This behaves as wanted:
double t = r[1][0] * .5;
But this doesn't:
double t = ((1/2)*r[1][0]);
r is a 2-D Vector.
Just thought of a possibility. Is it because (1/2) is considered an int and (1/2) == 0?
Is it because (1/2) is considered an int and (1/2) == 0?
Yes, both of those literals are of type int, therefore the result will be of type int, and that result is 0.
Instead, make one of those literals a float or double and you'll end up with the floating point result of 0.5, ie:
double t = ((1.0/2)*r[1][0]);
Because 1.0 is of type double, the int 2 will be promoted to a double and the result will be a double.
Write this instead:
double t = ((1/2.0)*r[1][0]);
1 / 2 is an integer division and the result is 0.
1 / 2.0 is a floating point division (with double values after the usual arithmetic conversions) and its result is 0.5.
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