im doing simple genetic algorithm uniform crossover operation . for that im using two arrays as parent and mother.i want concatenate the childs for getting the offsprings(childs).
i have problem in adding the arrays .any help plssss.i did it ubuntu
#include<iostream>
#include <fstream>
#include <algorithm>
#include<vector>
using namespace std;
int main()
{
int P[ ]={3,7,6,5,2,4,1,8};
int N[ ]={8,6,7,2,5,3,4,1};
int r= (sizeof(P)/sizeof(*P)) ;
int s= (sizeof(N)/sizeof(*N));
int val=r/2 ;
int t1[val],t2[val],t3[val],t4[val],n=0,p=0;
for(int m=0;m< val;m++)
{
t1[n]=P[m];
t2[n]=N[m];
n++;
}
for(int x=val;x< r;x++)
{
t3[p]=P[x];
t4[p]=N[x];
n++;
}
int* child=new int [val+val];
copy(t1,t1+val,child);
copy(t3,t3+val,child+val);
cout << child;
}
return 0;
}
This part is wrong:
int t1[val], t2[val], t3[val], t4[val]
You can only use constant values to declare the size of arrays.
You can either use a std::vector or dynamically allocate memory for the t-arrays.
std::vector<int> t1(val);
std::vector<int> t2(val);
for(int m = 0; m < val; m++)
{
t1[n] = P[m];
t2[n] = N[m];
n++;
}
There seem to be multiple errors in your code.
Variable length arrays are at present not supported in C++.
int val=r/2 ;
int t1[val]; // Not OK
In the second for loop I guess you meant p++ instead of n++;
Instead of manually doing all the memory allocation - deallocation, you should use std::vectors
cout << child; // This outputs the address of the pointer, not the entire array.
Related
In this below program, I'm trying to marge 2 arrays into a single vector, but while returning the function I'm getting additional garbage values along with it.
Please anyone suggest me how to remove those!
#include <bits/stdc++.h>
#include <vector>
#include <string>
using namespace std;
vector <int> merge(int a[],int b[]){
vector <int> marr1;
marr1.clear();
int i=0,j=0;
while(i+j <= ((*(&a+1)-a)+(*(&b+1)-b)))
{
if ((i<= *(&a+1)-a)){
marr1.push_back(a[i]);
i++;
}
else{
marr1.push_back(b[j]);
j++;
}
}
sort(marr1.begin(),marr1.end());
return marr1;
}
int main(){
//array imlementation
int arr1[] = {5,7,4,5},arr2[] = {8,3,7,1,9};
vector <int> ans;
ans.clear();
ans = merge(arr1,arr2);
for (auto i=ans.begin();i<ans.end();++i){
cout<<*i<<"\t";
}
}
output produced:
0 0 0 0 1 3 4 5 5 7 7 8 9 32614 32766 4207952 1400400592
You want something like this:
include <iostream>
#include <vector>
#include <algorithm> // <<<< dont use #include <bits/stdc++.h>,
// but include the standard headers
using namespace std;
vector <int> mergeandsort(int a[], int lengtha, int b[], int lengthb) { // <<<< pass the lengths of the arrays
vector <int> marr1; // <<<< and use meaningful names
// marr1.clear(); <<<< not needed
for (int i = 0; i < lengtha; i++)
{
marr1.push_back(a[i]);
}
for (int i = 0; i < lengthb; i++)
{
marr1.push_back(b[i]);
}
sort(marr1.begin(), marr1.end());
return marr1;
}
int main() {
int arr1[] = { 5,7,4,5 }, arr2[] = { 8,3,7,1,9 };
vector <int> ans;
// ans.clear(); <<<< not needed
ans = mergeandsort(arr1, 4, arr2, 5);
for (auto i = ans.begin(); i < ans.end(); ++i) {
cout << *i << "\t";
}
}
Look at the <<<< comments for explanations.
There is still room for improvement:
passing the hard coded lengths of the arrays in mergeandsort(arr1, 4, arr2, 5) is bad practice, if you add/remove element from the arrays, you need to change the lengths too.
you shouldn't use raw arrays in the first place but vectors like in vector<int> arr1[] = { 5,7,4,5 };, then you don't need to care about the sizes as a vectors knows it's own size. I leave this as an exercise for you.
Since you're not passing the length of the array, there is no way inside the merge function to know about their length. Your program seems to produce undefined behavior as can be seen here. If you execute this program again and again you'll notice that the output changes which is an indication of undefined behavior.
Secondly, you're using std::vector::clear when there is no need to use it in your program. I have commented it in the code example i have given below.
