I am trying to insert int array x to int *v. here is my code . please provide me with optimal solutions and the reason behind it.
there is an error in this line. Instead of copying array value its taking garbage value. line v1=x;
class vector
{
int *v;
int size;
public:
vector(int m)
{
v = new int[size = m];
for (int i = 0; i < size; i++)
v[i] = 0;
}
vector(int *a)
{
for (int i = 0; i < size; i++)
v[i] = a[i];
}
int operator *(vector &y)
{
int sum = 0;
for (int i = 0; i < size; i++)
sum += v[i] * y.v[i];
return sum;
}
void disp()
{
for (int i = 0; i < size; i++)
cout << v[i] << " ";
cout << "\n";
}
};
int main()
{
clrscr();
int x[3] = { 1,2,3 };
int y[3] = { 4,5,6 };
vector v1(3);
//v1.disp();
vector v2(3);
v2.disp();
v1 = x;
v1.disp();
//v2=y;
v2.disp();
int r = v1 * v2;
cout << "R = " << r;
getch();
return 0;
}
You forgot to add the assignment operator in your vector
class:
vector & operator=(int *a)
{
for (int i = 0; i < size; i++)
v[i] = a[i];
return *this;
}
In the the line
v1=x;
May be, you are expecting it to invoke the second constructor which takes int*
as argument. But it won't happen.
It can be seen as Type Conversion from Basic type to Class type . where we expect appropriate constructor will get invoked.
see http://www.hexainclude.com/basic-to-class-type-conversion/
But remember, Constructor will be invoked only once after the creation of object.
Here, in the line
vector v1(3);
the first constructor was already invoked. Then the line
v1=x;
won't invoke the second constructor now.
For every class, =
operator is default overloaded . That is the reason why we can easily assign objects to one another.
Therefore, the line v1=x
invokes default overloaded assignment =
operator. Here, it treats address of array x ie, &x[0]
as address of class object. As it is not address of vector class
object
=> it results a Segmentation fault .
YOUR ANSWER
To assign int
array to int pointer
ie, to the member variable int* v
of the vector class
,
=
inside the class . or
vector v1=x; // modify the class constructor to have size as a constant.
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