double

问题 As we know, because of the limited precision of double, the following two calculation may not give the exact the same value : A / B / C and A / ( B * C ) My question is even with the same two variable, A & B, can the compiler guarantee every time A / B yield the same value ? Or I should ask in the code, can we guarantee the following statement always return true: If ( A / B == A / B ) 回答1: A guarantee of behavior for a compiler requires some document specifying the behavior. The answer

问题 I really thought this would work, but when I ran it, I always get some random huge decimal and it makes no sense. The logic is right, so why doesn't the code work? do{ value = rnd.nextDouble(); }while(value>min && value<max); Keep in mind "min" and "max" are actually functions that do some simple arithmetic operations then return that value, but I just put min and max for simplicity so I don't have to post the entire class. 回答1: You can declare a start and end variables ( int or double ), and

问题 Prob = *((unsigned short*)(Prob)); Prob = (unsigned short)(Prob); This didn't work for me. How to convert it correctly? 回答1: The biggest issue you face is for those double values that exceed USHRT_MAX . Secondarily, you need to address rounding. A simple approach that addresses both and sets rounding at .5 could look something like the following where the function array_dbl2short returns the number of double values successfully converted to unsigned short : size_t array_dbl2short (unsigned

问题 I have a Fortran 77 code that shows a comment in the first column and written double precision in the same line. now my text editor (Notepad++) is showing that as a statement. I am confused if it is a statement or a comment. * double precision Ct,Jt,PDt,AeA0t,Zt,Cq,Jq,PDq,AeA0q,Zq, * &Kt,Kq,Eta,J,PD,AeA0 回答1: It's a comment. If there's anything other than a whitespace in the first column of a line in a fixed-form fortran (That is Fortran up to F77), it's a comment. These variables are not

问题 I got this question using read.table() with or without header=T , trying to extract a vector of doubles from the resulting data.frame with as.double(as.character()) (see ?factor ). But that's just how I realized that I don't understand R's logic. So you won't see e.g. read.table in the code below, only the necessary parts. Could you tell me what's the difference between the following options? With header=T equivalent: (a <- data.frame(array(c(0.5,0.5,0.5,0.5), c(1,4)))) as.character(a) # [1]

问题 i am trying to convert double to pascal real but when i convert 0.23 to real i got 0.23999999 real how can i truncate all 9999 to 0000. public static byte[] Double2Real48(double d) { byte[] r48 = new byte[6]; byte[] da = BitConverter.GetBytes(d); for (int i = 0; i < r48.Length; i++) r48[i] = 0; //Copy the negative flag r48[5] |= (byte)(da[7] & 0x80); //Get the expoent byte b1 = (byte)(da[7] & 0x7f); ushort n = (ushort)(b1 << 4); byte b2 = (byte)(da[6] & 0xf0); b2 >>= 4; n |= b2; if (n == 0)

问题 This question already has answers here : What is the difference between float and double? (11 answers) Closed last year . I have a sample program: #include <iostream> #include <stdio.h> using namespace std; int main() { float a = 33.30; double b = 33.30; char a1[1024]; char b1[1024]; sprintf(a1, "%0.6f", a); sprintf(b1, "%0.6lf", b); cout << a1 << endl; cout << b1 << endl; return 0; } The output I am getting is: 33.299999 33.300000 I am getting correct result for double and incorrect for

问题 I'm comparing simple floats and doubles in C, specifically the value 8.7 for both of them. Now I assign 8.7 to each variable, when I print I get a result of 8.7000 for both values. Why has the compiler added these zeros. And the main question I wanted to ask was is there any further numbers that I'm not seeing, as in hidden after the trailing zeros. I read that I shouldn't do comparisons like this with float because of a lack of precision, but I thought with such a small value surely it can

问题 I'm trying to generate a random double between but not including it's lower and upper bound (lower, upper). I've seen a lot of questions about generating a number from, and including it's lower bound up to, but not including it's upper bound [lower, upper), but they don't answer my question as they do not address this case. I've come up with two "solutions" to the problem, but am not really satisfied by either. First "solution" double myvalue; do { myvalue = myrandom.nextDouble() * (upper -

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 years ago . I'm working in Code::Blocks on a project in C. When I compile I get the error: "incompatible types when assigning to type 'double *' from type 'double'" on lines 81, 85, 90, 91. The project is to take a unit conversion tool and incorporate multiple functions instead of everything under the main(). http:/