c++ references array

If you are intending to pass the size of the array, then remove the reference

void f(int a[])

is equivalent to

void f(int* a)

so no copying will be done, if that is the concern.

If you want to take an array by reference, then you MUST specify the dimension. e.g.

void f(int (&a)[10])

Naturally, the best of the two is the third solution, which is to use std::vector's and pass them by reference, reference to const or by value if needed. HTH

Here is a slightly more C++ style of doing it:

#include <iostream>#include <vector>void writeTable(std::vector<int> &tab){    int val;    for (unsigned int i=0; i<tab.size(); i++)    {        std::cout << "Enter value " << i+1 << std::endl;        if (std::cin >> val)        {            tab[i] = val;        }    }}int main(){    int howMany;    std::cout << "How many elements?" << std::endl;    std::cin >> howMany;    std::vector<int> table(howMany);    writeTable(table);    return 0;}

You need not specify the dimension of the array if you make writeTable a function template.

template <typename T,size_t N>void writeTable(T (&tab)[N]) //Template argument deduction{    for(int i=0 ; i<N ; i++){       // code ....    }}

.

int table[howMany]; // C++ doesn't have Variable Length Arrays. `howMany` must be a constantwriteTable(table);  // type and size of `table` is automatically deduced