List to number base conversion

[Ed H]

The program takes the digits of a number from a list, and the base of the number from a variable, and returns the number.

Inputs:
B - the base of the input number
Ans - a list containing the digits of the input number in reverse order

Output:
Ans - the value of the number inputted

Example:
With 5 in B, and {1,2,3,4} in Ans, return 586 = 1 + 2*5 + 3*25 + 4*125.

Goal: 20 bytes

[bfr]

20 bytes:

[Ed H]

This is what I had:

[calc84maniac]

[Ed H]

Good catch, calcmaniac84.

[elfprince13]

17 bytes, assuming everything is a 1 byte token.

[bfr]

That returns a list, btw

EDIT: Nevermind, your new code doesn't return a list anymore. That still doesn't give the right answer, though, because you're doing: [digit]*B*1+[digit]*B*2... etc. instead of [digit]*B*0+[digit]*B*1

Without cumSum(, something like that would be a pretty nice solution though

[elfprince13]

my wireless pooped out right in the middle of fixing that with an edit or I would have beat you to it.

[Ed H]

[elfprince13]

haha, yeah, I just didn't have a calculator or emulator with me so I was doing that in my head, and I don't know how many bytes all the tokens I used are. I really just felt a need to exploit that cool u/v/w() thing DarkerLine posted.

also, you missed my edit with the fixed version.

[Ed H]

Using the sequence functions in that way only really helps if you're using seq( multiple times with the same expression but different start and end values. It is very cool, though - a useful substitute for seq( in sum situations.

However, in this situation, seq( is more efficient than the sequence functions. You could do

[DarkerLine]

In general, cumSum(Ans=Ans) is more effective than cumSum(Ans or 1); although it's the same size, it won't fail for complex inputs. I suppose it doesn't really matter here, though.