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-rw-r--r--drivers/media/usb/uvc/uvc_driver.c84
1 files changed, 0 insertions, 84 deletions
diff --git a/drivers/media/usb/uvc/uvc_driver.c b/drivers/media/usb/uvc/uvc_driver.c
index 9c05776f11d1..0f14dee4b6d7 100644
--- a/drivers/media/usb/uvc/uvc_driver.c
+++ b/drivers/media/usb/uvc/uvc_driver.c
@@ -329,90 +329,6 @@ static enum v4l2_ycbcr_encoding uvc_ycbcr_enc(const u8 matrix_coefficients)
return V4L2_YCBCR_ENC_DEFAULT; /* Reserved */
}
-/*
- * Simplify a fraction using a simple continued fraction decomposition. The
- * idea here is to convert fractions such as 333333/10000000 to 1/30 using
- * 32 bit arithmetic only. The algorithm is not perfect and relies upon two
- * arbitrary parameters to remove non-significative terms from the simple
- * continued fraction decomposition. Using 8 and 333 for n_terms and threshold
- * respectively seems to give nice results.
- */
-void uvc_simplify_fraction(u32 *numerator, u32 *denominator,
- unsigned int n_terms, unsigned int threshold)
-{
- u32 *an;
- u32 x, y, r;
- unsigned int i, n;
-
- an = kmalloc_array(n_terms, sizeof(*an), GFP_KERNEL);
- if (an == NULL)
- return;
-
- /*
- * Convert the fraction to a simple continued fraction. See
- * https://en.wikipedia.org/wiki/Continued_fraction
- * Stop if the current term is bigger than or equal to the given
- * threshold.
- */
- x = *numerator;
- y = *denominator;
-
- for (n = 0; n < n_terms && y != 0; ++n) {
- an[n] = x / y;
- if (an[n] >= threshold) {
- if (n < 2)
- n++;
- break;
- }
-
- r = x - an[n] * y;
- x = y;
- y = r;
- }
-
- /* Expand the simple continued fraction back to an integer fraction. */
- x = 0;
- y = 1;
-
- for (i = n; i > 0; --i) {
- r = y;
- y = an[i-1] * y + x;
- x = r;
- }
-
- *numerator = y;
- *denominator = x;
- kfree(an);
-}
-
-/*
- * Convert a fraction to a frame interval in 100ns multiples. The idea here is
- * to compute numerator / denominator * 10000000 using 32 bit fixed point
- * arithmetic only.
- */
-u32 uvc_fraction_to_interval(u32 numerator, u32 denominator)
-{
- u32 multiplier;
-
- /* Saturate the result if the operation would overflow. */
- if (denominator == 0 ||
- numerator/denominator >= ((u32)-1)/10000000)
- return (u32)-1;
-
- /*
- * Divide both the denominator and the multiplier by two until
- * numerator * multiplier doesn't overflow. If anyone knows a better
- * algorithm please let me know.
- */
- multiplier = 10000000;
- while (numerator > ((u32)-1)/multiplier) {
- multiplier /= 2;
- denominator /= 2;
- }
-
- return denominator ? numerator * multiplier / denominator : 0;
-}
-
/* ------------------------------------------------------------------------
* Terminal and unit management
*/