int c = (a * 10) + b;

Hmm I guess that is the way to do it if you want to go mathematically .. I should have thought of that... so dumb..lol

Thanks Ancient


I will use that but just to check if anyone knows of another way to do it please do post.

What I posted is also the method used to convert a string of digits to an integer if you don't want to use a standard C or C++ convertion function.

Here's aother way:
int c = (a << 3) + (a << 1) + b;

Edit: Wrong on my part, sorry.

Edit2: Actually... I think I got it.

a << 3 is the same as a * (2 ^ 3)
a >> 3 is the same as a / (2 ^ 3)

Correct me if I'm wrong on this.

Point made: it's obfuscatory and should only be used in the archaic systems, from whence it came, that lack the appropriate machine commands and/or a supporting assembly code library.

which way is "obfuscatory?"

Is there a specific command that has not been mentioned yet which is now available that will do the same thing?


Not that I have any problem with the way described

Well, you are given two ways of doing the same thing:

One of them is straight-forward, one is not. The one that is not is "obfuscatory" (i.e. more complicated and harder to understand for no reason other than to add confusion).

You're missing the point. When AD posted this:


his point was along the lines of "I've given you a perfectly good solution, yet you ask for another one with no explanation of why the first solution doesn't meet your needs." Then he gave you a solution that does the exact same thing as the first solution, but in an overly-complicated way that no one would ever use instead of the first solution. Why don't you like this?


I can't think of a command that will do this any better.

Why are you asking for a different way then? I'm sure we could come up with even more confusing mathematical formulas that do the exact same thing, but that's probably not what you are looking for.

I laughed so hard when I read this...

winner