Check if element found in array c++

Question

How can I check if my array has an element I'm looking for?

In Java, I would do something like this:

Foo someObject = new Foo(someParameter);
Foo foo;
//search through Foo[] arr
for(int i = 0; i < arr.length; i++){
  if arr[i].equals(someObject)
    foo = arr[i];
}
if (foo == null)
  System.out.println("Not found!");
else
  System.out.println("Found!");

But in C++ I don't think I'm allowed to search if an Object is null so what would be the C++ solution?

Answer 1

Here is a simple generic C++11 function contains which works for both arrays and containers:

using namespace std;

template<class C, typename T>
bool contains(C&& c, T e) { return find(begin(c), end(c), e) != end(c); };

Simple usage contains(arr, el) is somewhat similar to in keyword semantics in Python.

Here is a complete demo:

#include <algorithm>
#include <array>
#include <string>
#include <vector>
#include <iostream>

template<typename C, typename T>
bool contains(C&& c, T e) { 
    return std::find(std::begin(c), std::end(c), e) != std::end(c);
};

template<typename C, typename T>
void check(C&& c, T e) {
    std::cout << e << (contains(c,e) ? "" : " not") <<  " found\n";
}

int main() {
    int a[] = { 10, 15, 20 };
    std::array<int, 3> b { 10, 10, 10 };
    std::vector<int> v { 10, 20, 30 };
    std::string s { "Hello, Stack Overflow" };
    
    check(a, 10);
    check(b, 15);
    check(v, 20);
    check(s, 'Z');

    return 0;
}

Output:

10 found
15 not found
20 found
Z not found

Answer 2

You can do it in a beginners style by using control statements and loops..

#include <iostream>
using namespace std;
int main(){
    int arr[] = {10,20,30,40,50}, toFind= 10, notFound = -1;
    for(int i = 0; i<=sizeof(arr); i++){
        if(arr[i] == toFind){   
            cout<< "Element is found at " <<i <<" index" <<endl;
            return 0;
        }   
    }
    cout<<notFound<<endl;
}

Answer 3

One wants this to be done tersely. Nothing makes code more unreadable then spending 10 lines to achieve something elementary. In C++ (and other languages) we have all and any which help us to achieve terseness in this case. I want to check whether a function parameter is valid, meaning equal to one of a number of values. Naively and wrongly, I would first write

if (!any_of({ DNS_TYPE_A, DNS_TYPE_MX }, wtype) return false;

a second attempt could be

if (!any_of({ DNS_TYPE_A, DNS_TYPE_MX }, [&wtype](const int elem) { return elem == wtype; })) return false;

Less incorrect, but looses some terseness. However, this is still not correct because C++ insists in this case (and many others) that I specify both start and end iterators and cannot use the whole container as a default for both. So, in the end:

const vector validvalues{ DNS_TYPE_A, DNS_TYPE_MX };
if (!any_of(validvalues.cbegin(),  validvalues.cend(), [&wtype](const int elem) { return elem == wtype; })) return false;

which sort of defeats the terseness, but I don't know a better alternative... Thank you for not pointing out that in the case of 2 values I could just have just if ( || ). The best approach here (if possible) is to use a case structure with a default where not only the values are checked, but also the appropriate actions are done. The default case can be used for signalling an invalid value.

Answer 4

If you were originally looking for the answer to this question (int value in sorted (Ascending) int array), then you can use the following code that performs a binary search (fastest result):

static inline bool exists(int ints[], int size, int k) // array, array's size, searched value
{
    if (size <= 0)      // check that array size is not null or negative
         return false;
    // sort(ints, ints + size); // uncomment this line if array wasn't previously sorted
    return (std::binary_search(ints, ints + size, k));
}

edit: Also works for unsorted int array if uncommenting sort.

Answer 5

You can use old C-style programming to do the job. This will require little knowledge about C++. Good for beginners.

For modern C++ language you usually accomplish this through lambda, function objects, ... or algorithm: find, find_if, any_of, for_each, or the new for (auto& v : container) { } syntax. find class algorithm takes more lines of code. You may also write you own template find function for your particular need.

Here is my sample code

#include <iostream>
#include <functional>
#include <algorithm>
#include <vector>

using namespace std;

/**
 * This is old C-like style.  It is mostly gong from 
 * modern C++ programming.  You can still use this
 * since you need to know very little about C++.
 * @param storeSize you have to know the size of store
 *    How many elements are in the array.
 * @return the index of the element in the array,
 *   if not found return -1
 */
int in_array(const int store[], const int storeSize, const int query) {
   for (size_t i=0; i<storeSize; ++i) {
      if (store[i] == query) {
         return i;
      }
   }
   return -1;
}

void testfind() {
   int iarr[] = { 3, 6, 8, 33, 77, 63, 7, 11 };

   // for beginners, it is good to practice a looping method
   int query = 7;
   if (in_array(iarr, 8, query) != -1) {
      cout << query << " is in the array\n";
   }

   // using vector or list, ... any container in C++
   vector<int> vecint{ 3, 6, 8, 33, 77, 63, 7, 11 };
   auto it=find(vecint.begin(), vecint.end(), query);
   cout << "using find()\n";
   if (it != vecint.end()) {
      cout << "found " << query << " in the container\n";
   }
   else {
      cout << "your query: " << query << " is not inside the container\n";
   }

   using namespace std::placeholders;
   // here the query variable is bound to the `equal_to` function 
   // object (defined in std)
   cout << "using any_of\n";
   if (any_of(vecint.begin(), vecint.end(), bind(equal_to<int>(), _1, query))) {
      cout << "found " << query << " in the container\n";
   }
   else {
      cout << "your query: " << query << " is not inside the container\n";
   }

   // using lambda, here I am capturing the query variable
   // into the lambda function
   cout << "using any_of with lambda:\n";
   if (any_of(vecint.begin(), vecint.end(),
            [query](int val)->bool{ return val==query; })) {
      cout << "found " << query << " in the container\n";
   }
   else {
      cout << "your query: " << query << " is not inside the container\n";
   }
}

int main(int argc, char* argv[]) {
   testfind();

   return 0;
}

Say this file is named 'testalgorithm.cpp' you need to compile it with

g++ -std=c++11 -o testalgorithm testalgorithm.cpp

Hope this will help. Please update or add if I have made any mistake.

Answer 6

In C++ you would use std::find, and check if the resultant pointer points to the end of the range, like this:

Foo array[10];
... // Init the array here
Foo *foo = std::find(std::begin(array), std::end(array), someObject);
// When the element is not found, std::find returns the end of the range
if (foo != std::end(array)) {
    cerr << "Found at position " << std::distance(array, foo) << endl;
} else {
    cerr << "Not found" << endl;
}

Answer 7

There are many ways...one is to use the std::find() algorithm, e.g.

#include <algorithm>

int myArray[] = { 3, 2, 1, 0, 1, 2, 3 };
size_t myArraySize = sizeof(myArray) / sizeof(int);
int *end = myArray + myArraySize;
// find the value 0:
int *result = std::find(myArray, end, 0);
if (result != end) {
  // found value at "result" pointer location...
}

Answer 8

You would just do the same thing, looping through the array to search for the term you want. Of course if it's a sorted array this would be much faster, so something similar to prehaps:

for(int i = 0; i < arraySize; i++){
     if(array[i] == itemToFind){
         break;
     }
}

Answer 9

C++ has NULL as well, often the same as 0 (pointer to address 0x00000000).

Do you use NULL or 0 (zero) for pointers in C++?

So in C++ that null check would be:

 if (!foo)
    cout << "not found";