I am using string.Replace to replace substring A
func removeIP(text string) string {
text = strings.Replace(text, "someWord", "**NewWord**", -1)
return text
}
func removeIPUsingRegex(text string) string {
var re = regexp.MustCompile(`\b` + "someWord" + `\b`) // I want to replace whole word only
text = re.ReplaceAllString(text, "**NewWord**")
}
The Issue I am facing here is, I want to replace a whole word only if is not supported by string replace.
As I have to replace for very very huge strings may be in GBs. Regex is very slow compare to string replace.
eg: text: "abcdef defgh /def/ .def/ =def= def xxxy" -> Replace def with DEF
output: "abcdef defgh /DEF/ .DEF/ =DEF= DEF xxxy" //Notice only whole words have been replaced.
Regex bumps the time by almost 100 times (https://medium.com/codezillas/golang-replace-vs-regexp-de4e48482f53). Any Ideas will be much appreciated.
KMP ALGorithm used
// ReplaceWholeWord ...
func ReplaceWholeWord(text string, oldWord string, newWord string) string {
var patternLength = len(oldWord)
var textLength = len(text)
var copyIndex = 0
var textIndex = 0
var patternIndex = 0
var newString strings.Builder
var lps = computeLPSArray(oldWord)
for textIndex < textLength {
if oldWord[patternIndex] == text[textIndex] {
patternIndex++
textIndex++
}
if patternIndex == patternLength {
startIndex := textIndex - patternIndex
endIndex := textIndex - patternIndex + patternLength - 1
if checkIfWholeWord(text, startIndex, endIndex) {
if copyIndex != startIndex {
newString.WriteString(text[copyIndex:startIndex])
}
newString.WriteString(newWord)
copyIndex = endIndex + 1
}
patternIndex = 0
textIndex = endIndex + 1
} else if textIndex < textLength && oldWord[patternIndex] != text[textIndex] {
if patternIndex != 0 {
patternIndex = lps[patternIndex-1]
} else {
textIndex = textIndex + 1
}
}
}
newString.WriteString(text[copyIndex:])
return newString.String()
}
func computeLPSArray(pattern string) []int {
var length = 0
var i = 1
var patternLength = len(pattern)
var lps = make([]int, patternLength)
lps[0] = 0
for i = 1; i < patternLength; {
if pattern[i] == pattern[length] {
length++
lps[i] = length
i++
} else {
if length != 0 {
length = lps[length-1]
} else {
lps[i] = length
i++
}
}
}
return lps
}
func checkIfWholeWord(text string, startIndex int, endIndex int) bool {
startIndex = startIndex - 1
endIndex = endIndex + 1
if (startIndex < 0 && endIndex >= len(text)) ||
(startIndex < 0 && endIndex < len(text) && isNonWord(text[endIndex])) ||
(startIndex >= 0 && endIndex >= len(text) && isNonWord(text[startIndex])) ||
(startIndex >= 0 && endIndex < len(text) && isNonWord(text[startIndex]) && isNonWord(text[endIndex])) {
return true
}
return false
}
func isNonWord(c byte) bool {
return !((c >= '0' && c <= '9') || (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c == '_'))
}
Related
Given an arbitrary matrix, how do I find the co-ordinates that surrounds each city accurately?
E.g. City 1 has surrounding matrix of (0, 0), (0, 3), (1, 0), (1, 3), (2,0), (2, 1), (2, 2), (2, 3).
I have tried using a hardcoded method. Which is loop through each city's co-ordinate, however there are still inaccuracy in this method.
E.g. (0, 1) and from there check all 8 directions, up, down, left, right, upper left, upper right, bottom left, bottom right.
And if the char value is ' ', it is not a city which means it is part of a surrounding.
