Slashdot is powered by your submissions, so send in your scoop

 



Forgot your password?
typodupeerror
Check out the new SourceForge HTML5 internet speed test! No Flash necessary and runs on all devices. ×

Comment Re:Leaky abstractions (Score 1) 175

Well, thats because python, to save memory and processing time with lists, passes by reference, not by copy. The way to do what you want is this: >>> a=[[5]] >>> b=[a[0][:]] >>> a[0][0]=3 >>> b [[5]] However, if you are going to be passing around lists, and want copies to modify, then do this: b=a[:] to do a one-level copy of the list's contents. Any lists in the list, as above, will also be passed by reference, not copy. If you do have nested lists, then take a look at the copy module.

Slashdot Top Deals

"Be there. Aloha." -- Steve McGarret, _Hawaii Five-Oh_

Working...