So basically I have an array which looks like this:
char array[25];
I have an 'X' put in there, and 24 'O's. I am trying to create a function where I can find the location of 'X' in the array, so then I can move it whenever a control is typed in. Here is some code to get an idea on what I am thinking, I just can't put it together.
int findX(){
//make this find the location of X out of the 25 char array.
//then return the slot of X as a number, like if it's in the 20th slot, then it will return 20 ?
//return(locationOfX);
}
for(int i = 0; i <= array.Length; i++)
{
if(array[i] == 'X')
{
return i;
}
}
This should do it, I had no chance testing it, but it should work
I think Ron's answer is more correct because actually I think in C there isn't a function like "XXX.length()", so you need to pass the length of the array to the function.
Pass in the array and the number of elements.
1. The manual approach:
int findX(char arr[], int count){
for (int i = 0; i < count; i++){
if (arr[i] == 'X'){
return i;
}
}
return -1;
}
2. Use std::find function:
int findX2(char arr[], int count){
return std::find(arr, arr + count, 'X') - arr;
}
Slight twist with the std::distance function:
int findX2(char arr[], int count){
return std::distance(arr, std::find(arr, arr + count, 'X'));
}
3. Pass in the beginning and the end of the array:
int findX3(char* arrbegin, char* arrend){
return std::distance(arrbegin, std::find(arrbegin, arrend, 'X'));
}
and use like:
std::cout << findX3(std::begin(arr), std::end(arr));
4. Template function left as an exercise.
That being said, prefer std::vector or std::array to raw arrays.
In language C you can use strstr function.
#include "string.h"
int findX(char array[])
{
char *result = strstr(array, "s");
return result - array;
}
int main()
{
char array[25];
int index = findX(array);
}
in C++ use std::find() or just change array to std::string and use string::find
here is the code:
#include <string>
int main()
{
std::string array = "oooooooooosoooooooooooooo";
int index = array.find("s");
return 0;
}
Related
I am trying to reverse a char which has been provided in input from an user. I am having issues with the reverse function, particularly the loop. I can't get it to work- can I get advice?
#include <iostream>
using namespace std;
#include <cstring>
char* reverse(char* input) {
int len = strlen(input);
char temp[len];
for(int i=len; i>len; --i) {
temp[i]+=input[i];
}
return temp;
}
int main()
{
char input[100];
while(cin>>input) {
cout << reverse(input);
}
return 0;
}
Your Program has few issues
You're trying to return local variable address i.e. temp array address. The Function will return the address to main function. Since memory might get cleaned so it will print garbage value present at the address.
As Rohan Bari mentioned variable length array might cause undefined behavior. There for you can create a constant length array i.e.
char temp[100];
or you can dynamically allocate array on heap. Memory allocated on heap do not get cleared after termination of block but we have to manually delete it.
char* temp = new char[len];
As array start from 0 it goes till len-1 so loop condition should start from len-1 and has to go till 0 to reverse.
+ operator do not work's with array or char even if you are trying to add just char it preforms normal integer addition of their ASCII value.
Here is improved version of your code
#include<iostream>
using namespace std;
#include <cstring>
char* reverse(char* input) {
int len = strlen(input);
char* temp = new char [len]; // or you can use char temp[100];
int j = 0; //temp variable to enter values from 0th index if we use same as loop it just enter in the same order as original char array.
for(int i=len-1; i>=0; --i) {
temp[j++] = input[i];
}
temp[j] = '\0';
return temp;
}
You have got several errors in the program.
The variable-length arrays are used here:
char temp[len];
This should not be applied in C++ since this invokes undefined-behavior. Note that this is a valid statement in the C99 standard.
There is a better alternative to this. That is to take the std::string built-in type in use.
In the following line:
temp[i] += input[i];
You are not sequentially adding one character after another, but the values of them in a single integer. This could be not a problem if temp was of the type std::string.
