Segmentation Fault in Vectors - c++

So, I'm doing the famous "The Blocks Problem" from UVa Online Judge.
My approach is quite stupid, and that's because I wanted to play with vectors. So, I got vectors for pointers to each blocks in piles, and those vectors are stored in a vector called collection.
In order to find all the blocks, I have a vector called blockCollection, where pointers to all the blocks are stored inside.
The code has passed the samples provided. I will try to edit and provide comments later.
Full Source:
#include <iostream>
#include <vector>
struct Block
{
int id;
std::vector<Block*>* where;
};
int positionInVector(Block* b);
int main(int argc, const char * argv[])
{
std::vector<std::vector<Block*>*> collection;
std::vector<Block*> blockCollection;
std::string command = "", command2 = "";
int blockCount = 0, k = 0, A = 0, B = 0;
while ( std::cin >> blockCount )
{
std::vector<Block*>* vectors = new std::vector<Block*>[blockCount];
Block* blocks = new Block[blockCount];
for ( k = 0 ; k < blockCount ; ++ k)
{
blocks[k].id = k;
blocks[k].where = &vectors[k];
vectors[k].push_back(&blocks[k]);
blockCollection.push_back(&blocks[k]);
collection.push_back(&vectors[k]);
}
std::cin >> std::ws;
while ( std::cin >> command )
{
if ( command == "quit" ) break;
std::cin >> A >> command2 >> B;
Block* blockA = blockCollection[A];
Block* blockB = blockCollection[B];
std::vector<Block*>* vectorA = blockA -> where;
std::vector<Block*>* vectorB = blockB -> where;
//exception handle
if ( A > blockCount || B > blockCount ) continue;
if ( A == B ) continue;
if ( vectorA == vectorB ) continue;
if ( command == "move" )
{
//move anything on top of A to its original position
int positionOfBlockAInVectorA = positionInVector(blockA);
for ( int i = positionOfBlockAInVectorA + 1 ; i < vectorA -> size() ; ++ i )
{
Block* blockToBeMoved = *(vectorA -> begin() + i);
std::vector<Block*>* destinationVector = collection[blockToBeMoved -> id];
blockToBeMoved -> where = destinationVector;
destinationVector -> push_back(blockToBeMoved);
}
vectorA -> erase(vectorA -> begin() + positionOfBlockAInVectorA + 1, vectorA -> end());
}
if ( command2 == "onto" )
{
//move anything on top of B to its original position
int positionOfBlockBInVectorB = positionInVector(blockB);
for ( int i = positionOfBlockBInVectorB + 1 ; i < vectorB -> size() ; ++ i )
{
Block* blockToBeMoved = *(vectorB -> begin() + i);
std::vector<Block*>* destinationVector = collection[blockToBeMoved -> id];
blockToBeMoved -> where = destinationVector;
destinationVector -> push_back(blockToBeMoved);
}
if (positionOfBlockBInVectorB + 1 > vectorB -> size()) vectorA -> erase(vectorB -> begin() + positionOfBlockBInVectorB + 1, vectorB -> end());
}
if ( command == "move" )
{
//move block a to the pile containing block b
vectorA -> pop_back();
blockA -> where = vectorB;
vectorB -> push_back(blockA);
}
else
{
//move block a and those on top of it to the pile containing block b
std::vector<Block*> temperaryVector;
int positionOfBlockAInVectorA = positionInVector(blockA);
for ( int i = (int)vectorA -> size() - 1 ; i >= positionOfBlockAInVectorA ; -- i )
{
temperaryVector.push_back(vectorA -> at(i));
vectorA -> erase(vectorA -> begin() + i);
}
for ( int i = (int)temperaryVector.size() - 1 ; i >= 0 ; -- i )
{
temperaryVector[i] -> where = vectorB;
vectorB -> push_back(temperaryVector[i]);
}
}
}
for ( k = 0 ; k < blockCount ; ++ k )
{
std::vector<Block*>* vector = collection[k];
std::cout << k << ":";
if ( !vector -> empty() )
{
for ( int i = 0 ; i < vector -> size() ; ++ i )
{
std::cout << " " << vector -> at(i) -> id;
}
}
std::cout << std::endl;
}
delete [] blocks;
delete [] vectors;
}
return 0;
}
int positionInVector(Block* block)
{
std::vector<Block*> vector = *block -> where;
for ( int i = 0 ; i < vector.size() ; ++ i )
{
if ( vector[i] == block ) return i;
}
return -1;
}
Thanks!

