Dynamically create an array of strings with malloc

Question

I am trying to create an array of strings in C using malloc. The number of strings that the array will hold can change at run time, but the length of the string

Answer 1

Given that your strings are all fixed-length (presumably at compile-time?), you can do the following:

char (*orderedIds)[ID_LEN+1]
    = malloc(variableNumberOfElements * sizeof(*orderedIds));

// Clear-up
free(orderedIds);

A more cumbersome, but more general, solution, is to assign an array of pointers, and psuedo-initialising them to point at elements of a raw backing array:

char *raw = malloc(variableNumberOfElements * (ID_LEN + 1));
char **orderedIds = malloc(sizeof(*orderedIds) * variableNumberOfElements);

// Set each pointer to the start of its corresponding section of the raw buffer.
for (i = 0; i < variableNumberOfElements; i++)
{
    orderedIds[i] = &raw[i * (ID_LEN+1)];
}

...

// Clear-up pointer array
free(orderedIds);
// Clear-up raw array
free(raw);

Answer 2

char **orderIds;

orderIds = malloc(variableNumberOfElements * sizeof(char*));

for(int i = 0; i < variableNumberOfElements; i++) {
  orderIds[i] = malloc((ID_LEN + 1) * sizeof(char));
  strcpy(orderIds[i], your_string[i]);
}

Answer 3

#define ID_LEN 5
char **orderedIds;
int i;
int variableNumberOfElements = 5; /* Hard coded here */

orderedIds = (char **)malloc(variableNumberOfElements * (ID_LEN + 1) * sizeof(char));

..

Answer 4

You should assign an array of char pointers, and then, for each pointer assign enough memory for the string:

char **orderedIds;

orderedIds = malloc(variableNumberOfElements * sizeof(char*));
for (int i = 0; i < variableNumberOfElements; i++)
    orderedIds[i] = malloc((ID_LEN+1) * sizeof(char)); // yeah, I know sizeof(char) is 1, but to make it clear...

Seems like a good way to me. Although you perform many mallocs, you clearly assign memory for a specific string, and you can free one block of memory without freeing the whole "string array"