Pointer Array Malloc (C Programming)

Question

I found this one line that I don't understand. It says

char (*storage)[15] = malloc(sizeof *storage * 8)

Does anybody know what this is mean? why I see a lot of *?

I don't get it because why he/she multiplied by 8 (it seems like that) but then declared it was [15] too?

Correct me if I'm wrong.

cdecl.ridiculousfish.com/?q=char+%28*storage%29%5B15%5D (bit.ly/1WpCOWp) — ForceBru
– ForceBru, Commented Oct 31, 2015 at 15:48
In other words, storage is now an array of 8 elements, each of which is an array of 15 chars. — user4520
– user4520, Commented Oct 31, 2015 at 15:51
@ForceBru & Erik: It says syntax error bad character "=" :( — John
– John, Commented Oct 31, 2015 at 15:56

ameyCU · Accepted Answer · 2015-10-31 16:13:11Z

3

char (*storage)[15] = malloc(sizeof *storage * 8)

It allocates memory for 8 character arrays of size 15 .

For all * you ask-

char (*storage)[15]              // pointer to array of 15 chars

and this -

sizeof *storage * 8   // this is 8 times sizeof type to which storage points

edited Oct 31, 2015 at 16:13

answered Oct 31, 2015 at 15:50

ameyCU

16.7k2 gold badges28 silver badges41 bronze badges

Sign up to request clarification or add additional context in comments.

10 Comments

John Over a year ago

Ahh I see. Is there any alternative (much simpler) way to define it?

ameyCU Over a year ago

@John Yes , you can write like this -char storage[8][15];.

John Over a year ago

Now that seems familiar to me! Thank you.

Mike32ab Over a year ago

@ameyCU And how do you free() that?

alk Over a year ago

char *storage = malloc(15*8) is not the same as char (*storage)[15] = malloc(sizeof *storage * 8). Although the same amount of memory is allocated, storage behaves completely different in case pointer arithmetics are applied to it.

|

MikeCAT · Accepted Answer · 2015-10-31 15:47:46Z

1

char (*storage)[15] is a pointer to 15-element array of char.

sizeof *storage * 8 is 8 times the size of the type at which storage points.

answered Oct 31, 2015 at 15:47

MikeCAT

75.1k11 gold badges50 silver badges71 bronze badges

1 Comment

John Over a year ago

Ah I see. Is there any alternative (much simpler) way to define it?

John Bode · Accepted Answer · 2015-10-31 18:38:23Z

char (*storage)[15] = malloc(sizeof *storage * 8);

dynamically allocates an 8x15 array of char. This is roughly similar to writing

char storage[8][15];

except that

The lifetime of a dynamically-allocated object lasts until it is explicitly deallocated with free, whereas the lifetime of an automatically-allocated object lasts until the program exits the object's enclosing scope;
The storage for a dynamically-allocated object is (usually) taken from a different memory segment than automatically-allocated objects;
You can extend the size of a dynamically-allocated array as necessary. IOW, if you realized you needed two more rows, you could write
```
size_t curRows = 8;
char (*tmp)[15] = realloc( storage, sizeof *storage * (curRows + 2));
if ( tmp )
{
  storage = tmp;
  curRows += 2;
}
```
You can't do that with an automatically-allocated array.

An alternate way of dynamically allocating an 8x15 array is something like this:

char **storage = malloc( sizeof *storage * 8 );
if ( storage )
{
  for ( size_t i = 0; i < 8; i++ )
  {
    storage[i] = malloc( sizeof *storage[i] * 15 );
  }
}

Note that all three versions can be indexed as storage[i][j].

The main advantage of the last method is that each row can be different lengths if you want them to be. The main disadvantages are that the rows won't necessarily be adjacent in memory, and you have to free each storage[i] before you can free storage.

Collectives™ on Stack Overflow

Pointer Array Malloc (C Programming)

3 Answers 3

10 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

10 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related