The reverse function should look like this:
const char *reverse(char *input) {
int len = strlen(input);
std::string temp;
while (len--)
temp += input[len];
return temp.c_str();
}
len should actually be (len-1) and i should be >= 0 not len, so from (len-1) to 0 your loop should run.
for(int i = len-1; i >= 0; i--){}
You have to allocate the new array with the new keyword if you don't want to use a string. The following code does what you need:
char* reverse(char* input)
{
int len = strlen(input);
char* temp = new char[len + 1];
for (int i = len; i >= 0; --i)
{
temp[len-i-1] = input[i];
}
temp[len] = '\0';
return temp;
}
You could use a std::stack to reverse your input:
std::stack<char> s;
char c;
while (std::cin >> c)
{
s.push(c);
}
while (!s.empty())
{
std::cout << s.top();
s.pop();
}
It's 2021. Use the STL. If your instructor isn't aware of it or doesn't allow you to use it, your instructor is not keeping up-to-date and you should fire your instructor.
#include <algorithm>
#include <iostream>
#include <string>
int main() {
std::string input{};
while(std::getline(std::cin, input)) {
std::reverse(std::begin(input), std::end(input));
std::cout << input << '\n';
}
return 0;
}
There's quite many things wrong with the code as many people have already mentioned! Since you want to implement this without using STL it can be done this way,
#include <iostream>
using namespace std;
#include <cstring>
void reverse(char* input,int len) { //added len as argument
char temp[len];
for(int i=len-1; i>=0; --i) {
temp[len-i-1]=input[i];
cout<<temp[len-i-1]; //printing while reversing
}
cout<<endl;
}
int main()
{
char input[100];
int len=0;
//using do while since it has to run atleast once
do{
cin.getline(input,100);
len=strlen(input);
input[len]='\0';
if(len!=0)
reverse(input,len);
}while(len!=0) ;
return 0;
}
I am making a program to identify whether a 5 card ( user input ) array is a certain hand value. Pair, two pair, three of a kind, straight, full house, four of a kind ( all card values are ranked 2-9, no face cards, no suit ). I am trying to do this without sorting the array. I am currently using this to look through the array and identify if two elements are equal to each other
bool pair(const int array[])
{
for (int i = 0; i < array; i++)
{
if (array[i]==aray[i+1])
{
return true;
}
else
return false;
}
Does this section of code only evaluate whether the first two elements are the same, or will it return true if any two elements are the same? I.E if the hand entered were 2,3,2,4,5 would this return false, where 2,2,3,4,5 would return true? If so, how do I see if any two elements are equal, regardless of order, without sorting the array?
edit: please forgive the typos, I'm leaving the original post intact, so as not to create confusion.
I was not trying to compile the code, for the record.
It will do neither:
i < array will not work, array is an array not an int. You need something like int arraySize as a second argument to the function.
Even if you fix that then this; array[i]==aray[i+1] will cause undefined behaviour because you will access 1 past the end of the array. Use the for loop condition i < arraySize - 1.
If you fix both of those things then what you are checking is if 2 consecutive cards are equal which will only work if the array is sorted.
If you really cant sort the array (which would be so easy with std::sort) then you can do this:
const int NumCards = 9; // If this is a constant, this definition should occur somewhere.
bool hasPair(const int array[], const int arraySize) {
int possibleCards[NumCards] = {0}; // Initialize an array to represent the cards. Set
// the array elements to 0.
// Iterate over all of the cards in your hand.
for (int i = 0; i < arraySize; i++) {
int myCurrentCard = array[i]; // Get the current card number.
// Increment it in the array.
possibleCards[myCurrentCard] = possibleCards[myCurrentCard] + 1;
// Or the equivalent to the above in a single line.
possibleCards[array[i]]++; // Increment the card so that you
// count how many of each card is in your hand.
