Module RF 433 Mhz Library

Projects Programming
1

Hi , i need help please i have code . they work perfect but when i put a the library of Modul RF 433 Mhz #include <RH_ASK.h> .. i don't get the result so where is the problem

#include <RH_ASK.h>
#include <SPI.h>
RH_ASK rf_driver;
float x[10] ;
byte b[160] ;
float a ;
int r = 4 ;
int i ;
int L ;
int bprime[3][3];
int fprime;
char f = 'A' ;
int k1 = 23;
int k2 = 84 ;
int k3=233;
int k4=172;
int vector[9];
void setup() {
  Serial.begin(9600);    
   rf_driver.init();


    x[1]= 0.789632145698;
    for (  int i = 2; i <=10 ; i++) {
      x[i] = r * (x[i-1]*(1 - x[i-1])); 
}
    // partie de binairisation 
   for (  int u = 1; u <= 10 ; u++) {
        a=x[u];
        i=u-1 ;
     for (int k = 1 ; k <= 16; k++){
             if (a>=1/pow(2,k)){
                 a=a-(1/pow(2,k));
                 b[16*i+k]=1 ;
                         }  
                else  {
                a=a;
               b[16*i+k]= 0; 
               //Serial.println(a) ;
                }
          //   Serial.println(b[16*i+k]);   
              }
  } 
// matrix to array 
       for (  int i = 1; i <= 3 ; i++){
              for (int j = 1; j <= 3 ; j++){
               L = i*3+j;
           bprime[i][j]= 2*b[L]+b[L+1];
          //Serial.println(bprime[i][j]);
              }
       }
       
    // cryptage  
          for ( int k = 0 ; k <9  ; k++){
      
                if (vector[k]=0){
                    fprime= f^k1;}             
                   
                 if (vector[k]=1){
                    fprime= f^k3;} 
                      
                 if (vector[k]=2){
                    fprime= f^k2;} 
                    
                 if (vector[k]=3){
                       fprime= f^k4; }
                      }
                      Serial.println(fprime); 
   }
void loop() {

         }

and when i use my code without the library i get the result ... so this is the code without library


float x[10] ;
byte b[160] ;
float a ;
int r = 4 ;
int i ;
int L ;
int bprime[3][3];
int fprime;
char f = 'A' ;
int k1 = 23;
int k2 = 84 ;
int k3=233;
int k4=172;
int vector[9];
void setup() {
  Serial.begin(9600);    
   


    x[1]= 0.789632145698;
    for (  int i = 2; i <=10 ; i++) {
      x[i] = r * (x[i-1]*(1 - x[i-1])); 
}
    // partie de binairisation 
   for (  int u = 1; u <= 10 ; u++) {
        a=x[u];
        i=u-1 ;
     for (int k = 1 ; k <= 16; k++){
             if (a>=1/pow(2,k)){
                 a=a-(1/pow(2,k));
                 b[16*i+k]=1 ;
                         }  
                else  {
                a=a;
               b[16*i+k]= 0; 
               //Serial.println(a) ;
                }
          //   Serial.println(b[16*i+k]);   
              }
  } 
// matrix to array 
       for (  int i = 1; i <= 3 ; i++){
              for (int j = 1; j <= 3 ; j++){
               L = i*3+j;
           bprime[i][j]= 2*b[L]+b[L+1];
          //Serial.println(bprime[i][j]);
              }
       }
       
    // cryptage  
          for ( int k = 0 ; k <9  ; k++){
      
                if (vector[k]=0){
                    fprime= f^k1;}             
                   
                 if (vector[k]=1){
                    fprime= f^k3;} 
                      
                 if (vector[k]=2){
                    fprime= f^k2;} 
                    
                 if (vector[k]=3){
                       fprime= f^k4; }
                      }
                      Serial.println(fprime); 
   }
void loop() {

         }

And this is the result of my code

Capture d’écran 2021-06-03 132753
Capture d’écran 2021-06-03 132753791×419 17 KB

2 
int bprime[3][3];

...
...
// matrix to array 
       for (  int i = 1; i <= 3 ; i++){
              for (int j = 1; j <= 3 ; j++){
               L = i*3+j;
           bprime[i][j]= 2*b[L]+b[L+1];
          //Serial.println(bprime[i][j]);
              }
       }

Oops

 for (  int i = 2; i <=10 ; i++) {
      x[i]

Also oops

3 
// matrix to array

Comment is just wrong

4

I mostly ignore comments.
There are bigger problems.

(Five days ago I suggested you sorted out your understanding of arrays and their indices)

5

so what !

6

So, x[10] doesn't belong to you.

7

yes thank you I remember ... but the code give the right result .. the same result in matlab and the same in python

8

But you don't own that memory.
It's not yours to write to.
Same with bprime.

Until you fix your array accesses, nothing can be taken at face value.

9

X[10] It is an array of 10 elements.. my teacher forced me to use it He said x[10] in order to save the results for the next operation as an array of 10 elements

the next operation is

for (  int i = 2; i <=10 ; i++) {
      x[i] = r * (x[i-1]*(1 - x[i-1])); 
}
10

And they are numbered x[0] to x[9], so x[10] is memory outside of that allocated for the array.

11

ok thank you i changed it to x[9] because i start from 2 to 10 and i need 9 elements

12

Could you maybe involve your teacher in the discussion about bprime?
It is possible that s/he doesn't understand the issues here either.

13

I wish but I can't...they give me the algorithm and went and didn't reply in my emails so.. I asked helpe :sweat_smile: :sweat_smile:

14

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