I have just started c++ and in the algorythm course there is a pseudo-code and ı have tried it to c++ but ı have failed how can ı convert c++ any clue helps me thanks.
+about multiplication arrays ,this array's elements are constant of x to 1 2 .... n with same size. a1 and b1 is default zero.
+ı write same code in python easily but c++ ı don t know how to break type safety rules.
int* Mult2(int arr1[], int arr2[],int a1,int b1, int size)
{
int* R = new int[ 2* size - 1];
if ( size == 1 )
{
R[0] = arr1[a1] + arr2 [b1];
return R;
}
R = Mult2(arr1, arr2, a1, b1, size/2);
R = Mult2(arr1, arr2, a1+ size/2 , b1+ size/2 , size/2) ;
int* D0E1 = Mult2(arr1, arr2, a1, b1+size/2, size/2);
int* D1E0 = Mult2(arr1, arr2, a1+size/2 , b1, size/2);
R += D0E1 + D1E0;
return R;
}
error: invalid operands of types 'int*' and 'int*' to binary 'operator+' R += D0E1 + D1E0;
ı can t add two pointer. so
int* Mult2(int arr1[], int arr2[],int a1,int b1, int size)
{
int* R = new int[ 2* size - 1];
if ( size == 1 )
{
R[0] = arr1[a1] + arr2 [b1];
return R;
}
R = Mult2(arr1, arr2, a1, b1, size/2);
R = Mult2(arr1, arr2, a1+ size/2 , b1+ size/2 , size/2) ;
int* D0E1 = Mult2(arr1, arr2, a1, b1+size/2, size/2);
int* D1E0 = Mult2(arr1, arr2, a1+size/2 , b1, size/2);
for( int i = 0; i < size ; i++)
{
R[i] = D0E1[i] + D1E0[i];
}
return R;
}