Expected "," or "..." before numeric constant - arduino

I´ve tried looking up an answer but I was unable to deduct from any of those what exactly the issue here is. When I try to verify it says "Expected "," or "..." before numeric constant" and highlights "#define Odvesna"
#define A1 35
#define B1 15
#define Odvesna 10
#define Prepona 30
int O(int A1, int B1) {
int result;
result = 2*(A1+B1);
return result;
}
int S(int A1,int B1) {
int result;
result = A1*B1;
return result;
}
int St(int Odvesna,int Prepona) {
int result;
result = Prepona*Odvesna/2;
return result;
}
int Ot(int Odvesna,int Prepona) {
int result;
result = sqrt(pow(Odvesna)+pow(Prepona))+ Odvesna+Prepona;
return result;
}
void Vystup(O,S,St) {
Serial.print("O = ");Serial.println(O);
Serial.print("S = ");Serial.println(S);
Serial.print("St = ");Serial.println(St);
Serial.print("Ot = ");Serial.println(Ot);
/*******************SETUP**********************************/
void setup() {
Serial.begin(115200);
Vystup(O,S,St);
}
/**************MAIN PROGRAM********************************/
void loop() {
}

There is a } missing at the end of the Vystup() function.
You can't reuse #define names as variable names.
In this call: Vystup(O,S,St); you are passing function pointers as parameters.

int St(int Odvesna,int Prepona) {} is turned into int St(int 10,int 30) {} by the preprocessor, and the compiler doesn't understand what you mean by that.
You have to use other variable names when you define your functions.
Also, void Vystup(O,S,St) {} won't work; you are passing functions as parameters, not ints as you maybe think.

Related

vector<vector<vector>> v , v2.assign(v.begin(),v.end()) Does this operation work normally?

struct C
{
int c;
};
struct B
{
int b;
std::vector<c> v1;
};
struct A
{
int a;
std::vector<b> v;
};
int main()
{
std::vector<A> vecA;
//vecA.push_back(...) insert value
std::vector<A> vecB;
vecB.assign(vecA.begin(), vecA.end()); //<---
}
/*If I use assign, do vecA and vecB values ​​have the same value?
The result of my test has the same value, but I wonder if there is an exception case.
*/

how to return a pointer in a function returning char *?

I'm implementing my own strrchr - it searches for the last occurrence of the character c (an unsigned char) in the string pointed to by the argument str.
example:
Input: f("abcabc" , "b")
Output: "bc"
the function should return char *. How can i return a pointer to the char array in the function?
#include <stdio.h>
#include <string.h>
char* my_strrchr(char* param_1, char param_2)
{
int len = strlen(param_1);
char res ;
for(int i = len; i >= 0 ;i-- ){
if (param_1[i] == param_2){
res = param_1[i];
return *(res);
//here i tried return (char) res, (char*) res; nothing works
}
}
return NULL;
}
int main(){
char *d = "abcabc";
char r = 'b';
my_strrchr(d, r);
return 0 ;
}
You're trying to return value, not a pointer. Operator * means to get value by a pointer, when res isn't a pointer. You should make it a pointer an then return:
char* my_strrchr(char* param_1, char param_2)
{
int len = strlen(param_1);
char *res ; //make it a pointer
for(int i = len; i >= 0 ;i-- ){
if (param_1[i] == param_2){
res = &param_1[i]; //store address of an element to a pointer
return res; //return a pointer
}
}
return NULL;
}
Your variable res is of type char. To get the reference use the reference operator & (See Meaning of "referencing" and "dereferencing" in C):
return &res
However this will result in the address of your variable res and not in the address within your param_1 array. Have a look at Alex' answer to get the correct reference address: https://stackoverflow.com/a/61930295/6669161