Is there any way which is much more efficient and more accurate in finding the surrounding?
void surroundings(int x, int y) {
// Initiate the first city struct information
citySummInfo.cityId = cityLocList[0].cityId;
citySummInfo.cityName = cityLocList[0].cityName;
citySummInfoList.push_back(citySummInfo);
// Add unique cityID & cityName into vector
for (size_t i = 0; i < cityLocList.size(); i++) {
for (size_t j = 0; j < citySummInfoList.size(); j++) {
if (cityLocList[i].cityId == citySummInfoList[j].cityId && cityLocList[i].cityName == citySummInfoList[j].cityName) {
break;
}
else if (j == citySummInfoList.size() - 1) {
citySummInfo.cityId = cityLocList[i].cityId;
citySummInfo.cityName = cityLocList[i].cityName;
citySummInfoList.push_back(citySummInfo);
}
}
}
// To populate the entire matrix with city ID
for (int i = 0; i < x; i++) {
for (int j = 0; j < y; j++) {
for (size_t k = 0; k < cityLocList.size(); k++) {
if (cityLocList[k].xGrid == i && cityLocList[k].yGrid == j)
mapPtr[j][i] = cityLocList[k].cityId + '0';
}
}
}
// Main process of getting the surrounding
for (size_t i = 0; i < citySummInfoList.size(); i++) {
for (size_t j = 0; j < cityLocList.size(); j++) {
if (citySummInfoList[i].cityId == cityLocList[j].cityId)
citySummInfoList[i].coOrdinates.push_back(to_string(cityLocList[j].xGrid) + "." + to_string(cityLocList[j].yGrid));
}
}
for (size_t i = 0; i < citySummInfoList.size(); i++) {
vector<string> temp;
for (size_t j = 0; j < citySummInfoList[i].coOrdinates.size(); j++) {
char cityId = citySummInfoList[i].cityId + '0';
char delim[] = { '.' };
vector<string> tempAxis = tokenizer(citySummInfoList[i].coOrdinates[j], delim, 1);
int xAxis = stoi(tempAxis[0]);
int yAxis = stoi(tempAxis[1]);
if (xAxis - 1 < 0 || yAxis - 1 < 0) {
continue;
}
if (mapPtr[xAxis - 1][yAxis + 1] != cityId) {
if (xAxis + 1 == x || yAxis + 1 == y || xAxis - 1 < 0 || yAxis - 1 < 0)
continue;
string coOrd = to_string(xAxis - 1) + "." + to_string(yAxis + 1);
if (find(temp.begin(), temp.end(), coOrd) == temp.end()) {
temp.push_back(coOrd);
}
}
if (mapPtr[xAxis - 1][yAxis] != cityId) {
if (xAxis + 1 == x || yAxis + 1 == y || xAxis - 1 < 0 || yAxis - 1 < 0)
continue;
string coOrd = to_string(xAxis - 1) + "." + to_string(yAxis);
if (find(temp.begin(), temp.end(), coOrd) == temp.end()) {
temp.push_back(coOrd);
}
}
if (mapPtr[xAxis - 1][yAxis - 1] != cityId) {
if (xAxis + 1 == x || yAxis + 1 == y || xAxis - 1 < 0 || yAxis - 1 < 0)
continue;
string coOrd = to_string(xAxis - 1) + "." + to_string(yAxis - 1);
if (find(temp.begin(), temp.end(), coOrd) == temp.end()) {
temp.push_back(coOrd);
}
}
if (mapPtr[xAxis][yAxis + 1] != cityId) {
if (xAxis + 1 == x || yAxis + 1 == y || xAxis - 1 < 0 || yAxis - 1 < 0)
continue;
string coOrd = to_string(xAxis) + "." + to_string(yAxis + 1);
if (find(temp.begin(), temp.end(), coOrd) == temp.end()) {
temp.push_back(coOrd);
}
}
if (mapPtr[xAxis][yAxis - 1] != cityId) {
if (xAxis + 1 == x || yAxis + 1 == y || xAxis - 1 < 0 || yAxis - 1 < 0)
continue;
string coOrd = to_string(xAxis) + "." + to_string(yAxis - 1);
if (find(temp.begin(), temp.end(), coOrd) == temp.end()) {
temp.push_back(coOrd);
}
}
if (mapPtr[xAxis + 1][yAxis + 1] != cityId) {
if (xAxis + 1 == x || yAxis + 1 == y || xAxis - 1 < 0 || yAxis - 1 < 0)
continue;
string coOrd = to_string(xAxis + 1) + "." + to_string(yAxis + 1);
if (find(temp.begin(), temp.end(), coOrd) == temp.end()) {
temp.push_back(coOrd);
}
}
if (mapPtr[xAxis + 1][yAxis] != cityId) {
if (xAxis + 1 == x || yAxis + 1 == y || xAxis - 1 < 0 || yAxis - 1 < 0)
continue;
string coOrd = to_string(xAxis + 1) + "." + to_string(yAxis);
if (find(temp.begin(), temp.end(), coOrd) == temp.end()) {
temp.push_back(coOrd);
}
}
if (mapPtr[xAxis + 1][yAxis - 1] != cityId) {
if (xAxis + 1 == x || yAxis + 1 == y || xAxis - 1 < 0 || yAxis - 1 < 0)
continue;
string coOrd = to_string(xAxis + 1) + "." + to_string(yAxis - 1);
if (find(temp.begin(), temp.end(), coOrd) == temp.end()) {
temp.push_back(coOrd);
}
}
}
citySummInfoList[i].coOrdinates.reserve(temp.size());
citySummInfoList[i].coOrdinates.insert(citySummInfoList[i].coOrdinates.end(), temp.begin(), temp.end());
}
}
Also, is there a possibility that my print function may cause such unreliability?
