double

问题 Using Node.js, I'm evaluating the expression: 0 < Number.MIN_VALUE To my surprise, this returns true . Why is that? And: How can I get the smallest number available for which the comparison works as expected? 回答1: Number.MIN_VALUE is 5e-324 , i.e. the smallest positive number that can be represented within float precision, i.e. that's as close as you can get to zero. It defines the best resolution floats give you. Now the overall smallest value is Number.NEGATIVE_INFINITY although that's not

问题 Using Node.js, I'm evaluating the expression: 0 < Number.MIN_VALUE To my surprise, this returns true . Why is that? And: How can I get the smallest number available for which the comparison works as expected? 回答1: Number.MIN_VALUE is 5e-324 , i.e. the smallest positive number that can be represented within float precision, i.e. that's as close as you can get to zero. It defines the best resolution floats give you. Now the overall smallest value is Number.NEGATIVE_INFINITY although that's not

问题 Whenever I input a number in this program the program return a value which is 1 less than the actual result ... What is the problem here?? #include<stdio.h> #include<math.h> int main(void) { int a,b,c,n; scanf("%d",&n); c=pow((5),(n)); printf("%d",c); } 回答1: pow() returns a double , the implicit conversion from double to int is "rounding towards zero". So it depends on the behavior of the pow() function. If it's perfect then no problem, the conversion is exact. If not: 1) the result is

问题 Whenever I input a number in this program the program return a value which is 1 less than the actual result ... What is the problem here?? #include<stdio.h> #include<math.h> int main(void) { int a,b,c,n; scanf("%d",&n); c=pow((5),(n)); printf("%d",c); } 回答1: pow() returns a double , the implicit conversion from double to int is "rounding towards zero". So it depends on the behavior of the pow() function. If it's perfect then no problem, the conversion is exact. If not: 1) the result is

问题 folks! I've encountered a little problem: I'm doing a simple addition with three double values. The result has a smaller precision than the used values. double minutes = 3; minutes = minutes / (24.0*60.0); // contains 0.00208333 double hours = 3; hours = hours / 24.0; // contains 0.125 double days = 3; // contains 3 double age = days + hours + minutes; // result is 3.12708 I found no way to avoid this behaviour. 回答1: Nothing seems to be wrong with the calculation as what the comments on your

问题 I'm working with C, I have to do an exercise in which I have to print the value of long double min and long double max . I used float.h as header, but these two macros ( LDBL_MIN/MAX ) give me the same value as if it was just a double. I'm using Visual Studio 2015 and if I hover the mouse on LDBL MIN it says #define LDBL_MIN DBL_MIN . Is that why it prints dbl_min instead of ldbl_min ? How can I fix this problem? printf("Type: Long Double Value: %lf Min: %e Max: %e Memory:%lu\n", val10, LDBL

问题 In Java, I have defined k as double k=0.0; I am taking data from database and adding the same using while loop, while(rst.next()) { k = k + Double.parseDouble(rst.getString(5)); } NOTE: In database, I have values as 125.23, 458.45, 665.99 (all two decimals) When I display k, I get value as k = 6034.299999999992 Hence I introduced BigDecimal and changed code to below BigDecimal bd = new BigDecimal(k); bd = bd.setScale(2,BigDecimal.ROUND_UP); Now I get new total as bd=6034.30 which is correct.

问题 I have this column in my Oracle 11g mapped as NUMBER (21,20), which is mapped in Hibernate as: @Column(name = "PESO", precision = 21, scale = 20, nullable = false) public BigDecimal getWeight() { return weight; } For a particular record for which the value of the column is 0.493 I get a BigDecimal whose value is 0.49299999999. It seems that somewhere there is a loss of precision due (maybe) to a Double or Float conversion, but I couldn't track it down with a simple unit test like this: Double

问题 I have 2 nullable doubles, an expected value and an actual value (let's call them value and valueExpected). A percentage is found using 100 * (value / valueExpected). However, if valueExpected is zero, it returns NaN. Everything good so far. Now, what do I do if I need to check the value, to see if it is NaN? Normally one could use: if (!Double.IsNaN(myDouble)) But this doesn't work with nullable values (IsNaN only works with non-nullable variables). I have changed my code to do the check

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: Float to binary in C++ I have a very small double var, and when I print it I get -0. (using C++). Now in order to get better precision I tried using cout.precision(18); \\i think 18 is the max precision i can get. cout.setf(ios::fixed,ios::floatfield); cout<<var;\\var is a double. but it just writes -0.00000000000... I want to see the exact binary representation of the var. In other words I want to see what