three dimensional character array in C - c

Question

I want to declare a three dimensional array of strings
where each row has two strings.
Here I am able to declare it:
char *szArray[][2] = {
{"string1", "string2"},
{"string3", "string4"},
{"string5", "string6"},
{0, 0}
};
I want to do the same thing but number of rows are dynamically allocated.
Here you can assume 1st dimension is dynamically decided.
2nd dimension is 2,
3rd dimension is 25.
Tried in many ways no luck :(.
Thanks in advance !

Answer 1

First what you have in your question is not a 3-dimensional array of char but a 2-dimensional array of pointers to char. In C, pointers and arrays are not the same thing.
To answer your question, the simplest way in modern C, C99, is to use variable length arrays, VLA, for your purpose. For a 3-dimensional array you'd do
char szData[x][y][z];
where x, y and z are variables or expressions that are only determined at run time when you hit that declaration.
The inconvenience of VLA are that you have to be careful that they don't become too large for your stack, and that you have to initialize them by assignment to the individual entries (here by a nested for-loop).
To stay with your example of 2-d array of strings
char* szData[x][y];
and then you'd have to assign either individual strings to each of the pointers or to set them to 0.
for (size_t i = 0; i < x; ++i)
for (size_t j = 0; j < y; ++j)
szData[i][j] = 0;

Answer 2

int n = 0 ;
printf("Enter the number of rows\n");
scanf("%d",&n);
char *** szData = (char *** )malloc(n * sizeof(char ** )) ;
//Allocate memroy for each row
for(int i = 0 ; i < n ; i++ )
{
szData[i] = (char ** ) malloc(2 * sizeof(char * )) ;
for ( int j = 0 ; j < 2 ; j++ )
{
szData[i][j] = (char *) malloc (25 * sizeof(char));
}
}
//Assign some data
for( int i = 0 ; i < n ; i++ )
{
sprintf(szData[i][0],"string%d", i);
sprintf(szData[i][1],"string1%d", i);
}
//print all the elements
for( int i = 0 ; i < n ; i++ )
{
printf("%s\n",szData[i][0]);
printf("%s\n",szData[i][1]);
}
//free memory here
for(int i = 0 ; i < n ; i++ )
{
for ( int j = 0 ; j < 2 ; j++ )
{
delete szData[i][j];
}
}
for(int i = 0 ; i < n ; i++ )
{
delete szData[i];
}
delete szData;

Answer 3

I didnt get any error:
#include <stdio.h>
int main(){
char *szArray[][2] = { {"string1", "string2"}, {"string3", "string4"}, {"string5", "string6"}, {0, 0} };
printf("%s\n", szArray[0][0]);
printf("%s\n", szArray[2][0]);
}
Here is the output:
$ gcc test.c
$ ./a.exe
string1
string5
But you cannot print szArray[3][0][0] because it is 0, if you want to initialize the value to whatever, you can set to "\0" instead of just 0

Answer 4

Ok, no compiler here to double check, but you can do this a number of ways. The most straight forward would be to declare a Vector of char[25][3] and let c++ do it for you. Something like Vector foo and then just push_back your dynamic elements. Without a compiler though, I'm not 100% certain that this would work as there are funky rules when it comes to multi-dimensional arrays. You can always undimensionalize your array as a first pass just to get it working as well - something like a Vector and start adding how ever you want. You can also do the vector of vector of char approach and have jagged arrays of chars where and of the 3 dimensions can be dynamic, and this would probably even be more memory efficent. Lots of choices here.