Expand an non-pointer array

Question

let's say I have an array in C:

char *list[3] = {"Hello", "world", "!"};

And I want to expand it. If I daclare that array as:

char **list = (char **) malloc(3 * sizeof(char *));  // Or sth. like that...

I can resize it:

realloc(list, 5 * sizeof(char *));  // Not sure now if I should use `char *` or `char **`

If I try this:

char *list[3] = {"Hello", "world", "!"};
realloc(list, 5 * sizeof(char *));  // Not sure now if I should use `char *` or `char **`

It says that it can't resize memory that wasn't allocated.
Ok, but how can I then resize an array like this?

You're using sizeof(char *) correctly. A better and safer way would be to use sizeof *list instead. — Some programmer dude
– Some programmer dude, Commented Apr 25, 2016 at 13:18
It's simple logic: either you have an array of known size, or you have an array of unknown size. You can't have an array which is both of known and unknown size at once. — Lundin
– Lundin, Commented Apr 25, 2016 at 14:07

sepp2k · Accepted Answer · 2016-04-25 13:19:13Z

2

You can't. If you have a statically-sized array, you simply can't change its size. If you need to be able to change the size, don't use a statically-sized array.

PS: You should not ignore the return value of realloc. There's no guarantee that realloc is able to grow the given memory chunk, so it may return a pointer to an entirely new chunk of memory (freeing the old memory). So you should always use the pointer returned by realloc and never assume that the old pointer is still valid after calling realloc.

answered Apr 25, 2016 at 13:19

sepp2k

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Some programmer dude · Accepted Answer · 2016-04-25 13:22:26Z

The simple answer is that you can't. You can either have a fixed-size array, or a dynamically allocated array, but you can't mix them.

The reason for this is that fixed-sized arrays are created by the compiler at the time of compilation¹. It simply sets aside a bit of the memory used by the program to store the array. If it's a local variable that memory is on the stack, if it's a global variable it's in the global data space. The dynamic allocation function like malloc and realloc allocates from a different pool of memory, and these different pools of memory are not compatible.

¹Variable-length arrays are not allocated by the compiler at the time of compilation, or they would not be variable-length arrays. How the compiler implements them doesn't matter, they are still "fixed-sized" arrays and can not be reallocated once created.

DarkDust · Accepted Answer · 2016-04-25 13:26:25Z

0

When you're using the following statement:

char *list[3] = {"Hello", "world", "!"};

… the compiler is generating the array for you. But malloc doesn't know about it. You need to allocate with malloc, then copy your static array into the dynamic region. Only then can you change its size with realloc.

char *staticList[3] = {"Hello", "world", "!"};
// Allocate dynamically.
char **list = malloc(sizeof(staticList));
// Copy static array.
memcpy(list, staticList, sizeof(staticList));
// Change size to 5 entries.
list = realloc(list, 5 * sizeof(list[0]));

edited Apr 25, 2016 at 13:26

answered Apr 25, 2016 at 13:20

DarkDust

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2 Comments

Some programmer dude Over a year ago

That initialization of list will not work very well.

DarkDust Over a year ago

@JoachimPileborg: Thanks, fixed.

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