Question

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?

Answer 1

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.