I am working with some C code and I'm totally stuck in this function. It should compare two buffers with some deviator. For example if EEPROM_buffer[1] = 80, so TxBuffer values from 78 to 82 should be correct!
So the problem is that it always returns -1. I checked both buffers, data is correct and they should match, but won't. Program just runs while until reach i = 3 and returns -1..
I compile with atmel studio 6.1, atmel32A4U microcontroller..
int8_t CheckMatching(t_IrBuff * tx_buffer, t_IrBuff * tpool)
{
uint8_t i = 0;
uint16_t * TxBuffer = (uint16_t*) tx_buffer->data;
while((TxBuffer->state != Data_match) || (i != (SavedBuff_count))) // Data_match = 7;
{
uint16_t * EEPROM_buffer = (uint16_t*) tpool[i].data;
for(uint16_t j = 0; j < tpool[i].usedSize; j++) // tpool[i].usedSize = 67;
{
if(abs(TxBuffer[j] - EEPROM_buffer[j]) > 3)
{
i++;
continue;
}
}
i++;
TxBuffer->state = Data_match; // state value before Data_match equal 6!
}
tx_buffer->state = Buffer_empty;
if(i == (SavedBuff_count)) // SavedBuff_count = 3;
{
return -1;
}
return i;
}
Both your TxBuffer elements and EEPROM_buffer elements are uint16_t. When deducting 81 from 80 as uint16_t it would give 0xffff, with no chance of abs to help you. Do a typecast to int32_t and you will be better off.
Related
I am using Debian (8.3.0-6) on an embedded custom board and working on the dht11 sensor.
Briefly,
I need to read 40 bits from a GPIO pin and each bit can take a max 70 microseconds. When bit-level is high for max 28us or 70us, it means is logic 0 or 1, respectively. (So I have a timeout controller for each bit and if a bit takes more than 80us, I need to stop the process.).
In my situation, sometimes I can read all 40 bits correctly but sometimes I can't do it and the function of libgpiod gpiod_line_get_value(line); is missing the bit (my code is below). I am trying to figure out that why I cant read and lose a bit, what is the reason for that. But I haven't found a sensible answer yet. So I was wondering What am I missing out?, What is the proper way to GPIO programming?
Here is what I wanted to show you, How do I understand what I am missing a bit? Whenever I catch a bit, I am setting and resetting another GPIO pin at the rising and falling edge of a bit (So I can see which bit missing). Moreover, as far as I see I am always missing the two edges on one bit or one edge on two bits consecutively (rising and falling or falling and rising). In the first picture, you can see which bit I missed, the second is when I read all bits correctly.
Here it is my code:
//********************************************************* Start reading data bit by low level (50us) ***************************
for (int i = 0; i < DHT_DATA_BYTE_COUNT ; i++) //DHT_DATA_BYTE_COUNT = 5
{
for (int J = 7; J > -1; J--)
{
GPIO_SetOutPutPin(testPin); //gpiod_line_set_value(testPin, 1);
int ret;
start = micros();
do
{
ret = GPIO_IsInputPinSet(dht11pin);//gpiod_line_get_value(dht11pin);
delta = micros() - start;
if(ret == -1)
{
err_step.step = 9;
err_step.ret_val = -1;
return -1;
}
if(delta > DHT_START_BIT_TIMEOUT_US) //80us
{
err_step.step = 10;
err_step.ret_val = -2;
err_step.timestamp[is] = delta;
err_step.indx[is].i = i;
err_step.indx[is++].j = J;
GPIO_ResetOutPutPin(testPin);
return -2;
}
}while(ret == 0);
GPIO_ResetOutPutPin(testPin);
err_step.ret_val = 10;
GPIO_SetOutPutPin(testPin);
start = micros();
do
{
ret = GPIO_IsInputPinSet(dht11pin);
delta = micros() - start;
if(ret == -1)
{
err_step.step = 11;
err_step.ret_val = -1;
return -1;
}
if(delta > DHT_BEGIN_RESPONSE_TIMEOUT_US) //80us
{
err_step.step = 12;
err_step.ret_val = -2;
err_step.timestamp[is] = delta;
err_step.indx[is].i = i;
err_step.indx[is++].j = J;
return -2;
}
}while(ret == 1);
err_step.timestamp[is] = delta;
err_step.indx[is].i = i;
err_step.indx[is++].j = J;
GPIO_ResetOutPutPin(testPin);
err_step.ret_val = 10;
(delta > DHT_BIT_SET_DATA_DETECT_TIME_US) ? bitWrite(dht11_byte[i],J,1) : bitWrite(dht11_byte[i],J,0);
}
}
I have build a led cube using transistors, shift registers and an arduino nano (and it kinda works). I know shift registers may be a poor design choice but I have to work with what I got so please don't get stuck on that in your answers.
