#include <iostream>
using namespace std;
class my_array_over
{
size_t len = 1;
int *a = new int[1];
public:
my_array_over() { a[0] = 0; }
my_array_over(size_t ln, const int *o) : len(ln), a(new int[ln])
{
for (size_t n = 0; n < ln; n)
a[n] = o[n];
}
~my_array_over() { delete[] a; }
size_t get_length() const
{
return len;
}
int get(size_t n) const
{
return a[n];
}
int set(size_t n, int v)
{
int tmp = a[n];
a[n] = v;
return tmp;
}
};
void foo(const my_array_over &a2, size_t i)
{
if (i < a2.get_length())
std::cout << a2[i] << std::endl;
}
Been trying to fix this code but kept getting an error saying "no match for 'operator[]'enter code here
(operand types are 'const my_array_over' and 'size_t' {aka 'long unsigned int'})" on
std::cout << a2[i] << std::endl;
CodePudding user response:
In this statement
std::cout << a2[i] << std::endl;
there is used the subscript operator for an object of the type my_array_over
that (the subscript operator) is not defined within the class. It seems you mean
std::cout << a2.get( i ) << std::endl;
Otherwise you need to define the subscript operator within the class definition. For example
const int & operator []( size_t n ) const
{
return a[n];
}
int & operator []( size_t n )
{
return a[n];
}