void print(int x, int y) {
for (int i = 0; i <= x + 2; i++) {
if (i == 0 || i >= x + 1) // Indentation for 1st and last row of non city locations
cout << setw(4) << " ";
for (int j = 0; j <= y + 2; j++) {
if ((i == 0 || i == x + 1) && j <= y + 1) { // Layout for first and last row
cout << "# ";
}
else if ((j == 0 && (i != 0 || i <= x))) { // Numbering for each row
if (x - i >= 0) {
cout << setw(3) << left << x - i << " ";
}
else {
cout << " "; // Indentation for last row of #s
}
}
else if (j == 1 && i < x + 2) { // Layout for first column
cout << "#";
}
else if (j == y + 2 && i != 0 && i < x + 1) { // Layout for last column
cout << " #";
}
else if (i == x + 2 && j < y + 1) { // Numbering for each column
cout << j - 1 << " ";
}
else if ((i != 0 && i != x + 2 && j != y + 2)) {
cout << " " << mapPtr[x - i][j - 2]; // Plot map value
}
}
cout << endl;
}
cout << endl;
}
This is an O(n) answer for your problem. The idea behind it is to find all points that are edges (a point is an edge if it is adjacent to anything which is not its own city).
After finding all edge points, loop through each of them and add all points adjacent to them which are whitespace characters.
#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef pair<int, int> point;
string m[] = {
" ",
" 555 ",
" 555 ",
" 222 555 ",
" 222 ",
" 222 ",
" 222 ",
" ",
"11 33 ",
"11 44",
" 44"
};
//hash function for pairs
struct hash_pair {
template <class T1, class T2>
size_t operator()(const pair<T1, T2>& p) const {
auto hash1 = hash<T1>{}(p.first);
auto hash2 = hash<T2>{}(p.second);
return hash1 ^ hash2;
}
};
bool insideBounds(int x, int y, point &size){
if(x < 0 || x >= size.x || y < 0 || y >= size.y) return false;
return true;
}
bool isEdge(int x, int y, point &size){
for(int i = -1; i < 2; ++i){
for(int j = -1; j < 2; ++j){
if(!insideBounds(x + j, y + i, size)) return true;
if(m[y + i][x + j] == ' ') return true;
}
}
return false;
}
void FindSurrounding(int x, int y){
//size of map
point size = make_pair(11, 11);
//current city id
char city = m[y][x];
//finding a point in edge of the city
point edge = make_pair(x, y);
for(int i = x - 1; i >= 0; --i)
if(m[y][i] == city) edge = make_pair(i, y);
//find all edge points
unordered_set<point, hash_pair> visited;
stack<point> toVisit;
toVisit.push(edge);
while(toVisit.size()){
point current = toVisit.top();
visited.insert(current);
toVisit.pop();
for(int i = -1; i < 2; ++i){
for(int j = -1; j < 2; ++j){
point p = make_pair(current.x + j, current.y + i);
if(!insideBounds(p.x, p.y, size) || m[p.y][p.x] != city) continue;
if(visited.find(p) == visited.end() && isEdge(p.x, p.y, size)){
toVisit.push(p);
}
}
}
}
//find surrounding slots
unordered_set<point, hash_pair> surrounding;
for (const auto& p: visited) {
for(int i = -1; i < 2; ++i){
for(int j = -1; j < 2; ++j){
point curr = make_pair(p.x + j, p.y + i);
if(insideBounds(curr.x, curr.y, size) && m[curr.y][curr.x] == ' '){
surrounding.insert(curr);
}
}
}
}
//print answer
for (const auto& p: surrounding) {
cout<<"("<<p.x<<", "<<p.y<<"), ";
}
}
int main()
{
FindSurrounding(0, 8);
return 0;
}
OUTPUT: (2, 10), (1, 10), (2, 9), (0, 10), (2, 8), (2, 7), (1, 7), (0, 7),
I have a assingment were I need to code and decode txt files, for example: hello how are you? has to be coded as hel2o how are you? and aaaaaaaaaajkle as a10jkle.
