divide-by-zero

问题 This question already has answers here : Inconsistency in divide-by-zero behavior between different value types (5 answers) Closed 2 years ago . We get compile time error when integer is divided by zero whereas in case of double there is no compilation error but at run-time we get infinity/NaN as the result. Any idea why int & double have different behavior when it comes to divide by zero exception? void Main() { int number = 20; var result1 = number/0; // Divide by zero compile time

问题 This question already has answers here : Inconsistency in divide-by-zero behavior between different value types (5 answers) Closed 2 years ago . We get compile time error when integer is divided by zero whereas in case of double there is no compilation error but at run-time we get infinity/NaN as the result. Any idea why int & double have different behavior when it comes to divide by zero exception? void Main() { int number = 20; var result1 = number/0; // Divide by zero compile time

问题 I wrote a simulator that has some collision detection code and does a good bit of math on each object when it detects collisions. If these two objects are at the exact same location or in some rare other cases, I'm getting NaN (not a number) as their location somewhere along the line and I'd like to know where. Normally, the program would crash if I did these operations on integers but because + and - infinity are part of the floating point spec, it is allowed. So, somewhere along the line I

问题 I know in standard IEEE 754 division by zero is allowed. I want to know how it's represented in binary. For example, 0.25 in decimal is 0 01111101 00000000000000000000000 in binary. What about 5.0/0.0 or 0.0/0.0 do they have represenation in binary, and are they same? Thanks. 回答1: When you divide a finite number by zero you'll get an infinity with the sign of the number you tried to divide. So 5.0/0.0 is +inf but 0.0/0.0 returns something called a QNaN indefinite. Let’s say we are dividing

问题 Consider #include <iostream> int main() { double a = 1.0 / 0; double b = -1.0 / 0; double c = 0.0 / 0; std::cout << a << b << c; // to stop compilers from optimising out the code. } I have always thought that a will be +Inf, b will be -Inf, and c will be NaN. But I also hear rumours that strictly speaking the behaviour of floating point division by zero is undefined and therefore the above code cannot considered to be portable C++. (That theoretically obliterates the integrity of my million

问题 I am using this code to remove the lines from the following image: I don't know the reason, but it gives me as output ZeroDivisionError: division by zero error on line 34 - x0, x1, y0, y1 = (0, im_wb.shape[1], sum(y0_list)/len(y0_list), sum(y1_list)/len(y1_list)) . What's the reason ? How can I fix it ? import cv2 import numpy as np img = cv2.imread('lines.png',0) # Applies threshold and inverts the image colors (thresh, im_bw) = cv2.threshold(img, 128, 255, cv2.THRESH_BINARY | cv2.THRESH

问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Why doesn't Java throw an Exception when dividing by 0.0? Why the following statement in Java will not report an ArithmeticException? double d = 1.0/0; 回答1: In short: floating point numbers can represent infinity (or even operations that yield values which aren't numbers) so an operation that results in this (e.g. dividing by 0) is valid. Expanding upon Mohammod Hossain's answer, as well as this question and its

问题 If I divide by 0, I get either a ZeroDivisionError, Infinity or NaN depending on what is divided. ruby-1.9.2-p180 :018 > 0.0 / 0 => NaN ruby-1.9.2-p180 :020 > 3.0 / 0 => Infinity ruby-1.9.2-p180 :021 > 3 / 0 ZeroDivisionError: divided by 0 I understand that 0.0 / 0 is not an Infinity (in math terms), while 3.0 / 0 is but why then isn't 3 / 0 an Infinity? Why dividing an integer throws an exception but dividing a float doesn't? 回答1: In Ruby, not all numbers are created equal (pun intended).

问题 I am dividing two numpy arrays: >>> import numpy as np >>> a1 = np.array([[ 0, 3], [ 0, 2]]) >>> a2 = np.array([[ 0, 3], [ 0, 1]]) >>> d = a1/a2 >>> d array([[ nan, 1.], [ nan, 2.]]) >>> where_are_NaNs = np.isnan(d) >>> d[where_are_NaNs] = 0 >>> d >>> array([[ 0., 1.], [ 0., 2.]]) I am looking for a way to get 0 instead of Nan without using for loops? Does numpy have a similar function to fillna() in pandas? 回答1: This below should work and convert all NANs to 0 d[np.isnan(d)] = 0 If you want

问题 I have some places in my code where I want to assure that a division of 2 arbitrary floating point numbers (32 bit single precision) won't overflow. The target/compiler does not guarantee (explicitly enough) nice handling of -INF/INF and (does not fully guarantees IEEE 754 for the exceptional values - (possibly undefined) - and target might change). Also I cannot make save assumtions on the inputs for this few special places and I am bound to C90 standard libraries. I have read What Every