copying array to a new one with memcpy

Question

I have an issue where i'm using memcpy() to copy an array to a new array with dynamic memory. And my question is why there is 3 zeros between the numbers? like my original array is a1[] ={1, 2, 3, 4, 5} and when I use memcpy(mem1,a1,n1) then my mem1 will be 1 0 0 0 2 ? Here's my code:

int join_arrays(unsigned int n1, int *a1, unsigned int n2, int *a2, unsigned 
int n3, int *a3)
{
    /*Here I just print the original a1 just to make sure it's correct*/
    for (int j = 0; j < 5; j++) {
        printf("%d  ", a1[j]);
    }
    /*I allocate the memory for the new array*/
    char *mem1;
    mem1 = malloc(n1*sizeof(int));

    /*checking if the allocation succeeded*/
    if (!mem1) {
    printf("Memory allocation failed\n");
    exit(-1);
    }

    /*Using memcpy() to copy the original array to the new one*/
    memcpy(mem1, a1, n1);

    /*Printing the new array and this print gives me "1 0 0 0 2"
      and it should give me "1 2 3 4 5"*/
    printf("\n");
    for (int i = 0; i < 5; i++) {
        printf("%d  ", mem1[i]);
    }
    return 0;
}
int main(void)
{
    /* these are the original arrays which I need to put together to a single array */
    int a1[] = { 1, 2, 3, 4, 5 };
    int a2[] = { 10, 11, 12, 13, 14, 15, 16, 17 };
    int a3[] = { 20, 21, 22 };

    /*The number of elements are before the array itself*/
    join_arrays(5, a1, 8, a2, 3, a3);


    return 0;
}

chux · Accepted Answer · 2018-02-21 21:12:11Z

3

Instead of allocating the memory and assigning it to char* take a int* and work with it.

int *mem1;
mem1 = malloc(n1*sizeof(int)); // malloc(n1 * sizeof *mem1);
..
memcpy(mem1,a1,n1*sizeof(a1[0]));

Also check whether malloc failed or not - but add a proper error message:-

if (!mem1) {
   perror("Memory allocation failed\n");
   exit(EXIT_FAILURE);
}

Don't forget to free the dynamically allocated memory, in your case

free(mem1);

As you have said, that you need to return it from the function then you would do something like this

int *join_arrays(..){

  return mem1;
}
int main(void){

  int *p = join_arrays(..);
  /* work with it */

  free(p);
}

edited Feb 21, 2018 at 21:12

chux

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answered Feb 21, 2018 at 20:15

user2736738

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4 Comments

michael Over a year ago

This worked thanks, but my function have to return a pointer to that new array mem1 so do I have to free the memory still?

user2736738 Over a year ago

@michael.: Will add this to answer..wait.

user2736738 Over a year ago

@michael.: Edited.

michael Over a year ago

Okay yea that's what I thought!

N.Douss · Accepted Answer · 2018-02-21 20:22:48Z

1

mem1 here is a char pointer not an int pointer.

Therefore when you try to print mem1[i] it will actually print the byte that is stored at address mem1+i instead of 4 bytes. Apparently the integer 1 is stored like this on your machine :

 00000001 00000000 00000000 00000000

that is why you get 3 zeros.

Try changing your variable type to int* like this :

int *mem1;
mem1 = malloc(n1*sizeof(int));
memcpy(mem1,a1,n1*sizeof(int));

answered Feb 21, 2018 at 20:22

N.Douss

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