You can use pass the length of the arrays as arguments to the merge function. Below is the complete working example:
#include <bits/stdc++.h>
#include <vector>
#include <string>
using namespace std;
vector<int> merge(int a[], int lengthA, int b[], int lengthB){
vector <int> marr1;
//marr1.clear();//no need for this since the vector is empty at this point
for(int i = 0; i< lengthA; ++i)
{
//std::cout<<"adding: "<<a[i]<<std::endl;
marr1.push_back(a[i]);
}
for(int i = 0; i< lengthB; ++i)
{
//std::cout<<"adding: "<<b[i]<<std::endl;
marr1.push_back(b[i]);
}
sort(marr1.begin(),marr1.end());
return marr1;
}
int main(){
//array imlementation
int arr1[] = {5,7,4,5},arr2[] = {8,3,7,1,9};
vector <int> ans;
//ans.clear();//no need for this since the vector is empty at this point
ans = merge(arr1,4, arr2, 5);
for (auto i=ans.begin();i<ans.end();++i){
cout<<*i<<"\t";
}
}
You pass two int[] which degrade to pointers. This means you cannot tell the number of elements which you attempt to do with i+j <= ((*(&a+1)-a)+(*(&b+1)-b)). Either pass in a length of each array, or even better (C++) pass in two vectors instead. Also, if you don't know the STL has a merge() function in <algorithm>.
I'm learning C++ and I'm wondering if anyone can explain some strange behaviour I'm seeing.
I'm currently learning memory management and have been playing around with the following code:
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
// pass back by pointer (old C++)
const int array_size = 1e6; // determines size of the random number array
vector<int> *RandomNumbers1()
{
vector<int> *random_numbers = new vector<int>[array_size]; // allocate memory on the heap...
for (int i = 0; i < array_size; i++)
{
int b = rand();
(*random_numbers).push_back(b); // ...and fill it with random numbers
}
return random_numbers; // return pointer to heap memory
}
int main (){
vector<int> *random_numbers = RandomNumbers1();
for (int i = 0; i < (*random_numbers).size(); i++){
cout << (*random_numbers)[i] + "\n";
}
delete random_numbers;
}
What I'm trying to do is get a pointer to a vector containing random integers by calling the RandomNumbers1() function, and then print each random number on a new line.
However, when I run the above code, instead of printing out a random number, I get all sorts of random information. It seems as though the code is accessing random places in memory and printing out the contents.
Now I know that I'm doing something stupid here - I have an int and I am adding the string "\n" to it. If I change the code in main() to the following, it works fine:
int main (){
vector<int> *random_numbers = RandomNumbers1();
for (int i = 0; i < (*random_numbers).size(); i++){
cout << to_string((*random_numbers)[i]) + "\n";
}
}
However I just can't understand the behaviour I'm getting with the "wrong" code - i.e. how adding the string "\n" to (*random_numbers)[i]
causes the program to access random areas of memory, instead of where my pointer is pointing to. Surely I have de-referenced the pointer and accessed the element at position i before "adding" "\n" to it? So how is the program instead accessing a totally different memory address?
"\n" is a string literal. It is an array and it is converted to a pointer pointing at its first element in your expression.
(*random_numbers)[i] is an integer.
Adding a pointer to an integer means that advance the pointer by the integer.
This will drive the pointer to out-of-range because "\n" has only 2 elements ('\n' and '\0') but the numbers returnd from the rand() function are likely to be larger than 2.
There are several issues with your code.
you are using delete instead of delete[] to free the array allocated with new[].
you are creating an array of 1000000 vectors, but populating only the 1st vector with 1000000 integers. You probably meant to create just 1 vector instead.
you can and should use the -> operator when accessing an object's members via a pointer. Using the * and . operators will also work, but is more verbose and harder to read/code for.
you are trying to print a "\n" after each number, but you are using the + operator when you should be using the << operator instead. You can't append a string literal to an integer (well, you can, but it will invoke pointer arithmetic and thus the result will not be what you want, as you have seen).