while ( ! invoer.eof ( ) ) {
if (kar >= '0' && kar <= '9') {
counter = kar-48;
while (counter > 1){
uitvoer.put(vorigeKar);
counter--;
}
}else if (kar == '/'){
kar = invoer.get();
uitvoer.put(kar);
}else{
uitvoer.put(kar);
}
vorigeKar = kar;
kar = invoer.get ( );
}
but the problem I have is if need to decode a12bhr, the answer is aaaaaaaaaaaabhr but I can't seem to get the 12 as number without problems, I also can't use any strings.
try to put a repeated character when next is not numeric or end of string.
For prepare this, it needs to make number by parsing string.
about this, I recommend you to find how to convert string to integer in real time at C++.
bool isNumeric(char ch) {
return '0' <= ch && ch <= '9';
}
string decode(const string& s) {
int counter = 0;
string result;
char prevCh;
for (int i = 0; i < s.length(); i++) {
if (isNumeric(s[i])) { // update counter
counter = counter * 10 + (s[i] - '0');
if (isNumeric(s[i + 1]) == false || i + 1 == s.length()) {
// now, put previous character stacked
while (counter-- > 1) {
result.push_back(prevCh);
}
counter = 0;
}
}
else {
result.push_back(s[i]);
prevCh = s[i];
}
}
return result;
}
now, decode("a12bhr3") returns aaaaaaaaaaaabhrrr. it works well.
this code attempts to solve the 4 queens problem, placing 4 queens on a 4*4 chessboard without any of them being able to capture eachother
#include <iostream>
using namespace std;
int Place(int Chess[][4], int collumn, int i);
bool Check(int Chess[][4], int collumn, int i);
int findrow(int Chess[][4], int collumn);
const int size = 3;
int main()
{
int Chess[4][4];
int collumn;
int i = 0;
collumn = 0;
for(int s = 0; s < 4; s++)
{
for(int j = 0; j < 4; j ++)
{
Chess[s][j] = 0;
}
}
//Chess[0][0] = 1;
//Chess[3][3] = 1;
//if(Check(Chess, 3, 3) == false)
Place(Chess, collumn, i);
for(int z = 0; z < 4; z++)
{
for(int a = 0; a < 4; a++)
{
if(Chess[z][a] == 1)
cout<<"Row: "<<z<<"Collumn: "<<a<<"."<<endl;
}
cout<<endl;
}
system("pause");
return 0;
}
int Place(int Chess[][4], int collumn, int i)
{
if(collumn > size)
return 0;
while(i <= size)
{
if(Check(Chess, collumn, i) == true)
{
//cout<<"hi"<<endl;
Chess[collumn][i] = 1;
return(Place(Chess, (collumn + 1), i));
}
i ++;
}
if(i>= size)
{
//cout<<"hilo"<<endl;
return Place(Chess, collumn-1, findrow(Chess, collumn-1));
}
}
bool Check(int Chess[][4], int collumn, int i)//checks to see if it can be captured
{// very inneficitnt
int x = collumn;// this is so we can now work in terms of x and y
int y = i;
bool found = true;
// checks all the diagonal captures