}
for (int i = 0; i < NumCards; ++i) {
// If you want to check for a pair or above.
if (possibleCards[i] >= 2) { return true; }
// If you want to check for exactly a pair.
if (possibleCards[i] == 2) { return true; }
}
return false;
}
This algorithm is actually called the Bucket Sort and is really still sorting the array, its just not doing it in place.
do you know the meaning of return keyword? return means reaching the end of function, so in your code if two adjacent values are equal it immediately exits the function; if you want to continue checking for other equality possibilities then don't use return but you can store indexes of equal values in an array
#include <iostream>
using namespace std;
int* CheckForPairs(int[], int, int&);
int main()
{
int array[ ]= {2, 5, 5, 7, 7};
int nPairsFound = 0;
int* ptrPairs = CheckForPairs(array, 5, nPairsFound);
for(int i(0); i < nPairsFound; i++)
{
cout << ptrPairs[i] << endl;
}
if(ptrPairs)
{
delete[] ptrPairs;
ptrPairs = NULL;
}
return 0;
}
int* CheckForPairs(int array[] , int size, int& nPairsFound)
{
int *temp = NULL;
nPairsFound = 0;
int j = 0;
for(int i(0); i < size; i++)
{
if(array[i] == array[i + 1])
nPairsFound++;
}
temp = new int[nPairsFound];
for(int i(0); i < size; i++)
{
if(array[i] == array[i + 1])
{
temp[j] = i;
j++;
}
}
return temp;
}
You could use a std::unordered_set for a O(n) solution:
#include <unordered_set>
using namespace std;
bool hasMatchingElements(const int array[], int arraySize) {
unordered_set<int> seen;
for (int i = 0; i < arraySize; i++) {
int t = array[i];
if (seen.count(t)) {
return true;
} else {
seen.insert(t);
}
}
return false;
}
for (int i = 0; i < array; i++)
{
if (array[i]==aray[i+1])
{
return true;
}
else
return false;
This loop will only compare two adjacent values so the loop will return false for array[] = {2,3,2,4,5}.
You need a nested for loop:
#include <stdio.h>
#include <stdbool.h>
int main()
{
int unsortedArray[] = {2,3,2,4,5};
int size = 5;
for(int i=0;i<size-1;i++)
{ for(int j=i+1;j<size;j++)
{ if(unsortedArray[i]==unsortedArray[j])
{ printf("matching cards found\n");
return 0;
}
}
}
printf("matching cards not found\n");
return 0;
}
----EDIT------
Like Ben said, I should mention the function above will only find the first instance of 2 matching cards but it can't count how many cards match or if there are different cards matching. You could do something like below to count all the number of matching cards in the unsortedArray and save those values into a separate array. It's messier than the implementation above:
#include <iostream>
#include <stdio.h>
#include <stdbool.h>
#defin NUM_CARDS 52;
using namespace std;
int main()
{
int T;
cin>>T;
while(T--)
{
int N,i,j;
cin>>N;
int unsortedArray[N];
for(int i=0;i<N;i++)
cin>>unsortedArray[i];
int count[NUM_CARDS]={0};
int cnt = 0;
for( i=0;i<N-1;i++)
{
for( j=i+1;j<N;j++)
{ if(unsortedArray[i]==-1)
break;
if(unsortedArray[i]==unsortedArray[j])
{
unsortedArray[j]=-1;
cnt++;
}
}
if(unsortedArray[i]!=-1)
{
count[unsortedArray[i]]=cnt; //in case you need to store the number of each cards to
// determine the poker hand.
if(cnt==1)
cout<<" 2 matching cards of "<<unsortedArray[i]<<" was found"<<endl;
else if(cnt>=2)
cout<<" more than 2 matching cards of "<<unsortedArray[i]<<" was found"<<endl;
else
cout<<" no matching cards of "<<unsortedArray[i]<<" was found"<<endl;
cnt = 0;
}
}
This question already has answers here:
Can a local variable's memory be accessed outside its scope?
(20 answers)
Closed 8 years ago.
I'm at college and we're learning pointers. Our job was to input a char, compare it to an array and return a pointer to the first reference of that char in the array. But, as I don't like easy things, I've asked my teacher what about having that char more than once in the array.
That's where my headache begins.
