Merge "bridge_softmix: use a float type to store the internal REMB bitrate" into 16
This commit is contained in:
Joshua Colp
Gerrit Code Review
commit
4f0b8c3ed3
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@ -33,6 +33,8 @@
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#include "asterisk.h"
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#include <math.h>
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#include "asterisk/stream.h"
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#include "asterisk/test.h"
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#include "asterisk/vector.h"
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@ -75,8 +77,8 @@ struct softmix_remb_collector {
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struct ast_frame frame;
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/*! The REMB to send to the source which is collecting REMB reports */
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struct ast_rtp_rtcp_feedback feedback;
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/*! The maximum bitrate */
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unsigned int bitrate;
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/*! The maximum bitrate (A single precision floating point is big enough) */
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float bitrate;
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};
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struct softmix_stats {
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@ -1334,7 +1336,7 @@ static void remb_collect_report(struct ast_bridge *bridge, struct ast_bridge_cha
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struct softmix_bridge_data *softmix_data, struct softmix_channel *sc)
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{
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int i;
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unsigned int bitrate;
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float bitrate;
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/* If there are no video sources that we are a receiver of then we have noone to
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* report REMB to.
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@ -1391,6 +1393,7 @@ static void remb_collect_report(struct ast_bridge *bridge, struct ast_bridge_cha
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static void remb_send_report(struct ast_bridge_channel *bridge_channel, struct softmix_channel *sc)
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{
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int i;
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int exp;
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if (!sc->remb_collector) {
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return;
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@ -1398,17 +1401,52 @@ static void remb_send_report(struct ast_bridge_channel *bridge_channel, struct s
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/* We always do this calculation as even when the bitrate is zero the browser
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* still prefers it to be accurate instead of lying.
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*
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* The mantissa only has 18 bits available, so make sure it fits. Adjust the
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* value and exponent for those values that don't.
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*
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* For example given the following:
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*
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* bitrate = 123456789.0
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* frexp(bitrate, &exp);
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*
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* 'exp' should now equal 27 (number of bits needed to represent the value). Since
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* the mantissa must fit into an 18-bit unsigned integer, and the given bitrate is
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* too large to fit, we must subtract 18 from the exponent in order to get the
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* number of times the bitrate will fit into that size integer.
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*
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* exp -= 18;
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*
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* 'exp' is now equal to 9. Now we can get the mantissa that fits into an 18-bit
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* unsigned integer by dividing the bitrate by 2^exp:
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*
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* mantissa = 123456789.0 / 2^9
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*
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* This makes the final mantissa equal to 241126 (implicitly cast), which is less
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* than 262143 (the max value that can be put into an unsigned 18-bit integer).
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* So now we have the following:
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*
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* exp = 9;
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* mantissa = 241126;
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*
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* If we multiply that back we should come up with something close to the original
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* bit rate:
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*
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* 241126 * 2^9 = 123456512
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*
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* Precision is lost due to the nature of floating point values. Easier to why from
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* the binary:
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*
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* 241126 * 2^9 = 241126 << 9 = 111010110111100110 << 9 = 111010110111100110000000000
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*
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* Precision on the "lower" end is lost due to zeros being shifted in. This loss is
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* both expected and acceptable.
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*/
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sc->remb_collector->feedback.remb.br_mantissa = sc->remb_collector->bitrate;
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sc->remb_collector->feedback.remb.br_exp = 0;
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frexp(sc->remb_collector->bitrate, &exp);
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exp = exp > 18 ? exp - 18 : 0;
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/* The mantissa only has 18 bits available, so while it exceeds them we bump
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* up the exp.