if(Chess[x -1 ][y -1]== 1&& x>=1 && y >=1 )
found = false;
if(Chess[x -2 ][y - 2]== 1&& x>=2 && y>=2 )
found = false;
if(Chess[x - 3][y - 3]== 1 && x>=3 && y>=3 )
found = false;
if(Chess[x + 1][y - 1] == 1&& x<=2 && y>=1 )
found = false;
if(Chess[x + 2][y -2] == 1&& x<=1 && y>=2)
found = false;
if(Chess[x + 3][y - 3] == 1 && x<=0 && y>=3)
found = false;
if(Chess[x + 1][y + 1] == 1 && x<=2 && y<=2)
found = false;
if(Chess[x + 2][y + 2] == 1&& x<=1 && y<=1)
found = false;
if(Chess[x + 3][y + 3] == 1 && x<=0 && y<=0 )
found = false;
if(Chess[x -1 ][y + 1]== 1 && x>=1 && y<=2 )
found = false;
if(Chess[x - 2][y + 2] == 1&& x>=2 && y<=1 )
found = false;
if(Chess[x - 3][y + 3] == 1&& x>=3 && y<=0)
found = false;
//checks all the horizontal captures. We don't need to check for vertical captures
if(Chess[x + 1][y] == 1 && x<=2)
found = false;
if(Chess[x + 2][y] == 1&& x<=1 )
found = false;
if(Chess[x+3][y] == 1 && x<=0)
found = false;
if(Chess[x -1 ][y] == 1&& x>=1)
found = false;
if(Chess[x-2][y] == 1&& x>=2 )
found = false;
if(Chess[x-3][y] == 1 && x>=3)
found = false;
if(found == false)
return false;
if(found == true)
return true;
}
int findrow(int Chess[][4], int collumn)
{
for(int z = 0; z < 4; z++)
{
if(Chess[collumn][z] == 1)
{
Chess[collumn][z] = 0;
return z;
}
}
}
The first thing I see is a probable out-of-bounds access:
if(Chess[x -1 ][y -1]== 1&& x>=1 && y >=1 )
What if the value of x is 0? You are accessing Chess[-1][y], which is out of bounds. Your if statement does not stop this, even with the x>=1 condition.
The if will first test the Chess[x-1][y-1]==1 condition. If you want this to not happen, place the test for x>=1 before Chess[x-1][y-1]==1.
But even with this, that entire section of code looks suspicious. I wouldn't be surprised if there were more out-of-bounds accesses.
Problem Statement :-
A number is given, N, which is given in binary notation, and it
contains atmost 1000000 bits. You have to calculate the sum of LUCKY
FACTOR in range from 1 to N (decimal notation).
Here, LUCKY FACTOR means, (after converting into binary representation) if
rightmost or leftmost 1's neighbour is either 0 or nothing(for
boundary bit).
EDITED :-
Means if rightmost one's left neighbour is 0, means it count as a
LUCKY FACTOR, simlarly in the left side also
Example,
5 == 101, LUCKY FACTOR = 2.
7 == 111, LUCKY FACTOR = 0.
13 == 1101, LUCKY FACTOR = 1.
16 == 1110, LUCKY FACTOR = 0.
0 == 0, LUCKY FACTOR = 0.
Answer must be in binary form
I am totally stuck, give me a hint.
