How to fill a dynamically allocated 2D array? for example:
#define n 100
char **array;
array = new char*[n];
for(int i=0; i<n; i++)
{
array[i] = new char[n];
}
How to fill this 2D array with * value?
-
You can always use a loop inside a loop.Roy Avidan– Roy Avidan2020-08-15 09:35:36 +00:00Commented Aug 15, 2020 at 9:35
-
1you do not have a 2D array but an array of string, this is not the same thingbruno– bruno2020-08-15 09:43:45 +00:00Commented Aug 15, 2020 at 9:43
-
@bruno is right. array of string or vector of string.Amir Kadyrov– Amir Kadyrov2020-08-15 09:50:53 +00:00Commented Aug 15, 2020 at 9:50
Add a comment
|
3 Answers
- Your code came very close to achieve what you wanted.
- Just like you wrote,
array[i]is annlength character array. - To write
'*'to itsj-th entry all you have to do is:
for (int i=0;i<n;i++)
{
for (int j=0;j<n;j++)
{
array[i][j] = '*';
}
}
4 Comments
bruno
'a' is not '*'as expected by the OP, and why not using memset ?
OrenIshShalom
@bruno I prefer a double nested loop, but of course there are other ways too ...
bruno
memset is faster, may be propose both ?
bruno
You must also warn the OP array is not a 2D array
Both the answers you got so far are correct. You can see it as an 2D array or as an array of strings. You can store data on it char by char or directly writing a string. Maybe using constants instead of define would be a better practise.
const int n = 100;
char **array;
array = (char **)malloc(sizeof(char *) * n);
bool storeString = true;
for(int i=0; i<n; i++)
{
array[i] = (char *)malloc(sizeof(char) * n);
if(storeString)
array[i] = "Im striiinging in the rain";
else
for(int j = 0; j < n-1; j++)
{
array[i][j] = '0';
}
}
I'm using C type memory allocation but that shouldn't be an issue, I just don't remember how to do it precisely on C++ type, just import <cstdlib>
Comments
As @Bruno mentioned in comments, it is not 2D array but an array of strings. You can directly store the string with 1 loop. Either assign string by using = operator, or use memset function.
char **array;
array = new char*[n];
for(int i=0; i<n; i++)
{
array[i] = "whatever";
}