Converting ASCII to int in Arduino

I am trying to get user input from the serial monitor to turn a stepper motor according to the input. However my code returns the ASCII value rather than the original input.
#include <Stepper.h>
Stepper small_stepper(steps_per_motor_revolution, 8, 10, 9, 11);
void setup() {
// Put your setup code here, to run once:
Serial.begin(9600);
Serial.println("Ready");
}
void loop() {
// Put your main code here, to run repeatedly:
int Steps2Take = Serial.read();
Serial.println(Steps2Take); // Printing
if (Steps2Take == -1)
Steps2Take = 0;
else {
small_stepper.setSpeed(1000); // Setting speed
if (Steps2Take > 0)
small_stepper.step(Steps2Take * 32);
else
small_stepper.step(-Steps2Take * 32);
delay(2);
}
}
Just use the .toInt() function.
You should read the string from your serial and after that convert it to integer.
Serial.print(Serial.readString().toInt());
You could do this three ways! Notice, if the number is greater than 65535 then you have to use a long variable. With decimals use float variable.
You can use the toInt(), or toFloat() which require a String type variable. Heads up as the toFloat() is very time consuming.
// CODE:
String _int = "00254";
String _float = "002.54";
int value1 = _int.toInt();
float value2 = _float.toFloat();
Serial.println(value1);
Serial.println(value2);
// OUTPUT:
254
2.54
You could use the atoi. The function accepts a character array and then converts it to an integer.
// CODE:
// For some reason you have to have +1 your final size; if you don't you will get zeros.
char output[5] = {'1', '.', '2', '3'};
int value1 = atoi(output);
float value2 = atof(output);
Serial.print(value1);
Serial.print(value2);
// OUTPUT:
1
1.23
If you had a Character Array and and wanted to convert it to a string because you didn't know the length...like a message buffer or something, I dunno. You could use this below to change it to a string and then implement the toInt() or toFloat().
// CODE:
char _int[8];
String data = "";
for(int i = 0; i < 8; i++){
data += (char)_int[i];
}
char buf[data.length()+1];
data.toCharArray(buf, data.length()+1);
If it is just a "type-conversion" problem, you can use something like this:
int a_as_int = (int)'a';
or
#include <stdlib.h>
int num = atoi("23"); //atoi = ascii to integer
as it was point out here.
Does it solve the problem?

RSA encryption, decryption functions for Arduino

I need encrypt data with RSA using public key and decrypt using private key. Public and private keys generation not needed. I know that algorithm require computing resources, but I use only 32-128-bit keys or smaller. I can't find RSA lib for Arduino in internet.
I found cryptography AVR-Crypto-Lib library for AVR microcontrollers. How to use this library with Arduino? I think Arduino compatible, because it use ATmega328P, ATmega2560 microcontrollers etc. How include this library in Arduino IDE. It would be good that someone will write simple hello world example. Or can suggest other library that are implemented RSA algorithm.
I have developed the following code using random() of Arduino. But the code function once in 10 iterations that too at 8bit key.
#include "pRNG.h"
#include <Arduino.h>
unsigned int Firstprime, Secondprime, Privatekey, Publickey;
unsigned int Field, phin, Enc, Dec;
bool Hasrun= false;
unsigned int Text = 12;
pRNG prn;
unsigned int modMult(unsigned int a, unsigned int b, unsigned int mod) //modulo multiplication function
{
unsigned int res = 0; // Initialize result
a = a % mod;
while (b > 0)
{
// If b is odd, add 'a' to result
if (b % 2 == 1)
res = (res + a) % mod;
// Multiply 'a' with 2
a = (a * 2) % mod;
// Divide b by 2
b /= 2;
}
// Return result
return res % mod;
}
bool prime(unsigned int number) //primality check for prime numbers
{
for (unsigned int i = 2; i <=sqrt(number); ++i)
{
if (number % i == 0)
{
return false;
}
}
return true;
}
unsigned int PRN() //generation of a prime random number
{
unsigned int n1;
do
{
n1= prn.getRndByte();
//n1= random(100);
}while(n1==0||prime(n1)==false);
return n1;
}
unsigned int gcd(unsigned int a, unsigned int b) //function to check GCD
{
unsigned int temp;
while (1)
{
temp = a%b;
if (temp == 0)
return b;
a = b;
b= temp;
}
}
unsigned int E_gen(unsigned int n, unsigned int phi) //publickey generation e
{
for(unsigned int i=2; i<n; i++)
{
if(gcd(i,n)==1 && gcd(i,phi)==1)
{
return i;
//break;
}
}
Serial.println("Public key generated");
}
unsigned int D_gen(unsigned int en, unsigned int phi) //privatekey generation d
{
for(unsigned int i=2; i<phi; i++)
{
if(modMult(en,i,phi)==1)
{
return i;
//break;
}
}
Serial.println("Private key generated");
}
unsigned int power(unsigned int base, unsigned int expo, unsigned int mod)
{
unsigned int test;
for(test = 1; expo; expo >>= 1)
{
if (expo & 1)
test = (test * base) % mod;
base = (base * base) % mod;
}
return test;
}
/*unsigned int keygen()
{
}*/
void setup()
{
Serial.begin(9600);
// randomSeed(analogRead(A0));
Firstprime=PRN();
Serial.println(Firstprime);
do
{
Secondprime=PRN();
Serial.println(Secondprime);
}while(Firstprime==Secondprime);
Field=Firstprime*Secondprime;
phin=(Firstprime-1)*(Secondprime-1);
Publickey=E_gen(Field, phin);
Privatekey=D_gen(Publickey,phin);
}
void loop()
{
if(Hasrun==false)
{
//Serial.println(prn.getRndunsigned int());
Serial.print("Public key is:");
Serial.println(Publickey);
Serial.print("Private key is:");
Serial.println(Privatekey);
Serial.println("Encrypting....");
Enc= power(Text,Publickey, Field);
Serial.println(Enc);
Serial.println("Decrypting...");
Dec=power(Enc,Privatekey, Field);
Serial.println(Dec);
/*Serial.println(p);
Serial.println(q);*/
Hasrun=true;
}
}

