get the last # from a 2 digit sum

[Chris Losinger]

#include < math.h>
int a = 8;
int b = 4;
int lastDigit = (a + b) - (int)pow(10, (int)log10(a + b)));

-c

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When history comes, it always takes you by surprise.

Bobber!

[Nitron]

int num = 4 + 8;
CString szNumber = "";
szNumber.Format("%d", num);
int ones_col = atoi(szNumber.Right(1));

- Nitron

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"Those that say a task is impossible shouldn't interrupt the ones who are doing it." - Chinese Proverb

[Nitron]

Chris Losinger wrote: int lastDigit = (a + b) - (int)pow(10, (int)log10(a + b))); :omg: I guess I didn't think of it that way ;P - Nitron

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"Those that say a task is impossible shouldn't interrupt the ones who are doing it." - Chinese Proverb

[Chris Losinger]

:) i like a solution that's more complicated than the problem it solves. -c

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When history comes, it always takes you by surprise.

Bobber!

[Roger Allen]

Surely you can use the MOD operator?

int b = 4;
int c = 8;
int a = ((b + c) % 10);

Roger Allen Sonork 100.10016 Were you different as a kid? Did you ever say "Ooohhh, shiny red" even once? - Paul Watson 11-February-2003

[Joao Paulo Figueira]

Try:

int a = 4;
int b = 8;
int c = (a + b) % 10;

I think it is the fastest.

[Tim Smith]

I'm keeping my mouth shut. :laugh: Tim Smith I'm going to patent thought. I have yet to see any prior art.

[mijarral]

And also it is the best way to do this.

[Chris Losinger]

come on, spit it out. i'd hate for you to have to hold back. -c

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[Tim Smith]

Nah, was just going to be my same rant about some of the overly complex solutions to simple problems. It is the same rant I have always had. I wanted to spare people the misery. :laugh: Tim Smith I'm going to patent thought. I have yet to see any prior art.