So I have this code. The idea is: create a function that compares the input char to the entire array, get the pointers of the references and save them in an array and return that array.
Unfortunately it's not working as I wish :(
What can be wrong?
#include<iostream>
#include<cstdlib>
using namespace std;
char list [10];
int main()
{
initialize();
show();
cout<<search('1');
}
void initialize()
{
int i;
for(i=0; i<10;i++)
{
list[i]='1';
}
}
void show()
{
int i;
for(i=0; i<10;i++)
{
cout<<list[i];
}
}
int* search(char* input)
{
int* result[10];
int i;
char *temp;
for (i=0; i<10; i++)
{
*temp=list[i];
if(strcmp(temp, input) != NULL)
{
result[i]=i;
}
}
return result[];
}
I'm on a mobile device so I can't go into huge detail unfortunately, but you are returning a pointer to an array that you create in the function which goes out of scope at the end of the function.
My massive edit:
As everyone has already stated, a C++ array is actually only a pointer to the first element in the array. As a result, if you return a pointer to an array created in the scope of the function, you are returning a pointer to garbage. If I were doing this I would use a vector, but if I were to be forced into using an array, I would use something like the code below. Hope this helps!
#include <iostream>
#include <cstdlib>
void initialize(char* list) {
for(int i = 0; i < 10; ++i) {
if(i < 4) {
list[i] = '2';
} else {
list[i] = '1';
}
}
}
void show(char *list) {
for(int i = 0; i < 10; ++i) {
std::cout << list[i] << ' ';
}
std::cout << std::endl;
}
// Note the function requires an additional argument that is a pointer
// this is how you can avoid returning a pointer
int search(char input, char* list, char* result) {
int j = 0;
for(int i = 0; i < 10; ++i) {
// comparing characters can be done via ==
if(input == list[i]) {
*(result + j) = list[i];
// You could use result[j], but I used this to show that
// result really is just a pointer to the first element
// of the array. As a result saying result[j] is the same
// as saying I want to deference j elements past the first
// element or *(result + j)
++j; // increment j
}
}
// return how many elements matched
return(j);
}
int main(int argc, char *argv[]) {
char list[10];
char temp[10];
initialize(list);
show(list);
int size = search('1', list, temp);
// create a dynamically sized array containing space for each match
// because we don't know the size at compile time we must use
// a library type or a dynamically sized array
char *result = new char[size];
for(int i = 0; i < size; ++i) {
result[i] = temp[i];
// again you could use result[i]
std::cout << *(result + i) << std::endl;
}
delete[] result; // otherwise you'll get a memory leak
return(0);
}
For example, I have an array:
int Arr[10]={1,2,3,4,5,6,7,8,9,10};
How to change its order of elements using a pointer to receive the following array:
Arr={10,9,8,7,6,5,4,3,2,1}
to change the order odd and even using a pointer I've found this:
But I need only to reverse an array (without replacing odd and even)
#include <iostream>
using namespace std;
int main (const int& Argc, const char* Argv[]){
const int Nelem=10;
int Arr[]={1,2,3,4,5,6,7,8,9,10};
int *begAr=&Arr[0];
int *endAr=&Arr[Nelem];
int *itrAr=begAr;
int *tmpValAr=new int(0);
cout<<"Before\t1 2 3 4 5 6 7 8 9 10"<<endl;
while(itrAr<endAr){
*tmpValAr=*itrAr;
*itrAr=*(itrAr+1);
*(itrAr+1)=*tmpValAr;
itrAr+=2;
}
cout<<"After\t";
for(int i=0; i<Nelem; ++i)cout<<Arr[i]<<" ";
cout<<endl;
system("pause");
return 0;
}
Ok, a C-style approach using pointers to reverse an array? That shouldn't be too hard to figure out. Here's one approach:
int main ( void )
{
int i,//temp var
arr[10]= {1,2,3,4,5,6,7,8,9,10};//the array
int *start = &arr[0],//pointer to the start of the array
*end = &arr[9];//pointer to the last elem in array
//print out current arr values
for (i=0;i<10;++i)
printf("arr[%d] = %d\n", i, arr[i]);
do
{//simple loop
i = *start;//assign whatever start points to to i
*start = *end;//assign value of *end to *start
*end = i;//assign initial value of *start (stored in i) to *end
} while ( ++start < --end);//make sure start is < end, increment start and decrement end
//check output:
for (i=0;i<10;++i)
printf("arr[%d] = %d\n", i, arr[i]);
return 0;
}
As you can see here, this reverses the array just fine.