There is this piece of code:
bool input[32];
void resetLeds()
{
input[R1] = 0;
input[G1] = 0;
input[B1] = 0;
input[R2] = 0;
input[G2] = 0;
input[B2] = 0;
input[R3] = 0;
input[G3] = 0;
input[B3] = 0;
input[R4] = 0;
input[G4] = 0;
input[B4] = 0;
input[X1Z1] = 1;
input[X1Z2] = 1;
input[X1Z3] = 1;
input[X1Z4] = 1;
input[X2Z1] = 1;
input[X2Z2] = 1;
input[X2Z3] = 1;
input[X2Z4] = 1;
input[X3Z1] = 1;
input[X3Z2] = 1;
input[X3Z3] = 1;
input[X3Z4] = 1;
input[X4Z1] = 1;
input[X4Z2] = 1;
input[X4Z3] = 1;
input[X4Z4] = 1;
}
void loop()
{
T = micros();
for(int I = 0; I < 100; I++)
{
counter++;
if(counter >= 256 / DIVIDER) counter = 0;
for(int i = 0; i < 64; i += 4)
{
x = i / 16;
z = (i % 16) / 4;
resetLeds();
input[XZ[x][z]] = 0;
for(y = 0; y < 4; y++)
{
index = i + y;
if(counter < xyz[index][0]) input[Y[y][RED]] = 1;
if(counter < xyz[index][1]) input[Y[y][GREEN]] = 1;
if(counter < xyz[index][2]) input[Y[y][BLUE]] = 1;
}
PORTB = 0;
for(int j = 0; j < 32; j++)
{
bitWrite(OUT_PORT, 4, 0);
bitWrite(OUT_PORT, 3, input[j]);
PORTB = OUT_PORT;
bitWrite(OUT_PORT, 4, 1);
PORTB = OUT_PORT;
}
bitWrite(OUT_PORT, 0, 1);
PORTB = OUT_PORT;
}
}
T = micros() - T;
Serial.println(T / 100);
}
The runtime of a single iteration is reported to be 1274 microseconds, but I need it to be even lower to build a pwm function of sorts (manually turning a transistor on and off through a shift register). While optimizing I found this strange behavior I cannot explain. There is this line in the code:
bitWrite(OUT_PORT, 3, input[j]);
When I remove this line or change input[j] to 0 the runtime is halved. Apparently, an array lookup takes about 20 microseconds. But I find it very weird since I am indexing this array in more places in the code (when writing) and there it takes 23 microseconds for 28 writes.
Can somebody please explain to me what is going on and/or how to make this piece of code run faster? I guess you can do writes in a pipelined manner but a read stalls the code completely since you cannot continue before you receive the value from cache. But then again, I hardly doubt a read from cache should take 23 whole microseconds.
[EDIT 27/10/2020 16:55]
The part of the code that writes to the shiftregisters has a lot of bitWrites which are time consuming. Instead of writing the bits every time I implemented preconfigured options for the byte to write to PORTB:
PORTB = LATCH_LOW;
for(int j = 0; j < 32; j++)
{
if(input[j] == 0)
{
PORTB = CLOCK_OFF_DATA_0;
PORTB = CLOCK_ON_DATA_0;
}
else
{
PORTB = CLOCK_OFF_DATA_1;
PORTB = CLOCK_ON_DATA_1;
}
}
PORTB = LATCH_HIGH;
This cuts my running time roughly in half, which is kind of fast enough but I wonder I could get it to run even faster. When I remove everything from the loop except for the writing to the shift registers and I remove the input[j] read, I get a runtime of 200 microseconds. This means that if I could remove dependence on the input[j] read and compute its value inline I should be able to get at least another 2 times speed up. To achieve 8 bit PWM I calculated I need the running time to be 40 microseconds or less so I am going to stick with 16 (instead of 256) brightness levels for now to prevent flicker.