Are there any example of Mutual recursion?

Are there any examples for a recursive function that calls an other function which calls the first one too ?
Example :
function1()
{
//do something
function2();
//do something
}
function2()
{
//do something
function1();
//do something
}
Mutual recursion is common in code that parses mathematical expressions (and other grammars). A recursive descent parser based on the grammar below will naturally contain mutual recursion: expression-terms-term-factor-primary-expression.
expression
+ terms
- terms
terms
terms
term + terms
term - terms
term
factor
factor * term
factor / term
factor
primary
primary ^ factor
primary
( expression )
number
name
name ( expression )
The proper term for this is Mutual Recursion.
http://en.wikipedia.org/wiki/Mutual_recursion
There's an example on that page, I'll reproduce here in Java:
boolean even( int number )
{
if( number == 0 )
return true;
else
return odd(abs(number)-1)
}
boolean odd( int number )
{
if( number == 0 )
return false;
else
return even(abs(number)-1);
}
Where abs( n ) means return the absolute value of a number.
Clearly this is not efficient, just to demonstrate a point.
An example might be the minmax algorithm commonly used in game programs such as chess. Starting at the top of the game tree, the goal is to find the maximum value of all the nodes at the level below, whose values are defined as the minimum of the values of the nodes below that, whose values are defines as the maximum of the values below that, whose values ...
I can think of two common sources of mutual recursion.
Functions dealing with mutually recursive types
Consider an Abstract Syntax Tree (AST) that keeps position information in every node. The type might look like this:
type Expr =
| Int of int
| Var of string
| Add of ExprAux * ExprAux
and ExprAux = Expr of int * Expr
The easiest way to write functions that manipulate values of these types is to write mutually recursive functions. For example, a function to find the set of free variables:
let rec freeVariables = function
| Int n -> Set.empty
| Var x -> Set.singleton x
| Add(f, g) -> Set.union (freeVariablesAux f) (freeVariablesAux g)
and freeVariablesAux (Expr(loc, e)) =
freeVariables e
State machines
Consider a state machine that is either on, off or paused with instructions to start, stop, pause and resume (F# code):
type Instruction = Start | Stop | Pause | Resume
The state machine might be written as mutually recursive functions with one function for each state:
type State = State of (Instruction -> State)
let rec isOff = function
| Start -> State isOn
| _ -> State isOff
and isOn = function
| Stop -> State isOff
| Pause -> State isPaused
| _ -> State isOn
and isPaused = function
| Stop -> State isOff
| Resume -> State isOn
| _ -> State isPaused
It's a bit contrived and not very efficient, but you could do this with a function to calculate Fibbonacci numbers as in:
fib2(n) { return fib(n-2); }
fib1(n) { return fib(n-1); }
fib(n)
{
if (n>1)
return fib1(n) + fib2(n);
else
return 1;
}
In this case its efficiency can be dramatically enhanced if the language supports memoization
In a language with proper tail calls, Mutual Tail Recursion is a very natural way of implementing automata.
Here is my coded solution. For a calculator app that performs *,/,- operations using mutual recursion. It also checks for brackets (()) to decide the order of precedence.
Flow:: expression -> term -> factor -> expression
Calculator.h
#ifndef CALCULATOR_H_
#define CALCULATOR_H_
#include <string>
using namespace std;
