I would like to calculate (1.0-p)^n where p is a double be开发者_如何学Ctween 0 and 1 (often very close to 0) and n is a positive integer that might be on the order of hundreds or thousands (perhaps l
I\'m building a loc-based app, using CLLocationManager and locationManager:didUpdateToLocation: fromLocation:
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical andcannot be reasonably answered in its current form. For help clari
In MATLAB (r2009b) I have a uint32 variable containing the value 2147484101. This number (its 4-bytes) has been extracted from a digital machine-vision camera in a grabbing process. According to what
Every time I start a new project an开发者_Python百科d when I need to compare some float or double variables I write the code like this one:
Here is the code NSDate* d = [NSDate dateWithTimeIntervalSince1970:32.4560]; doub开发者_运维技巧le ti = [d timeIntervalSince1970];
I\'m working on porting the sqrt function (for 64-bit doubles) from fdlibm to a model-checker tool I\'m using at the moment (cbmc).
For storing large numbers I am refering to the next: 2.718281828459045534884808148490265011787414550开发者_C百科7812500
I noticed that when I store a double value such as e.g. x = 0.56657011973046234 in an sqlite database, and then retrieve it later, I get y = 0.56657011973046201. According to the sqlite spec and the .
I\'m writing a function that calculates the value of PI, and returns it as a double. So far so good. But once the function gets to 14 digits after the decimal place, it can\'t hold an开发者_如何学Pyth