My code
#include<stdio.h>
#include<string>
#include<string.h>
#include<vector>
//#include<iostream>
using namespace std;
vector<string> pp(10000001);
string add(string a, string b) {
if(b == "") return a;
string answer = "";
int c = 0;
int szeA = a.size() - 1;
int szeB = b.size() - 1;
while(szeA >= 0 || szeB >= 0) {
answer = (char)( ( ( ( (szeA >= 0) ? (a[szeA] - 48) : 0 ) ^ ( (szeB >= 0) ? (b[szeB] - 48) : 0 ) ) ^ (c) ) + 48 ) + answer;
c = ( ( ( (szeA >= 0) ? (a[szeA] - 48) : 0 ) & ( (szeB >= 0) ? (b[szeB] - 48) : 0 ) ) | ( ( (szeA >= 0) ? (a[szeA] - 48) : 0 ) & (c) ) | ( ( (szeB >= 0) ? (b[szeB] - 48) : 0 ) & (c) ) );
szeA--;
szeB--;
}
if(c) answer = '1' + answer;
return answer;
}
string subtract(string a, string b) {
int sze = a.size() - b.size();
while(sze--) b = '0' + b;
sze = a.size();
for(int i = 0; i < sze; i++) {
if(b[i] == '1') b[i] = '0';
else b[i] = '1';
}
if(b[sze-1] == '0') {
b[sze-1] = '1';
}
else {
int i = sze-1;
while(i >= 0 && b[i] == '1') {
b[i] = '0';
i--;
}
if(i >= 0) b[i] = '1';
else b = '1' + b;
}
b = add(a, b);
b.erase(b.begin() + 0);
//b[0] = '0';
while(b[0] == '0') b.erase(b.begin() + 0);
return b;
}
string power(int index) {
if(index < 0) return "";
string answer = "";
while(index--) {
answer = '0' + answer;
}
answer = '1' + answer;
return answer;
}
string convert(long long int val) {
int divisionStore=0;
int modStore=0;
string mainVector = "";
do {
modStore=val%2;
val=val/2;
mainVector = (char)(modStore+48) + mainVector;
}while(val!=0);
return mainVector;
}
string increment(string s) {
int sze = s.size()-1;
if(s[sze] == '0') {
s[sze] = '1';
return s;
}
while(sze >= 0 && s[sze] == '1') {
s[sze] = '0';
sze--;
}
if(sze >= 0) s[sze] = '1';
else s = '1' + s;
return s;
}
main() {
int T;
char s[1000001];
string answer;
scanf("%d", &T);
for(int t = 1; t <= T; t++) {
int num;
answer = "1";
int bitComeEver = 0;
int lastBit = 0;
scanf("%s", s);
int sze = strlen(s);
// I used below block because to avoid TLE.
if(sze > 3300) {
printf( "Case #%d\n", t);
for(int i = 0; i < sze; i++) printf("%c", '1');
printf("\n");
//continue;
}
else {
if(pp[sze-1] != "") answer = pp[sze-1];
else {
pp[sze-1] = power(sze-1);
answer = pp[sze-1];
}
answer = subtract(answer, convert(sze-1));
////////////////////////////
//cout << answer << endl;
for(int i = 1; i < sze; i++) {
if(s[i] == '1') {
if(s[1] == '0') {
num = sze-i-1;
if(num > 0) {
if( pp[num-1] == "") {
pp[num-1] = power(num-1);
}
if(pp[num+1] == "") {
pp[num+1] = power(num+1);
}
answer = add(answer, subtract(pp[num+1], pp[num-1]));
if(lastBit) answer = add(answer, "1");
//else answer = increment(answer);
//cout << "\t\t" << answer << endl;
}
else{
int inc;
if(lastBit) inc = 2; //answer = add(answer, "10");
else inc = 1; //answer = increment(answer);
if(s[i-1] == '0') lastBit = 1;
else lastBit = 0;
if(lastBit) inc += 2;
else inc += 1;
if(inc == 2) answer = add(answer, "10");
else if(inc == 3) answer = add(answer, "11");
else answer = add(answer, "100");
}
}
else {
if(num > 0) {
if(pp[num-1] != "") pp[num-1] = power(num-1);
answer = add(answer, pp[num-1]);
}
else {
int inc = 0;
if(lastBit) inc = 1; //answer = increment(answer);
if(s[i-1] == '0') lastBit = 1;
else lastBit = 0;
if(lastBit) inc += 1;
answer = add(answer, convert(inc));
}
}
if(s[i-1] == '0') lastBit = 1;
else lastBit = 0;
}
}
if(s[sze-1] == '0') {
if(lastBit) {
if(s[1] == '0') {
answer = add(answer, "10");
}
else answer = increment(answer);
}
else if(s[1] == '0'){
answer = increment(answer);
}
}
printf( "Case #%d\n", t);
for(int i = 0; i < sze; i++) printf("%c", answer[i]);
printf("\n");
}
}
return 0;
}
If a number has k bits, then calculate the number of such numbers having a LUCKY FACTOR of 2:
10.............01
Hence in this the 1st two and last two digits are fixed, the remaining k-4 digits can have any value. The number of such numbers = 2^(k-4).
So you can easily calculate the sum of lucky factors of such numbers = lucky_factor x 2^(k-4)
(ofcourse this is assuming k >= 4)
What's more, you do not need to calculate this number since it will be of the form 10000000.
