How to fill a char array without loop

Question

I need help with filling up a char array in c++ with a letter and without using a loop. This is what i have so far.

int n, m;
    cin >> n >> m;
    char canvas[n][m] = {{'B'}};

This however won't work. Does anyone know how I can do this?

You are trying to create a VLA, which is not included in the C++ standard (but is available as an extension in some implementations) — Ted Lyngmo, Feb 17 '21 at 21:39
Hint 2: VLAs like `char canvas[n][m]` are [not part of the C++ standard](https://stackoverflow.com/questions/1887097/why-arent-variable-length-arrays-part-of-the-c-standard) and have a tendency of blowing up the stack. `std::vector` will solve both of your problems. — Lukas-T, Feb 17 '21 at 21:39
@Shadow_Walker If you're writing values to many different locations, somebody has to do the loop. All you can hope for is that somebody writes the loop better than you do. — user4581301, Feb 17 '21 at 21:40

score 4 · Accepted Answer · answered Feb 17 '21 at 21:39

4

You can do this pretty easily with a 2D std::vector:

int n, m;
std::cin >> n >> m;
std::vector<std::vector<char>> canvas(n, std::vector<char>(m, 'B'));

answered Feb 17 '21 at 21:39

scohe001

2

Entry #9,582,108 in our ongoing series of "Why you should be using `std::vector`" – tadman Feb 17 '21 at 21:40
A 1D flat vector would be even better than a 2D sparse vector: `std::vector canvas(n*m, 'B');` Though, the code is dealing with `char`, so there is always `std::string` instead: `std::string canvas(n*m, 'B');` – Remy Lebeau Feb 17 '21 at 22:48

1 Answers1