[EDIT 28/10/2020 19:38]
I went into the platform.txt and change the optimization flag to -Ofast. This got my iteration time down another 200 microseconds!
I try to send a maximum of 8 bytes of data. The first 4 bytes are always the same and involve defined commands and an address. The last 4 bytes should be variable.
So far I'm using this approach. Unfortunatly I was told to not use any for loops in this case.
// Construct data
local_transmit_buffer[0] = EEPROM_CMD_WREN;
local_transmit_buffer[1] = EEPROM_CMD_WRITE;
local_transmit_buffer[2] = High(MSQ_Buffer.address);
local_transmit_buffer[3] = Low(MSQ_Buffer.address);
uint_fast8_t i = 0;
for(i = 0; i < MSQ_Buffer.byte_lenght || i < 4; i++){ // assign data
local_transmit_buffer[i + 4] = MSQ_Buffer.dataPointer[i];
}
This is some test code I'm trying to solve my problem:
#include <stdio.h>
__UINT_FAST8_TYPE__ local_transmit_buffer[8];
__UINT_FAST8_TYPE__ MSQ_Buffer_data[8];
void print_local(){
for (int i = 0; i < 8; i++)
{
printf("%d ", local_transmit_buffer[i]);
}
printf("\n");
}
void print_msg(){
for (int i = 0; i < 8; i++)
{
printf("%d ", MSQ_Buffer_data[i]);
}
printf("\n");
}
int main(){
// assign all local values to 0
for (int i = 0; i < 8; i++)
{
local_transmit_buffer[i] = 0;
} print_local();
// assign all msg values to 1
for (int i = 0; i < 8; i++)
{
MSQ_Buffer_data[i] = i + 1;
} print_msg();
*(local_transmit_buffer + 3) = (__UINT_FAST32_TYPE__)MSQ_Buffer_data;
printf("\n");
print_local();
return 0;
}
The first loops fills up the local_transmit_buffer with 0's and the MSQ_Buffer with 0,1,2,...
local_transmit_buffer -> 0 0 0 0 0 0 0 0
MSQ_Buffer_data -> 1 2 3 4 5 6 7 8
Now i want to assign the first 4 values of MSQ_Buffer_data to local_transmit_buffer like this:
local_transmit_buffer -> 0 0 0 0 1 2 3 4
Is there another way of solving this problem without using for loops or a bit_field?
Solved:
I used the memcpy function to solve my problem
// uint_fast8_t i = 0;
// for(i = 0; i < MSQ_Buffer.byte_lenght || i < 4; i++){ // assign data
// local_transmit_buffer[i + 4] = MSQ_Buffer.dataPointer[i];
// }
// copy a defined number data from the message to the local buffer to send
memcpy(&local_transmit_buffer[4], &MSQ_Buffer.dataPointer, local_save_data_length);
Either just unroll the loop manually by typing out each line, or simply use memcpy. In this case there's no reason why you need abstraction layers, so I'd write the sanest possible code, which is just manual unrolling (and get rid of icky macros):
uint8_t local_transmit_buffer [8];
...
local_transmit_buffer[0] = EEPROM_CMD_WREN;
local_transmit_buffer[1] = EEPROM_CMD_WRITE;
local_transmit_buffer[2] = (uint8_t) ((MSQ_Buffer.address >> 8) & 0xFFu);
local_transmit_buffer[3] = (uint8_t) (MSQ_Buffer.address & 0xFFu);
local_transmit_buffer[4] = MSQ_Buffer.dataPointer[0];
local_transmit_buffer[5] = MSQ_Buffer.dataPointer[1];
local_transmit_buffer[6] = MSQ_Buffer.dataPointer[2];
local_transmit_buffer[7] = MSQ_Buffer.dataPointer[3];
It is not obvious why you can't use a loop though, this doesn't look like the actual EEPROM programming (where overhead code might cause hiccups), but just preparations for it. Start to question such requirements.