With that said, try something more like this:
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
const int array_size = 1e6; // determines size of the random number array
vector<int>* RandomNumbers1()
{
vector<int> *random_numbers = new vector<int>;
random_numbers->reserve(array_size);
for (int i = 0; i < array_size; ++i)
{
int b = rand();
random_numbers->push_back(b);
}
return random_numbers;
}
int main (){
vector<int> *random_numbers = RandomNumbers1();
for (size_t i = 0; i < random_numbers->size(); ++i){
cout << (*random_numbers)[i] << "\n";
}
/* alternatively:
for (int number : *random_numbers){
cout << number << "\n";
}
*/
delete[] random_numbers;
}
However, if you are going to return a pointer to dynamic memory, you really should wrap it inside a smart pointer like std::unique_ptr or std::shared_ptr, and let it deal with the delete for you:
#include <iostream>
#include <vector>
#include <cmath>
#include <memory>
using namespace std;
const int array_size = 1e6; // determines size of the random number array
unique_ptr<vector<int>> RandomNumbers1()
{
auto random_numbers = make_unique<vector<int>>();
// or: unique_ptr<vector<int>> random_numbers(new vector<int>);
random_numbers->reserve(array_size);
for (int i = 0; i < array_size; ++i)
{
int b = rand();
random_numbers->push_back(b);
}
return random_numbers;
}
int main (){
auto random_numbers = RandomNumbers1();
for (size_t i = 0; i < random_numbers->size(); ++i){
cout << (*random_numbers)[i] << "\n";
}
/* alternatively:
for (int number : *random_numbers){
cout << number << "\n";
}
*/
}
Though, in this case, there is really no good reason to create the vector dynamically at all. 99% of the time, it is unnecessary and unwanted to use standard containers like that. Since the vector manages dynamic memory internally, there is no reason for the vector itself to also be created in dynamic memory. Return the vector by value instead, and let the compiler optimize the return for you.
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
const int array_size = 1e6; // determines size of the random number array
vector<int> RandomNumbers1()
{
vector<int> random_numbers;
random_numbers.reserve(array_size);
for (int i = 0; i < array_size; ++i)
{
int b = rand();
random_numbers.push_back(b);
}
return random_numbers;
}
int main (){
vector<int> random_numbers = RandomNumbers1();
for (size_t i = 0; i < random_numbers.size(); ++i){
cout << random_numbers[i] << "\n";
}
/* alternatively:
for (int number : random_numbers){
cout << number << "\n";
}
*/
}
What am I doing wrong ???
It is giving me an error of segmentation fault .
I don't know what memory i'm accessing ??
#include <iostream>
using namespace std;
int main() {
int t;
int n;
int arr[n];
cin>>t;
cin>>n;
// taking array elements from user
for(int i=0; i<n;i++)
{
cin>>arr[i];
}
// reverse of the array
for(int i=n-1; i>=0;i--)
{
cout<<arr[i];
}
//code
return 0;
}
int n;
You've default initialised this variable. Thus, the value is indeterminate.
int arr[n];
Here, you use that indeterminate value. Thus, the behaviour of the program is undefined.
There are "data flow" languages where using a variable will stop execution waiting for you to initialise it later and continue. C++ isn't such language. You must initialise everything before using the value.
Besides that, n isn't a compile time constant expression. Because the size of the array variable isn't compile time conastant, the progarm is ill-formed in C++.
If you want an array to have a dynamic size, you can use dynamic storage. Simplest way to create a dynamic array is to use std::vector.
For starters you are trying to declare a variable length array
int n;
int arr[n];
Variable length arrays is not a standard C++ feature. Moreover you are using an uninitialized variable n as the size of the array.
Either declare the array with an expected maximum size or use standard container std::vector<int>. At least you should write provided that the compiler supports variable length arrays
int t;
int n = 1;
cin>>t;
cin>>n;
if ( n < 1 ) n = 1;
int arr[n];
//...
Also you are not reversing an array. You are trying to output an array in the reverse order.
To reverse an array you could use standard algorithm std::reverse or you can write an appropriate loop yourself as for example
for ( int i = 0; i < n / 2; i++ )
{
// or use std::swap( arr[i], arr[n-i-1] );
int tmp = arr[i];
arr[i] = arr[n-i-1];
arr[n-i-1] = tmp;
}
and then you can output the reversed array.
You are not allowed to define array with Unknown size in C++ so int arr[n]; is Wrong! and if you have to use arrays and not know the size of it , you should use dynamic array with Pointers like this : int* a = new int[n] and also Deallocate Heap memory with Delete []array_name at end of your program and if it is possible for you not use arrays It's better for you use vectors because the size of it is dynamic.
