Dynamically allocate memory for array in C

Question

I allocated memory to an array (using malloc) but what if it needs more space, is it possible to expand the array later in the program? Or maybe create a new array and have the last element in the first array point to the new array?
I know that realloc would be much easier to use but I am trying to do this only using malloc.

Take a look at the man page for memcpy, instead of copying elements in the for loop. But yes, this is the right idea. — Don Shankin, Oct 15 '14 at 02:15
I'm confused about the line `newArray=new array(array1)` You're using malloc, so I'm not sure what that is meant to be — Don Shankin, Oct 15 '14 at 02:18
Also, don't cast the void pointer returned from malloc. If you absolutely must, then make sure that you `#include ` — Don Shankin, Oct 15 '14 at 02:19
I wasn't sure how to implement what Dwayne explained. The current array = newly created array (the one you copied into) — user2737810, Oct 15 '14 at 02:22
You need to check whether `num` is out of range *before* you write to `array1[num]`, not afterwards. In your current code you'll write to `array1[100]` before you figure out that that's past the end of your array. — Crowman, Oct 15 '14 at 02:59

Dwayne Towell · Accepted Answer · 2014-10-15T02:14:54.637

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The general algorithm is

allocate array of 100
while more input
    if no-room-in-array
        allocate another array 100 bigger than the current array
        copy values from current array into newly created array
        free(current array)
        current array = newly created array (the one you copied into)
    // now there will be room, so
    put input in array

edited Oct 15 '14 at 02:14

answered Oct 15 '14 at 01:57

Dwayne Towell

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Don't change your question. If you need to add something add it. – Dwayne Towell Oct 15 '14 at 02:15
No it doesn't, my implementation of current array = newly created array (the one you copied into) is wrong. Not really sure why – user2737810 Oct 15 '14 at 02:33

score 1 · Answer 2 · edited May 23 '17 at 12:29

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Yes, you can use realloc(). Be careful to check the return value before you assign it to the original pointer though. See here: https://stackoverflow.com/a/1986572/4323

edited May 23 '17 at 12:29

Community

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answered Oct 15 '14 at 01:50

John Zwinck

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Like I said I cant use realloc – user2737810 Oct 15 '14 at 01:51
@user2737810: what platform are you on that you can't use realloc()? – John Zwinck Oct 15 '14 at 01:53
The program specs dont allow realloc – user2737810 Oct 15 '14 at 01:54

chux - Reinstate Monica · Answer 3 · 2014-10-15T03:18:46.120

Wrong size passed to malloc().

Rather than passing n bytes, code should pass n * sizeof(int).

// int *array1=(int *)malloc(100);
int *array1 = malloc(100 * sizeof *array1);

// int *newArray=(int *)malloc(size+100);
int *newArray =  malloc((size+100) * szeof *newArray);

Other ideas include

1) No need to cast

    int *array1 = (int *) malloc(...;
    int *array1 = malloc(...);

2) Simplify with memcpy()

    // for(i=0; i<size; i++) newArray[i]=array1[i];
    memcpy(newArray, array, size * sizeof *newArray);

3) Be sure to free()

4) new is a C++ operator, this is C, use malloc().

5) Use size_t rather than int for size.

6) Grow exponentially, rather than linearly

// int *newArray=(int *)malloc(size+100);
size_t newsize = size*3/2;
int *newArray = malloc(newsize);

7) Check for malloc() failure

int *newArray = malloc(newsize);
if (newArray == NULL && newsize > 0) Handle_Failure();

Dynamically allocate memory for array in C

3 Answers3