declare an array with value

Question

I'm writing a program in C++, and I'm trying to declare an array with value but only this value is set only to the first member in the array.

char mines[100][100] = {'0'};

I'm not really sure what you're trying to ask. Are you trying to set every member of the mines array to 0, or are you trying to set the 100th member of the 100th element to 0? — alexyorke, Jul 17 '12 at 14:06
Initializer lists used in this manner will specifically set values for all included, everything not included will be initialized to 0. — Chad, Jul 17 '12 at 14:09

score 2 · Answer 1 · answered Jul 17 '12 at 14:09

Use std::fill on each dimension:

for (int i = 0; i < 100; ++i)
    std::fill (mines[i], mines[i] + 100, '0');

Or in C++11:

for (auto &chars : mines)
    std::fill (std::begin (chars), std::end (chars), '0');

Even easier, using the fact that arrays are contiguous, you can save yourself the loop:

std::fill (mines, mines + 100*100, '0');

To eliminate the magic number there, as Shahbaz points out, replace mines + 100*100 with sizeof(mines).

score 2 · Answer 2 · edited Jul 17 '12 at 14:15

2

If you don't want to write or generate an initializer list of 10000 elements, you could do:

char mines[100][100];
memset(mines, '0', sizeof(mines));

edited Jul 17 '12 at 14:15

jsist

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answered Jul 17 '12 at 14:09

Luchian Grigore

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Good point, I keep forgetting these things are contiguous for some reason. – chris Jul 17 '12 at 14:09
Well, they're 0, not '0'. Maybe that's it. – chris Jul 17 '12 at 14:10
@chris no, it's because I used memcpy instead of memset :) huh... good thing I noticed it quickly. – Luchian Grigore Jul 17 '12 at 14:11
@Rob thanks for the edit, I thought he wanted the array zeroed. – Luchian Grigore Jul 17 '12 at 14:13
Yes, but if Luna wanted the array zeroed, there wouldn't have been a question at all since that's what happens automatically for all array elements after the last initialized element. – Rob Kennedy Jul 17 '12 at 14:14
I tried to use "memset" before, but the problem is that I want to submit my code to uva.onlinejudge.org and they considered this as undeclared – Luna28 Jul 17 '12 at 14:16
@LunaAlRawas even if you `#include `? – Luchian Grigore Jul 17 '12 at 14:17
unfortunately yes, still being unknown to uva cause I'm getting a compilation error – Luna28 Jul 17 '12 at 14:37

score 0 · Answer 3 · answered Jul 17 '12 at 14:07

0

I usually use a for loop to initialize all values. As far as I know, you can only initalize by a list, not a default value for all elements.

answered Jul 17 '12 at 14:07

Kyle Preiksa

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yes exactly, I'm trying to do the same as this question, and using loops will make the program run slower – Luna28 Jul 17 '12 at 14:12

score 0 · Answer 4 · answered Jul 17 '12 at 14:10

0

You will only init the first element by doing

char mines[100][100] = {'0'};

The rest will be initialized with 0 (the value not the character) you will need a loop. Like this:

int row, column;
for (row = 0; row < 100; row++)
    for (column = 0; column < 100; column++)
        mines[row][column] = '0';

answered Jul 17 '12 at 14:10

qwertz

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I know this solution, but it will make the program slower – Luna28 Jul 17 '12 at 14:18
Why? This should be one of fastest ways to do it. – qwertz Jul 17 '12 at 15:17

score 0 · Answer 5 · edited Jun 20 '20 at 09:12

Luna, this is called zero-initializer and only works to initialize all values to zero corresponding value: null pointers, zero. If you try to initialize char to 0 it will work, but with any other value will not.

More information here.

#include <iostream>

int main(int argc, char *argv[])
{
    int value[2][2] = {0};

    for(int i = 0; i < 2; ++i)
       for (int j = 0; j < 2; ++j)
           std::cout << (value[i][j] == 0) << std::endl;

    return 0;
}

Output:

1 1 1 1

If you need to initialize with a specific value, in my opinion Luchian has the best solution with memset.

declare an array with value

5 Answers5