For this to work:
A = (int)command[5] - 48;
B = (int)command[12] - 48;
we are need to ensure that string is 5/12 characters long, and that there is a digit in those positions. The code should add checks for the length of input and validity of the digits in those places.

Every time you add or delete a Block to your blockCollection, every pointer you hold to any Block in the collection may be invalidated.
I think that's all I need to say initially...

Related

Product of all the nodes on the path of a tree

I was learning MO's Algorithm. In that I found a question. In which we have to make a program to take input n for n nodes of a tree then n-1 pairs of u and v denoting the connection between node u and node v. After that giving the n node values.
Then we will ask q queries. For each query we take input of k and l which denote the two nodes of that tree. Now we have to find the product of all the nodes in the path of k and l (including k and l).
I want to use MO's algorithm. https://codeforces.com/blog/entry/43230
But I am unable to make the code. Can anybody help me out in this.
The basic code for that would be:
int n, q;
int nxt[ N ], to[ N ], hd[ N ];
struct Que{
int u, v, id;
} que[ N ];
void init() {
// read how many nodes and how many queries
cin >> n >> q;
// read the edge of tree
for ( int i = 1 ; i < n ; ++ i ) {
int u, v; cin >> u >> v;
// save the tree using adjacency list
nxt[ i << 1 | 0 ] = hd[ u ];
to[ i << 1 | 0 ] = v;
hd[ u ] = i << 1 | 0;
nxt[ i << 1 | 1 ] = hd[ v ];
to[ i << 1 | 1 ] = u;
hd[ v ] = i << 1 | 1;
}
for ( int i = 0 ; i < q ; ++ i ) {
// read queries
cin >> que[ i ].u >> que[ i ].v;
que[ i ].id = i;
}
}
int dfn[ N ], dfn_, block_id[ N ], block_;
int stk[ N ], stk_;
void dfs( int u, int f ) {
dfn[ u ] = dfn_++;
int saved_rbp = stk_;
for ( int v_ = hd[ u ] ; v_ ; v_ = nxt[ v_ ] ) {
if ( to[ v_ ] == f ) continue;
dfs( to[ v_ ], u );
if ( stk_ - saved_rbp < SQRT_N ) continue;
for ( ++ block_ ; stk_ != saved_rbp ; )
block_id[ stk[ -- stk_ ] ] = block_;
}
stk[ stk_ ++ ] = u;
}
bool inPath[ N ];
void SymmetricDifference( int u ) {
if ( inPath[ u ] ) {
// remove this edge
} else {
// add this edge
}
inPath[ u ] ^= 1;
}
void traverse( int& origin_u, int u ) {
for ( int g = lca( origin_u, u ) ; origin_u != g ; origin_u = parent_of[ origin_u ] )
SymmetricDifference( origin_u );
for ( int v = u ; v != origin_u ; v = parent_of[ v ] )
SymmetricDifference( v );
origin_u = u;
}
void solve() {
// construct blocks using dfs
dfs( 1, 1 );
while ( stk_ ) block_id[ stk[ -- stk_ ] ] = block_;
// re-order our queries
sort( que, que + q, [] ( const Que& x, const Que& y ) {
return tie( block_id[ x.u ], dfn[ x.v ] ) < tie( block_id[ y.u ], dfn[ y.v ] );
} );
// apply mo's algorithm on tree
int U = 1, V = 1;
for ( int i = 0 ; i < q ; ++ i ) {
pass( U, que[ i ].u );
pass( V, que[ i ].v );
// we could our answer of que[ i ].id
}
}
This problem is a slight modification of the blog that you have shared.
