Ah, that's `lc` the lower-case function, not `1c` the hex representation of 28.
But I still don't understand why you can have the following pop without a delimiter after the assignment statement. I think the purpose of the pop is to subtract one from the result. | [reply] [d/l] [select] |

I may be mistaken, but I read it the way, that the pop delivers the scalar parameter for lc, (replace it with shift, the result is the same).
the basic is that you get a list of all values between a and the parameter, evaluated in scalar-context (I had this wrong at first :-), you get the length of the list, which is of course the decimal representation.
regards,
tomte
**Hlade's Law:**
If you have a difficult task, give it to a lazy person --
they will find an easier way to do it.
| [reply] |

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