I have a char array of char array like so:
char my_test[2][10];
As you can see I have a length of 2 and then 10. If I need to increase the first char array (2), how can this be done dynamically?
For example, half way through my application char[2] might be in use so therefore I need to use position 3 in the char array. I would then end up with this:
char store[3][10];
But keeping the data originally store in:
char store[0][10];
char store[1][10];
char store[2][10];
You should dynamically allocate memory for the array using standard C functions malloc and realloc declared in header <stdlib.h>.
Here is a demonstrative program that shows how the memory can be allocated.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 10
int main(void)
{
size_t n = 2;
char ( *my_test )[N] = malloc( n * sizeof( char[N] ) );
strcpy( my_test[0], "first" );
strcpy( my_test[1], "second" );
for ( size_t i = 0; i < n; i++ ) puts( my_test[i] );
putchar( '\n' );
char ( *tmp )[N] = realloc( my_test, ( n + 1 ) * sizeof( char[N] ) );
if ( tmp != NULL )
{
my_test = tmp;
strcpy( my_test[n++], "third" );
}
for ( size_t i = 0; i < n; i++ ) puts( my_test[i] );
free( my_test );
return 0;
}
The program output is
first
second
first
second
third
char my_test[2][10];
is compile-time constant, which means that the required memory to use that array is carved to the stone already before your application starts. So you will never be able to change it's size.
You'll have to use DYNAMIC allocation. Check for something called malloc and free in case you are really working with C, or with C++ new and delete are what you need. You'll also need to learn about pointers.
mallocandreallocfunctions to create a resizeable array.realloc().210character arrays. A 2D array is an array of arrays. Here, you have and array of 2 - 10-char arrays. Arrays cannot be resized. Blocks of memory can. You can declarechar (*my_test)[10];to create a pointer to an array of 10-chars and then allocate/reallocate the number of 10-char arrays in that block. Or, you can decalrechar **my_test;and allocate and reallocate each individual block each pointer points to.int a = 5;(aholds the immediate value5as its value),int *b;declares a pointer to typeint. Meaning it will hold the address of where an integer is stored as its value. (e.g.b = &a;) would assign the address whereais stored to the pointerb). Nothing else special about them.