1) Consider the array declaration, int x[20];. There is no memory allocated for

Question

1) Consider the array declaration, int x[20];. There is no memory allocated for x[20].

2) Given the two C++ array declarations:

int a[10], b[10];

You can successfully compute one array, say a, then assign b to a:

a = b;

3) Give the syntax of an array declaration. Mention the base type and declared size.

4) The dot operator is used between an object and a data member or between a calling object and a call to a member function from the class of the object.

5) You can change the behavior of + for the int type using operator overloading

Explanation / Answer

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Soution 1:

The given statement is false because whenever we declare any array with its size or statically it will automatically gets memory .

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Solution 2:

#include <iostream>

using namespace std;

int main()
{
  
int a[10],b[10];
int i;
//Values assign to array a
  
cout<<"Enter the values of Array a ";
for( i=0;i<10;i++)
{
cin>>a[i];
}

//Values assign to array b
  
cout<<" Enter the values of Array b ";
for( i=0;i<10;i++)
{
cin>>b[i];
}
//Printing values
cout<<" Printing array a";
for(i=0;i<10;i++)
{
cout<<a[i]<<" ";
}

  
cout<<" Printing array b";
for(i=0;i<10;i++)
{
cout<<b[i]<<" ";
}

//copying values of b to a array
for(i=0;i<10;i++)
{
a[i]=b[i];
}
cout<<" After copying the new 'a array': ";
for(i=0;i<10;i++)
{
cout<<a[i]<<" ";
}

return 0;
}

Output:

Enter the values of Array a 1 2 3 4 5 6 7 8 9 10

Enter the values of Array b 11 22 33 44 55 66 77 88 99 01

Printing array a 1 2 3 4 5 6 7 8 9 10

Printing array b 11 22 33 44 55 66 77 88 99 01

After copying the new 'a array':11 22 33 44 55 66 77 88 99 01

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Soution 3:

Array declaeation is:  

float arr[10];

here base type is float and size of array is 10.

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Solution 4:

Yes, the dot operator is used between an object and a data member or between a calling object and a call to a member function from the class of the object. Example:

#include <iostream>

using namespace std;

class Test
{
private:
int int_num;
float float_num;

public:
  
int test_num=100;
void insertIntegerData(int d)
{
int_num =d;
cout << " Number: " << int_num;
}

float insertFloatData()
{
cout << " Enter float number: ";
cin >> float_num;
return float_num;
}
};
int main()
{
  
Test obj; //object of test class
float float_data;
  
cout<<" Value of test_num is : "<<obj.test_num; //calling data member of test class using object

obj.insertIntegerData(12); //calling function of test class using object
float_data = obj.insertFloatData();//assign value using object

cout << " You entered : " << float_data;

return 0;
}

Output:

Value of test_num is : 100

Number: 12

Enter float number: 56.7

You entered : 56.7

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Solution 5:

No, we can not change the behavior of + for the int type using operator overoading. When we perform operator overloading on any operator at least one of the arguments to the operator must be of a class type.

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