int *tab1[5]; but dynamically

Question

I need a help with c++.

int *tab1[5];
int tab2[] = {3, 4, 5};
tab1[0] = tab2;

It works but I want to have a variable instead of 5 in first line. Any ideas? Thanks you in advance.

[`std::vector`](http://en.cppreference.com/w/cpp/container/vector). Always `std::vector`. — Some programmer dude, Mar 08 '18 at 11:59
arrays should ALWAYS be compile time constants (even if your compiler has an extension to allow you to use variables). You can get away with `constexpr unsigned int arraySize=5` but that's not a variable in the way I think you want it to be. — UKMonkey, Mar 08 '18 at 12:00
Possible duplicate of [How to create a dynamic array of integers](https://stackoverflow.com/questions/4029870/how-to-create-a-dynamic-array-of-integers) — alseether, Mar 08 '18 at 12:04

score 1 · Accepted Answer · answered Mar 08 '18 at 12:18

1

Since you found the "wrong" answer I'll show you the "correct":

#include <iostream>
#include <vector>

int main()
{
    std::cout << "How many lines do you want? ";
    unsigned lines;
    if (!(std::cin >> lines))
    {
        std::cout << "Invalid input\n";
        return 1;
    }

    std::vector<std::vector<int>> tab(lines);

    if (lines > 0)
    {
        tab[0] = { 3, 4, 5 };
    }
}

answered Mar 08 '18 at 12:18

Some programmer dude

400,186
35
402
621

3

Ew, vectors of vectors – Lightness Races in Orbit Mar 08 '18 at 12:27

empthy · Answer 2 · 2018-03-08T12:20:14.527

0

I found an answer,

int n = 5;
int **tab1 = new int*[n];
int tab2[] = {3, 4, 5};
tab1[0] = tab2;

Thanks for your attention.

edited Mar 08 '18 at 12:20

answered Mar 08 '18 at 12:13

empthy

19
4

4

Careful: that assignment doesn't do what you [probably] think it does. – Lightness Races in Orbit Mar 08 '18 at 12:27

int *tab1[5]; but dynamically

2 Answers2