Also note that you should not use __UINT_FAST8_TYPE__ but uint8_t. Never use homebrewed types but always stdint.h. But you should not be using fast types for a RAM buffer used for EEPROM programming, because it cannot be allowed to contain padding, ever. This is a bug.
I'm trying to implement the Join Five game. It is a game where, given a grid and a starting configuration of dots, you have to add dots in free crossings, so that each dot that you add forms a 5-dot line with those already in the grid. Two lines may only have 1 dot in common (they may cross or touch end to end)
My game grid is an int array that contains 0 or 1. 1 if there is a dot, 0 if there isn't.
I'm doing kinda well in the implementation, but I'd like to display all the possibles moves.
I made a very long and ugly function that is available here : https://pastebin.com/tw9RdNgi (it was way too long for my post i'm sorry)
here is a code snippet :
if(jeu->plat[i][j] == 0) // if we're on a empty spot
{
for(k = 0; k < lineSize; k++) // for each direction
{
//NORTH
if(jeu->plat[i-1-k][j] == 1) // if there is a dot north
{
n++; // we count it
}
else
{
break; //we change direction
}
} //
This code repeats itself 7 other times changing directions and if n or any other variable reaches 4 we count the x and y as a possible move.
And it's not even treating all the cases, if the available spot is between 2 and 2 dots it will not count it. same for 3 and 1 and 1 and 3.
But I don't think the way I started doing it is the best one. I'm pretty sure there is an easier and more optimized way but i can't figure it out.
So my question is: could somebody help me figure out how to find all the possible 5-dot alignments, or tell me if there is a better way of doing it?
Ok, the problem is more difficult than it appears, and a lot of code is required. Everything would have been simpler if you posted all of the necessary code to run it, that is a Minimal, Complete, and Verifiable Example. Anyway, I resorted to putting together a structure for the problem which allows to test it.
The piece which answers your question is the following one:
typedef struct board {
int side_;
char **dots_;
} board;
void board_set_possible_moves(board *b)
{
/* Directions
012
7 3
654 */
static int dr[8] = { -1,-1,-1, 0, 1, 1, 1, 0 };
static int dc[8] = { -1, 0, 1, 1, 1, 0,-1,-1 };
int side_ = b->side_;
char **dots_ = b->dots_;
for (int r = 0; r < side_; ++r) {
for (int c = 0; c < side_; ++c) {
// The place already has a dot
if (dots_[r][c] == 1)
continue;
// Count up to 4 dots in the 8 directions from current position
int ndots[8] = { 0 };
for (int d = 0; d < 8; ++d) {
for (int i = 1; i <= 4; ++i) {
int nr = r + dr[d] * i;
int nc = c + dc[d] * i;
if (nr < 0 || nc < 0 || nr >= side_ || nc >= side_ || dots_[nr][nc] != 1)
break;
++ndots[d];
}
}
// Decide if the position is a valid one
for (int d = 0; d < 4; ++d) {
if (ndots[d] + ndots[d + 4] >= 4)
dots_[r][c] = 2;
}
}
}
}
Note that I defined a square board with a pointer to pointers to chars, one per place. If there is a 0 in one of the places, then there is no dot and the place is not a valid move; if there is a 1, then there is a dot; if there is a 2, then the place has no dot, but it is a valid move. Valid here means that there are at least 4 dots aligned with the current one.
You can model the directions with a number from 0 to 7 (start from NW, move clockwise). Each direction has an associated movement expressed as dr and dc. Moving in every direction I count how many dots are there (up to 4, and stopping as soon as I find a non dot), and later I can sum opposite directions to obtain the total number of aligned points.
Of course these move are not necessarily valid, because we are missing the definition of lines already drawn and so we cannot check for them.
Here you can find a test for the function.