Problem Tags:- MO's Algorithm, Trees, LCA, Binary Lifting, Sieve, Precomputation, Prime Factors
Precomputations:- Just we need to do some precomputations with seiveOfErothenesis to store the highest prime factor of each element possible in input constraints. Then using this we will store all the prime factors and their powers for each element in input array in another matrix.
Observation:- with the constraints you can see the there can be very few such primes possible for each element. For an element (10^6) there can be a maximum of 7 prime factors possible.
Modify MO Algo Given in blog:- Now in our compute method we just need to maintain a map that will store the current count of the prime factor. While adding or subtracting each element in solving the queries we will iterate on the prime factors of that element and divide our result(storing total no. of factors) with the old count of that prime and then update the count of that prime and the multiple our result with the new count.(This will be O(7) max for each addition/subtraction).
Complexity:- O(T * ((N + Q) * sqrt(N) * F)) where F is 7 in our case. F is the complexity of your check method().
T - no of test cases in input file.
N - the size of your input array.
Q - No. of queries.
Below is an implementation of the above approach in JAVA. computePrimePowers() and check() are the methods you would be interested in.
import java.util.*;
import java.io.*;
public class Main {
static int BLOCK_SIZE;
static int ar[];
static ArrayList<Integer> graph[];
static StringBuffer sb = new StringBuffer();
static boolean notPrime[] = new boolean[1000001];
static int hpf[] = new int[1000001];
static void seive(){
notPrime[0] = true;
notPrime[1] = true;
for(int i = 2; i < 1000001; i++){
if(!notPrime[i]){
hpf[i] = i;
for(int j = 2 * i; j < 1000001; j += i){
notPrime[j] = true;
hpf[j] = i;
}
}
}
}
static long modI[] = new long[1000001];
static void computeModI() {
for(int i = 1; i < 1000001; i++) {
modI[i] = pow(i, 1000000005);
}
}
static long pow(long x, long y) {
if (y == 0)
return 1;
long p = pow(x, y / 2);
p = (p >= 1000000007) ? p % 1000000007 : p;
p = p * p;
p = (p >= 1000000007) ? p % 1000000007 : p;
if ((y & 1) == 0)
return p;
else {
long tt = x * p;
return (tt >= 1000000007) ? tt % 1000000007 : tt;
}
}
public static void main(String[] args) throws Exception {
Reader s = new Reader();
int test = s.nextInt();
seive();
computeModI();
for(int ii = 0; ii < test; ii++){
int n = s.nextInt();
lcaTable = new int[19][n + 1];
graph = new ArrayList[n + 1];
arrPrimes = new int[n + 1][7][2];
primeCnt = new int[1000001];
visited = new int[n + 1];
ar = new int[n + 1];
for(int i = 0; i < graph.length; i++) graph[i] = new ArrayList<>();
for(int i = 1; i < n; i++){
int u = s.nextInt(), v = s.nextInt();
graph[u].add(v);
graph[v].add(u);
}
int ip = 1; while(ip <= n) ar[ip++] = s.nextInt();
computePrimePowers();
int q = s.nextInt();
LVL = new int[n + 1];
dfsTime = 0;
dfs(1, -1);
BLOCK_SIZE = (int) Math.sqrt(dfsTime);