Use reverse found in <algorithm>:
std::reverse(Arr, Arr+10);
It will reverse a set of data like you are requesting.
This is the approximate implementation of the function, which you could adapt if necessary to your needs if you would like to write the loop yourself:
template <class BidirectionalIterator>
void reverse (BidirectionalIterator first, BidirectionalIterator last)
{
while ((first!=last)&&(first!=--last)) {
std::iter_swap (first,last);
++first;
}
}
If you are in C or would like a less general solution, do something like this:
int i = 0; j = 9;
for(;i<j;++i;--j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = arr[i];
}
Take two pointers begAr pointing at arr[0] and endAr pointing at arr[9]. Traverse the array from both sides and swap *begAr with *endAr until begAr > endAr.
int tempValAr;
while(endAr >= begAr )
{
tempValAr = *begAr;
*begAr++ = *endAr;
*endAr-- = tempValAr;
}
See the test program.
I want to use the code below and I want to use it for "unknown size of input". For example there is an array int cac[1000][1000]. I can use vector<vector<int> > array;, then how can i initialize it with -1 ? Any suggestions?
#include <sstream>
#include <iostream>
#include <vector>
#include <cstdlib>
#include <memory.h>
using namespace std;
int cac[1000][1000];
string res[1000][1000];
vector<string> words;
int M;
int go(int a, int b){
if(cac[a][b]>= 0) return cac[a][b];
if(a == b) return 0;
int csum = -1;
for(int i=a; i<b; ++i){
csum += words[i].size() + 1;
}
if(csum <= M || a == b-1){
string sep = "";
for(int i=a; i<b; ++i){
res[a][b].append(sep);
res[a][b].append(words[i]);
sep = " ";
}
return cac[a][b] = (M-csum)*(M-csum);
}
int ret = 1000000000;
int best_sp = -1;
for(int sp=a+1; sp<b; ++sp){
int cur = go(a, sp) + go(sp,b);
if(cur <= ret){
ret = cur;
best_sp = sp;
}
}
res[a][b] = res[a][best_sp] + "\n" + res[best_sp][b];
return cac[a][b] = ret;
}
int main(int argc, char ** argv){
memset(cac, -1, sizeof(cac));
M = atoi(argv[1]);
string word;
while(cin >> word) words.push_back(word);
go(0, words.size());
cout << res[0][words.size()] << endl;
}
What you can do is to use a associative array, where the key is a pair (rowPosition, ColumnPosition). When you want to set array[i][j] you just add or update the value assoArray[Pair(i,j)]. You can assume that any element which is not in the associative array has the initial value.
In general infinite multidimensional arrays are used for theoretical purpose.I hope i didn't misunderstood the question.
Using std::vector from the STL is much more straightforward than the following solution, which was pointed out in the comments for this post. I find that this site explains that topic effectively: http://www.learncpp.com/cpp-programming/16-2-stl-containers-overview/
An array of infinite size is not actually possible. However, you can achieve basically that effect using dynamic allocation. Here's some sample code:
int counter = 0;
int* myArray = new int[1000];
Fill the array with data, incrementing counter each time you add a value. When counter reaches 1000, do the following:
int* largerArray = new int[2000];
for( int i = 0; i < 1000; i++ )
{
largerArray[i] = myArray[i];
}
delete[] myArray;
myArray = largerArray;
With this method, you create the closest thing possible to an infinitely sized array, and I don't believe performance will be an issue with the copy piece