从':: size_t'到'int'的转换需要缩小的转换

Question

I'm trying to implement a container class using the code given in the Chapter 18 of 'Programming Principles and Practices using C++'. And when I write the code for the initialization using initializer_list I get this error: 'conversion from '::size_t' to 'int' requires a narrowing conversion'.

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>


class vector
{
public:
    vector(int s) 
        :sz{ s },
        elem{ new double[sz] }
    {
        for (int i = 0; i < sz; i++)
            elem[i] = 0.0;
    }

    vector(std::initializer_list<double>lst)
        :sz{ lst.size() }, elem{ new double[sz] }
    {                                  //compiler points to here for the error
        std::copy(lst.begin(), lst.end(), elem);
    }

    ~vector() { delete[] elem; }

    int size() const { return sz; }

    double get(int n) const { return elem[n]; }
    void set(int n, double d) { elem[n] = d; }
private:
    int sz;
    double* elem;
};

Answer 1

You're trying to convert a size_t into an int . Assuming 32-bits, size_t can hold up to 2^32 because it's unsigned, whereas int can only hold 2^31 (because it can have negative values also).

When you do this:

vector(std::initializer_list<double>lst)
    :sz{ lst.size() }

If lst.size() is greater than the value stored by an int , then the value can't properly be stored. The solution would be to just use std::vector internally, and get rid of the sz member, but seeing as though it looks as if you're trying to make your own vector class, you should just make sz a size_t , since it better represents what you're trying to do. After all, it's not as if your vector can have < 0 elements in it.