int Q[][] = new int[q][4];
int i = 0;
while(q-- > 0) {
int u = s.nextInt(), v = s.nextInt();
Q[i][0] = lca(u, v);
if (l[u] > l[v]) {
int temp = u; u = v; v = temp;
}
if (Q[i][0] == u) {
Q[i][1] = l[u];
Q[i][2] = l[v];
}
else {
Q[i][1] = r[u]; // left at col1 in query
Q[i][2] = l[v]; // right at col2
}
Q[i][3] = i;
i++;
}
Arrays.sort(Q, new Comparator<int[]>() {
#Override
public int compare(int[] x, int[] y) {
int block_x = (x[1] - 1) / (BLOCK_SIZE + 1);
int block_y = (y[1] - 1) / (BLOCK_SIZE + 1);
if(block_x != block_y)
return block_x - block_y;
return x[2] - y[2];
}
});
solveQueries(Q);
}
System.out.println(sb);
}
static long res;
private static void solveQueries(int [][] Q) {
int M = Q.length;
long results[] = new long[M];
res = 1;
int curL = Q[0][1], curR = Q[0][1] - 1;
int i = 0;
while(i < M){
while (curL < Q[i][1]) check(ID[curL++]);
while (curL > Q[i][1]) check(ID[--curL]);
while (curR < Q[i][2]) check(ID[++curR]);
while (curR > Q[i][2]) check(ID[curR--]);
int u = ID[curL], v = ID[curR];
if (Q[i][0] != u && Q[i][0] != v) check(Q[i][0]);
results[Q[i][3]] = res;
if (Q[i][0] != u && Q[i][0] != v) check(Q[i][0]);
i++;
}
i = 0;
while(i < M) sb.append(results[i++] + "\n");
}
static int visited[];
static int primeCnt[];
private static void check(int x) {
if(visited[x] == 1){
for(int i = 0; i < 7; i++) {
int c = arrPrimes[x][i][1];
int pp = arrPrimes[x][i][0];
if(pp == 0) break;
long tem = res * modI[primeCnt[pp] + 1];
res = (tem >= 1000000007) ? tem % 1000000007 : tem;
primeCnt[pp] -= c;
tem = res * (primeCnt[pp] + 1);
res = (tem >= 1000000007) ? tem % 1000000007 : tem;
}
}
else if(visited[x] == 0){
for(int i = 0; i < 7; i++) {
int c = arrPrimes[x][i][1];
int pp = arrPrimes[x][i][0];
if(pp == 0) break;
long tem = res * modI[primeCnt[pp] + 1];
res = (tem >= 1000000007) ? tem % 1000000007 : tem;
primeCnt[pp] += c;
tem = res * (primeCnt[pp] + 1);
res = (tem >= 1000000007) ? tem % 1000000007 : tem;
}
}
visited[x] ^= 1;
}
static int arrPrimes[][][];
static void computePrimePowers() {
int n = arrPrimes.length;
int i = 0;
while(i < n) {
int ele = ar[i];
int k = 0;
while(ele > 1) {
int c = 0;
int pp = hpf[ele];
while(hpf[ele] == pp) {
c++; ele /= pp;
}
arrPrimes[i][k][0] = pp;
arrPrimes[i][k][1] = c;
k++;
}
i++;
}
}
static int dfsTime;
static int l[] = new int[1000001], r[] = new int[1000001], ID[] = new int[1000001], LVL[], lcaTable[][];
static void dfs(int u, int p){
l[u] = ++dfsTime;
ID[dfsTime] = u;
int i = 1;
while(i < 19) {
lcaTable[i][u] = lcaTable[i - 1][lcaTable[i - 1][u]];
i++;
}
i = 0;
while(i < graph[u].size()){
int v = graph[u].get(i);
i++;
if (v == p) continue;
LVL[v] = LVL[u] + 1;
lcaTable[0][v] = u;
dfs(v, u);
}
r[u] = ++dfsTime;
ID[dfsTime] = u;
}
static int lca(int u, int v){
if (LVL[u] > LVL[v]) {
int temp = u;
u = v; v = temp;
}
int i = 18;
while(i >= 0) {
if (LVL[v] - (1 << i) >= LVL[u]) v = lcaTable[i][v];
i--;
}
if (u == v) return u;
i = 18;
while(i >= 0){
if (lcaTable[i][u] != lcaTable[i][v]){
