I have to store some char arrays in an array. But I don't know how many of them I will have to store.
What would be the best: initializing my array with a small size (like 1) and then realloc everything? How am I supposed to use realloc or malloc?
I cannot use vectors nor stl containers nor strings unfortunately. Increasing the size of a vector is very easy and I tried to understand malloc and realloc but I don't...
char ** array=(char**)malloc(10*sizeof(char*));
for (int i=0;i<10;i++)
array[i]="10";
array=(char **)realloc(array,(sizeof(array)+1)*sizeof(char*));
array[10]="12";
I understood the basic principle yes. Is it in this way?
Well, it seems you can not use vectors but normally that's exactly their purpose.
The problem with your solution is it's more C than C++, you use a char array whereas in C++ you should use a string, and you use malloc and realloc whereas in C++ you should use new.
Furthermore, you need to allocate memory for every level of indirection you have, so for a char ** you need at least 2 calls to malloc. Here is a corrected solution (it's still almost C, not really C++, notice you don't use a single C++ header):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
//allocate a buffer of 10 pointers
size_t array_size = 10;
char ** array=(char **)malloc(array_size * sizeof(*array) );
if (!array)
return -1;
//now allocate the 10 pointers with a buffer to hold 42 characters
for (int i = 0; i < array_size; i++) {
array[i]= (char *)malloc( 42 * sizeof (*array[i]) );
if (!array[i])
return -1;
strcpy(array[i], "10");
}
//for some reason we need to increase array size by 1
++array_size;
array = (char **)realloc(array, array_size * sizeof(*array) );
array[array_size-1] = (char *)malloc(42 * sizeof(*array[array_size-1]) );
if (!array[array_size-1])
return -1;
strcpy(array[array_size-1], "12");
}