How do I avoid getting -0 when dividing in c++

Question

I have a script in which I want to find the chunk my player is in. Simplified version:

float x = -5
float y = -15
int chunkSize = 16

int player_chunk_x = int(x / chunkSize)
int player_chunk_y = int(y / chunkSize)

This gives the chunk the player is in, but when x or y is negative but not less than the chunkSize (-16), player_chunk_x or player_chunk_y is still 0 or '-0' when I need -1

Of course I can just do this:

if (x < 0) x--
if (y < 0) y--

But I was wondering if there is a better solution to my problem.

Thanks in advance.

If `x` is negative, `std::floor(x / chunkSize)`? If `x` is positive use `std::ceil` instead? — Some programmer dude, Aug 26 '22 at 07:10
I don't believe you. Normally only floating-point types can be `-0`. — HolyBlackCat, Aug 26 '22 at 07:17
I dont actually get -0 but if the number after the division is between -0.0000001 and -0.9999 i will get 0 when i want -1. I was wondering if I could do this without using if statements — Sandu Chicu, Aug 26 '22 at 07:18
Conversion from floating point to integer *always* truncates the decimals. And there's no way to get consistent results of rounding without any kind of conditional check for negative or non-negative values. — Some programmer dude, Aug 26 '22 at 07:24
Your actual problem seems related to [cast-to-int-vs-floor](https://stackoverflow.com/questions/3300290/cast-to-int-vs-floor). — Jarod42, Aug 26 '22 at 07:37
As a possible workaround, get the absolute value of the division and always use `std::ceil`? — Some programmer dude, Aug 26 '22 at 07:49

score 4 · Answer 1 · answered Aug 26 '22 at 07:15

Since C++20 it's impossible to get an integral type signed negative zero, and was only possible in a rare (but by no means extinct) situation where your platform had 1's complement int. It's still possible in C (although rare), and adding 0 to the result will remove it.

It's possible though to have a floating point signed negative zero. For that, adding 0.0 will remove it.

Note that for an integral -0, subtracting 1 will yield -1.

score 3 · Accepted Answer · answered Aug 26 '22 at 08:38

3

Your issue is that you are casting a floating point value to an integer value.

This rounds to zero by default.

If you want consistent round down, you first have to floor your value:

int player_chunk_x = int(std::floor(x / chunkSize);

answered Aug 26 '22 at 08:38

Guillaume Gris

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score 1 · Answer 3 · answered Aug 26 '22 at 07:35

1

If you don't like negative numbers then don't use them:

int player_chunk_x = (x - min_x) / chunkSize;
int player_chunk_y = (y - min_y) / chunkSize;

answered Aug 26 '22 at 07:35

463035818_is_not_an_ai

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ADITYA AHLAWAT · Answer 4 · 2022-08-26T09:12:40.450

1

If you want integer, in this case -1 on ( -5%16 or anything like it ) then this is possible using a math function:

Possible Ways :

using floor ->

float x = -5;
float y = -15;
int chunkSize = 16;

int player_chunk_x = floor(x / chunkSize) 
// will give -1 for (-5 % 16);
// 0 for (5%16) 
// 1 for any value between 1 & 2 and so on 

int player_chunk_y = floor(y / chunkSize);

edited Aug 26 '22 at 09:12

answered Aug 26 '22 at 08:25

ADITYA AHLAWAT

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How do I avoid getting -0 when dividing in c++

4 Answers4