How do I divide an odd number to leave two integers?
"less than half" int x
is equal x/2
. "Bigger half" - x/2 + x%2
or x - x/2
.
Note that the "less than" and "greater" refers to the absolute value, so in the case of negative x
, bigger < smaller
.
Of course, if x
it is always odd and positive, then it x%2
will 1
, and the greater half can also be calculated as x/2 + 1
.
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For people who use microcontrollers where /
u %
are terrible expenses :-)
This shows an alternative method using shift >>
and &
, which are sometimes cheaper:
#include <stdio.h>
int main (int argc, const char * argv[]) {
const int iplus = 9;
const int iminus = -9;
printf("iplus=%d iminus=%d\n", iplus, iminus);
printf("(iplus >> 1)=%d ((iplus >> 1) + (iplus & 1))=%d\n", iplus >> 1, (iplus >> 1) + (iplus & 1));
printf("(iminus >> 1)=%d ((iminus >> 1) + (iminus & 1))=%d\n", iminus >> 1, (iminus >> 1) + (iminus & 1));
return 0;
}
Output:
iplus=9 iminus=-9
(iplus >> 1)=4 ((iplus >> 1) + (iplus & 1))=5
(iminus >> 1)=-5 ((iminus >> 1) + (iminus & 1))=-4
Accordingly, Does ANSI C or ISO C indicate -5% 10?
There is a difference in terms of behavior /
between C89 and C99, and in particular C89 '/
with one negative number may return positive or negative, but C99 is negative.
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This would be my recommended way:
int low = floor(x / 2.0f);
int high = ceil(x / 2.0f);
I find this to be more concise than the version x/2 + x%2
. This version also benefits from the fact that the result is correct if you performed it using an even number.
EDIT:
People seemed to be complaining to me about using floating point numbers for integers, and here is the bit-based version:
int a = 9;
int b = a >> 1;
int c = b | (a & 0x1);
C # 2's only caveat is that if the input is negative, the results will not be as expected.
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I thought the accepted answer was in the stadium but unclear. If you want to copy and paste the code, this would be the best solution in my opinion
var number = 11;
var halfRoundedUp = (number % 2) ? number/2 + .5 : number/2;
var halfRoundedDown = (number % 2) ? number/2 - .5 : number/2;
alert(halfRoundedUp +" "+ halfRoundedDown);
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