u = lcaTable[i][u];
v = lcaTable[i][v];
}
i--;
}
return lcaTable[0][u];
}
}
// SIMILAR SOLUTION FOR FINDING NUMBER OF DISTINCT ELEMENTS FROM U TO V
// USING MO's ALGORITHM
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 40005;
const int MAXM = 100005;
const int LN = 19;
int N, M, K, cur, A[MAXN], LVL[MAXN], DP[LN][MAXN];
int BL[MAXN << 1], ID[MAXN << 1], VAL[MAXN], ANS[MAXM];
int d[MAXN], l[MAXN], r[MAXN];
bool VIS[MAXN];
vector < int > adjList[MAXN];
struct query{
int id, l, r, lc;
bool operator < (const query& rhs){
return (BL[l] == BL[rhs.l]) ? (r < rhs.r) : (BL[l] < BL[rhs.l]);
}
}Q[MAXM];
// Set up Stuff
void dfs(int u, int par){
l[u] = ++cur;
ID[cur] = u;
for (int i = 1; i < LN; i++) DP[i][u] = DP[i - 1][DP[i - 1][u]];
for (int i = 0; i < adjList[u].size(); i++){
int v = adjList[u][i];
if (v == par) continue;
LVL[v] = LVL[u] + 1;
DP[0][v] = u;
dfs(v, u);
}
r[u] = ++cur; ID[cur] = u;
}
// Function returns lca of (u) and (v)
inline int lca(int u, int v){
if (LVL[u] > LVL[v]) swap(u, v);
for (int i = LN - 1; i >= 0; i--)
if (LVL[v] - (1 << i) >= LVL[u]) v = DP[i][v];
if (u == v) return u;
for (int i = LN - 1; i >= 0; i--){
if (DP[i][u] != DP[i][v]){
u = DP[i][u];
v = DP[i][v];
}
}
return DP[0][u];
}
inline void check(int x, int& res){
// If (x) occurs twice, then don't consider it's value
if ( (VIS[x]) and (--VAL[A[x]] == 0) ) res--;
else if ( (!VIS[x]) and (VAL[A[x]]++ == 0) ) res++;
VIS[x] ^= 1;
}
void compute(){
// Perform standard Mo's Algorithm
int curL = Q[0].l, curR = Q[0].l - 1, res = 0;
for (int i = 0; i < M; i++){
while (curL < Q[i].l) check(ID[curL++], res);
while (curL > Q[i].l) check(ID[--curL], res);
while (curR < Q[i].r) check(ID[++curR], res);
while (curR > Q[i].r) check(ID[curR--], res);
int u = ID[curL], v = ID[curR];
// Case 2
if (Q[i].lc != u and Q[i].lc != v) check(Q[i].lc, res);
ANS[Q[i].id] = res;
if (Q[i].lc != u and Q[i].lc != v) check(Q[i].lc, res);
}
for (int i = 0; i < M; i++) printf("%d\n", ANS[i]);
}
int main(){
int u, v, x;
while (scanf("%d %d", &N, &M) != EOF){
// Cleanup
cur = 0;
memset(VIS, 0, sizeof(VIS));
memset(VAL, 0, sizeof(VAL));
for (int i = 1; i <= N; i++) adjList[i].clear();
// Inputting Values
for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
memcpy(d + 1, A + 1, sizeof(int) * N);
// Compressing Coordinates
sort(d + 1, d + N + 1);
K = unique(d + 1, d + N + 1) - d - 1;
for (int i = 1; i <= N; i++) A[i] = lower_bound(d + 1, d + K + 1, A[i]) - d;
// Inputting Tree
for (int i = 1; i < N; i++){
scanf("%d %d", &u, &v);
adjList[u].push_back(v);
adjList[v].push_back(u);
}
// Preprocess
DP[0][1] = 1;
dfs(1, -1);
int size = sqrt(cur);
for (int i = 1; i <= cur; i++) BL[i] = (i - 1) / size + 1;
for (int i = 0; i < M; i++){
scanf("%d %d", &u, &v);
Q[i].lc = lca(u, v);
if (l[u] > l[v]) swap(u, v);
if (Q[i].lc == u) Q[i].l = l[u], Q[i].r = l[v];
else Q[i].l = r[u], Q[i].r = l[v];
Q[i].id = i;
}
sort(Q, Q + M);
compute();
}
}
Demo

Binary search for insert char string. Where is the bug?

