Thread: finding number 7 in array using recursion

i want to find a number 7 in a array using recursion
here is my programme

Code:

bool iscontainseven(int a[],int size){
    if(size == 0)
    {
        a[0] == 7;
        return true;
    }


    if(size == 1)
    {
        if((a[0]||a[1]) == 7)
            return true;
    }


      int smallerarray[100];
      for(int i = 1;i < size ;i++)
      {


          smallerarray[i-1] = a[ i];
      }
      bool issmallerarraycontain7 = iscontainseven(smallerarray,size - 1);
      if(issmallerarraycontain7)
        return true;
      else
        return false;
}



int main()
{
cout<<"enter no of elements"<<endl;
    int n;
    int a[50];
    cin>>n;
    for(int i =  0;i < n;i++)
    {
        cout<<"enter next element"<<endl;
        cin>>a[i];
    }
    cout<<endl;
    cout<<"printing an array"<<endl;
    for(int i = 0;i < n;i++)
    {
        cout<<a[i]<<" ";
    }
    cout<<endl;
  
 
    cout<<"is contain seven"<<endl;
   // here i call a function
    cout<<iscontainseven(a,n);
    cout<<endl;

}

but it gives error
could someone plz correct me

Originally Posted by Salem

> but it gives error
You should tell us what that error is.


Your size == 0 test is redundant. There are no elements in an array of size zero.

Your size == 1 test must ALWAYS return either true or false. At the moment, it goes onto make copies of size -1, size -2, size -3 etc.

whenever i give input for eg 1 2 3 4 5 6
output = 1
if i give input 1 2 3 4 1 7
output = 1
so could u please give me the code

I highly recommend either writing all your functions and variabes in_snake_case or youShouldUseSomethingLikeCamelCase.