/****** A Calculator Class holding expression, term, factor ********/
class Calculator
{
public:
/**Default Constructor*/
Calculator();
/** Parameterized Constructor common for all exception
* #aparam e exception value
* */
Calculator(char e);
/**
* Function to start computation
* #param input - input expression*/
void start(string input);
/**
* Evaluates Term*
* #param input string for term*/
double term(string& input);
/* Evaluates factor*
* #param input string for factor*/
double factor(string& input);
/* Evaluates Expression*
* #param input string for expression*/
double expression(string& input);
/* Evaluates number*
* #param input string for number*/
string number(string n);
/**
* Prints calculates value of the expression
* */
void print();
/**
* Converts char to double
* #param c input char
* */
double charTONum(const char* c);
/**
* Get error
*/
char get_value() const;
/** Reset all values*/
void reset();
private:
int lock;//set lock to check extra parenthesis
double result;// result
char error_msg;// error message
};
/**Error for unexpected string operation*/
class Unexpected_error:public Calculator
{
public:
Unexpected_error(char e):Calculator(e){};
};
/**Error for missing parenthesis*/
class Missing_parenthesis:public Calculator
{
public:
Missing_parenthesis(char e):Calculator(e){};
};
/**Error if divide by zeros*/
class DivideByZero:public Calculator{
public:
DivideByZero():Calculator(){};
};
#endif
===============================================================================
Calculator.cpp
//============================================================================
// Name : Calculator.cpp
// Author : Anurag
// Version :
// Copyright : Your copyright notice
// Description : Calculator using mutual recursion in C++, Ansi-style
//============================================================================
#include "Calculator.h"
#include <iostream>
#include <string>
#include <math.h>
#include <exception>
using namespace std;
Calculator::Calculator():lock(0),result(0),error_msg(' '){
}
Calculator::Calculator(char e):result(0), error_msg(e) {
}
char Calculator::get_value() const {
return this->error_msg;
}
void Calculator::start(string input) {
try{
result = expression(input);
print();
}catch (Unexpected_error e) {
cout<<result<<endl;
cout<<"***** Unexpected "<<e.get_value()<<endl;
}catch (Missing_parenthesis e) {
cout<<"***** Missing "<<e.get_value()<<endl;
}catch (DivideByZero e) {
cout<<"***** Division By Zero" << endl;
}
}
double Calculator::expression(string& input) {
double expression=0;
if(input.size()==0)
return 0;
expression = term(input);
if(input[0] == ' ')
input = input.substr(1);
if(input[0] == '+') {
input = input.substr(1);
expression += term(input);
}
else if(input[0] == '-') {
input = input.substr(1);
expression -= term(input);
}
if(input[0] == '%'){
result = expression;
throw Unexpected_error(input[0]);
}
if(input[0]==')' && lock<=0 )
throw Missing_parenthesis(')');
return expression;
}
double Calculator::term(string& input) {
if(input.size()==0)
return 1;
double term=1;
term = factor(input);
if(input[0] == ' ')
input = input.substr(1);
if(input[0] == '*') {
input = input.substr(1);
term = term * factor(input);
}
else if(input[0] == '/') {
input = input.substr(1);
double den = factor(input);
if(den==0) {
throw DivideByZero();
}
term = term / den;
}
return term;
}
double Calculator::factor(string& input) {
double factor=0;
if(input[0] == ' ') {
input = input.substr(1);
}
if(input[0] == '(') {
lock++;
input = input.substr(1);
factor = expression(input);
if(input[0]==')') {
lock--;
input = input.substr(1);