I have array of strings. I must find one char string in array of strings by binary search algoritm. If there is this one string then function must return position and return true otherwise this function must return position for insert string in array and false.
I have somewhere bug, but I dont know where ((
Example:
bool Binary_search ( char * arr_strings[], int & position, const char * search_string )
{
int start = 0 ;
int end = 10 - 1; // arr_strings [10]
int for_compare;
int middle;
while ( start <= end )
{
middle = ( start + end ) / 2;
for_compare = strcmp ( arr_strings[middle], search_string );
if ( for_compare > 0 )
{
start = middle + 1;
}
else if ( for_compare < 0 )
{
end = middle - 1;
}
else
{
// if search_string is found in array, then function return position in array of strings and return true
position = middle;
return true;
}
}
// if search_string is not found in array, then function must return position for insert string and return false
position = middle;
return false;
}
I think maybe it should be:
if ( for_compare > 0 )
{
end = middle - 1;
}
else if ( for_compare < 0 )
{
start = middle + 1;
}
The issue is that your insertion position is not right.
You should do something like the following:
bool Binary_search ( char * arr_strings[], const char * search_string )
{ //^^^you are not doing recursive, so you don't need position as parameter
int start = 0 ;
int end = 10 - 1; // arr_strings [10]
int for_compare;
int middle;
int position = -1;
while ( start <= end )
{
middle = ( start + end ) / 2;
for_compare = strcmp ( arr_strings[middle], search_string );
if ( for_compare > 0 )
{ //^^^here should switch the order
end = middle - 1;
}
else if ( for_compare < 0 )
{
start = middle + 1;
}
else
{
position = middle;
return true;
}
}
if (position == -1)
{
if(strcmp(arr_strings[middle],search_string) <0 )
{
position = middle + 1;
}else
{
position = middle;
}
}
return position;
}

Why did I get a Memory-related error here?

I had a problem from a website. Given a string s and st, I have to found all possible combination of st in s. For example,
s = "doomdogged"
st = "dg"
answer = 4
I can choose the d from 0 or 4, and g from 6 or 7. Which gives me 4 possible combinations.
Here's my code:
#include <iostream>
#include <vector>
using namespace std;
string s, st;
bool target[26];
vector<int> positions[26];
vector<vector<int>> possibleCombinations;
void DFS_Enumeration(int, vector<int>*);
int DFS_index_max = 0;
int main(int argc, char *argv[])
{
int answer = 0;
cin >> s; //Given a string s
cin >> st; //Given a string st
//Find all possible combination of st in s
for ( int i = 0 ; i < 26 ; ++ i )
target[i] = 0;
for ( int i = 0 ; i < st.length() ; ++ i )
target[st[i] - 97] = 1;
for ( int i = 0 ; i < 26 ; ++ i )
{
if ( target[i] == 0 ) continue;
for ( int j = 0 ; j < s.length() ; ++ j )
{
if ( s[j] == i + 97 ) positions[i].push_back(j);
}
}
DFS_index_max = st.length();
vector<int> trail(0);
DFS_Enumeration(0, &trail); //Here I got an runtime error
for ( vector<int> vi : possibleCombinations )
{
int currentMax = 0;
for ( int i = 0 ; i < vi.size() ; ++ i )
{
if ( vi[i] > currentMax )
{
if ( i == vi.size() - 1 ) ++ answer;
currentMax = vi[i];
continue;
}
else
break;
}
}
cout << answer;
}
void DFS_Enumeration(int index, vector<int>* trail)
{
if ( index == DFS_index_max )
{
possibleCombinations.push_back(*trail);
return;
}
for ( int i = 0 ; i < positions[st[index] - 97].size() ; ++ i )
{
trail -> push_back(positions[st[index] - 97][i]);
DFS_Enumeration(++index, trail);
trail -> pop_back();
}
return;
}
First I look for characters in st, and mark them as needed to found in my boolean array target.