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*/
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while (sc->remb_collector->feedback.remb.br_mantissa > 0x3ffff) {
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sc->remb_collector->feedback.remb.br_mantissa = sc->remb_collector->feedback.remb.br_mantissa >> 1;
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sc->remb_collector->feedback.remb.br_exp++;
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}
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sc->remb_collector->feedback.remb.br_mantissa = sc->remb_collector->bitrate / (1 << exp);
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sc->remb_collector->feedback.remb.br_exp = exp;
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for (i = 0; i < AST_VECTOR_SIZE(&bridge_channel->stream_map.to_bridge); ++i) {
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int bridge_num = AST_VECTOR_GET(&bridge_channel->stream_map.to_bridge, i);
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@ -31,6 +31,8 @@
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#include "asterisk.h"
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#include <math.h>
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#include "asterisk/module.h"
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#include "asterisk/cli.h"
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#include "asterisk/channel.h"
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@ -39,9 +41,9 @@
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struct remb_values {
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/*! \brief The amount of bitrate to use for REMB received from the channel */
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unsigned int receive_bitrate;
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float receive_bitrate;
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/*! \brief The amount of bitrate to use for REMB sent to the channel */
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unsigned int send_bitrate;
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float send_bitrate;
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};
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static void remb_values_free(void *data)
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@ -59,6 +61,8 @@ static struct ast_frame *remb_hook_event_cb(struct ast_channel *chan, struct ast
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struct ast_rtp_rtcp_feedback *feedback;
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struct ast_datastore *remb_store;
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struct remb_values *remb_values;
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int exp;
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float bitrate;
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if (!frame) {
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return NULL;
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@ -91,20 +95,57 @@ static struct ast_frame *remb_hook_event_cb(struct ast_channel *chan, struct ast
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/* If a bitrate override has been set apply it to the REMB Frame */
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if (event == AST_FRAMEHOOK_EVENT_READ && remb_values->receive_bitrate) {
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feedback->remb.br_mantissa = remb_values->receive_bitrate;
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feedback->remb.br_exp = 0;
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bitrate = remb_values->receive_bitrate;
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} else if (event == AST_FRAMEHOOK_EVENT_WRITE && remb_values->send_bitrate) {
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feedback->remb.br_mantissa = remb_values->send_bitrate;
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feedback->remb.br_exp = 0;
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bitrate = remb_values->send_bitrate;
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}
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/* The mantissa only has 18 bits available, so while it exceeds them we bump
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* up the exp.
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/*
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* The mantissa only has 18 bits available, so make sure it fits. Adjust the
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* value and exponent for those values that don't.
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*
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* For example given the following:
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*
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* bitrate = 123456789.0
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* frexp(bitrate, &exp);
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*
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* 'exp' should now equal 27 (number of bits needed to represent the value). Since
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* the mantissa must fit into an 18-bit unsigned integer, and the given bitrate is
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* too large to fit, we must subtract 18 from the exponent in order to get the
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* number of times the bitrate will fit into that size integer.
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*
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* exp -= 18;
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*
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* 'exp' is now equal to 9. Now we can get the mantissa that fits into an 18-bit
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* unsigned integer by dividing the bitrate by 2^exp:
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*
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* mantissa = 123456789.0 / 2^9
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*
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* This makes the final mantissa equal to 241126 (implicitly cast), which is less
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* than 262143 (the max value that can be put into an unsigned 18-bit integer).
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* So now we have the following:
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*
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* exp = 9;
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* mantissa = 241126;
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*
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* If we multiply that back we should come up with something close to the original
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* bit rate:
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*
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* 241126 * 2^9 = 123456512
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*
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* Precision is lost due to the nature of floating point values. Easier to why from
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* the binary:
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*
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* 241126 * 2^9 = 241126 << 9 = 111010110111100110 << 9 = 111010110111100110000000000
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*
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* Precision on the "lower" end is lost due to zeros being shifted in. This loss is
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* both expected and acceptable.
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*/
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while (feedback->remb.br_mantissa > 0x3ffff) {
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feedback->remb.br_mantissa = feedback->remb.br_mantissa >> 1;
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feedback->remb.br_exp++;
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}
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frexp(bitrate, &exp);
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exp = exp > 18 ? exp - 18 : 0;
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feedback->remb.br_mantissa = bitrate / (1 << exp);
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feedback->remb.br_exp = exp;
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return frame;
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}
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