If the number n is 11010010. Then 8 bit numbers less than n shall be of form:
10........ or 1100...... or 1101000_. If you see a pattern, then we have divided the calculation in terms of the number of 1s in the number n
.
I leave the rest for you.
Question: I'm new to C++ and after writing the following code seems like there should be a way to shorten it. Maybe by somehow matching the string? How would this be done?
The function takes a string message received via Serial port and sets the value of a particular element of the pinValues[] array depending on the message. The value that will be set is determined by the last character H or L just before the \n.
String pattern: (a number)(H or L)\n
Eg: message == "4H\n" will set the 5th element pinValues[4] to HIGH. The number at the start of the string can be 1 to 2 digits.
void setPinValues(String message) {
if( message == "1H\n" ) {
pinValues[1] = HIGH;
}
if( message == "1L\n" ) {
pinValues[1] = LOW;
}
if( message == "2H\n" ) {
pinValues[2] = HIGH;
}
if( message == "2L\n" ) {
pinValues[2] = LOW;
}
if( message == "3H\n" ) {
pinValues[3] = HIGH;
}
if( message == "3L\n" ) {
pinValues[3] = LOW;
}
if( message == "4H\n" ) {
pinValues[4] = HIGH;
}
if( message == "4L\n" ) {
pinValues[4] = LOW;
}
if( message == "5H\n" ) {
pinValues[5] = HIGH;
}
if( message == "5L\n" ) {
pinValues[5] = LOW;
}
if( message == "6H\n" ) {
pinValues[6] = HIGH;
}
if( message == "6L\n" ) {
pinValues[6] = LOW;
}
}
This is probably not the official "C++"-approved way of doing it, but you could do:
unsigned int pinNo = 0;
unsigned char level = 0;
int result = sscanf(message.c_str(), "%u%c", &pinNo, &level);
if (result < 2)
// it failed
if (pinNo > 6)
// bad data
levelVal = (level == 'H') ? HIGH : LOW;
I'd do some sanity checking on the string while extracting the key and value from the first two chars. If you don't need to sanity check the message, it could be as short as
void setPinValues(String message) {
pinValues[ message[0] - '0' ] = (message[1] == 'H') ? HIGH:LOW;
}
Although you may want to make that a little longer, i.e. check the string length, and that the 2 chars your checking are in the right range. i.e
void setPinValues(string message) {
if (
message.size() >= 2
and
message[0] >= '1' and message[0] <= '6'
and (message[1]=='H' or message[1]=='L')
) {
pinValues[ message[0] - '0' ] = (message[1] == 'H') ? HIGH:LOW;
}
}
EDIT: you could also extend that to checking two leading digits, i.e.
int n, off=0;
if ( s[off] <= '9' and s[off] >= '0')
{
n = s[off++] - '0';
}
if ( s[off] <= '9' and s[off] >= '0')
{
n = 10*n + s[off++] - '0';
}
if (off > 0 and (s[1]=='H' or s[1]=='L')) {
pinValues[ message[0] - '0' ] = (message[1] == 'H') ? HIGH:LOW;
}
Assuming String is actually a std::string or has an identical interface, and also assuming an ASCII-compatible character set...
void setPinValues(String message) {
const size_t sz = message.size();
// input validation, ignore the message if it doesn't fit the pattern
// you can remove this "if" block if the message has already been validated
if ( (sz < 3) || (sz > 4)
// note how message[0] will be checked twice if sz == 3
// once as message[0] and once as message[sz -3]
// but if sz == 4 we check message[0] and message[1]
|| (message[0] < '0') || (message[0] > '9')
|| (message[sz - 3] < '0') || (message[sz - 3] > '9')
|| ((message[sz - 2] != 'H') && (message[sz - 2] != 'L'))
|| (message[sz - 1] != '\n'))
return;
// convert the first or two characters to a number
int pinNumber = message[0] - '0';
if (sz == 4)
pinNumber = (pinNumber * 10) + (message[1] - '0');
// additional check to verify the pin number is in the correct range
if ((pinNumber < 1) || (pinNumber > 6))
return;
// apply
pinValues[pinNumber] = (message[sz - 2] == 'H' ? HIGH : LOW);
}