Then, I use DFS to enumerate all possible combinations. For the above example of "doomdogged" and "dg", d exists in 0, 4, 9. And g exist in 6, 7. I will get 06, 07, 46, 47, 96, 97.
Lastly, I count those which make sense, and output the answer. For some reason, my code doesn't work and generate an runtime error concerning memory at the line I've marked.
DFS_Enumeration might increment index any number of times, so st[index] could likely be past the end of the string st.

Sorting char arrays by swapping pointers, C++

I am trying to sort an array of char pointers (char * _string) by swapping pointers.
I have this method, and what I want to do is use the values I get from _string and sort them by not manipulating _string, but the empty helper array (char * _output) which I also hand over to the method.
Can anyone help me and tell me what I am doing wrong?
void sortAsc(char* _string, char* _output)
{
int length = strlen(_string);
// output and string now point to the same area in the memory
_output = _string;
for( int i = 0; i < length; i++) {
for( int j = 0; j < length; j++) {
if( *(_output) > (_output[j] ) ) {
// save the pointer
char* tmp = _output;
// now output points to the smaller value
_output = _output+j;
// move up the pointer to the smaller value
_output + j;
// now the pointer of the smaller value points to the higher value
_output = tmp;
// move down to where we were + 1
_output - j + 1;
}
}
}
//_output[length]='\0';
//delete chars;
}
In my main-Method, I do something like this:
char * string = {"bcdae"};
char * output = new char[5];
sortAsc(string, output);
After that code, I want the output array to contain the sorted values.
Let's do the selection sort for a 10 size int array using pointer notation, you can simply change it to an array list.
*---*---*---*---*---* ........
a[] = | 1 | 2 | 4 | 0 | 3 | ........
*---*---*---*---*---* ........
^--------We start here looking for the smaller numbers and sort the array.
for( i = 0; i < 10; i++ ){
k = i;
bypass = *( a + i );
for( j = i + 1; j < 10; j++ ){
/* To get Increasing order. */
if( bypass > *( a + j ) ){
bypass = *( a + j );
k = j;
}
}
if ( k != i ){
*( a + k ) = *( a + i );
*( a + i ) = bypass;
}
}
This sorts the string into an already allocated buffer, and if the buffer isn't large enough tells you how big it has to be:
std::size_t sortAsc(char const* string, char* dest, std::size_t dest_length) {
std::size_t str_length = strlen(string);
char const* str_end = string + str_length;
if (dest_length < str_length+1)
return str_length+1;
std::copy( string, str_end, output );
output[str_length] = '\0';
std::sort( output, output+strlen(output) );
return str_length+1;
}
This does the poor "allocate a new string" pattern, using the above implementation:
char* allocate_and_sortAsc(char const* string) {
std::size_t str_length = strlen(string);
char* retval = new char[str_length+1];
std::size_t count = sortAsc( string, retval, str_length+1);
ASSERT( count <= str_length );
return retval;
}
And don't use variable names that start with an _, it is a bad practice because it wanders really near compiler reserved names. _Capital is reserved everywhere, and _lower in global scope, and foo__bar everywhere.

