In C++ program which passes array in constructor, execution stops

Question

I am trying to set elements of a set at an index to 1 if an array contains those indices as its elements. Array size is 20 i.e. index 0 to 19

For ex -

 int myArray[5] = {1,4,2}; //user input or statically defined in driver(main)
 int set[] = {0,1,1,0,1}; //successfully set in constructor
 IntegerSet intObj(set);//At a point, program stops execution. Any idea why?

Here is the partial code

    //integerset.h 

    class IntegerSet{
          public :
              IntegerSet( int [] );
              .....
          private :
              int set[20];            
    };

  //integerset.cpp (header files included)

  IntegerSet :: IntegerSet( int arr[]){
       for(int i = 0; i <20; i++) //works fine (printed "test" in loop)
               set[i] = 0; //if not this then garbage elems set
       for ( int i = 0; arr[i] != '\0' ; i++ ) //works fine. (printed "test" in loop)
          set [arr[i]] = 1;        
       for ( int i = 0; i < 20; i++ ) //program stops execution here
           cout<<i<<"\t"<<set[i]<<"\n"; //tried to print "test" in loop,program crashed
  }

  //main.cpp (header files included)


int main(){
             int a[] = {1,3,0,12,14,15,'\0'}; // premature end??
             IntegerSet obj2(a);
             system("pause");
             return 0;
   }

doesn't change the program output/execution. set isn't a keyword. i changed it to setArray though set looked more intuitive but what about the answer to question? — ShivaniDhanked
– ShivaniDhanked, Commented May 24, 2013 at 3:54
@Sergey : in the for loop with != '\0' condition, if you print "test" it is printed 7 times. i found the answer on stackoverflow to the question ..how to find length of array when array is passed as argument, because sizeof operator in such case would return size of pointer which in my machine is 4 — ShivaniDhanked
– ShivaniDhanked, Commented May 24, 2013 at 3:58

Jesse Good · Accepted Answer · 2013-05-24 04:07:21Z

1

In arr[i] != '\0', your array does not have a null terminator, so the loop continues and indexes elements pass the end of the array.

It is best practice to use std::array or std::vector. Another option is the following fix to your code:

template <int N>
IntegerSet :: IntegerSet(int (&arr) [N]) : set() {
       for (int i = 0; i < N ; i++)
          set [arr[i]] = 1;        
       for (int i = 0; i < 20; i++)
           cout<<i<<"\t"<<set[i]<<"\n";
  }
}

edited May 24, 2013 at 4:07

answered May 24, 2013 at 3:54

Jesse Good

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7 Comments

ShivaniDhanked Over a year ago

int a[] = {1,3,4,12,14,15,'\0'} in main.cpp does the job! Thanks all!

Jesse Good Over a year ago

@ShivaniDhanked: That can cause problems, try the following array: int a[] = {1,3,4,0,14,15,'\0'};. The loop will stop prematurely.

ShivaniDhanked Over a year ago

oh yes, i soon realized it after posting this comment.. but how come the loop printed "test string" 7 times i.e. correctly parsed to length of array without using terminating null...isn't there a simpler way out than what you have just suggested. i am looking for something closest to my code., Thanks

Jesse Good Over a year ago

strings are special, they do have an implicit null terminator at the end. Another way to solve your problem is to pass the length of the array as the second parameter, IntegerSet :: IntegerSet( int arr[], int length) and loop until then: for (int i = 0; i < length ; i++).

ShivaniDhanked Over a year ago

Thanks but know it and can't do that, the set is the only private data member, so it alone can be passed to the constructor.

|

mecid · Accepted Answer · 2013-05-24 03:55:29Z

0

When you parse an array to a function as a parameter with its name, it refers to the address of the first element in the array. So in your case a is a type of int *. So you need to change your constuctor parameter as int *

answered May 24, 2013 at 3:55

mecid

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1 Comment

Jesse Good Over a year ago

int [] and int * mean the same thing as a parameter.

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