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 strings will always be consistent.

I've attempted this (see below), but am having trouble, any tips in the right direction will be much appreciated!

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

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

Ultimately I want to be able to use the array to do this:

strcpy(orderedIds[0], string1);
strcpy(orderedIds[1], string2);
/* etc */

Note that you shouldn't be using ints here. a) It's signed (and I doubt you want a -5 length string), and b) it's not guaranteed to be the right size to hold the values you need it to hold. Use the size_t type for storing array indices and object sizes. That's the type of the argument to malloc. — Chris Lutz
– Chris Lutz, Commented May 9, 2011 at 11:16
@Chris: that said, it's guaranteed to be big enough for 5. — Steve Jessop
– Steve Jessop, Commented May 9, 2011 at 11:37

MByD · Accepted Answer · 2013-07-25 04:51:17Z

102

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"

edited Jul 25, 2013 at 4:51

answered May 9, 2011 at 11:03

MByD

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2 Comments

Minh Tran Over a year ago

Do you have to use another for loop to free the memory you've allocated?

MByD Over a year ago

Yes, using loops for performing the same operation multiple time is a good and common practice, which becomes a necessity when performing it a variable number of times. (But each for loop may be implemented as a while loop as well.)

sahaj · Accepted Answer · 2011-05-09 11:11:27Z

10

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]);
}

answered May 9, 2011 at 11:11

sahaj

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Comments

Oliver Charlesworth · Accepted Answer · 2011-05-09 11:21:08Z

7

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);

edited May 9, 2011 at 11:21

answered May 9, 2011 at 11:04

Oliver Charlesworth

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4 Comments

MByD Over a year ago

Why would you assign everything to orderedIds[0]?

Oliver Charlesworth Over a year ago

@Chris: Yes, indeed. In fact, I've just made that change!

Chris Lutz Over a year ago

@Oli - So I saw. You might want to add a note about how one would go about properly freeing that first one, since it's rather unintuitive.

Dori Over a year ago

What does the first option's "malloc" call return? What type of pointer? If I try to cast it as anything (char*, char**...) I get errors

Roman · Accepted Answer · 2011-05-09 11:23:53Z

0


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

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

..

edited May 9, 2011 at 11:23

answered May 9, 2011 at 11:04

Roman

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3 Comments

Chris Lutz Over a year ago

It's not considered idiomatic C to cast the return value of malloc. Also, you're declaring orderedIds as an array rather than a pointer, which means you can't assign to it. Also, you're mallocing one large block for all the character data, then casting it to a char ** which is much different.

Roman Over a year ago

Changed *orderedIds[] to **orderedIds, though I believe that compiler wise it's the same, and the former is just more readable. Regarding the "one large block" - this is exactly what it is. Arrays are just memory blocks.

Chris Lutz Over a year ago

You're wrong. If you allocate it as a char[][ID_LEN + 1] you can't treat it as a char **. char ** expects you to have pointers to data that is stored elsewhere. It isn't stored in dynamic memory the same way a two-dimensional array is stored on the stack.

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