How to calculate bit transitions using bitset < >

I am new to C++. I want to calculate the no of transitions from 0 to 0, 0 to 1, 1 to 0 and 1 to 1 in a 9 bit sequence. I have written the following code;
int main {
srand((unsigned)time(0));
unsigned int x;
for (int i=0:i<=512;i++) // loop-1
{
x=rand()%512;
bitset<9>bitseq(x);
for(int j=0;j<=bitseq.size();j++) // loop-2
{
bool a= bitseq.test(j);
bool b= bitseq.test(j+1)
if ((a==0)&(b==0)==0)
{
transition0_0 = transition0_0 + 1; // transition from 0 to 0
}
else if ((a==0)&(b==1)==0)
{
transition0_1 = transition0_1 + 1;
else if ((a==1)&(b==0)==0)
{
transition1_0 = transition1_0 + 1;
else
{
transition1_1 = transition1_1 + 1;
cout<<transition0_0<<" "<<transition0_1<<endl;
cout<<transition1_0<<" "<<transition1_1<<endl;
}
}
Somebody please guide me on the following
how to save the last bit value in loop-2 to check the transition from last bit of the last bitset output to the 1st bit of the next bitset output?
If this does not work, How I can save it in vector and use iterators to check the transitions?
First of all, the loop index j is running past the end of the bitset. Indices go from 0 to bitseq.size()-1 (inclusive). If you're going to test j and j+1 the largest value j can take is bitseq.size()-2.
Second, the ==0 part that appears in your ifs is strange, you should just use
if( (a==0)&&(b==0) )
Notice the use of two &&. While a single & works for this code, I think it's better to use the operator that correctly conveys your intentions.
And then to answer your question, you can keep a "last bit" variable that is initially set to a sentinel value (indicating you're seeing the first bitseq just now) and compare it to bitseq[0] before the start of loop 2. Here's a modified version of your code that should do what you ask.
int main {
srand((unsigned)time(0));
unsigned int x;
int transition0_0 = 0,
transition0_1 = 0,
transition1_0 = 0,
transition1_1 = 0;
int prev = -1;
for (int i=0:i<=512;i++) // loop-1
{
x=rand()%512;
bitset<9> bitseq(x);
if( prev != -1 ) // don't check this on the first iteration
{
bool cur = bitseq.test(0);
if( !prev && !cur )
++transition0_0;
else if( !prev && cur )
++transition0_1;
else if( prev && !cur )
++transition1_0;
else
++transition1_1;
}
for(int j=0;j+1<bitseq.size();j++) // loop-2
{
bool a= bitseq.test(j);
bool b= bitseq.test(j+1)
if ((a==0)&&(b==0))
{
transition0_0 = transition0_0 + 1; // transition from 0 to 0
}
else if ((a==0)&&(b==1))
{
transition0_1 = transition0_1 + 1;
}
else if ((a==1)&&(b==0))
{
transition1_0 = transition1_0 + 1;
}
else
{
++transition1_1 = transition1_1 + 1;
}
} // for-2
prev = bitseq.test(bitseq.size()-1); // update prev for the next iteration
cout<<transition0_0<<" "<<transition0_1<<endl;
cout<<transition1_0<<" "<<transition1_1<<endl;
} // for-1
} // main
Would something like this be better for you? Use an array of 4 ints where [0] = 0->0, [1] = 0->1, [2] = 1->0, [3] = 1->1.
int main {
int nTransition[] = { 0,0,0,0 };
bool a,b;
unsigned int x;
int j;
srand ((unsigned)time(0));
for (int i = 0: i < 512; i++) {
x = rand () % 512;
bitset<9> bitseq(x);
if (i == 0) {
a = bitseq.test (0);
j = 1;
} else
j = 0;
for (; j < bitseq.size (); j++) {
b = bitseq.test(j);
int nPos = (a) ? ((b) ? 3 : 2) : ((b) ? 1 : 0);
nTransition[nPos]++;
a = b;
}
}
}