look at this with vectors :
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
//by : Salar Ashgi
int main()
{
int k;
cout<<"Enter number of elements ?\n";
cin>>k;
vector<int> v;
cout<<"--------------------\n";
int x;
for(int i=0;i<k;i++)
{
cout<<"Enter num "<<i+1<<" : ";
cin>>x;
v.push_back(x);
}
reverse(v.begin(),v.end());//algorithm.h
cout<<"Reverse : \n";
for(int i=0;i<k;i++)
{
cout<<v[i]<<" ";
}
}
I have written a code for solving magic square using Loubere's algorithm. I've created a function for this algorithm and I want to return a value using int** but how should i do that as I've used 2D array in the function body, also how should i store it in main() function
#include <iostream>
#include <cstdlib>
using namespace std ;
int ** magicSquare ( int size )
{
int number;
int sq=size*size;enter code here
int mid=size/2;
//for array
int m=0;
//rows
int n=mid; //columns
array[m][n]=n;
for(number=1; number<=sq; number++)
{
n++;
n=n-1;
m=m+1;
if(n==-1)
n=3;
if(m==3)
m=0;
if (array[n][m] !=0)
n++;
array[n][m]=number;
}
int* array_rows[size]=array[size][size];
int** p=array_rows;
return p;
}
int main()
{
int dimension;
cout<<"Enter dimension of the magic square : "; cin>>dimension;
magicSquare ( dimension );
}
As suggested, you could use std::vector<std::vector<int>> to store your data and return it properly. Something like this:
#include <vector>
#include <iostream>
using square = std::vector<std::vector<int>>;
square makeMagicSquare(size_t size) {
square magicSquare(size, std::vector(size, 0));
// <-- Your code here
// magicSquare[i][j] = ...;
return magicSquare;
}
int main()
{
square sq = makeMagicSquare(10);
// Print it
for (auto& row: sq) {
for (auto& cell: row) {
std::cout << cell << ' ';
}
std::cout << '\n';
}
}
If you want to store it as an array, you would need to change the method to return a int[][] instead of int**. They work very similarly, but there is a difference between a pointer and an array.
You can store it as a int[][] exactly as you would expect. Storing the magic square would like the following:
int[][] magicSquare = magicSquare(dimension);
This assigns the return value of your method to a 2D array that you will then have to free when your program ends.
Remember, the way an array works is to give the address of the head of a block of data. When you return an array, even if it is a int[][] you are referencing the address of a block of data. In this case, it is a block of int arrays. In order to access it later, you need to allocate that memory block before you can do anything with it. This means that in your magicSquare() method, you will need to create a int[][] that contains the information, and that you can pass to the int[][] to.
DISCLAIMER: you can definitely do this with pointers, but it would be easier and less likely to cause errors with std::vector<std::vector<int>>. Using a std::vector you don't have to allocate memory or free it, and it handles potentially undefined behavior such as reading past the end of the block of data.
#include <iostream>
using namespace std;
/*
*
*/
int main() {
int k, in[k],reversea[k],i,m,n;
cin>>k;
for (i=0;i<k;i++){
cin>>in[i];
}
for (m=k-1;m>=0;m--){
for (n=0;n<k;n++){
in[m]=reversea[n];
}
}
for(i=0;i<k;i++){
cout<<reversea[i];
}
return 0;
}
I have no idea why it says segmentation fault even before i start debugging it. I compile another one on calculating the frequency of 1, 5, and 10 in an array of k numbers, and it says the same thing...
Here is the other one:
#include <iostream>
using namespace std;
int main() {
int k,i,m,n,count5,count1,count10;
int input[k];
cin>>k;
for (i=0;i<k;i++){
cin>>input[i];
}//input all the numbers
for(i=0;i<k;i++){
if (input[i]=1){
count1++;
}
if (input[i]=5){
count5++;
}
if (input[i]=10){
count10++;
}
}
cout<<count1<<"\n"<<count5<<"\n"<<count10<<"\n";
return 0;
}
Please help me. Thanks.
On this line
int k, in[k],reversea[k]
How are you supposed to initialize an array with k elements if k isn't initialized? The size of an array must be known at compile time not run time. If k isn't know until run time, use a std::vector
int k;
std::cin >> k;
std::vector<int> in(k);
std::vector<int> reversea(k);
Both your programs have two major faults.
You need to know the size of an array while creating it. In your code, k is still uninitialized and you are using this value as the size of your array. Instead, change it to
int k,i,m,n;
cin >> k;
int in[k];
int reversea[k];
While reversing the array, you should be filling reversea using values from in, and not the other way round. Also, you don't need 2 for loops, just use 1 for loop.
for (m=k-1; m>=0; m--){
reversea[m] = in[k-1-m];
}
In the second program, you again need to get the value of k before creating the array input[k].
You are testing for equality with a = instead of == . Change your code from
if (input[i]=1){
to
if (input[i] == 1) {