四舍五入到最接近的整数 - 我是在作弊还是这已经足够了？

Question

Essentially, if the number generated is 2.3 then if I subtract .5 it will then be 1.8 but the rounding function will make it 2, which is what I want. Or if the answer is 2.99999 and I subtract .5, the answer is 2.49999 which should round down to 2 which is what I want. My question is if the answer is 2 even and I subtract .5, the answer is now 1.5, so will it still round up to 2.

temp1_1= Math.round(temp2_2/(360/temp_value)-.5);

this is my line of code for this.

Answer 1

There is already a function to do that. It's called floor :

double d = Math.floor(2.9999) //result: 2.0

Answer 2

Even simpler and potential faster

double d = 2.99999999;
long l = (long) d; // truncate to a whole number.

This will round towards 0. Math.floor() rounds towards negative infinity. Math.round(x - 0.5) also rounds towards negative infinity.

Answer 3

Everyone always wants to use fancy functions, but forgets about the humble modulus. My solution:

number = x-(x%1);

subtracts the remainder of division by one, so x = 2.999 will = 2, 3.111 will = 3 and so on. The cool thing about this is that you can round down the multiple of anything just by changing that 1 into whatever you like.