Camera UV coordinates — handle images with width or height of one

[ghost]

I'm not quite sure how the current equation was derived. But the case clearly fails for the case that the width and height are both equal to one, for example.

int x = 0;
int y = 0;
int width = 1;
int height = 1;
float u = (x / (width - 1)); // 0 / (1 - 1) = 0 / 0 = NaN
float v = (y / (height - 1));

Should be changed to:

float u = (x + 0.5f) / width;
float v = (y + 0.5f) / height;

The result can be seen clearly when you can a 2x2 image as an example.

The UV coordinates previously would be:

(0, 0) (1, 0)
(0, 1) (1, 1)

The correct result would be:

(0.25, 0.25) (0.75, 0.25)
(0.25, 0.75) (0.75, 0.75)

[hollasch]

Will add a special case to handle single-pixel images. I definitely do not want to promote the idea that a pixel is little square, or has any specific shape, or indeed even that it has a finite kernel.

I'll also revisit the text to ensure that we at least hint that square pixels are a poor (but cheap) model.

[trevordblack]

I forget where I read it, but the correct uv should be:

float u = floor((x + 0.5f) / width);
float v = floor((y + 0.5f) / height);

[tay10r]

@hollasch

This was my issue from a previous account. I wasn't proposing to tell readers that a pixel is a square, just offering a more simpler and commonly used expression of UV coordinates that does not require any special corner cases. You're free to reject it, that's totally fine.

[tay10r]

@trevordblack

Thanks for the suggestion! This:

float u = floor((x + 0.5f) / width);
float v = floor((y + 0.5f) / height);

Is not right because it will round to either 0 or 1 for all UV coordinates. So unless you have a 2x2 image, then you would get incorrect results. Using floor is more likely to be useful when converting from UV coordinates to image coordinates. For example:

int x = floor(uv_x * width);
int y = floor(uv_y * height);

But you can just use native floating-point to integer conversion instructions, which performs the truncation along with the conversion.

int x = uv_x * width;
int y = uv_y * height;

[hollasch]

Fixed in code in PR #1154 (text update pending).