I'm just revisiting C++, and I have a question about overloading of the [] operator, and more specifically why my program doesn't work.
Given the following code in vec.cpp:
double Vec::operator[](unsigned int i) const {
return this->values[i];
}
double & Vec::operator[](unsigned int i) {
return this->values[i];
}
These are defined in vec.h as methods to the Vec class, and if I do not use the operator in my main.cpp, all is fine and dandy. It compiles just as normal with no errors.
However once I do this in my main function (which is using std::cout and std::endl):
cout << a[0] << endl;
Things go wrong. The errors I get are a bunch of
candidate function template not viable: no known conversion from 'Vec' to 'char' for 2nd argument
operator<<(basic_ostream<_CharT, _Traits>& __os, char __cn)
where you can replace 'char' with any primitive data type.
Here is a working example
// In vec.h
#pragma once
#include <string>
#include <iostream>
class Vec {
private:
int dims;
double *values;
public:
Vec(int dims, double values[]);
double operator [](unsigned int i) const;
double& operator[](unsigned int i);
};
// In vec.cpp
#include <iostream>
#include <string>
#include <cmath>
#include "vec.h"
using std::cerr, std::endl, std::cout;
Vec::Vec(int dims, double values[]) {
this->dims = dims;
this->values = new double[dims];
for(int i = 0; i < dims; i++) {
this->values[i] = values[i];
}
}
double Vec::operator[](unsigned int i) const {
if(i >= this->dims) {
cerr << "Elem out of range" << endl;
}
return this->values[i];
}
double & Vec::operator[](unsigned int i) {
if(i >= this->dims) {
cerr << "Elem out of range" << endl;
}
return this->values[i];
}
// In main.cpp
#include <iostream>
#include <string>
#include "vec.h"
using std::cout, std::endl;
int main() {
double avals[2];
avals[0] = 1.0;
avals[1] = 2.0;
Vec *a = new Vec(2, avals);
cout << a[0] << endl; // Error occurs here
return 0;
}
Can anyone help me sort this out?
In this declaration
Vec *a = new Vec(2, avals);
there is declared a pointer of the type Vec *
. So an expression with the dereferenced pointer has the type Vec
.
So in this statement
cout << a[0] << endl;
the expression a[0]
has the type Vec
.
It seems you mean
( *a )[0]
or
a[0][0]