.123456789 Is there a way to set a "minimum" precision (a minimum number of decimal places), while retaining the

问题 Regarding my question I have seen a post on here but did not understand since i am new to C++. I wrote a small script which gets a number from user and script prints out the factorial of entered number. Once I entered bigger numbers like 30, script does not print out all the digits.Output is like 2.652528598 E+32 however What I want is exact number 265252859812191058636308480000000. Could someone explain how to get all digits in long double.Thanks in advance 回答1: You can set the precision of

问题 Regarding my question I have seen a post on here but did not understand since i am new to C++. I wrote a small script which gets a number from user and script prints out the factorial of entered number. Once I entered bigger numbers like 30, script does not print out all the digits.Output is like 2.652528598 E+32 however What I want is exact number 265252859812191058636308480000000. Could someone explain how to get all digits in long double.Thanks in advance 回答1: You can set the precision of

问题 If I run the statement Math.Exp(113.62826122038274).ToString("R") on a machine with .net 4.5.1 installed, then I get the answer 2.2290860617259248E+49 However, if I run the same command on a machine with .net framework 4.5.2 installed, then I get the answer 2.2290860617259246E+49 (i.e. the final digit changes) I realise that this is broadly insignificant in pure numeric terms, but does anyone know of any changes that have been made in .net 4.5.2 that would explain the change? (I don't prefer

问题 If I run the statement Math.Exp(113.62826122038274).ToString("R") on a machine with .net 4.5.1 installed, then I get the answer 2.2290860617259248E+49 However, if I run the same command on a machine with .net framework 4.5.2 installed, then I get the answer 2.2290860617259246E+49 (i.e. the final digit changes) I realise that this is broadly insignificant in pure numeric terms, but does anyone know of any changes that have been made in .net 4.5.2 that would explain the change? (I don't prefer

问题 I'm just trying to compute a good sigmoid function in C++ (and efficient). So i have to do something like: 1/(1 + exp(-x)) The problem is, when X becomes big (or even small), the result of 1 + e turns to be 0 or 1 For example, 1 + exp(-30) = 1 But this is incorrect... How can we add very small (or big) numbers easily and efficiently ? Datatype I am using : double Here is the code snippet : double Quaternion::sigmoidReal(double v){ return 1.0 / ( 1.0 + exp(-v) ) ; } Thanks ! 回答1: I think you

问题 I'm just trying to compute a good sigmoid function in C++ (and efficient). So i have to do something like: 1/(1 + exp(-x)) The problem is, when X becomes big (or even small), the result of 1 + e turns to be 0 or 1 For example, 1 + exp(-30) = 1 But this is incorrect... How can we add very small (or big) numbers easily and efficiently ? Datatype I am using : double Here is the code snippet : double Quaternion::sigmoidReal(double v){ return 1.0 / ( 1.0 + exp(-v) ) ; } Thanks ! 回答1: I think you

问题 This question already has answers here : Float and double datatype in Java (9 answers) Is floating point math broken? (31 answers) Closed 12 months ago . System.out.println(0.1F + 0.2F); // 0.3 System.out.println(0.1D + 0.2D); // 0.30000000000000004 I understood 0.1D + 0.2D ~= 0.30000000000000004. But I guessed these result are same, but it is not. Why result are different? 回答1: Why are the results different? In a general sense: Because the binary representations for float and double are