return factor;
}else{
throw Missing_parenthesis(')');
}
}
else if (input[0]>='0' && input[0]<='9'){
string nums = input.substr(0,1) + number(input.substr(1));
input = input.substr(nums.size());
return stod(nums);
}
else {
result = factor;
throw Unexpected_error(input[0]);
}
return factor;
}
string Calculator::number(string input) {
if(input.substr(0,2)=="E+" || input.substr(0,2)=="E-" || input.substr(0,2)=="e-" || input.substr(0,2)=="e-")
return input.substr(0,2) + number(input.substr(2));
else if((input[0]>='0' && input[0]<='9') || (input[0]=='.'))
return input.substr(0,1) + number(input.substr(1));
else
return "";
}
void Calculator::print() {
cout << result << endl;
}
void Calculator::reset(){
this->lock=0;
this->result=0;
}
int main() {
Calculator* cal = new Calculator;
string input;
cout<<"Expression? ";
getline(cin,input);
while(input!="."){
cal->start(input.substr(0,input.size()-2));
cout<<"Expression? ";
cal->reset();
getline(cin,input);
}
cout << "Done!" << endl;
return 0;
}
==============================================================
Sample input-> Expression? (42+8)*10 =
Output-> 500
Top down merge sort can use a pair of mutually recursive functions to alternate the direction of merge based on level of recursion.
For the example code below, a[] is the array to be sorted, b[] is a temporary working array. For a naive implementation of merge sort, each merge operation copies data from a[] to b[], then merges b[] back to a[], or it merges from a[] to b[], then copies from b[] back to a[]. This requires n · ceiling(log2(n)) copy operations. To eliminate the copy operations used for merging, the direction of merge can be alternated based on level of recursion, merge from a[] to b[], merge from b[] to a[], ..., and switch to in place insertion sort for small runs in a[], as insertion sort on small runs is faster than merge sort.
In this example, MergeSortAtoA() and MergeSortAtoB() are the mutually recursive functions.
Example java code:
static final int ISZ = 64; // use insertion sort if size <= ISZ
static void MergeSort(int a[])
{
int n = a.length;
if(n < 2)
return;
int [] b = new int[n];
MergeSortAtoA(a, b, 0, n);
}
static void MergeSortAtoA(int a[], int b[], int ll, int ee)
{
if ((ee - ll) <= ISZ){ // use insertion sort on small runs
InsertionSort(a, ll, ee);
return;
}
int rr = (ll + ee)>>1; // midpoint, start of right half
MergeSortAtoB(a, b, ll, rr);
MergeSortAtoB(a, b, rr, ee);
Merge(b, a, ll, rr, ee); // merge b to a
}
static void MergeSortAtoB(int a[], int b[], int ll, int ee)
{
int rr = (ll + ee)>>1; // midpoint, start of right half
MergeSortAtoA(a, b, ll, rr);
MergeSortAtoA(a, b, rr, ee);
Merge(a, b, ll, rr, ee); // merge a to b
}
static void Merge(int a[], int b[], int ll, int rr, int ee)
{
int o = ll; // b[] index
int l = ll; // a[] left index
int r = rr; // a[] right index
while(true){ // merge data
if(a[l] <= a[r]){ // if a[l] <= a[r]
b[o++] = a[l++]; // copy a[l]
if(l < rr) // if not end of left run
continue; // continue (back to while)
while(r < ee){ // else copy rest of right run
b[o++] = a[r++];
}
break; // and return
} else { // else a[l] > a[r]
b[o++] = a[r++]; // copy a[r]
if(r < ee) // if not end of right run
continue; // continue (back to while)
while(l < rr){ // else copy rest of left run
b[o++] = a[l++];
}
break; // and return
}
}
}
static void InsertionSort(int a[], int ll, int ee)
{
int i, j;
int t;
for (j = ll + 1; j < ee; j++) {
t = a[j];
i = j-1;
while(i >= ll && a[i] > t){
a[i+1] = a[i];
i--;
}
a[i+1] = t;
}
}

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