#include <stdio.h>
#include <stdlib.h>
board *board_init(board *b, int side) {
b->side_ = side;
b->dots_ = malloc(side * sizeof(char*));
b->dots_[0] = calloc(side*side, 1);
for (int r = 1; r < side; ++r) {
b->dots_[r] = b->dots_[r - 1] + side;
}
return b;
}
board *board_free(board *b) {
free(b->dots_[0]);
free(b->dots_);
return b;
}
void board_cross(board *b) {
board_init(b, 18);
for (int i = 0; i < 4; ++i) {
b->dots_[4][7 + i] = 1;
b->dots_[7][4 + i] = 1;
b->dots_[7][10 + i] = 1;
b->dots_[10][4 + i] = 1;
b->dots_[10][10 + i] = 1;
b->dots_[13][7 + i] = 1;
b->dots_[4 + i][7] = 1;
b->dots_[4 + i][10] = 1;
b->dots_[7 + i][4] = 1;
b->dots_[7 + i][13] = 1;
b->dots_[10 + i][7] = 1;
b->dots_[10 + i][10] = 1;
}
}
void board_print(const board *b, FILE *f)
{
int side_ = b->side_;
char **dots_ = b->dots_;
for (int r = 0; r < side_; ++r) {
for (int c = 0; c < side_; ++c) {
static char map[] = " oX";
fprintf(f, "%c%s", map[dots_[r][c]], c == side_ - 1 ? "" : " - ");
}
fprintf(f, "\n");
if (r < side_ - 1) {
for (int c = 0; c < side_; ++c) {
fprintf(f, "|%s", c == side_ - 1 ? "" : " ");
}
fprintf(f, "\n");
}
}
}
int main(void)
{
board b;
board_cross(&b);
board_set_possible_moves(&b);
board_print(&b, stdout);
board_free(&b);
return 0;
}
I have been trying to get KissFFT to work on a dsPIC, however after trying various different ways, the output is not what it should be. I was hoping to get some help to see if there are any configurations that I may be overlooking or if its just somthing i haven't thought of?
I am using a dsPIC33EP256MC202 with the XC16 compiler within MPLABX.
Declarations and memory assignment.
int readings[3] = {0, 0, 0};
kiss_fft_scalar zero;
memset(&zero,0,sizeof(zero));
int size = 128 * 2;
float fin[256];
kiss_fft_cpx in[size];
kiss_fft_cpx out[size];
for (i = 0; i < size; i++) {
in[i].r = zero;
in[i].i = zero;
out[i].r = zero;
out[i].i = zero;
}
kiss_fft_cfg mycfg = kiss_fft_alloc(size*2 ,0 ,NULL,NULL);
Get readings from an accellerometer on the breadboard and populate the float array (using pythagoras to consolidate the 3 axis' into one signal). The input XYZ value are scaled down as they come in anywhere between -2400 and 2400 on average.
while(1)
{
if(iii <= 1){
UART_Write_Text("Collecting...");
}
getOutput(readings);
X = (double)readings[0];
Y = (double)readings[1];
Z = (double)readings[2];
X = X / 50;
Y = Y / 50;
Z = Z / 50;
if(ii <= 256){
fin[ii] = sqrt(X*X + Y*Y + Z*Z);
ii++;
}
else{
i=0;
while(i<255){
fin[i] = fin[i+1];
i++;
}
fin[255] = sqrt(X*X + Y*Y + Z*Z);
}
Once the float array is full of values, populate the real component of the input complex array with the values in the float array. Then perform the Kiss FFT and populate a float array (arrayDFTOUT) with the absolute value of each real and imaginary value of the out array of Kiss FFT, the final loop makes any negative value positive.
if(iii == 255){
iii = 0;
UART_Write_Text("Processing...");
for (i = 0; i < size; i++) {
// samples are type of short
in[i].r = fin[i];
in[i].i = zero;
out[i].r = zero;
out[i].i = zero;
}
kiss_fft(mycfg, in, out);
for(i=0;i<128;i++){
arrayDFTOUT[i] = sqrt((out[i].r*out[i].r) + (out[i].i*out[i].i));
}
arrayDFTOUT[0] = 1;
for(i = 0; i<128; i++){
if(arrayDFTOUT[i] < 0){
arrayDFTOUT[i] = arrayDFTOUT[i] - (arrayDFTOUT[i]*2);
}
}
Finally display the output values through serial using the UART on the breadboard.
for(i = 0; i < 128; i++){
sprintf(temp, "%f,", arrayDFTOUT[i]);
UART_Write_Text(temp);
}
And are the results. All zero's aparet from the first value that was set to 1 after KissFFT had been performed. Any ideas?